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1,606 | Combinatorics | null | $\boxed{2}$ | [
"2"
] | false | null | Numerical | null | Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding 5000. Then she fixes 20 distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of positive integers not exceeding 20 , and she tells him back the set $\left\{a_{k}: k \in S\right\}$ without spelling out which number corresponds to which index. How many moves does Sergey need to determine for sure the number Xenia thought of? | [
"Sergey can determine Xenia's number in 2 but not fewer moves.\n\n\n\nWe first show that 2 moves are sufficient. Let Sergey provide the set $\\{17,18\\}$ on his first move, and the set $\\{18,19\\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able to identify $a_{17}, a_{18}$, and $a_{19}$, and thence the residue of $N$ modulo $17 \\cdot 18 \\cdot 19=5814>5000$, by the Chinese Remainder Theorem. This means that the given range contains a single number satisfying all congruences, and Sergey achieves his goal.\n\n\n\nTo show that 1 move is not sufficient, let $M=\\operatorname{lcm}(1,2, \\ldots, 10)=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7=2520$. Notice that $M$ is divisible by the greatest common divisor of every pair of distinct positive integers not exceeding 20. Let Sergey provide the set $S=\\left\\{s_{1}, s_{2}, \\ldots, s_{k}\\right\\}$. We show that there exist pairwise distinct positive integers $b_{1}, b_{2}, \\ldots, b_{k}$ such that $1 \\equiv b_{i}\\left(\\bmod s_{i}\\right)$ and $M+1 \\equiv b_{i-1}\\left(\\bmod s_{i}\\right)$ (indices are reduced modulo $k$ ). Thus, if in response Xenia provides the set $\\left\\{b_{1}, b_{2}, \\ldots, b_{k}\\right\\}$, then Sergey will be unable to distinguish 1 from $M+1$, as desired.\n\n\n\nTo this end, notice that, for each $i$, the numbers of the form $1+m s_{i}, m \\in \\mathbb{Z}$, cover all residues modulo $s_{i+1}$ which are congruent to $1(\\equiv M+1)$ modulo $\\operatorname{gcd}\\left(s_{i}, s_{i+1}\\right) \\mid M$. Xenia can therefore choose a positive integer $b_{i}$ such that $b_{i} \\equiv 1\\left(\\bmod s_{i}\\right)$ and $b_{i} \\equiv M+1\\left(\\bmod s_{i+1}\\right)$. Clearly, such choices can be performed so as to make the $b_{i}$ pairwise distinct, as required."
] | 2 |
1,610 | Algebra | null | $\boxed{\frac{1}{2 n+2}}$ | [
"$\\frac{1}{2 n+2}$"
] | false | null | Expression | null | Given a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4 n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\mu$. | [
"The required maximum is $\\frac{1}{2 n+2}$. To show that the condition in the statement is not met if $\\mu>\\frac{1}{2 n+2}$, let $U=(0,1) \\times(0,1)$, choose a small enough positive $\\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\\left(\\frac{i}{n+1} \\pm \\epsilon\\right) \\times\\left(\\frac{1}{2} \\pm \\epsilon\\right), i=1, \\ldots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at $\\operatorname{most}\\left(\\frac{1}{n+1}+\\epsilon\\right) \\cdot\\left(\\frac{1}{2}+\\epsilon\\right)<\\mu$ if $\\epsilon$ is small enough.\n\n\n\nWe now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu_{0}=\\frac{2}{|C|+4}$.\n\n\n\nTo prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.\n\n\n\nLemma 1. Let $k$ be a positive integer, and let $\\lambda<\\frac{1}{\\lfloor k / 2\\rfloor+1}$ be a positive real number. If $t_{1}, \\ldots, t_{k}$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$.\n\n\n\nLemma 2. Given an integer $k \\geq 2$ and positive integers $m_{1}, \\ldots, m_{k}$,\n\n\n\n$$\n\n\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+\\sum_{i=1}^{k}\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor \\leq \\sum_{i=1}^{k} m_{i}-k+2\n\n$$\n\n\n\nBack to the problem, let $U=(0,1) \\times(0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_{1}<\\cdots<x_{k}$ in the open unit interval $(0,1)$, let $\\ell_{i}$ be the vertical through $x_{i}$, and let $m_{i}=\\left|C \\cap \\ell_{i}\\right|, i=1, \\ldots, k$.\n\n\n\nSetting $x_{0}=0$ and $x_{k+1}=1$, assume that $x_{i+1}-x_{i-1}>\\left(\\left\\lfloor m_{i} / 2\\right\\rfloor+1\\right) \\mu_{0}$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \\cap \\ell_{i}$ from the other ones by an open subinterval $x_{i} \\times J$ of $x_{i} \\times(0,1)$ whose length is greater than or equal to $\\mu_{0} /\\left(x_{i+1}-x_{i-1}\\right)$. Consequently, $\\left(x_{i-1}, x_{i+1}\\right) \\times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\\mu_{0}$.\n\n\n\nNext, we rule out the case $x_{i+1}-x_{i-1} \\leq\\left(\\left\\lfloor m_{i} / 2\\right\\rfloor+1\\right) \\mu_{0}$ for all indices $i$. If this were the case, notice that necessarily $k>1$; also, $x_{1}-x_{0}<x_{2}-x_{0} \\leq\\left(\\left\\lfloor m_{1} / 2\\right\\rfloor+1\\right) \\mu_{0}$ and $x_{k+1}-x_{k}<$ $x_{k+1}-x_{k-1} \\leq\\left(\\left\\lfloor m_{k} / 2\\right\\rfloor+1\\right) \\mu_{0}$. With reference to Lemma 2 , write\n\n\n\n$$\n\n\\begin{aligned}\n\n2=2\\left(x_{k+1}-x_{0}\\right) & =\\left(x_{1}-x_{0}\\right)+\\sum_{i=1}^{k}\\left(x_{i+1}-x_{i-1}\\right)+\\left(x_{k+1}-x_{k}\\right) \\\\\n\n& <\\left(\\left(\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+1\\right)+\\sum_{i=1}^{k}\\left(\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+1\\right)+\\left(\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor+1\\right)\\right) \\cdot \\mu_{0} \\\\\n\n& \\leq\\left(\\sum_{i=1}^{k} m_{i}+4\\right) \\mu_{0}=(|C|+4) \\mu_{0}=2,\n\n\\end{aligned}\n\n$$\n\n\n\nand thereby reach a contradiction.\n\n\n\n\n\n\n\nFinally, we prove the two lemmas.\n\n\n\nProof of Lemma 1. Suppose, if possible, that no $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$. Without loss of generality, we may (and will) assume that $0=t_{0}<t_{1}<\\cdots<t_{k}<t_{k+1}=1$. Since the open interval $\\left(t_{i-1}, t_{i+1}\\right)$ isolates $t_{i}$ from the other $t_{j}$, its length, $t_{i+1}-t_{i-1}$, is less than $\\lambda$. Consequently, if $k$ is odd we have $1=\\sum_{i=0}^{(k-1) / 2}\\left(t_{2 i+2}-t_{2 i}\\right)<\\lambda\\left(1+\\frac{k-1}{2}\\right)<1$; if $k$ is even, we have $1<1+t_{k}-t_{k-1}=$ $\\sum_{i=0}^{k / 2-1}\\left(t_{2 i+2}-t_{2 i}\\right)+\\left(t_{k+1}-t_{k-1}\\right)<\\lambda\\left(1+\\frac{k}{2}\\right)<1$. A contradiction in either case.\n\n\n\nProof of Lemma 2. Let $I_{0}$, respectively $I_{1}$, be the set of all indices $i$ in the range $2, \\ldots, k-1$ such that $m_{i}$ is even, respectively odd. Clearly, $I_{0}$ and $I_{1}$ form a partition of that range. Since $m_{i} \\geq 2$ if $i$ is in $I_{0}$, and $m_{i} \\geq 1$ if $i$ is in $I_{1}$ (recall that the $m_{i}$ are positive integers),\n\n\n\n$$\n\n\\sum_{i=2}^{k-1} m_{i}=\\sum_{i \\in I_{0}} m_{i}+\\sum_{i \\in I_{1}} m_{i} \\geq 2\\left|I_{0}\\right|+\\left|I_{1}\\right|=2(k-2)-\\left|I_{1}\\right|, \\quad \\text { or } \\quad\\left|I_{1}\\right| \\geq 2(k-2)-\\sum_{i=2}^{k-1} m_{i}\n\n$$\n\n\n\nTherefore,\n\n\n\n$$\n\n\\begin{aligned}\n\n\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+\\sum_{i=1}^{k}\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor & \\leq m_{1}+\\left(\\sum_{i=2}^{k-1} \\frac{m_{i}}{2}-\\frac{\\left|I_{1}\\right|}{2}\\right)+m_{k} \\\\\n\n& \\leq m_{1}+\\left(\\frac{1}{2} \\sum_{i=2}^{k-1} m_{i}-(k-2)+\\frac{1}{2} \\sum_{i=2}^{k-1} m_{i}\\right)+m_{k} \\\\\n\n& =\\sum_{i=1}^{k} m_{i}-k+2 .\n\n\\end{aligned}\n\n$$"
] | $\frac{1}{2 n+2}$ |
1,612 | Number Theory | null | $\boxed{2^{1009}}$ | [
"$2^{1009}$"
] | false | null | Numerical | null |
Find (in closed form) the difference between the number of positive integers at most $2^{2017}$ with even weight and the number of positive integers at most $2^{2017}$ with odd weight. | [
"For every integer $M \\geq 0$, let $A_{M}=\\sum_{n=-2^{M}+1}^{0}(-1)^{w(n)}$ and let $B_{M}=$ $\\sum_{n=1}^{2^{M}}(-1)^{w(n)}$; thus, $B_{M}$ evaluates the difference of the number of even weight integers in the range 1 through $2^{M}$ and the number of odd weight integers in that range.\n\n\n\nNotice that\n\n\n\n$$\n\nw(n)= \\begin{cases}w\\left(n+2^{M}\\right)+1 & \\text { if }-2^{M}+1 \\leq n \\leq-2^{M-1} \\\\ w\\left(n-2^{M}\\right) & \\text { if } 2^{M-1}+1 \\leq n \\leq 2^{M}\\end{cases}\n\n$$\n\n\n\n\n\n\n\nto get\n\n\n\n$$\n\n\\begin{aligned}\n\n& A_{M}=-\\sum_{n=-2^{M}+1}^{-2^{M-1}}(-1)^{w\\left(n+2^{M}\\right)}+\\sum_{n=-2^{M-1}+1}^{0}(-1)^{w(n)}=-B_{M-1}+A_{M-1}, \\\\\n\n& B_{M}=\\sum_{n=1}^{2^{M-1}}(-1)^{w(n)}+\\sum_{n=2^{M-1}+1}^{2^{M}}(-1)^{w\\left(n-2^{M}\\right)}=B_{M-1}+A_{M-1} .\n\n\\end{aligned}\n\n$$\n\n\n\nIteration yields\n\n\n\n$$\n\n\\begin{aligned}\n\nB_{M} & =A_{M-1}+B_{M-1}=\\left(A_{M-2}-B_{M-2}\\right)+\\left(A_{M-2}+B_{M-2}\\right)=2 A_{M-2} \\\\\n\n& =2 A_{M-3}-2 B_{M-3}=2\\left(A_{M-4}-B_{M-4}\\right)-2\\left(A_{M-4}+B_{M-4}\\right)=-4 B_{M-4}\n\n\\end{aligned}\n\n$$\n\n\n\nThus, $B_{2017}=(-4)^{504} B_{1}=2^{1008} B_{1}$; since $B_{1}=(-1)^{w(1)}+(-1)^{w(2)}=2$, it follows that $B_{2017}=$ $2^{1009}$"
] | $2^{1009}$ |
1,613 | Algebra | null | $\boxed{2}$ | [
"2"
] | false | null | Numerical | null | Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \leq n$, and $k+1$ distinct integers $x_{1}, x_{2}, \ldots, x_{k+1}$ such that
$$
P\left(x_{1}\right)+P\left(x_{2}\right)+\cdots+P\left(x_{k}\right)=P\left(x_{k+1}\right) .
$$
Note. A polynomial is monic if the coefficient of the highest power is one. | [
"There is only one such integer, namely, $n=2$. In this case, if $P$ is a constant polynomial, the required condition is clearly satisfied; if $P=X+c$, then $P(c-1)+P(c+1)=$ $P(3 c)$; and if $P=X^{2}+q X+r$, then $P(X)=P(-X-q)$.\n\n\n\nTo rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of degree at most $n$ with integer coefficients, whose restriction to the integers is injective, and $P(x) \\equiv 1$ $(\\bmod\\ n)$ for all integers $x$. This is easily seen by reading the relation in the statement modulo $n$, to deduce that $k \\equiv 1(\\bmod\\ n)$, so $k=1$, since $1 \\leq k \\leq n$; hence $P\\left(x_{1}\\right)=P\\left(x_{2}\\right)$ for some distinct integers $x_{1}$ and $x_{2}$, which contradicts injectivity.\n\n\n\nIf $n=1$, let $P=X$, and if $n=4$, let $P=X^{4}+7 X^{2}+4 X+1$. In the latter case, clearly, $P(x) \\equiv 1(\\bmod\\ 4)$ for all integers $x$; and $P$ is injective on the integers, since $P(x)-P(y)=$ $(x-y)\\left((x+y)\\left(x^{2}+y^{2}+7\\right)+4\\right)$, and the absolute value of $(x+y)\\left(x^{2}+y^{2}+7\\right)$ is either 0 or at least 7 for integral $x$ and $y$.\n\n\n\nAssume henceforth $n \\geq 3, n \\neq 4$, and let $f_{n}=(X-1)(X-2) \\cdots(X-n)$. Clearly, $f_{n}(x) \\equiv$ $0(\\bmod n)$ for all integers $x$. If $n$ is odd, then $f_{n}$ is non-decreasing on the integers; and if, in addition, $n>3$, then $f_{n}(x) \\equiv 0(\\bmod n+1)$ for all integers $x$, since $f_{n}(0)=-n !=-1 \\cdot 2 \\cdot \\cdots$. $\\frac{n+1}{2} \\cdot \\cdots \\cdot n \\equiv 0(\\bmod\\ n+1)$.\n\n\n\nFinally, let $P=f_{n}+n X+1$ if $n$ is odd, and let $P=f_{n-1}+n X+1$ if $n$ is even. In either case, $P$ is strictly increasing, hence injective, on the integers, and $P(x) \\equiv 1(\\bmod n)$ for all integers $x$."
] | 2 |
1,614 | Combinatorics | null | $\boxed{2n-2}$ | [
"$2n-2$"
] | false | null | Expression | null | Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.
Note. A subset $A$ of $X$ is proper if $A \neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection. | [
"The required maximum is $2 n-2$. To describe a $(2 n-2)$-element collection satisfying the required conditions, write $X=\\{1,2, \\ldots, n\\}$ and set $B_{k}=\\{1,2, \\ldots, k\\}$, $k=1,2, \\ldots, n-1$, and $B_{k}=\\{k-n+2, k-n+3, \\ldots, n\\}, k=n, n+1, \\ldots, 2 n-2$. To show that no subcollection of the $B_{k}$ is tight, consider a subcollection $\\mathcal{C}$ whose union $U$ is a proper subset of $X$, let $m$ be an element in $X \\backslash U$, and notice that $\\mathcal{C}$ is a subcollection of $\\left\\{B_{1}, \\ldots, B_{m-1}, B_{m+n-1}, \\ldots, B_{2 n-2}\\right\\}$, since the other $B$ 's are precisely those containing $m$. If $U$ contains elements less than $m$, let $k$ be the greatest such and notice that $B_{k}$ is the only member of $\\mathcal{C}$ containing $k$; and if $U$ contains elements greater than $m$, let $k$ be the least such and notice that $B_{k+n-2}$ is the only member of $\\mathcal{C}$ containing $k$. Consequently, $\\mathcal{C}$ is not tight.\n\n\n\nWe now proceed to show by induction on $n \\geq 2$ that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$. The base case $n=2$ is clear, so let $n>2$ and suppose, if possible, that $\\mathcal{B}$ is a collection of $2 n-1$ proper non-empty subsets of $X$ containing no tight subcollection.\n\n\n\nTo begin, notice that $\\mathcal{B}$ has an empty intersection: if the members of $\\mathcal{B}$ shared an element $x$, then $\\mathcal{B}^{\\prime}=\\{B \\backslash\\{x\\}: B \\in \\mathcal{B}, B \\neq\\{x\\}\\}$ would be a collection of at least $2 n-2$ proper non-empty subsets of $X \\backslash\\{x\\}$ containing no tight subcollection, and the induction hypothesis would be contradicted.\n\n\n\nNow, for every $x$ in $X$, let $\\mathcal{B}_{x}$ be the (non-empty) collection of all members of $\\mathcal{B}$ not containing $x$. Since no subcollection of $\\mathcal{B}$ is tight, $\\mathcal{B}_{x}$ is not tight, and since the union of $\\mathcal{B}_{x}$ does not contain $x$, some $x^{\\prime}$ in $X$ is covered by a single member of $\\mathcal{B}_{x}$. In other words, there is a single set in $\\mathcal{B}$ covering $x^{\\prime}$ but not $x$. In this case, draw an arrow from $x$ to $x^{\\prime}$. Since there is at least one arrow from each $x$ in $X$, some of these arrows form a (minimal) cycle $x_{1} \\rightarrow x_{2} \\rightarrow \\cdots \\rightarrow x_{k} \\rightarrow x_{k+1}=x_{1}$ for some suitable integer $k \\geq 2$. Let $A_{i}$ be the unique member of $\\mathcal{B}$ containing $x_{i+1}$ but not $x_{i}$, and let $X^{\\prime}=\\left\\{x_{1}, x_{2}, \\ldots, x_{k}\\right\\}$.\n\n\n\nRemove $A_{1}, A_{2}, \\ldots, A_{k}$ from $\\mathcal{B}$ to obtain a collection $\\mathcal{B}^{\\prime}$ each member of which either contains or is disjoint from $X^{\\prime}$ : for if a member $B$ of $\\mathcal{B}^{\\prime}$ contained some but not all elements of $X^{\\prime}$, then $B$ should contain $x_{i+1}$ but not $x_{i}$ for some $i$, and $B=A_{i}$, a contradiction. This rules out the case $k=n$, for otherwise $\\mathcal{B}=\\left\\{A_{1}, A_{2}, \\ldots, A_{n}\\right\\}$, so $|\\mathcal{B}|<2 n-1$.\n\n\n\nTo rule out the case $k<n$, consider an extra element $x^{*}$ outside $X$ and let\n\n\n\n$$\n\n\\mathcal{B}^{*}=\\left\\{B: B \\in \\mathcal{B}^{\\prime}, B \\cap X^{\\prime}=\\varnothing\\right\\} \\cup\\left\\{\\left(B \\backslash X^{\\prime}\\right) \\cup\\left\\{x^{*}\\right\\}: B \\in \\mathcal{B}^{\\prime}, X^{\\prime} \\subseteq B\\right\\}\n\n$$\n\n\n\nthus, in each member of $\\mathcal{B}^{\\prime}$ containing $X^{\\prime}$, the latter is collapsed to $\\operatorname{singleton} x^{*}$. Notice that $\\mathcal{B}^{*}$ is a collection of proper non-empty subsets of $X^{*}=\\left(X \\backslash X^{\\prime}\\right) \\cup\\left\\{x^{*}\\right\\}$, no subcollection of which is tight. By the induction hypothesis, $\\left|\\mathcal{B}^{\\prime}\\right|=\\left|\\mathcal{B}^{*}\\right| \\leq 2\\left|X^{*}\\right|-2=2(n-k)$, so $|\\mathcal{B}| \\leq 2(n-k)+k=$ $2 n-k<2 n-1$, a final contradiction.",
"Proceed again by induction on $n$ to show that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$.\n\n\n\nConsider any collection $\\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection (we call such collection good). Assume that there exist $M, N \\in \\mathcal{B}$ such that $M \\cup N$ is distinct from $M, N$, and $X$. In this case, we will show how to modify $\\mathcal{B}$ so that it remains good, contains the same number of sets, but the total number of elements in the sets of $\\mathcal{B}$ increases.\n\n\n\n\n\n\n\nConsider a maximal (relative to set-theoretic inclusion) subcollection $\\mathcal{C} \\subseteq \\mathcal{B}$ such that the set $C=\\bigcup_{A \\in \\mathcal{C}} A$ is distinct from $X$ and from all members of $\\mathcal{C}$. Notice here that the union of any subcollection $\\mathcal{D} \\subset \\mathcal{B}$ cannot coincide with any $K \\in \\mathcal{B} \\backslash \\mathcal{D}$, otherwise $\\{K\\} \\cup \\mathcal{D}$ would be tight. Surely, $\\mathcal{C}$ exists (since $\\{M, N\\}$ is an example of a collection satisfying the requirements on $\\mathcal{C}$, except for maximality); moreover, $C \\notin \\mathcal{B}$ by the above remark.\n\n\n\nSince $C \\neq X$, there exists an $L \\in \\mathcal{C}$ and $x \\in L$ such that $L$ is the unique set in $\\mathcal{C}$ containing $x$. Now replace in $\\mathcal{B}$ the set $L$ by $C$ in order to obtain a new collection $\\mathcal{B}^{\\prime}$ (then $\\left|\\mathcal{B}^{\\prime}\\right|=|\\mathcal{B}|$ ). We claim that $\\mathcal{B}^{\\prime}$ is good.\n\n\n\nAssume, to the contrary, that $\\mathcal{B}^{\\prime}$ contained a tight subcollection $\\mathcal{T}$; clearly, $C \\in \\mathcal{T}$, otherwise $\\mathcal{B}$ is not good. If $\\mathcal{T} \\subseteq \\mathcal{C} \\cup\\{C\\}$, then $C$ is the unique set in $\\mathcal{T}$ containing $x$ which is impossible. Therefore, there exists $P \\in \\mathcal{T} \\backslash(\\mathcal{C} \\cup\\{C\\})$. By maximality of $\\mathcal{C}$, the collection $\\mathcal{C} \\cup\\{P\\}$ does not satisfy the requirements imposed on $\\mathcal{C}$; since $P \\cup C \\neq X$, this may happen only if $C \\cup P=P$, i.e., if $C \\subset P$. But then $\\mathcal{G}=(\\mathcal{T} \\backslash\\{C\\}) \\cup \\mathcal{C}$ is a tight subcollection in $\\mathcal{B}$ : all elements of $C$ are covered by $\\mathcal{G}$ at least twice (by $P$ and an element of $\\mathcal{C}$ ), and all the rest elements are covered by $\\mathcal{G}$ the same number of times as by $\\mathcal{T}$. A contradiction. Thus $\\mathcal{B}^{\\prime}$ is good.\n\n\n\nSuch modifications may be performed finitely many times, since the total number of elements of sets in $\\mathcal{B}$ increases. Thus, at some moment we arrive at a good collection $\\mathcal{B}$ for which the procedure no longer applies. This means that for every $M, N \\in \\mathcal{B}$, either $M \\cup N=X$ or one of them is contained in the other.\n\n\n\nNow let $M$ be a minimal (with respect to inclusion) set in $\\mathcal{B}$. Then each set in $\\mathcal{B}$ either contains $M$ or forms $X$ in union with $M$ (i.e., contains $X \\backslash M$ ). Now one may easily see that the two collections\n\n\n\n$$\n\n\\mathcal{B}_{+}=\\{A \\backslash M: A \\in \\mathcal{B}, M \\subset A, A \\neq M\\}, \\quad \\mathcal{B}_{-}=\\{A \\cap M: A \\in \\mathcal{B}, X \\backslash M \\subset A, A \\neq X \\backslash M\\}\n\n$$\n\n\n\nare good as collections of subsets of $X \\backslash M$ and $M$, respectively; thus, by the induction hypothesis, we have $\\left|\\mathcal{B}_{+}\\right|+\\left|\\mathcal{B}_{-}\\right| \\leq 2 n-4$.\n\n\n\nFinally, each set $A \\in \\mathcal{B}$ either produces a set in one of the two new collections, or coincides with $M$ or $X \\backslash M$. Thus $|\\mathcal{B}| \\leq\\left|\\mathcal{B}_{+}\\right|+\\left|\\mathcal{B}_{-}\\right|+2 \\leq 2 n-2$, as required.",
"We provide yet another proof of the estimate $|\\mathcal{B}| \\leq 2 n-2$. Consider any collection $\\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection (we call such collection good). Arguing indirectly, we assume that there exists a good collection $\\mathcal{B}$ with $|\\mathcal{B}| \\geq 2 n-1$, and choose one such for the minimal possible value of $n$. Clearly, $n>2$.\n\n\n\nFirstly, we perform a different modification of $\\mathcal{B}$. Choose any $x \\in X$, and consider the subcollection $\\mathcal{B}_{x}=\\{B: B \\in \\mathcal{B}, x \\notin B\\}$. By our assumption, $\\mathcal{B}_{x}$ is not tight. As the union of sets in $\\mathcal{B}_{x}$ is distinct from $X$, either this collection is empty, or there exists an element $y \\in X$ contained in a unique member $A_{x}$ of $\\mathcal{B}_{x}$. In the former case, we add the set $B_{x}=X \\backslash\\{x\\}$ to $\\mathcal{B}$, and in the latter we replace $A_{x}$ by $B_{x}$, to form a new collection $\\mathcal{B}^{\\prime}$. (Notice that if $B_{x} \\in \\mathcal{B}$, then $B_{x} \\in \\mathcal{B}_{x}$ and $y \\in B_{x}$, so $B_{x}=A_{x}$.)\n\n\n\nWe claim that the collection $\\mathcal{B}^{\\prime}$ is also good. Indeed, if $\\mathcal{B}^{\\prime}$ has a tight subcollection $\\mathcal{T}$, then $B_{x}$ should lie in $\\mathcal{T}$. Then, as the union of the sets in $\\mathcal{T}$ is distinct from $X$, we should have $\\mathcal{T} \\subseteq \\mathcal{B}_{x} \\cup\\left\\{B_{x}\\right\\}$. But in this case an element $y$ is contained in a unique member of $\\mathcal{T}$, namely $B_{x}$, so $\\mathcal{T}$ is not tight - a contradiction.\n\n\n\nPerform this procedure for every $x \\in X$, to get a good collection $\\mathcal{B}$ containing the sets $B_{x}=X \\backslash\\{x\\}$ for all $x \\in X$. Consider now an element $x \\in X$ such that $\\left|\\mathcal{B}_{x}\\right|$ is maximal. As we have mentioned before, there exists an element $y \\in X$ belonging to a unique member (namely, $B_{x}$ ) of $\\mathcal{B}_{x}$. Thus, $\\mathcal{B}_{x} \\backslash\\left\\{B_{x}\\right\\} \\subset \\mathcal{B}_{y}$; also, $B_{y} \\in \\mathcal{B}_{y} \\backslash \\mathcal{B}_{x}$. Thus we get $\\left|\\mathcal{B}_{y}\\right| \\geq\\left|\\mathcal{B}_{x}\\right|$, which by the maximality assumption yields the equality, which in turn means that $\\mathcal{B}_{y}=\\left(\\mathcal{B}_{x} \\backslash\\left\\{B_{x}\\right\\}\\right) \\cup\\left\\{B_{y}\\right\\}$.\n\n\n\nTherefore, each set in $\\mathcal{B} \\backslash\\left\\{B_{x}, B_{y}\\right\\}$ contains either both $x$ and $y$, or none of them. Collapsing $\\{x, y\\}$ to singleton $x^{*}$, we get a new collection of $|\\mathcal{B}|-2$ subsets of $(X \\backslash\\{x, y\\}) \\cup\\left\\{x^{*}\\right\\}$ containing no tight subcollection. This contradicts minimality of $n$."
] | $2n-2$ |
1,618 | Number Theory | null | $\boxed{(1,8,19), (2,7,13), (4,5,7)}$ | [
"$(1,8,19), (2,7,13), (4,5,7)$"
] | true | null | Tuple | null | Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $x^{3}+y^{3}=$ $p(x y+p)$. | [
"Up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$. The verification is routine.\n\n\n\nSet $s=x+y$. Rewrite the equation in the form $s\\left(s^{2}-3 x y\\right)=p(p+x y)$, and express $x y$ :\n\n\n\n$$\n\nx y=\\frac{s^{3}-p^{2}}{3 s+p} \\tag{*}\n\n$$\n\n\n\nIn particular,\n\n\n\n$$\n\ns^{2} \\geq 4 x y=\\frac{4\\left(s^{3}-p^{2}\\right)}{3 s+p}\n\n$$\n\n\n\nor\n\n\n\n$$\n\n(s-2 p)\\left(s^{2}+s p+2 p^{2}\\right) \\leq p^{2}-p^{3}<0\n\n$$\n\n\n\nso $s<2 p$.\n\n\n\nIf $p \\mid s$, then $s=p$ and $x y=p(p-1) / 4$ which is impossible for $x+y=p$ (the equation $t^{2}-p t+p(p-1) / 4=0$ has no integer solutions).\n\n\n\nIf $p \\nmid s$, rewrite $(*)$ in the form\n\n\n\n$$\n\n27 x y=\\left(9 s^{2}-3 s p+p^{2}\\right)-\\frac{p^{2}(p+27)}{3 s+p} .\n\n$$\n\n\n\nSince $p \\nmid s$, this could be integer only if $3 s+p \\mid$ $p+27$, and hence $3 s+p \\mid 27-s$.\n\n\n\nIf $s \\neq 9$, then $|3 s-27| \\geq 3 s+p$, so $27-3 s \\geq$ $3 s+p$, or $27-p \\geq 6 s$, whence $s \\leq 4$. These cases are ruled out by hand.\n\n\n\nIf $s=x+y=9$, then $(*)$ yields $x y=27-p$. Up to a swap of $x$ and $y$, all such triples $(x, y, p)$ are $(1,8,19),(2,7,13)$, and $(4,5,7)$.",
"Set again $s=x+y$. It is readily checked that $s \\leq 8$ provides no solutions, so assume $s \\geq 9$. Notice that $x^{3}+y^{3}=s\\left(x^{2}-x y+y^{2}\\right) \\geq$ $\\frac{1}{4} s^{3}$ and $x y \\leq \\frac{1}{4} s^{2}$. The condition in the statement then implies $s^{2}(s-p) \\leq 4 p^{2}$, so $s<p+4$.\n\n\n\nNotice that $p$ divides one of $s$ and $x^{2}-x y+y^{2}$. The case $p \\mid s$ is easily ruled out by the condition $s<p+4$ : The latter forces $s=p$, so $x^{2}-x y+y^{2}=x y+p$, i. e., $(x-y)^{2}=p$, which is impossible.\n\n\n\nHence $p \\mid x^{2}-x y+y^{2}$, so $x^{2}-x y+y^{2}=k p$ and $x y+p=k s$ for some positive integer $k$, implying\n\n\n\n$$\n\ns^{2}+3 p=k(3 s+p) \\tag{**}\n\n$$\n\n\n\nRecall that $p \\nmid s$ to infer that $3 k \\equiv s(\\bmod p)$. We now present two approaches.\n\n\n\n1st Approach. Write $3 k=s+m p$ for some integer $m$ and plug $k=\\frac{1}{3}(s+m p)$ into $(* *)$ to get $s=(9-m p) /(3 m+1)$. The condition $s \\geq 9$ then forces $m=0$, so $s=9$, in which case, up to a swap of the first two entries, the solutions turn out to be $(x, y, p)=(1,8,19),(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$.\n\n\n\n2nd Approach. Notice that $k=\\frac{s^{2}+3 p}{3 s+p}=3+$ $\\frac{s(s-9)}{3 s+p} \\leq 3+\\frac{1}{3}(s-9)=\\frac{1}{3} s \\leq \\frac{1}{3}(p+3)$, since $s<p+4$. Hence $3 k \\leq p+3$, and the congruence $3 k \\equiv s$ $(\\bmod p)$ then forces either $3 k=s-p$ or $3 k=s$.\n\n\n\nThe case $3 k=s-p$ is easily ruled out: Otherwise, $(* *)$ boils down to $2 s+p+9=0$, which is clearly impossible.\n\n\n\nFinally, if $3 k=s$, then $(* *)$ reduces to $s=9$. In this case, up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$."
] | $(1,8,19), (2,7,13), (4,5,7)$ |
1,620 | Algebra | null | $\boxed{2n}$ | [
"$2n$"
] | false | null | Expression | null | Let $n \geqslant 2$ be an integer, and let $f$ be a $4 n$-variable polynomial with real coefficients. Assume that, for any $2 n$ points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{2 n}, y_{2 n}\right)$ in the plane, $f\left(x_{1}, y_{1}, \ldots, x_{2 n}, y_{2 n}\right)=0$ if and only if the points form the vertices of a regular $2 n$-gon in some order, or are all equal.
Determine the smallest possible degree of $f$. | [
"The smallest possible degree is $2 n$. In what follows, we will frequently write $A_{i}=$ $\\left(x_{i}, y_{i}\\right)$, and abbreviate $P\\left(x_{1}, y_{1}, \\ldots, x_{2 n}, y_{2 n}\\right)$ to $P\\left(A_{1}, \\ldots, A_{2 n}\\right)$ or as a function of any $2 n$ points.\n\n\n\nSuppose that $f$ is valid. First, we note a key property:\n\n\n\nClaim (Sign of $f$ ). $f$ attains wither only nonnegative values, or only nonpositive values.\n\n\n\nProof. This follows from the fact that the zero-set of $f$ is very sparse: if $f$ takes on a positive and a negative value, we can move $A_{1}, \\ldots, A_{2 n}$ from the negative value to the positive value without ever having them form a regular $2 n$-gon - a contradiction.\n\n\n\nThe strategy for showing $\\operatorname{deg} f \\geq 2 n$ is the following. We will animate the points $A_{1}, \\ldots, A_{2 n}$ linearly in a variable $t$; then $g(t)=f\\left(A_{1}, \\ldots, A_{2 n}\\right)$ will have degree at most $\\operatorname{deg} f$ (assuming it is not zero). The claim above then establishes that any root of $g$ must be a multiple root, so if we can show that there are at least $n$ roots, we will have shown $\\operatorname{deg} g \\geq 2 n$, and so $\\operatorname{deg} f \\geq 2 n$.\n\n\n\nGeometrically, our goal is to exhibit $2 n$ linearly moving points so that they form a regular $2 n$-gon a total of $n$ times, but not always form one.\n\n\n\nWe will do this as follows. Draw $n$ mirrors through the origin, as lines making angles of $\\frac{\\pi}{n}$ with each other. Then, any point $P$ has a total of $2 n$ reflections in the mirrors, as shown below for $n=5$. (Some of these reflections may overlap.)\n\n\n\nDraw the $n$ angle bisectors of adjacent mirrors. Observe that the reflections of $P$ form a regular $2 n$ gon if and only if $P$ lies on one of the bisectors.\n\n\n\nWe will animate $P$ on any line $\\ell$ which intersects all $n$ bisectors (but does not pass through the origin), and let $P_{1}, \\ldots, P_{2 n}$ be its reflections. Clearly, these are also all linearly animated, and because of the reasons above, they will form a regular $2 n$-gon exactly $n$ times, when $\\ell$ meets each bisector. So this establishes $\\operatorname{deg} f \\geq 2 n$ for the reasons described previously.\n\n\n\nNow we pass to constructing a polynomial $f$ of degree $2 n$ having the desired property. First of all, we will instead find a polynomial $g$ which has this property, but only when points with sum zero are input. This still solves the problem, because then we can choose\n\n\n\n$$\nf\\left(A_{1}, A_{2}, \\ldots, A_{2 n}\\right)=g\\left(A_{1}-\\bar{A}, \\ldots, A_{2 n}-\\bar{A}\\right)\n$$\n\n\n\nwhere $\\bar{A}$ is the centroid of $A_{1}, \\ldots, A_{2 n}$. This has the upshot that we can now always assume $A_{1}+\\cdots+A_{2 n}=0$, which will simplify the ensuing discussion.\n\n\n\n<img_3624>\n\n\n\nWe will now construct a suitable $g$ as a sum of squares. This means that, if we write $g=g_{1}^{2}+g_{2}^{2}+$ $\\cdots+g_{m}^{2}$, then $g=0$ if and only if $g_{1}=\\cdots=g_{m}=0$, and that if their degrees are $d_{1}, \\ldots, d_{m}$, then $g$ has degree at most $2 \\max \\left(d_{1}, \\ldots, d_{m}\\right)$.\n\n\n\nThus, it is sufficient to exhibit several polynomials, all of degree at most $n$, such that $2 n$ points with zero sum are the vertices of a regular $2 n$-gon if and only if the polynomials are all zero at those points.\n\n\n\n\n\n\n\nFirst, we will impose the constraints that all $\\left|A_{i}\\right|^{2}=x_{i}^{2}+y_{i}^{2}$ are equal. This uses multiple degree 2 constraints.\n\n\n\nNow, we may assume that the points $A_{1}, \\ldots, A_{2 n}$ all lie on a circle with centre 0 , and $A_{1}+\\cdots+A_{2 n}=0$. If this circle has radius 0 , then all $A_{i}$ coincide, and we may ignore this case.\n\n\n\nOtherwise, the circle has positive radius. We will use the following lemma.\n\n\n\nLemma. Suppose that $a_{1}, \\ldots, a_{2 n}$ are complex numbers of the same non-zero magnitude, and suppose that $a_{1}^{k}+\\cdots+a_{2 n}^{k}=0, k=1, \\ldots, n$. Then $a_{1}, \\ldots, a_{2 n}$ form a regular $2 n$-gon centred at the origin. (Conversely, this is easily seen to be sufficient.)\n\n\n\nProof. Since all the hypotheses are homogenous, we may assume (mostly for convenience) that $a_{1}, \\ldots, a_{2 n}$ lie on the unit circle. By Newton's sums, the $k$-th symmetric sums of $a_{1}, \\ldots, a_{2 n}$ are all zero for $k$ in the range $1, \\ldots, n$.\n\n\n\nTaking conjugates yields $a_{1}^{-k}+\\cdots+a_{2 n}^{-k}=0$, $k=1, \\ldots, n$. Thus, we can repeat the above logic to obtain that the $k$-th symmetric sums of $a_{1}^{-1}, \\ldots, a_{2 n}^{-1}$ are also all zero for $k=1, \\ldots, n$. However, these are simply the $(2 n-k)$-th symmetric sums of $a_{1}, \\ldots, a_{2 n}$ (divided by $a_{1} \\cdots a_{2 n}$ ), so the first $2 n-1$ symmetric sums of $a_{1}, \\ldots, a_{2 n}$ are all zero. This implies that $a_{1}, \\ldots, a_{2 n}$ form a regular $2 n$-gon centred at the origin.\n\n\n\nWe will encode all of these constraints into our polynomial. More explicitly, write $a_{r}=x_{r}+y_{r} i$; then the constraint $a_{1}^{k}+\\cdots+a_{2 n}^{k}=0$ can be expressed as $p_{k}+q_{k} i=0$, where $p_{k}$ and $q_{k}$ are real polynomials in the coordinates. To incorporate this, simply impose the constraints $p_{k}=0$ and $q_{k}=0$; these are conditions of degree $k \\leq n$, so their squares are all of degree at most $2 n$.\n\n\n\nTo recap, taking the sum of squares of all of these constraints gives a polynomial $f$ of degree at most $2 n$ which works whenever $A_{1}+\\cdots+A_{2 n}=0$. Finally, the centroid-shifting trick gives a polynomial which works in general, as wanted."
] | $2n$ |
1,631 | Algebra | null | $\boxed{2}$ | [
"2"
] | false | null | Numerical | null | For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime. | [
"The largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \\mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \\ldots, y_{k}$ are primes for some $k \\geq 1$ then $a=x_{1}$ is also prime.\n\n\n\nNow we claim that for every odd prime $a$ at least one of the numbers $y_{1}, y_{2}, y_{3}$ is composite (and thus $k<3$ ). Assume, to the contrary, that $y_{1}, y_{2}$, and $y_{3}$ are primes; then $x_{1}, x_{2}, x_{3}$ are primes as well. Since $x_{1} \\geq 3$ is odd, we have $x_{2}>3$ and $x_{2} \\equiv 3(\\bmod 4)$; consequently, $x_{3} \\equiv 7$ $(\\bmod 8)$. This implies that 2 is a quadratic residue modulo $p=x_{3}$, so $2 \\equiv s^{2}(\\bmod p)$ for some integer $s$, and hence $2^{x_{2}}=2^{(p-1) / 2} \\equiv s^{p-1} \\equiv 1(\\bmod p)$. This means that $p \\mid y_{2}$, thus $2^{x_{2}}-1=x_{3}=2 x_{2}+1$. But it is easy to show that $2^{t}-1>2 t+1$ for all integer $t>3$. A contradiction.\n\n\n\nFinally, if $a=2$, then the numbers $y_{1}=3$ and $y_{2}=31$ are primes, while $y_{3}=2^{11}-1$ is divisible by 23 ; in this case we may choose $k=2$ but not $k=3$."
] | 2 |
1,641 | Combinatorics | null | $\boxed{\binom{2n}{n}}$ | [
"$\\binom{2n}{n}$"
] | false | null | Expression | null | Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow{C D}$ such that $A B C D$ is a convex quadrangle oriented clockwise. Determine the number of good configurations. | [
"The required number is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. To prove this, trace the circumference counterclockwise to label the points $a_{1}, a_{2}, \\ldots, a_{2 n}$.\n\nLet $\\mathcal{C}$ be any good configuration and let $O(\\mathcal{C})$ be the set of all points from which arrows emerge. We claim that every $n$-element subset $S$ of $\\left\\{a_{1}, \\ldots, a_{2 n}\\right\\}$ is an $O$-image of a unique good configuration; clearly, this provides the answer.\n\nTo prove the claim induct on $n$. The base case $n=1$ is clear. For the induction step, consider any $n$-element subset $S$ of $\\left\\{a_{1}, \\ldots, a_{2 n}\\right\\}$, and assume that $S=O(\\mathcal{C})$ for some good configuration $\\mathcal{C}$. Take any index $k$ such that $a_{k} \\in S$ and $a_{k+1} \\notin S$ (assume throughout that indices are cyclic modulo $2 n$, i.e., $a_{2 n+1}=a_{1}$ etc.).\n\nIf the arrow from $a_{k}$ points to some $a_{\\ell}, k+1<\\ell(<2 n+k)$, then the arrow pointing to $a_{k+1}$ emerges from some $a_{m}, m$ in the range $k+2$ through $\\ell-1$, since these two arrows do not cross. Then the arrows $a_{k} \\rightarrow a_{\\ell}$ and $a_{m} \\rightarrow a_{k+1}$ form a prohibited quadrangle. Hence, $\\mathcal{C}$ contains an arrow $a_{k} \\rightarrow a_{k+1}$.\n\nOn the other hand, if any configuration $\\mathcal{C}$ contains the arrow $a_{k} \\rightarrow a_{k+1}$, then this arrow cannot cross other arrows, neither can it occur in prohibited quadrangles.\n\nThus, removing the points $a_{k}, a_{k+1}$ from $\\left\\{a_{1}, \\ldots, a_{2 n}\\right\\}$ and the point $a_{k}$ from $S$, we may apply the induction hypothesis to find a unique good configuration $\\mathcal{C}^{\\prime}$ on $2 n-2$ points compatible with the new set of sources (i.e., points from which arrows emerge). Adjunction of the arrow $a_{k} \\rightarrow a_{k+1}$ to $\\mathcal{C}^{\\prime}$ yields a unique good configuration on $2 n$ points, as required.",
"Use the counterclockwise labelling $a_{1}, a_{2}, \\ldots, a_{2 n}$ in the solution above.\n\nLetting $D_{n}$ be the number of good configurations on $2 n$ points, we establish a recurrence relation for the $D_{n}$. To this end, let $C_{n}=\\frac{(2 n) !}{n !(n+1) !}$ the $n$th Catalan number; it is well-known that $C_{n}$ is the number of ways to connect $2 n$ given points on the circumference by $n$ pairwise disjoint chords.\n\nSince no two arrows cross, in any good configuration the vertex $a_{1}$ is connected to some $a_{2 k}$. Fix $k$ in the range 1 through $n$ and count the number of good configurations containing the arrow $a_{1} \\rightarrow a_{2 k}$. Let $\\mathcal{C}$ be any such configuration.\n\nIn $\\mathcal{C}$, the vertices $a_{2}, \\ldots, a_{2 k-1}$ are paired off with one other, each arrow pointing from the smaller to the larger index, for otherwise it would form a prohibited quadrangle with $a_{1} \\rightarrow a_{2 k}$. Consequently, there are $C_{k-1}$ ways of drawing such arrows between $a_{2}, \\ldots, a_{2 k-1}$.\n\nOn the other hand, the arrows between $a_{2 k+1}, \\ldots, a_{2 n}$ also form a good configuration, which can be chosen in $D_{n-k}$ ways. Finally, it is easily seen that any configuration of the first kind and any configuration of the second kind combine together to yield an overall good configuration.\n\nThus the number of good configurations containing the arrow $a_{1} \\rightarrow a_{2 k}$ is $C_{k-1} D_{n-k}$. Clearly, this is also the number of good configurations containing the arrow $a_{2(n-k+1)} \\rightarrow a_{1}$, so\n\n$$\nD_{n}=2 \\sum_{k=1}^{n} C_{k-1} D_{n-k} \\tag{*}\n$$\n\nTo find an explicit formula for $D_{n}$, let $d(x)=\\sum_{n=0}^{\\infty} D_{n} x^{n}$ and let $c(x)=\\sum_{n=0}^{\\infty} C_{n} x^{n}=$ $\\frac{1-\\sqrt{1-4 x}}{2 x}$ be the generating functions of the $D_{n}$ and the $C_{n}$, respectively. Since $D_{0}=1$, relation $(*)$\n\n\n\nyields $d(x)=2 x c(x) d(x)+1$, so\n\n$$\n\\begin{aligned}\nd(x)=\\frac{1}{1-2 x c(x)}=(1-4 x)^{-1 / 2} & =\\sum_{n \\geq 0}\\left(-\\frac{1}{2}\\right)\\left(-\\frac{3}{2}\\right) \\ldots\\left(-\\frac{2 n-1}{2}\\right) \\frac{(-4 x)^{n}}{n !} \\\\\n& =\\sum_{n \\geq 0} \\frac{2^{n}(2 n-1) ! !}{n !} x^{n}=\\sum_{n \\geq 0}\\left(\\begin{array}{c}\n2 n \\\\\nn\n\\end{array}\\right) x^{n} .\n\\end{aligned}\n$$\n\nConsequently, $D_{n}=\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$.\n\n### solution_2\nLet $C_{n}=\\frac{1}{n+1}\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ denote the $n$th Catalan number and recall that there are exactly $C_{n}$ ways to join $2 n$ distinct points on a circumference by $n$ pairwise disjoint chords. Such a configuration of chords will be referred to as a Catalan n-configuration. An orientation of the chords in a Catalan configuration $\\mathcal{C}$ making it into a good configuration (in the sense defined in the statement of the problem) will be referred to as a good orientation for $\\mathcal{C}$.\n\nWe show by induction on $n$ that there are exactly $n+1$ good orientations for any Catalan $n$-configuration, so there are exactly $(n+1) C_{n}=\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ good configurations on $2 n$ points. The base case $n=1$ is clear.\n\nFor the induction step, let $n>1$, let $\\mathcal{C}$ be a Catalan $n$-configuration, and let $a b$ be a chord of minimal length in $\\mathcal{C}$. By minimality, the endpoints of the other chords in $\\mathcal{C}$ all lie on the major arc $a b$ of the circumference.\n\nLabel the $2 n$ endpoints $1,2, \\ldots, 2 n$ counterclockwise so that $\\{a, b\\}=\\{1,2\\}$, and notice that the good orientations for $\\mathcal{C}$ fall into two disjoint classes: Those containing the arrow $1 \\rightarrow 2$, and those containing the opposite arrow.\n\nSince the arrow $1 \\rightarrow 2$ cannot be involved in a prohibited quadrangle, the induction hypothesis applies to the Catalan $(n-1)$-configuration formed by the other chords to show that the first class contains exactly $n$ good orientations.\n\nFinally, the second class consists of a single orientation, namely, $2 \\rightarrow 1$, every other arrow emerging from the smaller endpoint of the respective chord; a routine verification shows that this is indeed a good orientation. This completes the induction step and ends the proof.\n\n### solution_3\nWe intend to count the number of good orientations of a Catalan $n$-configuration.\n\nFor each such configuration, we consider its dual graph $T$ whose vertices are finite regions bounded by chords and the circle, and an edge connects two regions sharing a boundary segment. This graph $T$ is a plane tree with $n$ edges and $n+1$ vertices.\n\nThere is a canonical bijection between orientations of chords and orientations of edges of $T$ in such a way that each chord crosses an edge of $T$ from the right to the left of the arrow on that edge. A good orientation of chords corresponds to an orientation of the tree containing no two edges oriented towards each other. Such an orientation is defined uniquely by its source vertex, i.e., the unique vertex having no in-arrows.\n\nTherefore, for each tree $T$ on $n+1$ vertices, there are exactly $n+1$ ways to orient it so that the source vertex is unique - one for each choice of the source. Thus, the answer is obtained in the same way as above."
] | $\binom{2n}{n}$ |
1,644 | Combinatorics | null | $\boxed{m n-\lfloor m / 2\rfloor}$ | [
"$m n-\\lfloor m / 2\\rfloor$"
] | false | null | Expression | null | Given positive integers $m$ and $n \geq m$, determine the largest number of dominoes $(1 \times 2$ or $2 \times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2 n$ columns consisting of cells $(1 \times 1$ squares $)$ so that:
(i) each domino covers exactly two adjacent cells of the board;
(ii) no two dominoes overlap;
(iii) no two form a $2 \times 2$ square; and
(iv) the bottom row of the board is completely covered by $n$ dominoes. | [
"The required maximum is $m n-\\lfloor m / 2\\rfloor$ and is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed on alternate rows, so that the bottom row of the board is completely covered by $n$ dominoes.\n\n\n\nTo show that the number of dominoes in an arrangement satisfying the conditions in the statement does not exceed $m n-\\lfloor m / 2\\rfloor$, label the rows upwards $0,1, \\ldots, m-1$, and, for each $i$ in this range, draw a vertically symmetric block of $n-i$ fictitious horizontal dominoes in the $i$-th row (so the block on the $i$-th row leaves out $i$ cells on either side) - Figure 4 illustrates the case $m=n=6$. A fictitious domino is good if it is completely covered by a domino in the arrangement; otherwise, it is bad.\n\n\n\nIf the fictitious dominoes are all good, then the dominoes in the arrangement that cover no fictitious domino, if any, all lie in two triangular regions of side-length $m-1$ at the upper-left and upper-right corners of the board. Colour the cells of the board chess-like and notice that in each of the two triangular regions the number of black cells and the number of white cells differ by $\\lfloor m / 2\\rfloor$. Since each domino covers two cells of different colours, at least $\\lfloor m / 2\\rfloor$ cells are not covered in each of these regions, and the conclusion follows.\n\n\n\n<img_3888>\n\n\n\nFig. 4\n\n<img_3590>\n\n\n\nFig. 5\n\n\n\nTo deal with the remaining case where bad fictitious dominoes are present, we show that an arrangement satisfying the conditions in the statement can be transformed into another such with at least as many dominoes, but fewer bad fictitious dominoes. A finite number of such transformations eventually leads to an arrangement of at least as many dominoes all of whose fictitious dominoes are good, and the conclusion follows by the preceding.\n\n\n\nConsider the row of minimal rank containing bad fictitious dominoes - this is certainly not the bottom row - and let $D$ be one such. Let $\\ell$, respectively $r$, be the left, respectively right, cell of $D$ and notice that the cell below $\\ell$, respectively $r$, is the right, respectively left, cell of a domino $D_{1}$, respectively $D_{2}$, in the arrangement.\n\n\n\nIf $\\ell$ is covered by a domino $D_{\\ell}$ in the arrangement, since $D$ is bad and no two dominoes in the arrangement form a square, it follows that $D_{\\ell}$ is vertical. If $r$ were also covered by a domino $D_{r}$ in the arrangement, then $D_{r}$ would also be vertical, and would therefore form a square with $D_{\\ell}-$ a contradiction. Hence $r$ is not covered, and there is room for $D_{\\ell}$ to be placed so as to cover $D$, to obtain a new arrangement satisfying the conditions in the statement; the latter has as many dominoes as the former, but fewer bad fictitious dominoes. The case where $r$ is covered is dealt with similarly.\n\n\n\nFinally, if neither cell of $D$ is covered, addition of an extra domino to cover $D$ and, if necessary, removal of the domino above $D$ to avoid formation of a square, yields a new arrangement satisfying the conditions in the statement; the latter has at least as many dominoes as the former, but fewer bad fictitious dominoes. (Figure 5 illustrates the two cases.)",
"We present an alternative proof of the bound.\n\n\n\nLabel the rows upwards $0,1, \\ldots, m-1$, and the columns from the left to the right by $0,1, \\ldots, 2 n-1$; label each cell by the pair of its column's and row's numbers, so that $(1,0)$ is the second left cell in the bottom row. Colour the cells chess-like so that $(0,0)$ is white. For $0 \\leq i \\leq n-1$, we say that the $i$ th white diagonal is the set of cells of the form $(2 i+k, k)$, where $k$ ranges over all appropriate indices. Similarly, the ith black diagonal is the set of cells of the form $(2 i+1-k, k)$. (Notice that the white cells in the upper-left corner and the black cells in the upper-right corner are not covered by these diagonals.)\n\n\n\nClaim. Assume that $K$ lowest cells of some white diagonal are all covered by dominoes. Then all these $K$ dominoes face right or up from the diagonal. (In other words, the black cell of any such\n\n\n\n\n\n\n\ndomino is to the right or to the top of its white cell.) Similarly, if $K$ lowest cells of some black diagonal are covered by dominoes, then all these dominoes face left or up from the diagonal.\n\n\n\nProof. By symmetry, it suffices to prove the first statement. Assume that $K$ lowest cells of the $i$ th white diagonal is completely covered. We prove by induction on $k<K$ that the required claim holds for the domino covering $(2 i+k, k)$. The base case $k=0$ holds due to the problem condition. To establish the step, one observes that if $(2 i+k, k)$ is covered by a domino facing up of right, while $(2 i+k+1, k+1)$ is covered by a domino facing down or left, then these two dominoes form a square.\n\n\n\nWe turn to the solution. We will prove that there are at least $d=\\lfloor m / 2\\rfloor$ empty white cells. Since each domino covers exactly one white cell, the required bound follows.\n\n\n\nIf each of the first $d$ white diagonals contains an empty cell, the result is clear. Otherwise, let $i<d$ be the least index of a completely covered white diagonal. We say that the dominoes covering our diagonal are distinguished. After removing the distinguished dominoes, the board splits into two parts; the left part $L$ contains $i$ empty white cells on the previous diagonals. So, it suffices to prove that the right part $R$ contains at least $d-i$ empty white cells.\n\n\n\nLet $j$ be the number of distinguished dominoes facing up. Then at least $j-i$ of these dominoes cover some cells of (distinct) black diagonals (the relation $m \\leq n$ is used). Each such domino faces down from the corresponding black diagonal - so, by the Claim, each such black diagonal contains an empty cell in $R$. Thus, $R$ contains at least $j-i$ empty black cells.\n\n\n\nNow, let $w$ be the number of white cells in $R$. Then the number of black cells in $R$ is $w-d+j$, and at least $i-j$ of those are empty. Thus, the number of dominoes in $R$ is at most $(w-d+j)-(j-i)=w-(d-i)$, so $R$ contains at least $d-i$ empty white cells, as we wanted to show."
] | $m n-\lfloor m / 2\rfloor$ |
1,645 | Algebra | null | $\boxed{0}$ | [
"0"
] | false | null | Numerical | null | A cubic sequence is a sequence of integers given by $a_{n}=n^{3}+b n^{2}+c n+d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.
Determine the possible values of $a_{2015} \cdot a_{2016}$ for a cubic sequence satisfying the condition in part (a). | [
"The only possible value of $a_{2015} \\cdot a_{2016}$ is 0 . For simplicity, by performing a translation of the sequence (which may change the defining constants $b, c$ and $d$ ), we may instead concern ourselves with the values $a_{0}$ and $a_{1}$, rather than $a_{2015}$ and $a_{2016}$.\n\n\n\nSuppose now that we have a cubic sequence $a_{n}$ with $a_{0}=p^{2}$ and $a_{1}=q^{2}$ square numbers. We will show that $p=0$ or $q=0$. Consider the line $y=(q-p) x+p$ passing through $(0, p)$ and $(1, q)$; the latter are two points the line under consideration and the cubic $y^{2}=x^{3}+b x^{2}+c x+d$ share. Hence the two must share a third point whose $x$-coordinate is the third root of the polynomial $t^{3}+\\left(b-(q-p)^{2}\\right) t^{2}+(c-2(q-p) p) t+\\left(d-p^{2}\\right)$ (it may well happen that this third point coincide with one of the other two points the line and the cubic share).\n\n\n\nNotice that the sum of the three roots is $(q-p)^{2}-b$, so the third intersection has integral $x$-coordinate $X=(q-p)^{2}-b-1$. Its $y$-coordinate $Y=(q-p) X+p$ is also an integer, and hence $a_{X}=X^{3}+b X^{2}+c X+d=Y^{2}$ is a square. This contradicts our assumption on the sequence unless $X=0$ or $X=1$, i.e. unless $(q-p)^{2}=b+1$ or $(q-p)^{2}=b+2$.\n\n\n\n\n\n\n\nApplying the same argument to the line through $(0,-p)$ and $(1, q)$, we find that $(q+p)^{2}=b+1$ or $b+2$ also. Since $(q-p)^{2}$ and $(q+p)^{2}$ have the same parity, they must be equal, and hence $p q=0$, as desired.\n\n\n\nIt remains to show that such sequences exist, say when $p=0$. Consider the sequence $a_{n}=$ $n^{3}+\\left(q^{2}-2\\right) n^{2}+n$, chosen to satisfy $a_{0}=0$ and $a_{1}=q^{2}$. We will show that when $q=1$, the only square terms of the sequence are $a_{0}=0$ and $a_{1}=1$. Indeed, suppose that $a_{n}=n\\left(n^{2}-n+1\\right)$ is square. Since the second factor is positive, and the two factors are coprime, both must be squares; in particular, $n \\geq 0$. The case $n=0$ is clear, so let $n \\geq 1$. Finally, if $n>1$, then $(n-1)^{2}<n^{2}-n+1<n^{2}$, so $n^{2}-n+1$ is not a square. Consequently, $n=0$ or $n=1$, and the conclusion follows."
] | 0 |
1,659 | Algebra | null | $\boxed{f(x)=2 x}$ | [
"$f(x)=2 x$"
] | false | null | Expression | null | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$
f(x+f(y))=f(x+y)+f(y)\tag{1}
$$
for all $x, y \in \mathbb{R}^{+}$. (Symbol $\mathbb{R}^{+}$denotes the set of all positive real numbers.) | [
"First we show that $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y) \\neq y$ immediately. If $f(y)<y$ for some $y$, then setting $x=y-f(y)$ we get\n\n$$\nf(y)=f((y-f(y))+f(y))=f((y-f(y))+y)+f(y)>f(y),\n$$\n\ncontradiction. Therefore $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$.\n\nFor $x \\in \\mathbb{R}^{+}$define $g(x)=f(x)-x$; then $f(x)=g(x)+x$ and, as we have seen, $g(x)>0$. Transforming (1) for function $g(x)$ and setting $t=x+y$,\n\n$$\n\\begin{aligned}\nf(t+g(y)) & =f(t)+f(y) \\\\\ng(t+g(y))+t+g(y) & =(g(t)+t)+(g(y)+y)\n\\end{aligned}\n$$\n\nand therefore\n\n$$\ng(t+g(y))=g(t)+y \\quad \\text { for all } t>y>0 \\tag{2}\n$$\n\nNext we prove that function $g(x)$ is injective. Suppose that $g\\left(y_{1}\\right)=g\\left(y_{2}\\right)$ for some numbers $y_{1}, y_{2} \\in \\mathbb{R}^{+}$. Then by $(2)$,\n\n$$\ng(t)+y_{1}=g\\left(t+g\\left(y_{1}\\right)\\right)=g\\left(t+g\\left(y_{2}\\right)\\right)=g(t)+y_{2}\n$$\n\nfor all $t>\\max \\left\\{y_{1}, y_{2}\\right\\}$. Hence, $g\\left(y_{1}\\right)=g\\left(y_{2}\\right)$ is possible only if $y_{1}=y_{2}$.\n\nNow let $u, v$ be arbitrary positive numbers and $t>u+v$. Applying (2) three times,\n\n$$\ng(t+g(u)+g(v))=g(t+g(u))+v=g(t)+u+v=g(t+g(u+v)) \\text {. }\n$$\n\nBy the injective property we conclude that $t+g(u)+g(v)=t+g(u+v)$, hence\n\n$$\ng(u)+g(v)=g(u+v)\\tag{3}\n$$\n\nSince function $g(v)$ is positive, equation (3) also shows that $g$ is an increasing function.\n\nFinally we prove that $g(x)=x$. Combining (2) and (3), we obtain\n\n$$\ng(t)+y=g(t+g(y))=g(t)+g(g(y))\n$$\n\nand hence\n\n$$\ng(g(y))=y\n$$\n\nSuppose that there exists an $x \\in \\mathbb{R}^{+}$such that $g(x) \\neq x$. By the monotonicity of $g$, if $x>g(x)$ then $g(x)>g(g(x))=x$. Similarly, if $x<g(x)$ then $g(x)<g(g(x))=x$. Both cases lead to contradiction, so there exists no such $x$.\n\nWe have proved that $g(x)=x$ and therefore $f(x)=g(x)+x=2 x$ for all $x \\in \\mathbb{R}^{+}$. This function indeed satisfies the functional equation (1).",
"We prove that $f(y)>y$ and introduce function $g(x)=f(x)-x>0$ in the same way as in Solution 1.\n\nFor arbitrary $t>y>0$, substitute $x=t-y$ into (1) to obtain\n\n$$\nf(t+g(y))=f(t)+f(y)\n$$\n\nwhich, by induction, implies\n\n$$\nf(t+n g(y))=f(t)+n f(y) \\quad \\text { for all } t>y>0, n \\in \\mathbb{N} \\tag{4}\n$$\n\nTake two arbitrary positive reals $y$ and $z$ and a third fixed number $t>\\max \\{y, z\\}$. For each positive integer $k$, let $\\ell_{k}=\\left\\lfloor k \\frac{g(y)}{g(z)}\\right\\rfloor$. Then $t+k g(y)-\\ell_{k} g(z) \\geq t>z$ and, applying (4) twice,\n\n$$\n\\begin{gathered}\nf\\left(t+k g(y)-\\ell_{k} g(z)\\right)+\\ell_{k} f(z)=f(t+k g(y))=f(t)+k f(y), \\\\\n0<\\frac{1}{k} f\\left(t+k g(y)-\\ell_{k} g(z)\\right)=\\frac{f(t)}{k}+f(y)-\\frac{\\ell_{k}}{k} f(z) .\n\\end{gathered}\n$$\n\nAs $k \\rightarrow \\infty$ we get\n\n$$\n0 \\leq \\lim _{k \\rightarrow \\infty}\\left(\\frac{f(t)}{k}+f(y)-\\frac{\\ell_{k}}{k} f(z)\\right)=f(y)-\\frac{g(y)}{g(z)} f(z)=f(y)-\\frac{f(y)-y}{f(z)-z} f(z)\n$$\n\nand therefore\n\n$$\n\\frac{f(y)}{y} \\leq \\frac{f(z)}{z}\n$$\n\nExchanging variables $y$ and $z$, we obtain the reverse inequality. Hence, $\\frac{f(y)}{y}=\\frac{f(z)}{z}$ for arbitrary $y$ and $z$; so function $\\frac{f(x)}{x}$ is constant, $f(x)=c x$.\n\nSubstituting back into (1), we find that $f(x)=c x$ is a solution if and only if $c=2$. So the only solution for the problem is $f(x)=2 x$."
] | $f(x)=2 x$ |
1,662 | Combinatorics | null | $\boxed{3 n}$ | [
"$3 n$"
] | false | null | Expression | null | Let $n>1$ be an integer. In the space, consider the set
$$
S=\{(x, y, z) \mid x, y, z \in\{0,1, \ldots, n\}, x+y+z>0\}
$$
Find the smallest number of planes that jointly contain all $(n+1)^{3}-1$ points of $S$ but none of them passes through the origin. | [
"It is easy to find $3 n$ such planes. For example, planes $x=i, y=i$ or $z=i$ $(i=1,2, \\ldots, n)$ cover the set $S$ but none of them contains the origin. Another such collection consists of all planes $x+y+z=k$ for $k=1,2, \\ldots, 3 n$.\n\nWe show that $3 n$ is the smallest possible number.\n\nLemma 1. Consider a nonzero polynomial $P\\left(x_{1}, \\ldots, x_{k}\\right)$ in $k$ variables. Suppose that $P$ vanishes at all points $\\left(x_{1}, \\ldots, x_{k}\\right)$ such that $x_{1}, \\ldots, x_{k} \\in\\{0,1, \\ldots, n\\}$ and $x_{1}+\\cdots+x_{k}>0$, while $P(0,0, \\ldots, 0) \\neq 0$. Then $\\operatorname{deg} P \\geq k n$.\n\nProof. We use induction on $k$. The base case $k=0$ is clear since $P \\neq 0$. Denote for clarity $y=x_{k}$.\n\nLet $R\\left(x_{1}, \\ldots, x_{k-1}, y\\right)$ be the residue of $P$ modulo $Q(y)=y(y-1) \\ldots(y-n)$. Polynomial $Q(y)$ vanishes at each $y=0,1, \\ldots, n$, hence $P\\left(x_{1}, \\ldots, x_{k-1}, y\\right)=R\\left(x_{1}, \\ldots, x_{k-1}, y\\right)$ for all $x_{1}, \\ldots, x_{k-1}, y \\in\\{0,1, \\ldots, n\\}$. Therefore, $R$ also satisfies the condition of the Lemma; moreover, $\\operatorname{deg}_{y} R \\leq n$. Clearly, $\\operatorname{deg} R \\leq \\operatorname{deg} P$, so it suffices to prove that $\\operatorname{deg} R \\geq n k$.\n\nNow, expand polynomial $R$ in the powers of $y$ :\n$$\nR\\left(x_{1}, \\ldots, x_{k-1}, y\\right)=R_{n}\\left(x_{1}, \\ldots, x_{k-1}\\right) y^{n}+R_{n-1}\\left(x_{1}, \\ldots, x_{k-1}\\right) y^{n-1}+\\cdots+R_{0}\\left(x_{1}, \\ldots, x_{k-1}\\right)\n$$\nWe show that polynomial $R_{n}\\left(x_{1}, \\ldots, x_{k-1}\\right)$ satisfies the condition of the induction hypothesis.\n\nConsider the polynomial $T(y)=R(0, \\ldots, 0, y)$ of degree $\\leq n$. This polynomial has $n$ roots $y=1, \\ldots, n$; on the other hand, $T(y) \\not \\equiv 0$ since $T(0) \\neq 0$. Hence $\\operatorname{deg} T=n$, and its leading coefficient is $R_{n}(0,0, \\ldots, 0) \\neq 0$. In particular, in the case $k=1$ we obtain that coefficient $R_{n}$ is nonzero.\n\nSimilarly, take any numbers $a_{1}, \\ldots, a_{k-1} \\in\\{0,1, \\ldots, n\\}$ with $a_{1}+\\cdots+a_{k-1}>0$. Substituting $x_{i}=a_{i}$ into $R\\left(x_{1}, \\ldots, x_{k-1}, y\\right)$, we get a polynomial in $y$ which vanishes at all points $y=0, \\ldots, n$ and has degree $\\leq n$. Therefore, this polynomial is null, hence $R_{i}\\left(a_{1}, \\ldots, a_{k-1}\\right)=0$ for all $i=0,1, \\ldots, n$. In particular, $R_{n}\\left(a_{1}, \\ldots, a_{k-1}\\right)=0$.\n\nThus, the polynomial $R_{n}\\left(x_{1}, \\ldots, x_{k-1}\\right)$ satisfies the condition of the induction hypothesis. So, we have $\\operatorname{deg} R_{n} \\geq(k-1) n$ and $\\operatorname{deg} P \\geq \\operatorname{deg} R \\geq \\operatorname{deg} R_{n}+n \\geq k n$.\n\nNow we can finish the solution. Suppose that there are $N$ planes covering all the points of $S$ but not containing the origin. Let their equations be $a_{i} x+b_{i} y+c_{i} z+d_{i}=0$. Consider the polynomial\n$$\nP(x, y, z)=\\prod_{i=1}^{N}\\left(a_{i} x+b_{i} y+c_{i} z+d_{i}\\right)\n$$\nIt has total degree $N$. This polynomial has the property that $P\\left(x_{0}, y_{0}, z_{0}\\right)=0$ for any $\\left(x_{0}, y_{0}, z_{0}\\right) \\in S$, while $P(0,0,0) \\neq 0$. Hence by Lemma 1 we get $N=\\operatorname{deg} P \\geq 3 n$, as desired.",
"We present a different proof of the main Lemma 1. Here we confine ourselves to the case $k=3$, which is applied in the solution, and denote the variables by $x, y$ and $z$. (The same proof works for the general statement as well.)\n\nThe following fact is known with various proofs; we provide one possible proof for the completeness.\n\nLemma 2. For arbitrary integers $0 \\leq m<n$ and for an arbitrary polynomial $P(x)$ of degree $m$,\n$$\n\\sum_{k=0}^{n}(-1)^{k}\\left(\\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right) P(k)=0\\tag{1}\n$$\nProof. We use an induction on $n$. If $n=1$, then $P(x)$ is a constant polynomial, hence $P(1)-P(0)=0$, and the base is proved.\n\nFor the induction step, define $P_{1}(x)=P(x+1)-P(x)$. Then clearly $\\operatorname{deg} P_{1}=\\operatorname{deg} P-1=$ $m-1<n-1$, hence by the induction hypothesis we get\n$$\n\\begin{aligned}\n0 & =-\\sum_{k=0}^{n-1}(-1)^{k}\\left(\\begin{array}{c}\nn-1 \\\\\nk\n\\end{array}\\right) P_{1}(k)=\\sum_{k=0}^{n-1}(-1)^{k}\\left(\\begin{array}{c}\nn-1 \\\\\nk\n\\end{array}\\right)(P(k)-P(k+1)) \\\\\n& =\\sum_{k=0}^{n-1}(-1)^{k}\\left(\\begin{array}{c}\nn-1 \\\\\nk\n\\end{array}\\right) P(k)-\\sum_{k=0}^{n-1}(-1)^{k}\\left(\\begin{array}{c}\nn-1 \\\\\nk\n\\end{array}\\right) P(k+1) \\\\\n& =\\sum_{k=0}^{n-1}(-1)^{k}\\left(\\begin{array}{c}\nn-1 \\\\\nk\n\\end{array}\\right) P(k)+\\sum_{k=1}^{n}(-1)^{k}\\left(\\begin{array}{c}\nn-1 \\\\\nk-1\n\\end{array}\\right) P(k) \\\\\n& =P(0)+\\sum_{k=1}^{n-1}(-1)^{k}\\left(\\left(\\begin{array}{c}\nn-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-1 \\\\\nk\n\\end{array}\\right)\\right) P(k)+(-1)^{n} P(n)=\\sum_{k=0}^{n}(-1)^{k}\\left(\\begin{array}{c}\nn \\\\\nk\n\\end{array}\\right) P(k) .\n\\end{aligned}\n$$\nNow return to the proof of Lemma 1. Suppose, to the contrary, that $\\operatorname{deg} P=N<3 n$. Consider the sum\n$$\n\\Sigma=\\sum_{i=0}^{n} \\sum_{j=0}^{n} \\sum_{k=0}^{n}(-1)^{i+j+k}\\left(\\begin{array}{c}\nn \\\\\ni\n\\end{array}\\right)\\left(\\begin{array}{l}\nn \\\\\nj\n\\end{array}\\right)\\left(\\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right) P(i, j, k)\n$$\nThe only nonzero term in this sum is $P(0,0,0)$ and its coefficient is $\\left(\\begin{array}{l}n \\\\ 0\\end{array}\\right)^{3}=1$; therefore $\\Sigma=P(0,0,0) \\neq 0$.\n\nOn the other hand, if $P(x, y, z)=\\sum_{\\alpha+\\beta+\\gamma \\leq N} p_{\\alpha, \\beta, \\gamma} x^{\\alpha} y^{\\beta} z^{\\gamma}$, then\n$$\n\\begin{aligned}\n\\Sigma & =\\sum_{i=0}^{n} \\sum_{j=0}^{n} \\sum_{k=0}^{n}(-1)^{i+j+k}\\left(\\begin{array}{c}\nn \\\\\ni\n\\end{array}\\right)\\left(\\begin{array}{l}\nn \\\\\nj\n\\end{array}\\right)\\left(\\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right) \\sum_{\\alpha+\\beta+\\gamma \\leq N} p_{\\alpha, \\beta, \\gamma} i^{\\alpha} j^{\\beta} k^{\\gamma} \\\\\n& =\\sum_{\\alpha+\\beta+\\gamma \\leq N} p_{\\alpha, \\beta, \\gamma}\\left(\\sum_{i=0}^{n}(-1)^{i}\\left(\\begin{array}{c}\nn \\\\\ni\n\\end{array}\\right) i^{\\alpha}\\right)\\left(\\sum_{j=0}^{n}(-1)^{j}\\left(\\begin{array}{c}\nn \\\\\nj\n\\end{array}\\right) j^{\\beta}\\right)\\left(\\sum_{k=0}^{n}(-1)^{k}\\left(\\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right) k^{\\gamma}\\right) .\n\\end{aligned}\n$$\nConsider an arbitrary term in this sum. We claim that it is zero. Since $N<3 n$, one of three inequalities $\\alpha<n, \\beta<n$ or $\\gamma<n$ is valid. For the convenience, suppose that $\\alpha<n$. Applying Lemma 2 to polynomial $x^{\\alpha}$, we get $\\sum_{i=0}^{n}(-1)^{i}\\left(\\begin{array}{c}n \\\\ i\\end{array}\\right) i^{\\alpha}=0$, hence the term is zero as required.\n\nThis yields $\\Sigma=0$ which is a contradiction. Therefore, $\\operatorname{deg} P \\geq 3 n$."
] | $3 n$ |
1,664 | Combinatorics | null | $\boxed{69$,$84}$ | [
"$69$,$84$"
] | true | null | Numerical | null | Find all positive integers $n$, for which the numbers in the set $S=\{1,2, \ldots, n\}$ can be colored red and blue, with the following condition being satisfied: the set $S \times S \times S$ contains exactly 2007 ordered triples $(x, y, z)$ such that (i) $x, y, z$ are of the same color and (ii) $x+y+z$ is divisible by $n$. | [
"Suppose that the numbers $1,2, \\ldots, n$ are colored red and blue. Denote by $R$ and $B$ the sets of red and blue numbers, respectively; let $|R|=r$ and $|B|=b=n-r$. Call a triple $(x, y, z) \\in S \\times S \\times S$ monochromatic if $x, y, z$ have the same color, and bichromatic otherwise. Call a triple $(x, y, z)$ divisible if $x+y+z$ is divisible by $n$. We claim that there are exactly $r^{2}-r b+b^{2}$ divisible monochromatic triples.\n\nFor any pair $(x, y) \\in S \\times S$ there exists a unique $z_{x, y} \\in S$ such that the triple $\\left(x, y, z_{x, y}\\right)$ is divisible; so there are exactly $n^{2}$ divisible triples. Furthermore, if a divisible triple $(x, y, z)$ is bichromatic, then among $x, y, z$ there are either one blue and two red numbers, or vice versa. In both cases, exactly one of the pairs $(x, y),(y, z)$ and $(z, x)$ belongs to the set $R \\times B$. Assign such pair to the triple $(x, y, z)$.\n\nConversely, consider any pair $(x, y) \\in R \\times B$, and denote $z=z_{x, y}$. Since $x \\neq y$, the triples $(x, y, z),(y, z, x)$ and $(z, x, y)$ are distinct, and $(x, y)$ is assigned to each of them. On the other hand, if $(x, y)$ is assigned to some triple, then this triple is clearly one of those mentioned above. So each pair in $R \\times B$ is assigned exactly three times.\n\nThus, the number of bichromatic divisible triples is three times the number of elements in $R \\times B$, and the number of monochromatic ones is $n^{2}-3 r b=(r+b)^{2}-3 r b=r^{2}-r b+b^{2}$, as claimed.\n\nSo, to find all values of $n$ for which the desired coloring is possible, we have to find all $n$, for which there exists a decomposition $n=r+b$ with $r^{2}-r b+b^{2}=2007$. Therefore, $9 \\mid r^{2}-r b+b^{2}=(r+b)^{2}-3 r b$. From this it consequently follows that $3|r+b, 3| r b$, and then $3|r, 3| b$. Set $r=3 s, b=3 c$. We can assume that $s \\geq c$. We have $s^{2}-s c+c^{2}=223$.\n\nFurthermore,\n$$\n892=4\\left(s^{2}-s c+c^{2}\\right)=(2 c-s)^{2}+3 s^{2} \\geq 3 s^{2} \\geq 3 s^{2}-3 c(s-c)=3\\left(s^{2}-s c+c^{2}\\right)=669\n$$\nso $297 \\geq s^{2} \\geq 223$ and $17 \\geq s \\geq 15$. If $s=15$ then\n$$\nc(15-c)=c(s-c)=s^{2}-\\left(s^{2}-s c+c^{2}\\right)=15^{2}-223=2\n$$\nwhich is impossible for an integer $c$. In a similar way, if $s=16$ then $c(16-c)=33$, which is also impossible. Finally, if $s=17$ then $c(17-c)=66$, and the solutions are $c=6$ and $c=11$. Hence, $(r, b)=(51,18)$ or $(r, b)=(51,33)$, and the possible values of $n$ are $n=51+18=69$ and $n=51+33=84$."
] | $69$,$84$ |
1,675 | Geometry | null | $\boxed{k=1}$ | [
"$k=1$"
] | false | null | Numerical | null | Determine the smallest positive real number $k$ with the following property.
Let $A B C D$ be a convex quadrilateral, and let points $A_{1}, B_{1}, C_{1}$ and $D_{1}$ lie on sides $A B, B C$, $C D$ and $D A$, respectively. Consider the areas of triangles $A A_{1} D_{1}, B B_{1} A_{1}, C C_{1} B_{1}$, and $D D_{1} C_{1}$; let $S$ be the sum of the two smallest ones, and let $S_{1}$ be the area of quadrilateral $A_{1} B_{1} C_{1} D_{1}$. Then we always have $k S_{1} \geq S$. | [
"Throughout the solution, triangles $A A_{1} D_{1}, B B_{1} A_{1}, C C_{1} B_{1}$, and $D D_{1} C_{1}$ will be referred to as border triangles. We will denote by $[\\mathcal{R}]$ the area of a region $\\mathcal{R}$.\n\nFirst, we show that $k \\geq 1$. Consider a triangle $A B C$ with unit area; let $A_{1}, B_{1}, K$ be the midpoints of its sides $A B, B C, A C$, respectively. Choose a point $D$ on the extension of $B K$, close to $K$. Take points $C_{1}$ and $D_{1}$ on sides $C D$ and $D A$ close to $D$ (see Figure 1). We have $\\left[B B_{1} A_{1}\\right]=\\frac{1}{4}$. Moreover, as $C_{1}, D_{1}, D \\rightarrow K$, we get $\\left[A_{1} B_{1} C_{1} D_{1}\\right] \\rightarrow\\left[A_{1} B_{1} K\\right]=\\frac{1}{4}$, $\\left[A A_{1} D_{1}\\right] \\rightarrow\\left[A A_{1} K\\right]=\\frac{1}{4},\\left[C C_{1} B_{1}\\right] \\rightarrow\\left[C K B_{1}\\right]=\\frac{1}{4}$ and $\\left[D D_{1} C_{1}\\right] \\rightarrow 0$. Hence, the sum of the two smallest areas of border triangles tends to $\\frac{1}{4}$, as well as $\\left[A_{1} B_{1} C_{1} D_{1}\\right]$; therefore, their ratio tends to 1 , and $k \\geq 1$.\n\nWe are left to prove that $k=1$ satisfies the desired property.\n\n<img_3730>\n\nFigure 1\n\n<img_3171>\n\nFigure 2\n\n<img_4006>\n\nFigure 3\n\nLemma. Let points $A_{1}, B_{1}, C_{1}$ lie respectively on sides $B C, C A, A B$ of a triangle $A B C$. Then $\\left[A_{1} B_{1} C_{1}\\right] \\geq \\min \\left\\{\\left[A C_{1} B_{1}\\right],\\left[B A_{1} C_{1}\\right],\\left[C B_{1} A_{1}\\right]\\right\\}$.\n\nProof. Let $A^{\\prime}, B^{\\prime}, C^{\\prime}$ be the midpoints of sides $B C, C A$ and $A B$, respectively.\n\nSuppose that two of points $A_{1}, B_{1}, C_{1}$ lie in one of triangles $A C^{\\prime} B^{\\prime}, B A^{\\prime} C^{\\prime}$ and $C B^{\\prime} A^{\\prime}$ (for convenience, let points $B_{1}$ and $C_{1}$ lie in triangle $A C^{\\prime} B^{\\prime}$; see Figure 2). Let segments $B_{1} C_{1}$ and $A A_{1}$ intersect at point $X$. Then $X$ also lies in triangle $A C^{\\prime} B^{\\prime}$. Hence $A_{1} X \\geq A X$, and we have\n$$\n\\frac{\\left[A_{1} B_{1} C_{1}\\right]}{\\left[A C_{1} B_{1}\\right]}=\\frac{\\frac{1}{2} A_{1} X \\cdot B_{1} C_{1} \\cdot \\sin \\angle A_{1} X C_{1}}{\\frac{1}{2} A X \\cdot B_{1} C_{1} \\cdot \\sin \\angle A X B_{1}}=\\frac{A_{1} X}{A X} \\geq 1\n$$\nas required.\n\nOtherwise, each one of triangles $A C^{\\prime} B^{\\prime}, B A^{\\prime} C^{\\prime}, C B^{\\prime} A^{\\prime}$ contains exactly one of points $A_{1}$, $B_{1}, C_{1}$, and we can assume that $B A_{1}<B A^{\\prime}, C B_{1}<C B^{\\prime}, A C_{1}<A C^{\\prime}$ (see Figure 3). Then lines $B_{1} A_{1}$ and $A B$ intersect at a point $Y$ on the extension of $A B$ beyond point $B$, hence $\\frac{\\left[A_{1} B_{1} C_{1}\\right]}{\\left[A_{1} B_{1} C^{\\prime}\\right]}=\\frac{C_{1} Y}{C^{\\prime} Y}>1$; also, lines $A_{1} C^{\\prime}$ and $C A$ intersect at a point $Z$ on the extension of $C A$ beyond point $A$, hence $\\frac{\\left[A_{1} B_{1} C^{\\prime}\\right]}{\\left[A_{1} B^{\\prime} C^{\\prime}\\right]}=\\frac{B_{1} Z}{B^{\\prime} Z}>1$. Finally, since $A_{1} A^{\\prime} \\| B^{\\prime} C^{\\prime}$, we have $\\left[A_{1} B_{1} C_{1}\\right]>\\left[A_{1} B_{1} C^{\\prime}\\right]>\\left[A_{1} B^{\\prime} C^{\\prime}\\right]=\\left[A^{\\prime} B^{\\prime} C^{\\prime}\\right]=\\frac{1}{4}[A B C]$.\n\n\n\nNow, from $\\left[A_{1} B_{1} C_{1}\\right]+\\left[A C_{1} B_{1}\\right]+\\left[B A_{1} C_{1}\\right]+\\left[C B_{1} A_{1}\\right]=[A B C]$ we obtain that one of the remaining triangles $A C_{1} B_{1}, B A_{1} C_{1}, C B_{1} A_{1}$ has an area less than $\\frac{1}{4}[A B C]$, so it is less than $\\left[A_{1} B_{1} C_{1}\\right]$.\n\nNow we return to the problem. We say that triangle $A_{1} B_{1} C_{1}$ is small if $\\left[A_{1} B_{1} C_{1}\\right]$ is less than each of $\\left[B B_{1} A_{1}\\right]$ and $\\left[C C_{1} B_{1}\\right]$; otherwise this triangle is big (the similar notion is introduced for triangles $B_{1} C_{1} D_{1}, C_{1} D_{1} A_{1}, D_{1} A_{1} B_{1}$ ). If both triangles $A_{1} B_{1} C_{1}$ and $C_{1} D_{1} A_{1}$ are big, then $\\left[A_{1} B_{1} C_{1}\\right]$ is not less than the area of some border triangle, and $\\left[C_{1} D_{1} A_{1}\\right]$ is not less than the area of another one; hence, $S_{1}=\\left[A_{1} B_{1} C_{1}\\right]+\\left[C_{1} D_{1} A_{1}\\right] \\geq S$. The same is valid for the pair of $B_{1} C_{1} D_{1}$ and $D_{1} A_{1} B_{1}$. So it is sufficient to prove that in one of these pairs both triangles are big.\n\nSuppose the contrary. Then there is a small triangle in each pair. Without loss of generality, assume that triangles $A_{1} B_{1} C_{1}$ and $D_{1} A_{1} B_{1}$ are small. We can assume also that $\\left[A_{1} B_{1} C_{1}\\right] \\leq$ $\\left[D_{1} A_{1} B_{1}\\right]$. Note that in this case ray $D_{1} C_{1}$ intersects line $B C$.\n\nConsider two cases.\n\n<img_3464>\n\nFigure 4\n\n<img_3580>\n\nFigure 5\n\nCase 1. Ray $C_{1} D_{1}$ intersects line $A B$ at some point $K$. Let ray $D_{1} C_{1}$ intersect line $B C$ at point $L$ (see Figure 4). Then we have $\\left[A_{1} B_{1} C_{1}\\right]<\\left[C C_{1} B_{1}\\right]<\\left[L C_{1} B_{1}\\right],\\left[A_{1} B_{1} C_{1}\\right]<\\left[B B_{1} A_{1}\\right]$ (both - since $\\left[A_{1} B_{1} C_{1}\\right]$ is small), and $\\left[A_{1} B_{1} C_{1}\\right] \\leq\\left[D_{1} A_{1} B_{1}\\right]<\\left[A A_{1} D_{1}\\right]<\\left[K A_{1} D_{1}\\right]<\\left[K A_{1} C_{1}\\right]$ (since triangle $D_{1} A_{1} B_{1}$ is small). This contradicts the Lemma, applied for triangle $A_{1} B_{1} C_{1}$ inside $L K B$.\n\nCase 2. Ray $C_{1} D_{1}$ does not intersect $A B$. Then choose a \"sufficiently far\" point $K$ on ray $B A$ such that $\\left[K A_{1} C_{1}\\right]>\\left[A_{1} B_{1} C_{1}\\right]$, and that ray $K C_{1}$ intersects line $B C$ at some point $L$ (see Figure 5). Since ray $C_{1} D_{1}$ does not intersect line $A B$, the points $A$ and $D_{1}$ are on different sides of $K L$; then $A$ and $D$ are also on different sides, and $C$ is on the same side as $A$ and $B$. Then analogously we have $\\left[A_{1} B_{1} C_{1}\\right]<\\left[C C_{1} B_{1}\\right]<\\left[L C_{1} B_{1}\\right]$ and $\\left[A_{1} B_{1} C_{1}\\right]<\\left[B B_{1} A_{1}\\right]$ since triangle $A_{1} B_{1} C_{1}$ is small. This (together with $\\left[A_{1} B_{1} C_{1}\\right]<\\left[K A_{1} C_{1}\\right]$ ) contradicts the Lemma again."
] | $k=1$ |
1,678 | Number Theory | null | $\boxed{(2,4)}$ | [
"$(2,4)$"
] | false | null | Tuple | null | Find all pairs $(k, n)$ of positive integers for which $7^{k}-3^{n}$ divides $k^{4}+n^{2}$. | [
"Suppose that a pair $(k, n)$ satisfies the condition of the problem. Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even, hence $k$ and $n$ have the same parity. If $k$ and $n$ are odd, then $k^{4}+n^{2} \\equiv 1+1=2(\\bmod 4)$, while $7^{k}-3^{n} \\equiv 7-3 \\equiv 0(\\bmod 4)$, so $k^{4}+n^{2}$ cannot be divisible by $7^{k}-3^{n}$. Hence, both $k$ and $n$ must be even.\n\nWrite $k=2 a, n=2 b$. Then $7^{k}-3^{n}=7^{2 a}-3^{2 b}=\\frac{7^{a}-3^{b}}{2} \\cdot 2\\left(7^{a}+3^{b}\\right)$, and both factors are integers. So $2\\left(7^{a}+3^{b}\\right) \\mid 7^{k}-3^{n}$ and $7^{k}-3^{n} \\mid k^{4}+n^{2}=2\\left(8 a^{4}+2 b^{2}\\right)$, hence\n$$\n7^{a}+3^{b} \\leq 8 a^{4}+2 b^{2}\n$$\nWe prove by induction that $8 a^{4}<7^{a}$ for $a \\geq 4,2 b^{2}<3^{b}$ for $b \\geq 1$ and $2 b^{2}+9 \\leq 3^{b}$ for $b \\geq 3$. In the initial cases $a=4, b=1, b=2$ and $b=3$ we have $8 \\cdot 4^{4}=2048<7^{4}=2401,2<3$, $2 \\cdot 2^{2}=8<3^{2}=9$ and $2 \\cdot 3^{2}+9=3^{3}=27$, respectively.\n\nIf $8 a^{4}<7^{a}(a \\geq 4)$ and $2 b^{2}+9 \\leq 3^{b}(b \\geq 3)$, then\n$$\n\\begin{aligned}\n8(a+1)^{4} & =8 a^{4}\\left(\\frac{a+1}{a}\\right)^{4}<7^{a}\\left(\\frac{5}{4}\\right)^{4}=7^{a} \\frac{625}{256}<7^{a+1} \\quad \\text { and } \\\\\n2(b+1)^{2}+9 & <\\left(2 b^{2}+9\\right)\\left(\\frac{b+1}{b}\\right)^{2} \\leq 3^{b}\\left(\\frac{4}{3}\\right)^{2}=3^{b} \\frac{16}{9}<3^{b+1},\n\\end{aligned}\n$$\nas desired.\n\nFor $a \\geq 4$ we obtain $7^{a}+3^{b}>8 a^{4}+2 b^{2}$ and inequality (1) cannot hold. Hence $a \\leq 3$, and three cases are possible.\n\nCase 1: $a=1$. Then $k=2$ and $8+2 b^{2} \\geq 7+3^{b}$, thus $2 b^{2}+1 \\geq 3^{b}$. This is possible only if $b \\leq 2$. If $b=1$ then $n=2$ and $\\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\\frac{2^{4}+2^{2}}{7^{2}-3^{2}}=\\frac{1}{2}$, which is not an integer. If $b=2$ then $n=4$ and $\\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\\frac{2^{4}+4^{2}}{7^{2}-3^{4}}=-1$, so $(k, n)=(2,4)$ is a solution.\n\nCase 2: $a=2$. Then $k=4$ and $k^{4}+n^{2}=256+4 b^{2} \\geq\\left|7^{4}-3^{n}\\right|=\\left|49-3^{b}\\right| \\cdot\\left(49+3^{b}\\right)$. The smallest value of the first factor is 22 , attained at $b=3$, so $128+2 b^{2} \\geq 11\\left(49+3^{b}\\right)$, which is impossible since $3^{b}>2 b^{2}$.\n\nCase 3: $a=3$. Then $k=6$ and $k^{4}+n^{2}=1296+4 b^{2} \\geq\\left|7^{6}-3^{n}\\right|=\\left|343-3^{b}\\right| \\cdot\\left(343+3^{b}\\right)$. Analogously, $\\left|343-3^{b}\\right| \\geq 100$ and we have $324+b^{2} \\geq 25\\left(343+3^{b}\\right)$, which is impossible again.\n\nWe find that there exists a unique solution $(k, n)=(2,4)$."
] | $(2,4)$ |
1,681 | Number Theory | null | $\boxed{f(n)=n}$ | [
"$f(n)=n$"
] | false | null | Expression | null | Find all surjective functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for every $m, n \in \mathbb{N}$ and every prime $p$, the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p$.
( $\mathbb{N}$ is the set of all positive integers.) | [
"Suppose that function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies the problem conditions.\n\nLemma. For any prime $p$ and any $x, y \\in \\mathbb{N}$, we have $x \\equiv y(\\bmod p)$ if and only if $f(x) \\equiv f(y)$ $(\\bmod p)$. Moreover, $p \\mid f(x)$ if and only if $p \\mid x$.\n\nProof. Consider an arbitrary prime $p$. Since $f$ is surjective, there exists some $x \\in \\mathbb{N}$ such that $p \\mid f(x)$. Let\n$$\nd=\\min \\{x \\in \\mathbb{N}: p \\mid f(x)\\}\n$$\nBy induction on $k$, we obtain that $p \\mid f(k d)$ for all $k \\in \\mathbb{N}$. The base is true since $p \\mid f(d)$. Moreover, if $p \\mid f(k d)$ and $p \\mid f(d)$ then, by the problem condition, $p \\mid f(k d+d)=f((k+1) d)$ as required.\n\nSuppose that there exists an $x \\in \\mathbb{N}$ such that $d \\not x$ but $p \\mid f(x)$. Let\n$$\ny=\\min \\{x \\in \\mathbb{N}: d \\nmid x, p \\mid f(x)\\} .\n$$\nBy the choice of $d$, we have $y>d$, and $y-d$ is a positive integer not divisible by $d$. Then $p \\nmid f(y-d)$, while $p \\mid f(d)$ and $p \\mid f(d+(y-d))=f(y)$. This contradicts the problem condition. Hence, there is no such $x$, and\n$$\np|f(x) \\Longleftrightarrow d| x .\\tag{1}\n$$\nTake arbitrary $x, y \\in \\mathbb{N}$ such that $x \\equiv y(\\bmod d)$. We have $p \\mid f(x+(2 x d-x))=f(2 x d)$; moreover, since $d \\mid 2 x d+(y-x)=y+(2 x d-x)$, we get $p \\mid f(y+(2 x d-x))$. Then by the problem condition $p|f(x)+f(2 x d-x), p| f(y)+f(2 x d-x)$, and hence $f(x) \\equiv-f(2 x d-x) \\equiv f(y)$ $(\\bmod p)$.\n\nOn the other hand, assume that $f(x) \\equiv f(y)(\\bmod p)$. Again we have $p \\mid f(x)+f(2 x d-x)$ which by our assumption implies that $p \\mid f(x)+f(2 x d-x)+(f(y)-f(x))=f(y)+f(2 x d-x)$. Hence by the problem condition $p \\mid f(y+(2 x d-x))$. Using (1) we get $0 \\equiv y+(2 x d-x) \\equiv y-x$ $(\\bmod d)$.\n\nThus, we have proved that\n$$\nx \\equiv y \\quad(\\bmod d) \\Longleftrightarrow f(x) \\equiv f(y) \\quad(\\bmod p)\\tag{2}\n$$\nWe are left to show that $p=d$ : in this case (1) and (2) provide the desired statements.\n\nThe numbers $1,2, \\ldots, d$ have distinct residues modulo $d$. By (2), numbers $f(1), f(2), \\ldots$, $f(d)$ have distinct residues modulo $p$; hence there are at least $d$ distinct residues, and $p \\geq d$. On the other hand, by the surjectivity of $f$, there exist $x_{1}, \\ldots, x_{p} \\in \\mathbb{N}$ such that $f\\left(x_{i}\\right)=i$ for any $i=1,2, \\ldots, p$. By (2), all these $x_{i}$ 's have distinct residues modulo $d$. For the same reasons, $d \\geq p$. Hence, $d=p$.\n\nNow we prove that $f(n)=n$ by induction on $n$. If $n=1$ then, by the Lemma, $p \\nmid f(1)$ for any prime $p$, so $f(1)=1$, and the base is established. Suppose that $n>1$ and denote $k=f(n)$. Note that there exists a prime $q \\mid n$, so by the Lemma $q \\mid k$ and $k>1$.\n\nIf $k>n$ then $k-n+1>1$, and there exists a prime $p \\mid k-n+1$; we have $k \\equiv n-1$ $(\\bmod p)$. By the induction hypothesis we have $f(n-1)=n-1 \\equiv k=f(n)(\\bmod p)$. Now, by the Lemma we obtain $n-1 \\equiv n(\\bmod p)$ which cannot be true.\n\n\n\nAnalogously, if $k<n$, then $f(k-1)=k-1$ by induction hypothesis. Moreover, $n-k+1>1$, so there exists a prime $p \\mid n-k+1$ and $n \\equiv k-1(\\bmod p)$. By the Lemma again, $k=f(n) \\equiv$ $f(k-1)=k-1(\\bmod p)$, which is also false. The only remaining case is $k=n$, so $f(n)=n$.\n\nFinally, the function $f(n)=n$ obviously satisfies the condition."
] | $f(n)=n$ |
1,687 | Algebra | null | $\boxed{f(n)=n$, $g(n)=1}$ | [
"$f(n)=n$, $g(n)=1$"
] | false | null | Expression | null | Determine all pairs $(f, g)$ of functions from the set of positive integers to itself that satisfy
$$
f^{g(n)+1}(n)+g^{f(n)}(n)=f(n+1)-g(n+1)+1
$$
for every positive integer $n$. Here, $f^{k}(n)$ means $\underbrace{f(f(\ldots f}_{k}(n) \ldots))$. | [
"The given relation implies\n\n$$\nf\\left(f^{g(n)}(n)\\right)<f(n+1) \\quad \\text { for all } n\n\\tag{1}\n$$\n\nwhich will turn out to be sufficient to determine $f$.\n\nLet $y_{1}<y_{2}<\\ldots$ be all the values attained by $f$ (this sequence might be either finite or infinite). We will prove that for every positive $n$ the function $f$ attains at least $n$ values, and we have (i) $)_{n}: f(x)=y_{n}$ if and only if $x=n$, and $(\\mathrm{ii})_{n}: y_{n}=n$. The proof will follow the scheme\n\n$$\n(\\mathrm{i})_{1},(\\mathrm{ii})_{1},(\\mathrm{i})_{2},(\\mathrm{ii})_{2}, \\ldots,(\\mathrm{i})_{n},(\\mathrm{ii})_{n}, \\ldots\n\\tag{2}\n$$\n\nTo start, consider any $x$ such that $f(x)=y_{1}$. If $x>1$, then (1) reads $f\\left(f^{g(x-1)}(x-1)\\right)<y_{1}$, contradicting the minimality of $y_{1}$. So we have that $f(x)=y_{1}$ is equivalent to $x=1$, establishing $(\\mathrm{i})_{1}$.\n\nNext, assume that for some $n$ statement $(\\mathrm{i})_{n}$ is established, as well as all the previous statements in (2). Note that these statements imply that for all $k \\geq 1$ and $a<n$ we have $f^{k}(x)=a$ if and only if $x=a$.\n\nNow, each value $y_{i}$ with $1 \\leq i \\leq n$ is attained at the unique integer $i$, so $y_{n+1}$ exists. Choose an arbitrary $x$ such that $f(x)=y_{n+1}$; we necessarily have $x>n$. Substituting $x-1$ into (1) we have $f\\left(f^{g(x-1)}(x-1)\\right)<y_{n+1}$, which implies\n\n$$\nf^{g(x-1)}(x-1) \\in\\{1, \\ldots, n\\}\n\\tag{3}\n$$\n\nSet $b=f^{g(x-1)}(x-1)$. If $b<n$ then we would have $x-1=b$ which contradicts $x>n$. So $b=n$, and hence $y_{n}=n$, which proves (ii) ${ }_{n}$. Next, from (i) ${ }_{n}$ we now get $f(k)=n \\Longleftrightarrow k=n$, so removing all the iterations of $f$ in (3) we obtain $x-1=b=n$, which proves $(\\mathrm{i})_{n+1}$.\n\nSo, all the statements in (2) are valid and hence $f(n)=n$ for all $n$. The given relation between $f$ and $g$ now reads $n+g^{n}(n)=n+1-g(n+1)+1$ or $g^{n}(n)+g(n+1)=2$, from which it immediately follows that we have $g(n)=1$ for all $n$.\n\n"
] | $f(n)=n$, $g(n)=1$ |
1,694 | Combinatorics | null | $\boxed{3}$ | [
"3"
] | false | null | Numerical | null | Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \ldots, A_{k}$ such that for all integers $n \geq 15$ and all $i \in\{1,2, \ldots, k\}$ there exist two distinct elements of $A_{i}$ whose sum is $n$. | [
"There are various examples showing that $k=3$ does indeed have the property under consideration. E.g. one can take\n\n$$\n\\begin{gathered}\nA_{1}=\\{1,2,3\\} \\cup\\{3 m \\mid m \\geq 4\\} \\\\\nA_{2}=\\{4,5,6\\} \\cup\\{3 m-1 \\mid m \\geq 4\\} \\\\\nA_{3}=\\{7,8,9\\} \\cup\\{3 m-2 \\mid m \\geq 4\\}\n\\end{gathered}\n$$\n\nTo check that this partition fits, we notice first that the sums of two distinct elements of $A_{i}$ obviously represent all numbers $n \\geq 1+12=13$ for $i=1$, all numbers $n \\geq 4+11=15$ for $i=2$, and all numbers $n \\geq 7+10=17$ for $i=3$. So, we are left to find representations of the numbers 15 and 16 as sums of two distinct elements of $A_{3}$. These are $15=7+8$ and $16=7+9$.\n\nLet us now suppose that for some $k \\geq 4$ there exist sets $A_{1}, A_{2}, \\ldots, A_{k}$ satisfying the given property. Obviously, the sets $A_{1}, A_{2}, A_{3}, A_{4} \\cup \\cdots \\cup A_{k}$ also satisfy the same property, so one may assume $k=4$.\n\nPut $B_{i}=A_{i} \\cap\\{1,2, \\ldots, 23\\}$ for $i=1,2,3,4$. Now for any index $i$ each of the ten numbers $15,16, \\ldots, 24$ can be written as sum of two distinct elements of $B_{i}$. Therefore this set needs to contain at least five elements. As we also have $\\left|B_{1}\\right|+\\left|B_{2}\\right|+\\left|B_{3}\\right|+\\left|B_{4}\\right|=23$, there has to be some index $j$ for which $\\left|B_{j}\\right|=5$. Let $B_{j}=\\left\\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\\right\\}$. Finally, now the sums of two distinct elements of $A_{j}$ representing the numbers $15,16, \\ldots, 24$ should be exactly all the pairwise sums of the elements of $B_{j}$. Calculating the sum of these numbers in two different ways, we reach\n\n$$\n4\\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\\right)=15+16+\\ldots+24=195\n$$\n\nThus the number 195 should be divisible by 4, which is false. This contradiction completes our solution.",
"Again we only prove that $k \\leq 3$. Assume that $A_{1}, A_{2}, \\ldots, A_{k}$ is a partition satisfying the given property. We construct a graph $\\mathcal{G}$ on the set $V=\\{1,2, \\ldots, 18\\}$ of vertices as follows. For each $i \\in\\{1,2, \\ldots, k\\}$ and each $d \\in\\{15,16,17,19\\}$ we choose one pair of distinct elements $a, b \\in A_{i}$ with $a+b=d$, and we draw an $e d g e$ in the $i^{\\text {th }}$ color connecting $a$ with $b$. By hypothesis, $\\mathcal{G}$ has exactly 4 edges of each color.\n\nClaim. The graph $\\mathcal{G}$ contains at most one circuit.\n\nProof. Note that all the connected components of $\\mathcal{G}$ are monochromatic and hence contain at most four edges. Thus also all circuits of $\\mathcal{G}$ are monochromatic and have length at most four. Moreover, each component contains at most one circuit since otherwise it should contain at least five edges.\n\nSuppose that there is a 4-cycle in $\\mathcal{G}$, say with vertices $a, b, c$, and $d$ in order. Then $\\{a+b, b+$ $c, c+d, d+a\\}=\\{15,16,17,19\\}$. Taking sums we get $2(a+b+c+d)=15+16+17+19$ which is impossible for parity reasons. Thus all circuits of $\\mathcal{G}$ are triangles.\n\nNow if the vertices $a, b$, and $c$ form such a triangle, then by a similar reasoning the set $\\{a+b, b+$ $c, c+a\\}$ coincides with either $\\{15,16,17\\}$, or $\\{15,16,19\\}$, or $\\{16,17,19\\}$, or $\\{15,17,19\\}$. The last of these alternatives can be excluded for parity reasons again, whilst in the first three cases the set $\\{a, b, c\\}$ appears to be either $\\{7,8,9\\}$, or $\\{6,9,10\\}$, or $\\{7,9,10\\}$, respectively. Thus, a component containing a circuit should contain 9 as a vertex. Therefore there is at most one such component and hence at most one circuit.\n\nBy now we know that $\\mathcal{G}$ is a graph with $4 k$ edges, at least $k$ components and at most one circuit. Consequently, $\\mathcal{G}$ must have at least $4 k+k-1$ vertices. Thus $5 k-1 \\leq 18$, and $k \\leq 3$."
] | 3 |
1,695 | Combinatorics | null | $\boxed{\frac{3 m}{2}-1}$ | [
"$\\frac{3 m}{2}-1$"
] | false | null | Expression | null | Let $m$ be a positive integer and consider a checkerboard consisting of $m$ by $m$ unit squares. At the midpoints of some of these unit squares there is an ant. At time 0, each ant starts moving with speed 1 parallel to some edge of the checkerboard. When two ants moving in opposite directions meet, they both turn $90^{\circ}$ clockwise and continue moving with speed 1 . When more than two ants meet, or when two ants moving in perpendicular directions meet, the ants continue moving in the same direction as before they met. When an ant reaches one of the edges of the checkerboard, it falls off and will not re-appear.
Considering all possible starting positions, determine the latest possible moment at which the last ant falls off the checkerboard or prove that such a moment does not necessarily exist. | [
"For $m=1$ the answer is clearly correct, so assume $m>1$. In the sequel, the word collision will be used to denote meeting of exactly two ants, moving in opposite directions.\n\nIf at the beginning we place an ant on the southwest corner square facing east and an ant on the southeast corner square facing west, then they will meet in the middle of the bottom row at time $\\frac{m-1}{2}$. After the collision, the ant that moves to the north will stay on the board for another $m-\\frac{1}{2}$ time units and thus we have established an example in which the last ant falls off at time $\\frac{m-1}{2}+m-\\frac{1}{2}=\\frac{3 m}{2}-1$. So, we are left to prove that this is the latest possible moment.\n\nConsider any collision of two ants $a$ and $a^{\\prime}$. Let us change the rule for this collision, and enforce these two ants to turn anticlockwise. Then the succeeding behavior of all the ants does not change; the only difference is that $a$ and $a^{\\prime}$ swap their positions. These arguments may be applied to any collision separately, so we may assume that at any collision, either both ants rotate clockwise or both of them rotate anticlockwise by our own choice.\n\nFor instance, we may assume that there are only two types of ants, depending on their initial direction: NE-ants, which move only north or east, and $S W$-ants, moving only south and west. Then we immediately obtain that all ants will have fallen off the board after $2 m-1$ time units. However, we can get a better bound by considering the last moment at which a given ant collides with another ant.\n\nChoose a coordinate system such that the corners of the checkerboard are $(0,0),(m, 0),(m, m)$ and $(0, m)$. At time $t$, there will be no NE-ants in the region $\\{(x, y): x+y<t+1\\}$ and no SW-ants in the region $\\{(x, y): x+y>2 m-t-1\\}$. So if two ants collide at $(x, y)$ at time $t$, we have\n\n$$\nt+1 \\leq x+y \\leq 2 m-t-1\n\\tag{1}\n$$\n\n\n\nAnalogously, we may change the rules so that each ant would move either alternatingly north and west, or alternatingly south and east. By doing so, we find that apart from (11) we also have $|x-y| \\leq m-t-1$ for each collision at point $(x, y)$ and time $t$.\n\nTo visualize this, put\n\n$$\nB(t)=\\left\\{(x, y) \\in[0, m]^{2}: t+1 \\leq x+y \\leq 2 m-t-1 \\text { and }|x-y| \\leq m-t-1\\right\\}\n$$\n\nAn ant can thus only collide with another ant at time $t$ if it happens to be in the region $B(t)$. The following figure displays $B(t)$ for $t=\\frac{1}{2}$ and $t=\\frac{7}{2}$ in the case $m=6$ :\n\n<img_3463>\n\nNow suppose that an NE-ant has its last collision at time $t$ and that it does so at the point $(x, y)$ (if the ant does not collide at all, it will fall off the board within $m-\\frac{1}{2}<\\frac{3 m}{2}-1$ time units, so this case can be ignored). Then we have $(x, y) \\in B(t)$ and thus $x+y \\geq t+1$ and $x-y \\geq-(m-t-1)$. So we get\n\n$$\nx \\geq \\frac{(t+1)-(m-t-1)}{2}=t+1-\\frac{m}{2}\n$$\n\nBy symmetry we also have $y \\geq t+1-\\frac{m}{2}$, and hence $\\min \\{x, y\\} \\geq t+1-\\frac{m}{2}$. After this collision, the ant will move directly to an edge, which will take at $\\operatorname{most} m-\\min \\{x, y\\}$ units of time. In sum, the total amount of time the ant stays on the board is at most\n\n$$\nt+(m-\\min \\{x, y\\}) \\leq t+m-\\left(t+1-\\frac{m}{2}\\right)=\\frac{3 m}{2}-1\n$$\n\nBy symmetry, the same bound holds for SW-ants as well."
] | $\frac{3 m}{2}-1$ |
1,697 | Combinatorics | null | $\boxed{3986729}$ | [
"3986729"
] | false | null | Numerical | null | On a square table of 2011 by 2011 cells we place a finite number of napkins that each cover a square of 52 by 52 cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is the largest value of $k$ ? | [
"Let $m=39$, then $2011=52 m-17$. We begin with an example showing that there can exist 3986729 cells carrying the same positive number.\n\n<img_3188>\n\nTo describe it, we number the columns from the left to the right and the rows from the bottom to the top by $1,2, \\ldots, 2011$. We will denote each napkin by the coordinates of its lowerleft cell. There are four kinds of napkins: first, we take all napkins $(52 i+36,52 j+1)$ with $0 \\leq j \\leq i \\leq m-2$; second, we use all napkins $(52 i+1,52 j+36)$ with $0 \\leq i \\leq j \\leq m-2$; third, we use all napkins $(52 i+36,52 i+36)$ with $0 \\leq i \\leq m-2$; and finally the napkin $(1,1)$. Different groups of napkins are shown by different types of hatchings in the picture.\n\nNow except for those squares that carry two or more different hatchings, all squares have the number 1 written into them. The number of these exceptional cells is easily computed to be $\\left(52^{2}-35^{2}\\right) m-17^{2}=57392$.\n\nWe are left to prove that 3986729 is an upper bound for the number of cells containing the same number. Consider any configuration of napkins and any positive integer $M$. Suppose there are $g$ cells with a number different from $M$. Then it suffices to show $g \\geq 57392$. Throughout the solution, a line will mean either a row or a column.\n\nConsider any line $\\ell$. Let $a_{1}, \\ldots, a_{52 m-17}$ be the numbers written into its consecutive cells. For $i=1,2, \\ldots, 52$, let $s_{i}=\\sum_{t \\equiv i(\\bmod 52)} a_{t}$. Note that $s_{1}, \\ldots, s_{35}$ have $m$ terms each, while $s_{36}, \\ldots, s_{52}$ have $m-1$ terms each. Every napkin intersecting $\\ell$ contributes exactly 1 to each $s_{i}$;\n\n\n\nhence the number $s$ of all those napkins satisfies $s_{1}=\\cdots=s_{52}=s$. Call the line $\\ell$ rich if $s>(m-1) M$ and poor otherwise.\n\nSuppose now that $\\ell$ is rich. Then in each of the sums $s_{36}, \\ldots, s_{52}$ there exists a term greater than $M$; consider all these terms and call the corresponding cells the rich bad cells for this line. So, each rich line contains at least 17 cells that are bad for this line.\n\nIf, on the other hand, $\\ell$ is poor, then certainly $s<m M$ so in each of the sums $s_{1}, \\ldots, s_{35}$ there exists a term less than $M$; consider all these terms and call the corresponding cells the poor bad cells for this line. So, each poor line contains at least 35 cells that are bad for this line.\n\nLet us call all indices congruent to $1,2, \\ldots$, or 35 modulo 52 small, and all other indices, i.e. those congruent to $36,37, \\ldots$, or 52 modulo 52 , big. Recall that we have numbered the columns from the left to the right and the rows from the bottom to the top using the numbers $1,2, \\ldots, 52 m-17$; we say that a line is big or small depending on whether its index is big or small. By definition, all rich bad cells for the rows belong to the big columns, while the poor ones belong to the small columns, and vice versa.\n\nIn each line, we put a strawberry on each cell that is bad for this line. In addition, for each small rich line we put an extra strawberry on each of its (rich) bad cells. A cell gets the strawberries from its row and its column independently.\n\nNotice now that a cell with a strawberry on it contains a number different from $M$. If this cell gets a strawberry by the extra rule, then it contains a number greater than $M$. Moreover, it is either in a small row and in a big column, or vice versa. Suppose that it is in a small row, then it is not bad for its column. So it has not more than two strawberries in this case. On the other hand, if the extra rule is not applied to some cell, then it also has not more than two strawberries. So, the total number $N$ of strawberries is at most $2 g$.\n\nWe shall now estimate $N$ in a different way. For each of the $2 \\cdot 35 \\mathrm{~m}$ small lines, we have introduced at least 34 strawberries if it is rich and at least 35 strawberries if it is poor, so at least 34 strawberries in any case. Similarly, for each of the $2 \\cdot 17(m-1)$ big lines, we put at least $\\min (17,35)=17$ strawberries. Summing over all lines we obtain\n\n$$\n2 g \\geq N \\geq 2(35 m \\cdot 34+17(m-1) \\cdot 17)=2(1479 m-289)=2 \\cdot 57392\n$$\n\nas desired.",
"We present a different proof of the estimate which is the hard part of the problem. Let $S=35, H=17, m=39$; so the table size is $2011=S m+H(m-1)$, and the napkin size is $52=S+H$. Fix any positive integer $M$ and call a cell vicious if it contains a number distinct\n\n\n\nfrom $M$. We will prove that there are at least $H^{2}(m-1)+2 S H m$ vicious cells.\n\nFirstly, we introduce some terminology. As in the previous solution, we number rows and columns and we use the same notions of small and big indices and lines; so, an index is small if it is congruent to one of the numbers $1,2, \\ldots, S$ modulo $(S+H)$. The numbers $1,2, \\ldots, S+H$ will be known as residues. For two residues $i$ and $j$, we say that a cell is of type $(i, j)$ if the index of its row is congruent to $i$ and the index of its column to $j$ modulo $(S+H)$. The number of vicious cells of this type is denoted by $v_{i j}$.\n\nLet $s, s^{\\prime}$ be two variables ranging over small residues and let $h, h^{\\prime}$ be two variables ranging over big residues. A cell is said to be of class $A, B, C$, or $D$ if its type is of shape $\\left(s, s^{\\prime}\\right),(s, h),(h, s)$, or $\\left(h, h^{\\prime}\\right)$, respectively. The numbers of vicious cells belonging to these classes are denoted in this order by $a, b, c$, and $d$. Observe that each cell belongs to exactly one class.\n\nClaim 1. We have\n\n$$\nm \\leq \\frac{a}{S^{2}}+\\frac{b+c}{2 S H}\n\\tag{1}\n$$\n\nProof. Consider an arbitrary small row $r$. Denote the numbers of vicious cells on $r$ belonging to the classes $A$ and $B$ by $\\alpha$ and $\\beta$, respectively. As in the previous solution, we obtain that $\\alpha \\geq S$ or $\\beta \\geq H$. So in each case we have $\\frac{\\alpha}{S}+\\frac{\\beta}{H} \\geq 1$.\n\nPerforming this argument separately for each small row and adding up all the obtained inequalities, we get $\\frac{a}{S}+\\frac{b}{H} \\geq m S$. Interchanging rows and columns we similarly get $\\frac{a}{S}+\\frac{c}{H} \\geq m S$. Summing these inequalities and dividing by $2 S$ we get what we have claimed.\n\nClaim 2. Fix two small residue $s, s^{\\prime}$ and two big residues $h, h^{\\prime}$. Then $2 m-1 \\leq v_{s s^{\\prime}}+v_{s h^{\\prime}}+v_{h h^{\\prime}}$. Proof. Each napkin covers exactly one cell of type $\\left(s, s^{\\prime}\\right)$. Removing all napkins covering a vicious cell of this type, we get another collection of napkins, which covers each cell of type $\\left(s, s^{\\prime}\\right)$ either 0 or $M$ times depending on whether the cell is vicious or not. Hence $\\left(m^{2}-v_{s s^{\\prime}}\\right) M$ napkins are left and throughout the proof of Claim 2 we will consider only these remaining napkins. Now, using a red pen, write in each cell the number of napkins covering it. Notice that a cell containing a red number greater than $M$ is surely vicious.\n\nWe call two cells neighbors if they can be simultaneously covered by some napkin. So, each cell of type $\\left(h, h^{\\prime}\\right)$ has not more than four neighbors of type $\\left(s, s^{\\prime}\\right)$, while each cell of type $\\left(s, h^{\\prime}\\right)$ has not more than two neighbors of each of the types $\\left(s, s^{\\prime}\\right)$ and $\\left(h, h^{\\prime}\\right)$. Therefore, each red number at a cell of type $\\left(h, h^{\\prime}\\right)$ does not exceed $4 M$, while each red number at a cell of type $\\left(s, h^{\\prime}\\right)$ does not exceed $2 M$.\n\nLet $x, y$, and $z$ be the numbers of cells of type $\\left(h, h^{\\prime}\\right)$ whose red number belongs to $(M, 2 M]$, $(2 M, 3 M]$, and $(3 M, 4 M]$, respectively. All these cells are vicious, hence $x+y+z \\leq v_{h h^{\\prime}}$. The red numbers appearing in cells of type $\\left(h, h^{\\prime}\\right)$ clearly sum up to $\\left(m^{2}-v_{s s^{\\prime}}\\right) M$. Bounding each of these numbers by a multiple of $M$ we get\n\n$$\n\\left(m^{2}-v_{s s^{\\prime}}\\right) M \\leq\\left((m-1)^{2}-(x+y+z)\\right) M+2 x M+3 y M+4 z M\n$$\n\n\n\ni.e.\n\n$$\n2 m-1 \\leq v_{s s^{\\prime}}+x+2 y+3 z \\leq v_{s s^{\\prime}}+v_{h h^{\\prime}}+y+2 z\n$$\n\nSo, to prove the claim it suffices to prove that $y+2 z \\leq v_{s h^{\\prime}}$.\n\nFor a cell $\\delta$ of type $\\left(h, h^{\\prime}\\right)$ and a cell $\\beta$ of type $\\left(s, h^{\\prime}\\right)$ we say that $\\delta$ forces $\\beta$ if there are more than $M$ napkins covering both of them. Since each red number in a cell of type $\\left(s, h^{\\prime}\\right)$ does not exceed $2 M$, it cannot be forced by more than one cell.\n\nOn the other hand, if a red number in a $\\left(h, h^{\\prime}\\right)$-cell belongs to $(2 M, 3 M]$, then it forces at least one of its neighbors of type $\\left(s, h^{\\prime}\\right)$ (since the sum of red numbers in their cells is greater than $2 M)$. Analogously, an $\\left(h, h^{\\prime}\\right)$-cell with the red number in $(3 M, 4 M]$ forces both its neighbors of type $\\left(s, h^{\\prime}\\right)$, since their red numbers do not exceed $2 M$. Therefore there are at least $y+2 z$ forced cells and clearly all of them are vicious, as desired.\n\nClaim 3. We have\n\n$$\n2 m-1 \\leq \\frac{a}{S^{2}}+\\frac{b+c}{2 S H}+\\frac{d}{H^{2}}\n\\tag{2}\n$$\n\nProof. Averaging the previous result over all $S^{2} H^{2}$ possibilities for the quadruple $\\left(s, s^{\\prime}, h, h^{\\prime}\\right)$, we get $2 m-1 \\leq \\frac{a}{S^{2}}+\\frac{b}{S H}+\\frac{d}{H^{2}}$. Due to the symmetry between rows and columns, the same estimate holds with $b$ replaced by $c$. Averaging these two inequalities we arrive at our claim.\n\nNow let us multiply (2) by $H^{2}$, multiply (II) by $\\left(2 S H-H^{2}\\right)$ and add them; we get\n\n$H^{2}(2 m-1)+\\left(2 S H-H^{2}\\right) m \\leq a \\cdot \\frac{H^{2}+2 S H-H^{2}}{S^{2}}+(b+c) \\frac{H^{2}+2 S H-H^{2}}{2 S H}+d=a \\cdot \\frac{2 H}{S}+b+c+d$.\n\nThe left-hand side is exactly $H^{2}(m-1)+2 S H m$, while the right-hand side does not exceed $a+b+c+d$ since $2 H \\leq S$. Hence we come to the desired inequality."
] | 3986729 |
1,709 | Number Theory | null | $\boxed{1,3,5}$ | [
"1,3,5"
] | true | null | Numerical | null | For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences
$$
t(n+a)-t(n), \quad t(n+a+1)-t(n+1), \quad \ldots, \quad t(n+2 a-1)-t(n+a-1)
$$
are divisible by 4 . | [
"A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs.\n\nNow suppose that $a$ is a positive integer not equal to 1,3 , and 5 . We will show that there are no winning pairs $(a, n)$ by distinguishing three cases.\n\nCase 1: $a$ is even. In this case we have $a=2^{\\alpha} d$ for some positive integer $\\alpha$ and some odd $d$. Since $a \\geq 2^{\\alpha}$, for each positive integer $n$ there exists an $i \\in\\{0,1, \\ldots, a-1\\}$ such that $n+i=2^{\\alpha-1} e$, where $e$ is some odd integer. Then we have $t(n+i)=t\\left(2^{\\alpha-1} e\\right)=e$ and\n\n$$\nt(n+a+i)=t\\left(2^{\\alpha} d+2^{\\alpha-1} e\\right)=2 d+e \\equiv e+2 \\quad(\\bmod 4) .\n$$\n\nSo we get $t(n+i)-t(n+a+i) \\equiv 2(\\bmod 4)$, and $(a, n)$ is not a winning pair.\n\nCase 2: $a$ is odd and $a>8$. For each positive integer $n$, there exists an $i \\in\\{0,1, \\ldots, a-5\\}$ such that $n+i=2 d$ for some odd $d$. We get\n\n$$\nt(n+i)=d \\not \\equiv d+2=t(n+i+4) \\quad(\\bmod 4)\n$$\n\nand\n\n$$\nt(n+a+i)=n+a+i \\equiv n+a+i+4=t(n+a+i+4) \\quad(\\bmod 4)\n$$\n\nTherefore, the integers $t(n+a+i)-t(n+i)$ and $t(n+a+i+4)-t(n+i+4)$ cannot be both divisible by 4 , and therefore there are no winning pairs in this case.\n\nCase 3: $a=7$. For each positive integer $n$, there exists an $i \\in\\{0,1, \\ldots, 6\\}$ such that $n+i$ is either of the form $8 k+3$ or of the form $8 k+6$, where $k$ is a nonnegative integer. But we have\n\n$$\nt(8 k+3) \\equiv 3 \\not \\equiv 1 \\equiv 4 k+5=t(8 k+3+7) \\quad(\\bmod 4)\n$$\n\nand\n\n$$\nt(8 k+6)=4 k+3 \\equiv 3 \\not \\equiv 1 \\equiv t(8 k+6+7) \\quad(\\bmod 4)\n$$\n\nHence, there are no winning pairs of the form $(7, n)$."
] | 1,3,5 |
1,716 | Algebra | null | $\boxed{\frac{25}{2}}$ | [
"$\\frac{25}{2}$"
] | false | null | Numerical | null | Let $x_{1}, \ldots, x_{100}$ be nonnegative real numbers such that $x_{i}+x_{i+1}+x_{i+2} \leq 1$ for all $i=1, \ldots, 100$ (we put $x_{101}=x_{1}, x_{102}=x_{2}$ ). Find the maximal possible value of the sum
$$
S=\sum_{i=1}^{100} x_{i} x_{i+2}
$$ | [
"Let $x_{2 i}=0, x_{2 i-1}=\\frac{1}{2}$ for all $i=1, \\ldots, 50$. Then we have $S=50 \\cdot\\left(\\frac{1}{2}\\right)^{2}=\\frac{25}{2}$. So, we are left to show that $S \\leq \\frac{25}{2}$ for all values of $x_{i}$ 's satisfying the problem conditions.\n\nConsider any $1 \\leq i \\leq 50$. By the problem condition, we get $x_{2 i-1} \\leq 1-x_{2 i}-x_{2 i+1}$ and $x_{2 i+2} \\leq 1-x_{2 i}-x_{2 i+1}$. Hence by the AM-GM inequality we get\n\n$$\n\\begin{aligned}\nx_{2 i-1} x_{2 i+1} & +x_{2 i} x_{2 i+2} \\leq\\left(1-x_{2 i}-x_{2 i+1}\\right) x_{2 i+1}+x_{2 i}\\left(1-x_{2 i}-x_{2 i+1}\\right) \\\\\n& =\\left(x_{2 i}+x_{2 i+1}\\right)\\left(1-x_{2 i}-x_{2 i+1}\\right) \\leq\\left(\\frac{\\left(x_{2 i}+x_{2 i+1}\\right)+\\left(1-x_{2 i}-x_{2 i+1}\\right)}{2}\\right)^{2}=\\frac{1}{4} .\n\\end{aligned}\n$$\n\nSumming up these inequalities for $i=1,2, \\ldots, 50$, we get the desired inequality\n\n$$\n\\sum_{i=1}^{50}\\left(x_{2 i-1} x_{2 i+1}+x_{2 i} x_{2 i+2}\\right) \\leq 50 \\cdot \\frac{1}{4}=\\frac{25}{2}\n$$",
"We present another proof of the estimate. From the problem condition, we get\n\n$$\n\\begin{aligned}\nS=\\sum_{i=1}^{100} x_{i} x_{i+2} \\leq \\sum_{i=1}^{100} x_{i}\\left(1-x_{i}-x_{i+1}\\right) & =\\sum_{i=1}^{100} x_{i}-\\sum_{i=1}^{100} x_{i}^{2}-\\sum_{i=1}^{100} x_{i} x_{i+1} \\\\\n& =\\sum_{i=1}^{100} x_{i}-\\frac{1}{2} \\sum_{i=1}^{100}\\left(x_{i}+x_{i+1}\\right)^{2}\n\\end{aligned}\n$$\n\nBy the AM-QM inequality, we have $\\sum\\left(x_{i}+x_{i+1}\\right)^{2} \\geq \\frac{1}{100}\\left(\\sum\\left(x_{i}+x_{i+1}\\right)\\right)^{2}$, so\n\n$$\n\\begin{aligned}\nS \\leq \\sum_{i=1}^{100} x_{i}-\\frac{1}{200}\\left(\\sum_{i=1}^{100}\\left(x_{i}+x_{i+1}\\right)\\right)^{2} & =\\sum_{i=1}^{100} x_{i}-\\frac{2}{100}\\left(\\sum_{i=1}^{100} x_{i}\\right)^{2} \\\\\n& =\\frac{2}{100}\\left(\\sum_{i=1}^{100} x_{i}\\right)\\left(\\frac{100}{2}-\\sum_{i=1}^{100} x_{i}\\right) .\n\\end{aligned}\n$$\n\nAnd finally, by the AM-GM inequality\n\n$$\nS \\leq \\frac{2}{100} \\cdot\\left(\\frac{1}{2}\\left(\\sum_{i=1}^{100} x_{i}+\\frac{100}{2}-\\sum_{i=1}^{100} x_{i}\\right)\\right)^{2}=\\frac{2}{100} \\cdot\\left(\\frac{100}{4}\\right)^{2}=\\frac{25}{2}\n$$"
] | $\frac{25}{2}$ |
1,718 | Algebra | null | $\boxed{f(x)=\frac{1}{x}}$ | [
"$f(x)=\\frac{1}{x}$"
] | false | null | Expression | null | Denote by $\mathbb{Q}^{+}$the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ which satisfy the following equation for all $x, y \in \mathbb{Q}^{+}$:
$$
f\left(f(x)^{2} y\right)=x^{3} f(x y)
\tag{1}
$$ | [
"By substituting $y=1$, we get\n\n$$\nf\\left(f(x)^{2}\\right)=x^{3} f(x)\\tag{2}\n$$\n\nThen, whenever $f(x)=f(y)$, we have\n\n$$\nx^{3}=\\frac{f\\left(f(x)^{2}\\right)}{f(x)}=\\frac{f\\left(f(y)^{2}\\right)}{f(y)}=y^{3}\n$$\n\nwhich implies $x=y$, so the function $f$ is injective.\n\nNow replace $x$ by $x y$ in (2), and apply (1) twice, second time to $\\left(y, f(x)^{2}\\right)$ instead of $(x, y)$ :\n\n$$\nf\\left(f(x y)^{2}\\right)=(x y)^{3} f(x y)=y^{3} f\\left(f(x)^{2} y\\right)=f\\left(f(x)^{2} f(y)^{2}\\right)\n$$\n\nSince $f$ is injective, we get\n\n$$\n\\begin{aligned}\nf(x y)^{2} & =f(x)^{2} f(y)^{2} \\\\\nf(x y) & =f(x) f(y) .\n\\end{aligned}\n$$\n\nTherefore, $f$ is multiplicative. This also implies $f(1)=1$ and $f\\left(x^{n}\\right)=f(x)^{n}$ for all integers $n$.\n\nThen the function equation (1) can be re-written as\n\n$$\n\\begin{aligned}\nf(f(x))^{2} f(y) & =x^{3} f(x) f(y), \\\\\nf(f(x)) & =\\sqrt{x^{3} f(x)} .\n\\end{aligned}\n\\tag{3}\n$$\n\nLet $g(x)=x f(x)$. Then, by (3), we have\n\n$$\n\\begin{aligned}\ng(g(x)) & =g(x f(x))=x f(x) \\cdot f(x f(x))=x f(x)^{2} f(f(x))= \\\\\n& =x f(x)^{2} \\sqrt{x^{3} f(x)}=(x f(x))^{5 / 2}=(g(x))^{5 / 2},\n\\end{aligned}\n$$\n\nand, by induction,\n\n$$\n\\underbrace{g(g(\\ldots g}_{n+1}(x) \\ldots))=(g(x))^{(5 / 2)^{n}}\n\\tag{4}\n$$\n\nfor every positive integer $n$.\n\nConsider (4) for a fixed $x$. The left-hand side is always rational, so $(g(x))^{(5 / 2)^{n}}$ must be rational for every $n$. We show that this is possible only if $g(x)=1$. Suppose that $g(x) \\neq 1$, and let the prime factorization of $g(x)$ be $g(x)=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}$ where $p_{1}, \\ldots, p_{k}$ are distinct primes and $\\alpha_{1}, \\ldots, \\alpha_{k}$ are nonzero integers. Then the unique prime factorization of (4) is\n\n$$\n\\underbrace{g(g(\\ldots g}_{n+1}(x) \\ldots))=(g(x))^{(5 / 2)^{n}}=p_{1}^{(5 / 2)^{n} \\alpha_{1}} \\ldots p_{k}^{(5 / 2)^{n} \\alpha_{k}}\n$$\n\n\n\nwhere the exponents should be integers. But this is not true for large values of $n$, for example $\\left(\\frac{5}{2}\\right)^{n} \\alpha_{1}$ cannot be a integer number when $2^{n} \\nmid \\alpha_{1}$. Therefore, $g(x) \\neq 1$ is impossible.\n\nHence, $g(x)=1$ and thus $f(x)=\\frac{1}{x}$ for all $x$.\n\nThe function $f(x)=\\frac{1}{x}$ satisfies the equation (1):\n\n$$\nf\\left(f(x)^{2} y\\right)=\\frac{1}{f(x)^{2} y}=\\frac{1}{\\left(\\frac{1}{x}\\right)^{2} y}=\\frac{x^{3}}{x y}=x^{3} f(x y)\n$$"
] | $f(x)=\frac{1}{x}$ |
1,723 | Combinatorics | null | $\boxed{M=2^{N-2}+1}$ | [
"$M=2^{N-2}+1$"
] | false | null | Expression | null | On some planet, there are $2^{N}$ countries $(N \geq 4)$. Each country has a flag $N$ units wide and one unit high composed of $N$ fields of size $1 \times 1$, each field being either yellow or blue. No two countries have the same flag.
We say that a set of $N$ flags is diverse if these flags can be arranged into an $N \times N$ square so that all $N$ fields on its main diagonal will have the same color. Determine the smallest positive integer $M$ such that among any $M$ distinct flags, there exist $N$ flags forming a diverse set. | [
"When speaking about the diagonal of a square, we will always mean the main diagonal.\n\nLet $M_{N}$ be the smallest positive integer satisfying the problem condition. First, we show that $M_{N}>2^{N-2}$. Consider the collection of all $2^{N-2}$ flags having yellow first squares and blue second ones. Obviously, both colors appear on the diagonal of each $N \\times N$ square formed by these flags.\n\nWe are left to show that $M_{N} \\leq 2^{N-2}+1$, thus obtaining the desired answer. We start with establishing this statement for $N=4$.\n\nSuppose that we have 5 flags of length 4 . We decompose each flag into two parts of 2 squares each; thereby, we denote each flag as $L R$, where the $2 \\times 1$ flags $L, R \\in \\mathcal{S}=\\{\\mathrm{BB}, \\mathrm{BY}, \\mathrm{YB}, \\mathrm{YY}\\}$ are its left and right parts, respectively. First, we make two easy observations on the flags $2 \\times 1$ which can be checked manually.\n\n(i) For each $A \\in \\mathcal{S}$, there exists only one $2 \\times 1$ flag $C \\in \\mathcal{S}$ (possibly $C=A$ ) such that $A$ and $C$ cannot form a $2 \\times 2$ square with monochrome diagonal (for part $\\mathrm{BB}$, that is $\\mathrm{YY}$, and for $\\mathrm{BY}$ that is $\\mathrm{YB)}$.\n\n(ii) Let $A_{1}, A_{2}, A_{3} \\in \\mathcal{S}$ be three distinct elements; then two of them can form a $2 \\times 2$ square with yellow diagonal, and two of them can form a $2 \\times 2$ square with blue diagonal (for all parts but $\\mathrm{BB}$, a pair (BY, YB) fits for both statements, while for all parts but BY, these pairs are $(\\mathrm{YB}, \\mathrm{YY})$ and $(\\mathrm{BB}, \\mathrm{YB}))$.\n\nNow, let $\\ell$ and $r$ be the numbers of distinct left and right parts of our 5 flags, respectively. The total number of flags is $5 \\leq r \\ell$, hence one of the factors (say, $r$ ) should be at least 3 . On the other hand, $\\ell, r \\leq 4$, so there are two flags with coinciding right part; let them be $L_{1} R_{1}$ and $L_{2} R_{1}\\left(L_{1} \\neq L_{2}\\right)$.\n\nNext, since $r \\geq 3$, there exist some flags $L_{3} R_{3}$ and $L_{4} R_{4}$ such that $R_{1}, R_{3}, R_{4}$ are distinct. Let $L^{\\prime} R^{\\prime}$ be the remaining flag. By (i), one of the pairs $\\left(L^{\\prime}, L_{1}\\right)$ and $\\left(L^{\\prime}, L_{2}\\right)$ can form a $2 \\times 2$ square with monochrome diagonal; we can assume that $L^{\\prime}, L_{2}$ form a square with a blue diagonal. Finally, the right parts of two of the flags $L_{1} R_{1}, L_{3} R_{3}, L_{4} R_{4}$ can also form a $2 \\times 2$ square with a blue diagonal by (ii). Putting these $2 \\times 2$ squares on the diagonal of a $4 \\times 4$ square, we find a desired arrangement of four flags.\n\nWe are ready to prove the problem statement by induction on $N$; actually, above we have proved the base case $N=4$. For the induction step, assume that $N>4$, consider any $2^{N-2}+1$ flags of length $N$, and arrange them into a large flag of size $\\left(2^{N-2}+1\\right) \\times N$. This flag contains a non-monochrome column since the flags are distinct; we may assume that this column is the first one. By the pigeonhole principle, this column contains at least $\\left\\lceil\\frac{2^{N-2}+1}{2}\\right\\rceil=2^{N-3}+1$ squares of one color (say, blue). We call the flags with a blue first square good.\n\nConsider all the good flags and remove the first square from each of them. We obtain at least $2^{N-3}+1 \\geq M_{N-1}$ flags of length $N-1$; by the induction hypothesis, $N-1$ of them\n\n\n\ncan form a square $Q$ with the monochrome diagonal. Now, returning the removed squares, we obtain a rectangle $(N-1) \\times N$, and our aim is to supplement it on the top by one more flag.\n\nIf $Q$ has a yellow diagonal, then we can take each flag with a yellow first square (it exists by a choice of the first column; moreover, it is not used in $Q$ ). Conversely, if the diagonal of $Q$ is blue then we can take any of the $\\geq 2^{N-3}+1-(N-1)>0$ remaining good flags. So, in both cases we get a desired $N \\times N$ square.",
"We present a different proof of the estimate $M_{N} \\leq 2^{N-2}+1$. We do not use the induction, involving Hall's lemma on matchings instead.\n\nConsider arbitrary $2^{N-2}+1$ distinct flags and arrange them into a large $\\left(2^{N-2}+1\\right) \\times N$ flag. Construct two bipartite graphs $G_{\\mathrm{y}}=\\left(V \\cup V^{\\prime}, E_{\\mathrm{y}}\\right)$ and $G_{\\mathrm{b}}=\\left(V \\cup V^{\\prime}, E_{\\mathrm{b}}\\right)$ with the common set of vertices as follows. Let $V$ and $V^{\\prime}$ be the set of columns and the set of flags under consideration, respectively. Next, let the edge $(c, f)$ appear in $E_{\\mathrm{y}}$ if the intersection of column $c$ and flag $f$ is yellow, and $(c, f) \\in E_{\\mathrm{b}}$ otherwise. Then we have to prove exactly that one of the graphs $G_{\\mathrm{y}}$ and $G_{\\mathrm{b}}$ contains a matching with all the vertices of $V$ involved.\n\nAssume that these matchings do not exist. By Hall's lemma, it means that there exist two sets of columns $S_{\\mathrm{y}}, S_{\\mathrm{b}} \\subset V$ such that $\\left|E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)\\right| \\leq\\left|S_{\\mathrm{y}}\\right|-1$ and $\\left|E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right| \\leq\\left|S_{\\mathrm{b}}\\right|-1$ (in the left-hand sides, $E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)$ and $E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)$ denote respectively the sets of all vertices connected to $S_{\\mathrm{y}}$ and $S_{\\mathrm{b}}$ in the corresponding graphs). Our aim is to prove that this is impossible. Note that $S_{\\mathrm{y}}, S_{\\mathrm{b}} \\neq V$ since $N \\leq 2^{N-2}+1$.\n\nFirst, suppose that $S_{\\mathrm{y}} \\cap S_{\\mathrm{b}} \\neq \\varnothing$, so there exists some $c \\in S_{\\mathrm{y}} \\cap S_{\\mathrm{b}}$. Note that each flag is connected to $c$ either in $G_{\\mathrm{y}}$ or in $G_{\\mathrm{b}}$, hence $E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right) \\cup E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)=V^{\\prime}$. Hence we have $2^{N-2}+1=\\left|V^{\\prime}\\right| \\leq\\left|E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)\\right|+\\left|E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right| \\leq\\left|S_{\\mathrm{y}}\\right|+\\left|S_{\\mathrm{b}}\\right|-2 \\leq 2 N-4$; this is impossible for $N \\geq 4$.\n\nSo, we have $S_{\\mathrm{y}} \\cap S_{\\mathrm{b}}=\\varnothing$. Let $y=\\left|S_{\\mathrm{y}}\\right|, b=\\left|S_{\\mathrm{b}}\\right|$. From the construction of our graph, we have that all the flags in the set $V^{\\prime \\prime}=V^{\\prime} \\backslash\\left(E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right) \\cup E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right)$ have blue squares in the columns of $S_{\\mathrm{y}}$ and yellow squares in the columns of $S_{\\mathrm{b}}$. Hence the only undetermined positions in these flags are the remaining $N-y-b$ ones, so $2^{N-y-b} \\geq\\left|V^{\\prime \\prime}\\right| \\geq\\left|V^{\\prime}\\right|-\\left|E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)\\right|-\\left|E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right| \\geq$ $2^{N-2}+1-(y-1)-(b-1)$, or, denoting $c=y+b, 2^{N-c}+c>2^{N-2}+2$. This is impossible since $N \\geq c \\geq 2$."
] | $M=2^{N-2}+1$ |
1,724 | Combinatorics | null | $\boxed{2}$ | [
"2"
] | false | null | Numerical | null | 2500 chess kings have to be placed on a $100 \times 100$ chessboard so that
(i) no king can capture any other one (i.e. no two kings are placed in two squares sharing a common vertex);
(ii) each row and each column contains exactly 25 kings.
Find the number of such arrangements. (Two arrangements differing by rotation or symmetry are supposed to be different.) | [
"Suppose that we have an arrangement satisfying the problem conditions. Divide the board into $2 \\times 2$ pieces; we call these pieces blocks. Each block can contain not more than one king (otherwise these two kings would attack each other); hence, by the pigeonhole principle each block must contain exactly one king.\n\nNow assign to each block a letter $\\mathrm{T}$ or $\\mathrm{B}$ if a king is placed in its top or bottom half, respectively. Similarly, assign to each block a letter $\\mathrm{L}$ or $\\mathrm{R}$ if a king stands in its left or right half. So we define T-blocks, B-blocks, L-blocks, and $R$-blocks. We also combine the letters; we call a block $a T L$-block if it is simultaneously T-block and L-block. Similarly we define TR-blocks, $B L$-blocks, and BR-blocks. The arrangement of blocks determines uniquely the arrangement of kings; so in the rest of the solution we consider the $50 \\times 50$ system of blocks (see Fig. 1). We identify the blocks by their coordinate pairs; the pair $(i, j)$, where $1 \\leq i, j \\leq 50$, refers to the $j$ th block in the $i$ th row (or the $i$ th block in the $j$ th column). The upper-left block is $(1,1)$.\n\nThe system of blocks has the following properties..\n\n$\\left(\\mathrm{i}^{\\prime}\\right)$ If $(i, j)$ is a B-block then $(i+1, j)$ is a B-block: otherwise the kings in these two blocks can take each other. Similarly: if $(i, j)$ is a T-block then $(i-1, j)$ is a T-block; if $(i, j)$ is an L-block then $(i, j-1)$ is an L-block; if $(i, j)$ is an R-block then $(i, j+1)$ is an R-block.\n\n(ii') Each column contains exactly 25 L-blocks and 25 R-blocks, and each row contains exactly 25 T-blocks and 25 B-blocks. In particular, the total number of L-blocks (or R-blocks, or T-blocks, or B-blocks) is equal to $25 \\cdot 50=1250$.\n\nConsider any B-block of the form $(1, j)$. By $\\left(\\mathrm{i}^{\\prime}\\right)$, all blocks in the $j$ th column are B-blocks; so we call such a column $B$-column. By (ii'), we have 25 B-blocks in the first row, so we obtain 25 B-columns. These $25 \\mathrm{~B}$-columns contain $1250 \\mathrm{~B}$-blocks, hence all blocks in the remaining columns are T-blocks, and we obtain 25 T-columns. Similarly, there are exactly 25 L-rows and exactly 25 -rows.\n\nNow consider an arbitrary pair of a T-column and a neighboring B-column (columns with numbers $j$ and $j+1$ ).\n\n<img_3973>\n\nFig. 1\n\n<img_3920>\n\nFig. 2\n\nCase 1. Suppose that the $j$ th column is a T-column, and the $(j+1)$ th column is a Bcolumn. Consider some index $i$ such that the $i$ th row is an L-row; then $(i, j+1)$ is a BL-block. Therefore, $(i+1, j)$ cannot be a TR-block (see Fig. 2), hence $(i+1, j)$ is a TL-block, thus the $(i+1)$ th row is an L-row. Now, choosing the $i$ th row to be the topmost L-row, we successively obtain that all rows from the $i$ th to the 50th are L-rows. Since we have exactly 25 L-rows, it follows that the rows from the 1st to the 25th are R-rows, and the rows from the 26th to the 50th are L-rows.\n\nNow consider the neighboring R-row and L-row (that are the rows with numbers 25 and 26). Replacing in the previous reasoning rows by columns and vice versa, the columns from the 1 st to the 25th are T-columns, and the columns from the 26th to the 50th are B-columns. So we have a unique arrangement of blocks that leads to the arrangement of kings satisfying the condition of the problem (see Fig. 3).\n\n<img_4043>\n\nFig. 3\n\n<img_3321>\n\nFig. 4\n\nCase 2. Suppose that the $j$ th column is a B-column, and the $(j+1)$ th column is a T-column. Repeating the arguments from Case 1, we obtain that the rows from the 1st to the 25th are L-rows (and all other rows are R-rows), the columns from the 1st to the 25th are B-columns (and all other columns are T-columns), so we find exactly one more arrangement of kings (see Fig. 4)."
] | 2 |
1,736 | Number Theory | null | $\boxed{39}$ | [
"39"
] | false | null | Numerical | null | Find the least positive integer $n$ for which there exists a set $\left\{s_{1}, s_{2}, \ldots, s_{n}\right\}$ consisting of $n$ distinct positive integers such that
$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010}
$$ | [
"Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}<s_{2}<\\cdots<s_{n}$. Surely $s_{1}>1$ since otherwise $1-\\frac{1}{s_{1}}=0$. So we have $2 \\leq s_{1} \\leq s_{2}-1 \\leq \\cdots \\leq s_{n}-(n-1)$, hence $s_{i} \\geq i+1$ for each $i=1, \\ldots, n$. Therefore\n\n$$\n\\begin{aligned}\n\\frac{51}{2010} & =\\left(1-\\frac{1}{s_{1}}\\right)\\left(1-\\frac{1}{s_{2}}\\right) \\ldots\\left(1-\\frac{1}{s_{n}}\\right) \\\\\n& \\geq\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\ldots\\left(1-\\frac{1}{n+1}\\right)=\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots \\frac{n}{n+1}=\\frac{1}{n+1}\n\\end{aligned}\n$$\n\nwhich implies\n\n$$\nn+1 \\geq \\frac{2010}{51}=\\frac{670}{17}>39\n$$\n\nso $n \\geq 39$.\n\nNow we are left to show that $n=39$ fits. Consider the set $\\{2,3, \\ldots, 33,35,36, \\ldots, 40,67\\}$ which contains exactly 39 numbers. We have\n\n$$\n\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots \\frac{32}{33} \\cdot \\frac{34}{35} \\cdots \\frac{39}{40} \\cdot \\frac{66}{67}=\\frac{1}{33} \\cdot \\frac{34}{40} \\cdot \\frac{66}{67}=\\frac{17}{670}=\\frac{51}{2010}\n\\tag{1}\n$$\n\nhence for $n=39$ there exists a desired example."
] | 39 |
1,737 | Number Theory | null | $\boxed{(6,3),(9,3),(9,5),(54,5)}$ | [
"$(6,3),(9,3),(9,5),(54,5)$"
] | true | null | Tuple | null | Find all pairs $(m, n)$ of nonnegative integers for which
$$
m^{2}+2 \cdot 3^{n}=m\left(2^{n+1}-1\right)
\tag{1}
$$ | [
"For fixed values of $n$, the equation (1) is a simple quadratic equation in $m$. For $n \\leq 5$ the solutions are listed in the following table.\n\n| case | equation | discriminant | integer roots |\n| :--- | :--- | :--- | :--- |\n| $n=0$ | $m^{2}-m+2=0$ | -7 | none |\n| $n=1$ | $m^{2}-3 m+6=0$ | -15 | none |\n| $n=2$ | $m^{2}-7 m+18=0$ | -23 | none |\n| $n=3$ | $m^{2}-15 m+54=0$ | 9 | $m=6$ and $m=9$ |\n| $n=4$ | $m^{2}-31 m+162=0$ | 313 | none |\n| $n=5$ | $m^{2}-63 m+486=0$ | $2025=45^{2}$ | $m=9$ and $m=54$ |\n\nWe prove that there is no solution for $n \\geq 6$.\n\nSuppose that $(m, n)$ satisfies (1) and $n \\geq 6$. Since $m \\mid 2 \\cdot 3^{n}=m\\left(2^{n+1}-1\\right)-m^{2}$, we have $m=3^{p}$ with some $0 \\leq p \\leq n$ or $m=2 \\cdot 3^{q}$ with some $0 \\leq q \\leq n$.\n\nIn the first case, let $q=n-p$; then\n\n$$\n2^{n+1}-1=m+\\frac{2 \\cdot 3^{n}}{m}=3^{p}+2 \\cdot 3^{q}\n$$\n\nIn the second case let $p=n-q$. Then\n\n$$\n2^{n+1}-1=m+\\frac{2 \\cdot 3^{n}}{m}=2 \\cdot 3^{q}+3^{p}\n$$\n\nHence, in both cases we need to find the nonnegative integer solutions of\n\n$$\n3^{p}+2 \\cdot 3^{q}=2^{n+1}-1, \\quad p+q=n .\n\\tag{2}\n$$\n\nNext, we prove bounds for $p, q$. From (2) we get\n\n$$\n3^{p}<2^{n+1}=8^{\\frac{n+1}{3}}<9^{\\frac{n+1}{3}}=3^{\\frac{2(n+1)}{3}}\n$$\n\nand\n\n$$\n2 \\cdot 3^{q}<2^{n+1}=2 \\cdot 8^{\\frac{n}{3}}<2 \\cdot 9^{\\frac{n}{3}}=2 \\cdot 3^{\\frac{2 n}{3}}<2 \\cdot 3^{\\frac{2(n+1)}{3}}\n$$\n\nso $p, q<\\frac{2(n+1)}{3}$. Combining these inequalities with $p+q=n$, we obtain\n\n$$\n\\frac{n-2}{3}<p, q<\\frac{2(n+1)}{3}\n\\tag{3}\n$$\n\nNow let $h=\\min (p, q)$. By (3) we have $h>\\frac{n-2}{3}$; in particular, we have $h>1$. On the left-hand side of (2), both terms are divisible by $3^{h}$, therefore $9\\left|3^{h}\\right| 2^{n+1}-1$. It is easy check that $\\operatorname{ord}_{9}(2)=6$, so $9 \\mid 2^{n+1}-1$ if and only if $6 \\mid n+1$. Therefore, $n+1=6 r$ for some positive integer $r$, and we can write\n\n$$\n2^{n+1}-1=4^{3 r}-1=\\left(4^{2 r}+4^{r}+1\\right)\\left(2^{r}-1\\right)\\left(2^{r}+1\\right)\n\\tag{4}\n$$\n\n\n\nNotice that the factor $4^{2 r}+4^{r}+1=\\left(4^{r}-1\\right)^{2}+3 \\cdot 4^{r}$ is divisible by 3 , but it is never divisible by 9 . The other two factors in (4), $2^{r}-1$ and $2^{r}+1$ are coprime: both are odd and their difference is 2 . Since the whole product is divisible by $3^{h}$, we have either $3^{h-1} \\mid 2^{r}-1$ or $3^{h-1} \\mid 2^{r}+1$. In any case, we have $3^{h-1} \\leq 2^{r}+1$. Then\n\n$$\n\\begin{gathered}\n3^{h-1} \\leq 2^{r}+1 \\leq 3^{r}=3^{\\frac{n+1}{6}} \\\\\n\\frac{n-2}{3}-1<h-1 \\leq \\frac{n+1}{6} \\\\\nn<11 .\n\\end{gathered}\n$$\n\nBut this is impossible since we assumed $n \\geq 6$, and we proved $6 \\mid n+1$."
] | $(6,3),(9,3),(9,5),(54,5)$ |
1,738 | Number Theory | null | $\boxed{5}$ | [
"5"
] | false | null | Numerical | null | Find the smallest number $n$ such that there exist polynomials $f_{1}, f_{2}, \ldots, f_{n}$ with rational coefficients satisfying
$$
x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+\cdots+f_{n}(x)^{2} .
$$ | [
"The equality $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$ shows that $n \\leq 5$. It remains to show that $x^{2}+7$ is not a sum of four (or less) squares of polynomials with rational coefficients.\n\nSuppose by way of contradiction that $x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+f_{3}(x)^{2}+f_{4}(x)^{2}$, where the coefficients of polynomials $f_{1}, f_{2}, f_{3}$ and $f_{4}$ are rational (some of these polynomials may be zero).\n\nClearly, the degrees of $f_{1}, f_{2}, f_{3}$ and $f_{4}$ are at most 1 . Thus $f_{i}(x)=a_{i} x+b_{i}$ for $i=1,2,3,4$ and some rationals $a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}, a_{4}, b_{4}$. It follows that $x^{2}+7=\\sum_{i=1}^{4}\\left(a_{i} x+b_{i}\\right)^{2}$ and hence\n\n$$\n\\sum_{i=1}^{4} a_{i}^{2}=1, \\quad \\sum_{i=1}^{4} a_{i} b_{i}=0, \\quad \\sum_{i=1}^{4} b_{i}^{2}=7\n\\tag{1}\n$$\n\nLet $p_{i}=a_{i}+b_{i}$ and $q_{i}=a_{i}-b_{i}$ for $i=1,2,3,4$. Then\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{4} p_{i}^{2} & =\\sum_{i=1}^{4} a_{i}^{2}+2 \\sum_{i=1}^{4} a_{i} b_{i}+\\sum_{i=1}^{4} b_{i}^{2}=8, \\\\\n\\sum_{i=1}^{4} q_{i}^{2} & =\\sum_{i=1}^{4} a_{i}^{2}-2 \\sum_{i=1}^{4} a_{i} b_{i}+\\sum_{i=1}^{4} b_{i}^{2}=8 \\\\\n\\text { and } \\sum_{i=1}^{4} p_{i} q_{i} & =\\sum_{i=1}^{4} a_{i}^{2}-\\sum_{i=1}^{4} b_{i}^{2}=-6,\n\\end{aligned}\n$$\n\nwhich means that there exist a solution in integers $x_{1}, y_{1}, x_{2}, y_{2}, x_{3}, y_{3}, x_{4}, y_{4}$ and $m>0$ of the system of equations\n(i) $\\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}$\n(ii) $\\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}$\n(iii) $\\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2}$.\n\nWe will show that such a solution does not exist.\n\nAssume the contrary and consider a solution with minimal $m$. Note that if an integer $x$ is odd then $x^{2} \\equiv 1(\\bmod 8)$. Otherwise (i.e., if $x$ is even) we have $x^{2} \\equiv 0(\\bmod 8)$ or $x^{2} \\equiv 4$ $(\\bmod 8)$. Hence, by $(\\mathrm{i})$, we get that $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are even. Similarly, by (ii), we get that $y_{1}, y_{2}, y_{3}$ and $y_{4}$ are even. Thus the LHS of (iii) is divisible by 4 and $m$ is also even. It follows that $\\left(\\frac{x_{1}}{2}, \\frac{y_{1}}{2}, \\frac{x_{2}}{2}, \\frac{y_{2}}{2}, \\frac{x_{3}}{2}, \\frac{y_{3}}{2}, \\frac{x_{4}}{2}, \\frac{y_{4}}{2}, \\frac{m}{2}\\right)$ is a solution of the system of equations (i), (ii) and (iii), which contradicts the minimality of $m$."
] | 5 |
1,747 | Algebra | null | $\boxed{M=\frac{9}{32} \sqrt{2}}$ | [
"$M=\\frac{9}{32} \\sqrt{2}$"
] | false | null | Numerical | null | Determine the smallest number $M$ such that the inequality
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}
$$
holds for all real numbers $a, b, c$. | [
"We first consider the cubic polynomial\n\n$$\nP(t)=t b\\left(t^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c t\\left(c^{2}-t^{2}\\right) .\n$$\n\nIt is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore\n\n$$\nP(t)=(b-c)(t-b)(t-c)(t+b+c)\n$$\n\nsince the cubic coefficient is $b-c$. The left-hand side of the proposed inequality can therefore be written in the form\n\n$$\n\\left|a b\\left(a^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c a\\left(c^{2}-a^{2}\\right)\\right|=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)| .\n$$\n\nThe problem comes down to finding the smallest number $M$ that satisfies the inequality\n\n$$\n|(b-c)(a-b)(a-c)(a+b+c)| \\leq M \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{2} . \\tag{1}\n$$\n\nNote that this expression is symmetric, and we can therefore assume $a \\leq b \\leq c$ without loss of generality. With this assumption,\n\n$$\n|(a-b)(b-c)|=(b-a)(c-b) \\leq\\left(\\frac{(b-a)+(c-b)}{2}\\right)^{2}=\\frac{(c-a)^{2}}{4} \\tag{2}\n$$\n\nwith equality if and only if $b-a=c-b$, i.e. $2 b=a+c$. Also\n\n$$\n\\left(\\frac{(c-b)+(b-a)}{2}\\right)^{2} \\leq \\frac{(c-b)^{2}+(b-a)^{2}}{2} \\tag{3}\n$$\n\nor equivalently,\n\n$$\n3(c-a)^{2} \\leq 2 \\cdot\\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\\right]\n$$\n\nagain with equality only for $2 b=a+c$. From (2) and (3) we get\n\n$$\n\\begin{aligned}\n& |(b-c)(a-b)(a-c)(a+b+c)| \\\\\n\\leq & \\frac{1}{4} \\cdot\\left|(c-a)^{3}(a+b+c)\\right| \\\\\n= & \\frac{1}{4} \\cdot \\sqrt{(c-a)^{6}(a+b+c)^{2}} \\\\\n\\leq & \\frac{1}{4} \\cdot \\sqrt{\\left(\\frac{2 \\cdot\\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\\right]}{3}\\right)^{3} \\cdot(a+b+c)^{2}} \\\\\n= & \\frac{\\sqrt{2}}{2} \\cdot\\left(\\sqrt[4]{\\left(\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\\right)^{3} \\cdot(a+b+c)^{2}}\\right)^{2} .\n\\end{aligned}\n$$\n\n\n\nBy the weighted AM-GM inequality this estimate continues as follows:\n\n$$\n\\begin{aligned}\n& |(b-c)(a-b)(a-c)(a+b+c)| \\\\\n\\leq & \\frac{\\sqrt{2}}{2} \\cdot\\left(\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\\right)^{2} \\\\\n= & \\frac{9 \\sqrt{2}}{32} \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{2} .\n\\end{aligned}\n$$\n\nWe see that the inequality (1) is satisfied for $M=\\frac{9}{32} \\sqrt{2}$, with equality if and only if $2 b=a+c$ and\n\n$$\n\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2}\n$$\n\nPlugging $b=(a+c) / 2$ into the last equation, we bring it to the equivalent form\n\n$$\n2(c-a)^{2}=9(a+c)^{2} .\n$$\n\nThe conditions for equality can now be restated as\n\n$$\n2 b=a+c \\quad \\text { and } \\quad(c-a)^{2}=18 b^{2} .\n$$\n\nSetting $b=1$ yields $a=1-\\frac{3}{2} \\sqrt{2}$ and $c=1+\\frac{3}{2} \\sqrt{2}$. We see that $M=\\frac{9}{32} \\sqrt{2}$ is indeed the smallest constant satisfying the inequality, with equality for any triple $(a, b, c)$ proportional to $\\left(1-\\frac{3}{2} \\sqrt{2}, 1,1+\\frac{3}{2} \\sqrt{2}\\right)$, up to permutation.\n\n"
] | $M=\frac{9}{32} \sqrt{2}$ |
1,750 | Combinatorics | null | $\boxed{1003}$ | [
"$1003$"
] | false | null | Numerical | null | A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals.
Suppose the 2006-gon has been dissected into triangles by 2003 nonintersecting diagonals. Find the maximum possible number of isosceles triangles with two odd sides. | [
"Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity.\n\nLemma. Let $A B$ be one of dissecting diagonals and let $\\mathcal{L}$ be the shorter part of the boundary of the 2006-gon with endpoints $A, B$. Suppose that $\\mathcal{L}$ consists of $n$ segments. Then the number of iso-odd triangles with vertices on $\\mathcal{L}$ does not exceed $n / 2$.\n\nProof. This is obvious for $n=2$. Take $n$ with $2<n \\leq 1003$ and assume the claim to be true for every $\\mathcal{L}$ of length less than $n$. Let now $\\mathcal{L}$ (endpoints $A, B$ ) consist of $n$ segments. Let $P Q$ be the longest diagonal which is a side of an iso-odd triangle $P Q S$ with all vertices on $\\mathcal{L}$ (if there is no such triangle, there is nothing to prove). Every triangle whose vertices lie on $\\mathcal{L}$ is obtuse or right-angled; thus $S$ is the summit of $P Q S$. We may assume that the five points $A, P, S, Q, B$ lie on $\\mathcal{L}$ in this order and partition $\\mathcal{L}$ into four pieces $\\mathcal{L}_{A P}, \\mathcal{L}_{P S}, \\mathcal{L}_{S Q}, \\mathcal{L}_{Q B}$ (the outer ones possibly reducing to a point).\n\nBy the definition of $P Q$, an iso-odd triangle cannot have vertices on both $\\mathcal{L}_{A P}$ and $\\mathcal{L}_{Q B}$. Therefore every iso-odd triangle within $\\mathcal{L}$ has all its vertices on just one of the four pieces. Applying to each of these pieces the induction hypothesis and adding the four inequalities we get that the number of iso-odd triangles within $\\mathcal{L}$ other than $P Q S$ does not exceed $n / 2$. And since each of $\\mathcal{L}_{P S}, \\mathcal{L}_{S Q}$ consists of an odd number of sides, the inequalities for these two pieces are actually strict, leaving a $1 / 2+1 / 2$ in excess. Hence the triangle $P S Q$ is also covered by the estimate $n / 2$. This concludes the induction step and proves the lemma.\n\nThe remaining part of the solution in fact repeats the argument from the above proof. Consider the longest dissecting diagonal $X Y$. Let $\\mathcal{L}_{X Y}$ be the shorter of the two parts of the boundary with endpoints $X, Y$ and let $X Y Z$ be the triangle in the dissection with vertex $Z$ not on $\\mathcal{L}_{X Y}$. Notice that $X Y Z$ is acute or right-angled, otherwise one of the segments $X Z, Y Z$ would be longer than $X Y$. Denoting by $\\mathcal{L}_{X Z}, \\mathcal{L}_{Y Z}$ the two pieces defined by $Z$ and applying the lemma to each of $\\mathcal{L}_{X Y}, \\mathcal{L}_{X Z}, \\mathcal{L}_{Y Z}$ we infer that there are no more than 2006/2 iso-odd triangles in all, unless $X Y Z$ is one of them. But in that case $X Z$ and $Y Z$ are odd diagonals and the corresponding inequalities are strict. This shows that also in this case the total number of iso-odd triangles in the dissection, including $X Y Z$, is not greater than 1003.\n\nThis bound can be achieved. For this to happen, it just suffices to select a vertex of the 2006-gon and draw a broken line joining every second vertex, starting from the selected one. Since 2006 is even, the line closes. This already gives us the required 1003 iso-odd triangles. Then we can complete the triangulation in an arbitrary fashion.",
"Let the terms odd triangle and iso-odd triangle have the same meaning as in the first solution.\n\nLet $A B C$ be an iso-odd triangle, with $A B$ and $B C$ odd sides. This means that there are an odd number of sides of the 2006-gon between $A$ and $B$ and also between $B$ and $C$. We say that these sides belong to the iso-odd triangle $A B C$.\n\nAt least one side in each of these groups does not belong to any other iso-odd triangle. This is so because any odd triangle whose vertices are among the points between $A$ and $B$ has two sides of equal length and therefore has an even number of sides belonging to it in total. Eliminating all sides belonging to any other iso-odd triangle in this area must therefore leave one side that belongs to no other iso-odd triangle. Let us assign these two sides (one in each group) to the triangle $A B C$.\n\nTo each iso-odd triangle we have thus assigned a pair of sides, with no two triangles sharing an assigned side. It follows that at most 1003 iso-odd triangles can appear in the dissection.\n\nThis value can be attained, as shows the example from the first solution."
] | $1003$ |
1,760 | Geometry | null | $\boxed{\angle B E A_{1}=90$,$\angle A E B_{1}=90}$ | [
"$\\angle B E A_{1}=90$,$\\angle A E B_{1}=90$"
] | true | ^{\circ} | Numerical | null | In triangle $A B C$, let $J$ be the centre of the excircle tangent to side $B C$ at $A_{1}$ and to the extensions of sides $A C$ and $A B$ at $B_{1}$ and $C_{1}$, respectively. Suppose that the lines $A_{1} B_{1}$ and $A B$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $D J$. Determine the angles $\angle B E A_{1}$ and $\angle A E B_{1}$. | [
"Let $K$ be the intersection point of lines $J C$ and $A_{1} B_{1}$. Obviously $J C \\perp A_{1} B_{1}$ and since $A_{1} B_{1} \\perp A B$, the lines $J K$ and $C_{1} D$ are parallel and equal. From the right triangle $B_{1} C J$ we obtain $J C_{1}^{2}=J B_{1}^{2}=J C \\cdot J K=J C \\cdot C_{1} D$ from which we infer that $D C_{1} / C_{1} J=C_{1} J / J C$ and the right triangles $D C_{1} J$ and $C_{1} J C$ are similar. Hence $\\angle C_{1} D J=\\angle J C_{1} C$, which implies that the lines $D J$ and $C_{1} C$ are perpendicular, i.e. the points $C_{1}, E, C$ are collinear.\n\n<img_3975>\n\nSince $\\angle C A_{1} J=\\angle C B_{1} J=\\angle C E J=90^{\\circ}$, points $A_{1}, B_{1}$ and $E$ lie on the circle of diameter $C J$. Then $\\angle D B A_{1}=\\angle A_{1} C J=\\angle D E A_{1}$, which implies that quadrilateral $B E A_{1} D$ is cyclic; therefore $\\angle A_{1} E B=90^{\\circ}$.\n\nQuadrilateral $A D E B_{1}$ is also cyclic because $\\angle E B_{1} A=\\angle E J C=\\angle E D C_{1}$, therefore we obtain $\\angle A E B_{1}=\\angle A D B=90^{\\circ}$.\n\n<img_3438>",
"Consider the circles $\\omega_{1}, \\omega_{2}$ and $\\omega_{3}$ of diameters $C_{1} D, A_{1} B$ and $A B_{1}$, respectively. Line segments $J C_{1}, J B_{1}$ and $J A_{1}$ are tangents to those circles and, due to the right angle at $D$, $\\omega_{2}$ and $\\omega_{3}$ pass through point $D$. Since $\\angle C_{1} E D$ is a right angle, point $E$ lies on circle $\\omega_{1}$, therefore\n\n$$\nJ C_{1}^{2}=J D \\cdot J E\n$$\n\nSince $J A_{1}=J B_{1}=J C_{1}$ are all radii of the excircle, we also have\n\n$$\nJ A_{1}^{2}=J D \\cdot J E \\quad \\text { and } \\quad J B_{1}^{2}=J D \\cdot J E .\n$$\n\nThese equalities show that $E$ lies on circles $\\omega_{2}$ and $\\omega_{3}$ as well, so $\\angle B E A_{1}=\\angle A E B_{1}=90^{\\circ}$.\n\n### solution_2\nFirst note that $A_{1} B_{1}$ is perpendicular to the external angle bisector $C J$ of $\\angle B C A$ and parallel to the internal angle bisector of that angle. Therefore, $A_{1} B_{1}$ is perpendicular to $A B$ if and only if triangle $A B C$ is isosceles, $A C=B C$. In that case the external bisector $C J$ is parallel to $A B$.\n\nTriangles $A B C$ and $B_{1} A_{1} J$ are similar, as their corresponding sides are perpendicular. In particular, we have $\\angle D A_{1} J=\\angle C_{1} B A_{1}$; moreover, from cyclic deltoid $J A_{1} B C_{1}$,\n\n$$\n\\angle C_{1} A_{1} J=\\angle C_{1} B J=\\frac{1}{2} \\angle C_{1} B A_{1}=\\frac{1}{2} \\angle D A_{1} J\n$$\n\nTherefore, $A_{1} C_{1}$ bisects angle $\\angle D A_{1} J$.\n\n<img_3155>\n\nIn triangle $B_{1} A_{1} J$, line $J C_{1}$ is the external bisector at vertex $J$. The point $C_{1}$ is the intersection of two external angle bisectors (at $A_{1}$ and $J$ ) so $C_{1}$ is the centre of the excircle $\\omega$, tangent to side $A_{1} J$, and to the extension of $B_{1} A_{1}$ at point $D$.\n\nNow consider the similarity transform $\\varphi$ which moves $B_{1}$ to $A, A_{1}$ to $B$ and $J$ to $C$. This similarity can be decomposed into a rotation by $90^{\\circ}$ around a certain point $O$ and a homothety from the same centre. This similarity moves point $C_{1}$ (the centre of excircle $\\omega$ ) to $J$ and moves $D$ (the point of tangency) to $C_{1}$.\n\nSince the rotation angle is $90^{\\circ}$, we have $\\angle X O \\varphi(X)=90^{\\circ}$ for an arbitrary point $X \\neq O$. For $X=D$ and $X=C_{1}$ we obtain $\\angle D O C_{1}=\\angle C_{1} O J=90^{\\circ}$. Therefore $O$ lies on line segment $D J$ and $C_{1} O$ is perpendicular to $D J$. This means that $O=E$.\n\nFor $X=A_{1}$ and $X=B_{1}$ we obtain $\\angle A_{1} O B=\\angle B_{1} O A=90^{\\circ}$, i.e.\n\n$$\n\\angle B E A_{1}=\\angle A E B_{1}=90^{\\circ} .\n$$"
] | $\angle B E A_{1}=90$,$\angle A E B_{1}=90$ |
1,766 | Number Theory | null | $\boxed{(0,2),(0,-2),(4,23),(4,-23)}$ | [
"$(0,2),(0,-2),(4,23),(4,-23)$"
] | true | null | Tuple | null | Determine all pairs $(x, y)$ of integers satisfying the equation
$$
1+2^{x}+2^{2 x+1}=y^{2}
$$ | [
"If $(x, y)$ is a solution then obviously $x \\geq 0$ and $(x,-y)$ is a solution too. For $x=0$ we get the two solutions $(0,2)$ and $(0,-2)$.\n\nNow let $(x, y)$ be a solution with $x>0$; without loss of generality confine attention to $y>0$. The equation rewritten as\n\n$$\n2^{x}\\left(1+2^{x+1}\\right)=(y-1)(y+1)\n$$\n\nshows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by 4 . Hence $x \\geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^{x}$. So\n\n$$\ny=2^{x-1} m+\\epsilon, \\quad m \\text { odd }, \\quad \\epsilon= \\pm 1\\tag{1}\n$$\n\nPlugging this into the original equation we obtain\n\n$$\n2^{x}\\left(1+2^{x+1}\\right)=\\left(2^{x-1} m+\\epsilon\\right)^{2}-1=2^{2 x-2} m^{2}+2^{x} m \\epsilon,\n$$\n\nor, equivalently\n\n$$\n1+2^{x+1}=2^{x-2} m^{2}+m \\epsilon .\n$$\n\nTherefore\n\n$$\n1-\\epsilon m=2^{x-2}\\left(m^{2}-8\\right) .\\tag{2}\n$$\n\nFor $\\epsilon=1$ this yields $m^{2}-8 \\leq 0$, i.e., $m=1$, which fails to satisfy (2).\n\nFor $\\epsilon=-1$ equation (2) gives us\n\n$$\n1+m=2^{x-2}\\left(m^{2}-8\\right) \\geq 2\\left(m^{2}-8\\right),\n$$\n\nimplying $2 m^{2}-m-17 \\leq 0$. Hence $m \\leq 3$; on the other hand $m$ cannot be 1 by $(2)$. Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From (1) we get $y=23$. These values indeed satisfy the given equation. Recall that then $y=-23$ is also good. Thus we have the complete list of solutions $(x, y):(0,2),(0,-2),(4,23),(4,-23)$."
] | $(0,2),(0,-2),(4,23),(4,-23)$ |
1,775 | Algebra | null | $\boxed{\left\lfloor\log _{2} n\right\rfloor+1}$ | [
"$\\left\\lfloor\\log _{2} n\\right\\rfloor+1$"
] | false | null | Expression | null | Given a positive integer $n$, find the smallest value of $\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{n}}{n}\right\rfloor$ over all permutations $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $(1,2, \ldots, n)$. | [
"Suppose that $2^{k} \\leqslant n<2^{k+1}$ with some nonnegative integer $k$. First we show a permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ such that $\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor=k+1$; then we will prove that $\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor \\geqslant k+1$ for every permutation. Hence, the minimal possible value will be $k+1$.\n\nI. Consider the permutation\n\n$$\n\\begin{gathered}\n\\left(a_{1}\\right)=(1), \\quad\\left(a_{2}, a_{3}\\right)=(3,2), \\quad\\left(a_{4}, a_{5}, a_{6}, a_{7}\\right)=(7,4,5,6), \\\\\n\\left(a_{2^{k-1}}, \\ldots, a_{2^{k}-1}\\right)=\\left(2^{k}-1,2^{k-1}, 2^{k-1}+1, \\ldots, 2^{k}-2\\right), \\\\\n\\left(a_{2^{k}}, \\ldots, a_{n}\\right)=\\left(n, 2^{k}, 2^{k}+1, \\ldots, n-1\\right) .\n\\end{gathered}\n$$\n\nThis permutation consists of $k+1$ cycles. In every cycle $\\left(a_{p}, \\ldots, a_{q}\\right)=(q, p, p+1, \\ldots, q-1)$ we have $q<2 p$, so\n\n$$\n\\sum_{i=p}^{q}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor=\\left\\lfloor\\frac{q}{p}\\right\\rfloor+\\sum_{i=p+1}^{q}\\left\\lfloor\\frac{i-1}{i}\\right\\rfloor=1\n$$\n\nThe total sum over all cycles is precisely $k+1$.\n\nII. In order to establish the lower bound, we prove a more general statement.\n\nClaim. If $b_{1}, \\ldots, b_{2^{k}}$ are distinct positive integers then\n\n$$\n\\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant k+1\n$$\n\nFrom the Claim it follows immediately that $\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant k+1$.\n\nProof of the Claim. Apply induction on $k$. For $k=1$ the claim is trivial, $\\left\\lfloor\\frac{b_{1}}{1}\\right\\rfloor \\geqslant 1$. Suppose the Claim holds true for some positive integer $k$, and consider $k+1$.\n\nIf there exists an index $j$ such that $2^{k}<j \\leqslant 2^{k+1}$ and $b_{j} \\geqslant j$ then\n\n$$\n\\sum_{i=1}^{2^{k+1}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor+\\left\\lfloor\\frac{b_{j}}{j}\\right\\rfloor \\geqslant(k+1)+1\n$$\n\nby the induction hypothesis, so the Claim is satisfied.\n\nOtherwise we have $b_{j}<j \\leqslant 2^{k+1}$ for every $2^{k}<j \\leqslant 2^{k+1}$. Among the $2^{k+1}$ distinct numbers $b_{1}, \\ldots, b_{2^{k+1}}$ there is some $b_{m}$ which is at least $2^{k+1}$; that number must be among $b_{1} \\ldots, b_{2^{k}}$. Hence, $1 \\leqslant m \\leqslant 2^{k}$ and $b_{m} \\geqslant 2^{k+1}$.\n\nWe will apply the induction hypothesis to the numbers\n\n$$\nc_{1}=b_{1}, \\ldots, c_{m-1}=b_{m-1}, \\quad c_{m}=b_{2^{k}+1}, \\quad c_{m+1}=b_{m+1}, \\ldots, c_{2^{k}}=b_{2^{k}}\n$$\n\nso take the first $2^{k}$ numbers but replace $b_{m}$ with $b_{2^{k}+1}$. Notice that\n\n$$\n\\left\\lfloor\\frac{b_{m}}{m}\\right\\rfloor \\geqslant\\left\\lfloor\\frac{2^{k+1}}{m}\\right\\rfloor=\\left\\lfloor\\frac{2^{k}+2^{k}}{m}\\right\\rfloor \\geqslant\\left\\lfloor\\frac{b_{2^{k}+1}+m}{m}\\right\\rfloor=\\left\\lfloor\\frac{c_{m}}{m}\\right\\rfloor+1\n$$\n\n\n\nFor the other indices $i$ with $1 \\leqslant i \\leqslant 2^{k}, i \\neq m$ we have $\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor=\\left\\lfloor\\frac{c_{i}}{i}\\right\\rfloor$, so\n\n$$\n\\sum_{i=1}^{2^{k+1}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor=\\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{c_{i}}{i}\\right\\rfloor+1 \\geqslant(k+1)+1\n$$\n\nThat proves the Claim and hence completes the solution.",
"Assume $2^{k} \\leqslant n<2^{k+1}$, and let $P=\\left\\{2^{0}, 2^{1}, \\ldots, 2^{k}\\right\\}$ be the set of powers of 2 among $1,2, \\ldots, n$. Call an integer $i \\in\\{1,2, \\ldots, n\\}$ and the interval $\\left[i, a_{i}\\right]$ good if $a_{i} \\geqslant i$.\n\nLemma 1. The good intervals cover the integers $1,2, \\ldots, n$.\n\nProof. Consider an arbitrary $x \\in\\{1,2 \\ldots, n\\}$; we want to find a good interval $\\left[i, a_{i}\\right]$ that covers $x$; i.e., $i \\leqslant x \\leqslant a_{i}$. Take the cycle of the permutation that contains $x$, that is $\\left(x, a_{x}, a_{a_{x}}, \\ldots\\right)$. In this cycle, let $i$ be the first element with $a_{i} \\geqslant x$; then $i \\leqslant x \\leqslant a_{i}$.\n\nLemma 2. If a good interval $\\left[i, a_{i}\\right]$ covers $p$ distinct powers of 2 then $\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant p$; more formally, $\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant\\left|\\left[i, a_{i}\\right] \\cap P\\right|$.\n\nProof. The ratio of the smallest and largest powers of 2 in the interval is at least $2^{p-1}$. By Bernoulli's inequality, $\\frac{a_{i}}{i} \\geqslant 2^{p-1} \\geqslant p$; that proves the lemma.\n\nNow, by Lemma 1, the good intervals cover $P$. By applying Lemma 2 as well, we obtain that\n\n$$\n\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor=\\sum_{i \\text { is good }}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i \\text { is good }}^{n}\\left|\\left[i, a_{i}\\right] \\cap P\\right| \\geqslant|P|=k+1\n$$",
"We show proof based on the following inequality.\n\nLemma 3. \n\n$$\n\\left\\lfloor\\frac{a}{b}\\right\\rfloor \\geqslant \\log _{2} \\frac{a+1}{b}\n$$\n\nfor every pair $a, b$ of positive integers.\n\nProof. Let $t=\\left\\lfloor\\frac{a}{b}\\right\\rfloor$, so $t \\leqslant \\frac{a}{b}$ and $\\frac{a+1}{b} \\leqslant t+1$. By applying the inequality $2^{t} \\geqslant t+1$, we obtain\n\n$$\n\\left\\lfloor\\frac{a}{b}\\right\\rfloor=t \\geqslant \\log _{2}(t+1) \\geqslant \\log _{2} \\frac{a+1}{b}\n$$\n\nBy applying the lemma to each term, we get\n\n$$\n\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{n} \\log _{2} \\frac{a_{i}+1}{i}=\\sum_{i=1}^{n} \\log _{2}\\left(a_{i}+1\\right)-\\sum_{i=1}^{n} \\log _{2} i\n$$\n\nNotice that the numbers $a_{1}+1, a_{2}+1, \\ldots, a_{n}+1$ form a permutation of $2,3, \\ldots, n+1$. Hence, in the last two sums all terms cancel out, except for $\\log _{2}(n+1)$ in the first sum and $\\log _{2} 1=0$ in the second sum. Therefore,\n\n$$\n\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\log _{2}(n+1)>k\n$$\n\nAs the left-hand side is an integer, it must be at least $k+1$."
] | $\left\lfloor\log _{2} n\right\rfloor+1$ |
1,782 | Combinatorics | null | $\boxed{m_{\max }=n^{2}-n-1}$ | [
"$m_{\\max }=n^{2}-n-1$"
] | false | null | Expression | null | Let $n \geqslant 3$ be an integer. An integer $m \geqslant n+1$ is called $n$-colourful if, given infinitely many marbles in each of $n$ colours $C_{1}, C_{2}, \ldots, C_{n}$, it is possible to place $m$ of them around a circle so that in any group of $n+1$ consecutive marbles there is at least one marble of colour $C_{i}$ for each $i=1, \ldots, n$.
Prove that there are only finitely many positive integers which are not $n$-colourful. Find the largest among them. | [
"First suppose that there are $n(n-1)-1$ marbles. Then for one of the colours, say blue, there are at most $n-2$ marbles, which partition the non-blue marbles into at most $n-2$ groups with at least $(n-1)^{2}>n(n-2)$ marbles in total. Thus one of these groups contains at least $n+1$ marbles and this group does not contain any blue marble.\n\nNow suppose that the total number of marbles is at least $n(n-1)$. Then we may write this total number as $n k+j$ with some $k \\geqslant n-1$ and with $0 \\leqslant j \\leqslant n-1$. We place around a circle $k-j$ copies of the colour sequence $[1,2,3, \\ldots, n]$ followed by $j$ copies of the colour sequence $[1,1,2,3, \\ldots, n]$."
] | $m_{\max }=n^{2}-n-1$ |
1,789 | Combinatorics | null | $\boxed{\frac{100!}{2^{50}}}$ | [
"$\\frac{100!}{2^{50}}$"
] | false | null | Expression | null | Determine the largest $N$ for which there exists a table $T$ of integers with $N$ rows and 100 columns that has the following properties:
(i) Every row contains the numbers 1,2, ., 100 in some order.
(ii) For any two distinct rows $r$ and $s$, there is a column $c$ such that $|T(r, c)-T(s, c)| \geqslant 2$.
Here $T(r, c)$ means the number at the intersection of the row $r$ and the column $c$. | [
"Non-existence of a larger table. Let us consider some fixed row in the table, and let us replace (for $k=1,2, \\ldots, 50$ ) each of two numbers $2 k-1$ and $2 k$ respectively by the symbol $x_{k}$. The resulting pattern is an arrangement of 50 symbols $x_{1}, x_{2}, \\ldots, x_{50}$, where every symbol occurs exactly twice. Note that there are $N=100 ! / 2^{50}$ distinct patterns $P_{1}, \\ldots, P_{N}$.\n\nIf two rows $r \\neq s$ in the table have the same pattern $P_{i}$, then $|T(r, c)-T(s, c)| \\leqslant 1$ holds for all columns $c$. As this violates property (ii) in the problem statement, different rows have different patterns. Hence there are at most $N=100 ! / 2^{50}$ rows.\n\nExistence of a table with $N$ rows. We construct the table by translating every pattern $P_{i}$ into a corresponding row with the numbers $1,2, \\ldots, 100$. We present a procedure that inductively replaces the symbols by numbers. The translation goes through steps $k=1,2, \\ldots, 50$ in increasing order and at step $k$ replaces the two occurrences of symbol $x_{k}$ by $2 k-1$ and $2 k$.\n\n- The left occurrence of $x_{1}$ is replaced by 1 , and its right occurrence is replaced by 2 .\n- For $k \\geqslant 2$, we already have the number $2 k-2$ somewhere in the row, and now we are looking for the places for $2 k-1$ and $2 k$. We make the three numbers $2 k-2,2 k-1,2 k$ show up (ordered from left to right) either in the order $2 k-2,2 k-1,2 k$, or as $2 k, 2 k-2,2 k-1$, or as $2 k-1,2 k, 2 k-2$. This is possible, since the number $2 k-2$ has been placed in the preceding step, and shows up before / between / after the two occurrences of the symbol $x_{k}$.\n\nWe claim that the $N$ rows that result from the $N$ patterns yield a table with the desired property (ii). Indeed, consider the $r$-th and the $s$-th row $(r \\neq s)$, which by construction result from patterns $P_{r}$ and $P_{s}$. Call a symbol $x_{i}$ aligned, if it occurs in the same two columns in $P_{r}$ and in $P_{s}$. Let $k$ be the largest index, for which symbol $x_{k}$ is not aligned. Note that $k \\geqslant 2$. Consider the column $c^{\\prime}$ with $T\\left(r, c^{\\prime}\\right)=2 k$ and the column $c^{\\prime \\prime}$ with $T\\left(s, c^{\\prime \\prime}\\right)=2 k$. Then $T\\left(r, c^{\\prime \\prime}\\right) \\leqslant 2 k$ and $T\\left(s, c^{\\prime}\\right) \\leqslant 2 k$, as all symbols $x_{i}$ with $i \\geqslant k+1$ are aligned.\n\n- If $T\\left(r, c^{\\prime \\prime}\\right) \\leqslant 2 k-2$, then $\\left|T\\left(r, c^{\\prime \\prime}\\right)-T\\left(s, c^{\\prime \\prime}\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(s, c^{\\prime}\\right) \\leqslant 2 k-2$, then $\\left|T\\left(r, c^{\\prime}\\right)-T\\left(s, c^{\\prime}\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(r, c^{\\prime \\prime}\\right)=2 k-1$ and $T\\left(s, c^{\\prime}\\right)=2 k-1$, then the symbol $x_{k}$ is aligned; contradiction.\n\nIn the only remaining case we have $c^{\\prime}=c^{\\prime \\prime}$, so that $T\\left(r, c^{\\prime}\\right)=T\\left(s, c^{\\prime}\\right)=2 k$ holds. Now let us consider the columns $d^{\\prime}$ and $d^{\\prime \\prime}$ with $T\\left(r, d^{\\prime}\\right)=2 k-1$ and $T\\left(s, d^{\\prime \\prime}\\right)=2 k-1$. Then $d \\neq d^{\\prime \\prime}$ (as the symbol $x_{k}$ is not aligned), and $T\\left(r, d^{\\prime \\prime}\\right) \\leqslant 2 k-2$ and $T\\left(s, d^{\\prime}\\right) \\leqslant 2 k-2$ (as all symbols $x_{i}$ with $i \\geqslant k+1$ are aligned).\n\n- If $T\\left(r, d^{\\prime \\prime}\\right) \\leqslant 2 k-3$, then $\\left|T\\left(r, d^{\\prime \\prime}\\right)-T\\left(s, d^{\\prime \\prime}\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(s, c^{\\prime}\\right) \\leqslant 2 k-3$, then $\\left|T\\left(r, d^{\\prime}\\right)-T\\left(s, d^{\\prime}\\right)\\right| \\geqslant 2$ as desired.\n\nIn the only remaining case we have $T\\left(r, d^{\\prime \\prime}\\right)=2 k-2$ and $T\\left(s, d^{\\prime}\\right)=2 k-2$. Now the row $r$ has the numbers $2 k-2,2 k-1,2 k$ in the three columns $d^{\\prime}, d^{\\prime \\prime}, c^{\\prime}$. As one of these triples violates the ordering property of $2 k-2,2 k-1,2 k$, we have the final contradiction."
] | $\frac{100!}{2^{50}}$ |
1,798 | Number Theory | null | $\boxed{2}$ | [
"2"
] | false | null | Numerical | null | Determine all integers $n \geqslant 1$ for which there exists a pair of positive integers $(a, b)$ such that no cube of a prime divides $a^{2}+b+3$ and
$$
\frac{a b+3 b+8}{a^{2}+b+3}=n
$$ | [
"As $b \\equiv-a^{2}-3\\left(\\bmod a^{2}+b+3\\right)$, the numerator of the given fraction satisfies\n\n$$\na b+3 b+8 \\equiv a\\left(-a^{2}-3\\right)+3\\left(-a^{2}-3\\right)+8 \\equiv-(a+1)^{3} \\quad\\left(\\bmod a^{2}+b+3\\right)\n$$\n\nAs $a^{2}+b+3$ is not divisible by $p^{3}$ for any prime $p$, if $a^{2}+b+3$ divides $(a+1)^{3}$ then it does also divide $(a+1)^{2}$. Since\n\n$$\n0<(a+1)^{2}<2\\left(a^{2}+b+3\\right)\n$$\n\nwe conclude $(a+1)^{2}=a^{2}+b+3$. This yields $b=2(a-1)$ and $n=2$. The choice $(a, b)=(2,2)$ with $a^{2}+b+3=9$ shows that $n=2$ indeed is a solution."
] | 2 |
1,800 | Number Theory | null | $\boxed{1,3}$ | [
"1,3"
] | true | null | Numerical | null | Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $\left(d_{1}, d_{2}, \ldots, d_{k}\right)$ such that for every $i=1,2, \ldots, k$, the number $d_{1}+\cdots+d_{i}$ is a perfect square. | [
"For $i=1,2, \\ldots, k$ let $d_{1}+\\ldots+d_{i}=s_{i}^{2}$, and define $s_{0}=0$ as well. Obviously $0=s_{0}<s_{1}<s_{2}<\\ldots<s_{k}$, so\n\n$$\ns_{i} \\geqslant i \\quad \\text { and } \\quad d_{i}=s_{i}^{2}-s_{i-1}^{2}=\\left(s_{i}+s_{i-1}\\right)\\left(s_{i}-s_{i-1}\\right) \\geqslant s_{i}+s_{i-1} \\geqslant 2 i-1\n\\tag{1}\n$$\n\nThe number 1 is one of the divisors $d_{1}, \\ldots, d_{k}$ but, due to $d_{i} \\geqslant 2 i-1$, the only possibility is $d_{1}=1$.\n\nNow consider $d_{2}$ and $s_{2} \\geqslant 2$. By definition, $d_{2}=s_{2}^{2}-1=\\left(s_{2}-1\\right)\\left(s_{2}+1\\right)$, so the numbers $s_{2}-1$ and $s_{2}+1$ are divisors of $n$. In particular, there is some index $j$ such that $d_{j}=s_{2}+1$.\n\nNotice that\n\n$$\ns_{2}+s_{1}=s_{2}+1=d_{j} \\geqslant s_{j}+s_{j-1} ;\n\\tag{2}\n$$\n\nsince the sequence $s_{0}<s_{1}<\\ldots<s_{k}$ increases, the index $j$ cannot be greater than 2. Hence, the divisors $s_{2}-1$ and $s_{2}+1$ are listed among $d_{1}$ and $d_{2}$. That means $s_{2}-1=d_{1}=1$ and $s_{2}+1=d_{2} ;$ therefore $s_{2}=2$ and $d_{2}=3$.\n\nWe can repeat the above process in general.\n\nClaim. $d_{i}=2 i-1$ and $s_{i}=i$ for $i=1,2, \\ldots, k$.\n\nProof. Apply induction on $i$. The Claim has been proved for $i=1,2$. Suppose that we have already proved $d=1, d_{2}=3, \\ldots, d_{i}=2 i-1$, and consider the next divisor $d_{i+1}$ :\n\n$$\nd_{i+1}=s_{i+1}^{2}-s_{i}^{2}=s_{i+1}^{2}-i^{2}=\\left(s_{i+1}-i\\right)\\left(s_{i+1}+i\\right)\n$$\n\nThe number $s_{i+1}+i$ is a divisor of $n$, so there is some index $j$ such that $d_{j}=s_{i+1}+i$.\n\nSimilarly to (2), by (1) we have\n\n$$\ns_{i+1}+s_{i}=s_{i+1}+i=d_{j} \\geqslant s_{j}+s_{j-1}\n\\tag{3}\n$$\n\nsince the sequence $s_{0}<s_{1}<\\ldots<s_{k}$ increases, (3) forces $j \\leqslant i+1$. On the other hand, $d_{j}=s_{i+1}+i>2 i>d_{i}>d_{i-1}>\\ldots>d_{1}$, so $j \\leqslant i$ is not possible. The only possibility is $j=i+1$.\n\nHence,\n\n$$\n\\begin{gathered}\ns_{i+1}+i=d_{i+1}=s_{i+1}^{2}-s_{i}^{2}=s_{i+1}^{2}-i^{2} \\\\\ns_{i+1}^{2}-s_{i+1}=i(i+1) .\n\\end{gathered}\n$$\n\nBy solving this equation we get $s_{i+1}=i+1$ and $d_{i+1}=2 i+1$, that finishes the proof.\n\nNow we know that the positive divisors of the number $n$ are $1,3,5, \\ldots, n-2, n$. The greatest divisor is $d_{k}=2 k-1=n$ itself, so $n$ must be odd. The second greatest divisor is $d_{k-1}=n-2$; then $n-2$ divides $n=(n-2)+2$, so $n-2$ divides 2 . Therefore, $n$ must be 1 or 3 .\n\nThe numbers $n=1$ and $n=3$ obviously satisfy the requirements: for $n=1$ we have $k=1$ and $d_{1}=1^{2}$; for $n=3$ we have $k=2, d_{1}=1^{2}$ and $d_{1}+d_{2}=1+3=2^{2}$."
] | 1,3 |
1,807 | Algebra | null | $\boxed{-2,0,2}$ | [
"$-2,0,2$"
] | true | null | Numerical | null | Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
- In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin.
- In the second line, Gugu writes down every number of the form $q a b$, where $a$ and $b$ are two (not necessarily distinct) numbers from the first line.
- In the third line, Gugu writes down every number of the form $a^{2}+b^{2}-c^{2}-d^{2}$, where $a, b, c, d$ are four (not necessarily distinct) numbers from the first line.
Determine all values of $q$ such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line. | [
"Call a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers 0 and \\pm 2 are good. The third line necessarily contains 0 , so 0 is good. For any two numbers $a, b$ in the first line, write $a=x-y$ and $b=u-v$, where $x, y, u, v$ are (not necessarily distinct) numbers on the napkin. We may now write\n\n$$\n2 a b=2(x-y)(u-v)=(x-v)^{2}+(y-u)^{2}-(x-u)^{2}-(y-v)^{2},\n$$\n\nwhich shows that 2 is good. By negating both sides of the above equation, we also see that -2 is good.\n\nWe now show that $-2,0$, and 2 are the only good numbers. Assume for sake of contradiction that $q$ is a good number, where $q \\notin\\{-2,0,2\\}$. We now consider some particular choices of numbers on Gugu's napkin to arrive at a contradiction.\n\nAssume that the napkin contains the integers $1,2, \\ldots, 10$. Then, the first line contains the integers $-9,-8, \\ldots, 9$. The second line then contains $q$ and $81 q$, so the third line must also contain both of them. But the third line only contains integers, so $q$ must be an integer. Furthermore, the third line contains no number greater than $162=9^{2}+9^{2}-0^{2}-0^{2}$ or less than -162 , so we must have $-162 \\leqslant 81 q \\leqslant 162$. This shows that the only possibilities for $q$ are \\pm 1 .\n\nNow assume that $q= \\pm 1$. Let the napkin contain $0,1,4,8,12,16,20,24,28,32$. The first line contains \\pm 1 and \\pm 4 , so the second line contains \\pm 4 . However, for every number $a$ in the first line, $a \\not \\equiv 2(\\bmod 4)$, so we may conclude that $a^{2} \\equiv 0,1(\\bmod 8)$. Consequently, every number in the third line must be congruent to $-2,-1,0,1,2(\\bmod 8)$; in particular, \\pm 4 cannot be in the third line, which is a contradiction.",
"Let $q$ be a good number, as defined in the first solution, and define the polynomial $P\\left(x_{1}, \\ldots, x_{10}\\right)$ as\n\n$$\n\\prod_{i<j}\\left(x_{i}-x_{j}\\right) \\prod_{a_{i} \\in S}\\left(q\\left(x_{1}-x_{2}\\right)\\left(x_{3}-x_{4}\\right)-\\left(a_{1}-a_{2}\\right)^{2}-\\left(a_{3}-a_{4}\\right)^{2}+\\left(a_{5}-a_{6}\\right)^{2}+\\left(a_{7}-a_{8}\\right)^{2}\\right)\n$$\n\nwhere $S=\\left\\{x_{1}, \\ldots, x_{10}\\right\\}$.\n\nWe claim that $P\\left(x_{1}, \\ldots, x_{10}\\right)=0$ for every choice of real numbers $\\left(x_{1}, \\ldots, x_{10}\\right)$. If any two of the $x_{i}$ are equal, then $P\\left(x_{1}, \\ldots, x_{10}\\right)=0$ trivially. If no two are equal, assume that Gugu has those ten numbers $x_{1}, \\ldots, x_{10}$ on his napkin. Then, the number $q\\left(x_{1}-x_{2}\\right)\\left(x_{3}-x_{4}\\right)$ is in the second line, so we must have some $a_{1}, \\ldots, a_{8}$ so that\n\n$$\nq\\left(x_{1}-x_{2}\\right)\\left(x_{3}-x_{4}\\right)-\\left(a_{1}-a_{2}\\right)^{2}-\\left(a_{3}-a_{4}\\right)^{2}+\\left(a_{5}-a_{6}\\right)^{2}+\\left(a_{7}-a_{8}\\right)^{2}=0\n$$\n\n\n\nand hence $P\\left(x_{1}, \\ldots, x_{10}\\right)=0$.\n\nSince every polynomial that evaluates to zero everywhere is the zero polynomial, and the product of two nonzero polynomials is necessarily nonzero, we may define $F$ such that\n\n$$\nF\\left(x_{1}, \\ldots, x_{10}\\right) \\equiv q\\left(x_{1}-x_{2}\\right)\\left(x_{3}-x_{4}\\right)-\\left(a_{1}-a_{2}\\right)^{2}-\\left(a_{3}-a_{4}\\right)^{2}+\\left(a_{5}-a_{6}\\right)^{2}+\\left(a_{7}-a_{8}\\right)^{2} \\equiv 0\n$$\n\nfor some particular choice $a_{i} \\in S$.\n\nEach of the sets $\\left\\{a_{1}, a_{2}\\right\\},\\left\\{a_{3}, a_{4}\\right\\},\\left\\{a_{5}, a_{6}\\right\\}$, and $\\left\\{a_{7}, a_{8}\\right\\}$ is equal to at most one of the four sets $\\left\\{x_{1}, x_{3}\\right\\},\\left\\{x_{2}, x_{3}\\right\\},\\left\\{x_{1}, x_{4}\\right\\}$, and $\\left\\{x_{2}, x_{4}\\right\\}$. Thus, without loss of generality, we may assume that at most one of the sets $\\left\\{a_{1}, a_{2}\\right\\},\\left\\{a_{3}, a_{4}\\right\\},\\left\\{a_{5}, a_{6}\\right\\}$, and $\\left\\{a_{7}, a_{8}\\right\\}$ is equal to $\\left\\{x_{1}, x_{3}\\right\\}$. Let $u_{1}, u_{3}, u_{5}, u_{7}$ be the indicator functions for this equality of sets: that is, $u_{i}=1$ if and only if $\\left\\{a_{i}, a_{i+1}\\right\\}=\\left\\{x_{1}, x_{3}\\right\\}$. By assumption, at least three of the $u_{i}$ are equal to 0 .\n\nWe now compute the coefficient of $x_{1} x_{3}$ in $F$. It is equal to $q+2\\left(u_{1}+u_{3}-u_{5}-u_{7}\\right)=0$, and since at least three of the $u_{i}$ are zero, we must have that $q \\in\\{-2,0,2\\}$, as desired."
] | $-2,0,2$ |
1,810 | Algebra | null | $\boxed{-(n-1) / 2}$ | [
"$-(n-1) / 2$"
] | false | null | Expression | null | An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have
$$
\sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1
$$
Find the largest constant $K=K(n)$ such that
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant K
$$
holds for every Shiny $n$-tuple $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$. | [
"First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take $x_{2}=x_{3}=\\cdots=t$ and $x_{1}=-1 /(2 t)$. Then, every product $x_{i} x_{j}(i \\neq j)$ is equal to either $t^{2}$ or $-1 / 2$. Hence, for every permutation $y_{i}$ of the $x_{i}$, we have\n\n$$\ny_{1} y_{2}+\\cdots+y_{n-1} y_{n} \\geqslant(n-3) t^{2}-1 \\geqslant-1\n$$\n\nThis justifies that the $n$-tuple $\\left(x_{1}, \\ldots, x_{n}\\right)$ is Shiny. Now, we have\n\n$$\n\\sum_{i<j} x_{i} x_{j}=-\\frac{n-1}{2}+\\frac{(n-1)(n-2)}{2} t^{2}\n$$\n\nThus, as $t$ approaches 0 from above, $\\sum_{i<j} x_{i} x_{j}$ gets arbitrarily close to $-(n-1) / 2$. This shows that we may not take $K$ any larger than $-(n-1) / 2$. It remains to show that $\\sum_{i<j} x_{i} x_{j} \\geqslant$ $-(n-1) / 2$ for any Shiny choice of the $x_{i}$.\n\nFrom now onward, assume that $\\left(x_{1}, \\ldots, x_{n}\\right)$ is a Shiny $n$-tuple. Let the $z_{i}(1 \\leqslant i \\leqslant n)$ be some permutation of the $x_{i}$ to be chosen later. The indices for $z_{i}$ will always be taken modulo $n$. We will first split up the sum $\\sum_{i<j} x_{i} x_{j}=\\sum_{i<j} z_{i} z_{j}$ into $\\lfloor(n-1) / 2\\rfloor$ expressions, each of the form $y_{1} y_{2}+\\cdots+y_{n-1} y_{n}$ for some permutation $y_{i}$ of the $z_{i}$, and some leftover terms. More specifically, write\n\n$$\n\\sum_{i<j} z_{i} z_{j}=\\sum_{q=0}^{n-1} \\sum_{\\substack{i+j \\equiv q \\\\ i \\neq j}} z_{i} z_{j}=\\sum_{\\substack{(\\bmod n)}}^{\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor} \\sum_{\\substack{i+j \\equiv 2 p-1,2 p(\\bmod n) \\\\ i \\neq j}} z_{i} z_{j}+L\n\\tag{1}\n$$\n\nwhere $L=z_{1} z_{-1}+z_{2} z_{-2}+\\cdots+z_{(n-1) / 2} z_{-(n-1) / 2}$ if $n$ is odd, and $L=z_{1} z_{-1}+z_{1} z_{-2}+z_{2} z_{-2}+$ $\\cdots+z_{(n-2) / 2} z_{-n / 2}$ if $n$ is even. We note that for each $p=1,2, \\ldots,\\lfloor(n-1) / 2\\rfloor$, there is some permutation $y_{i}$ of the $z_{i}$ such that\n\n$$\n\\sum_{\\substack{i+j \\equiv 2 p-1,2 p(\\bmod n) \\\\ i \\neq j}} z_{i} z_{j}=\\sum_{k=1}^{n-1} y_{k} y_{k+1}\n$$\n\nbecause we may choose $y_{2 i-1}=z_{i+p-1}$ for $1 \\leqslant i \\leqslant(n+1) / 2$ and $y_{2 i}=z_{p-i}$ for $1 \\leqslant i \\leqslant n / 2$.\n\nWe show (1) graphically for $n=6,7$ in the diagrams below. The edges of the graphs each represent a product $z_{i} z_{j}$, and the dashed and dotted series of lines represents the sum of the edges, which is of the form $y_{1} y_{2}+\\cdots+y_{n-1} y_{n}$ for some permutation $y_{i}$ of the $z_{i}$ precisely when the series of lines is a Hamiltonian path. The filled edges represent the summands of $L$.\n\n\n\n<img_3150>\n\nNow, because the $z_{i}$ are Shiny, we have that (1) yields the following bound:\n\n$$\n\\sum_{i<j} z_{i} z_{j} \\geqslant-\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor+L\n$$\n\nIt remains to show that, for each $n$, there exists some permutation $z_{i}$ of the $x_{i}$ such that $L \\geqslant 0$ when $n$ is odd, and $L \\geqslant-1 / 2$ when $n$ is even. We now split into cases based on the parity of $n$ and provide constructions of the permutations $z_{i}$.\n\nSince we have not made any assumptions yet about the $x_{i}$, we may now assume without loss of generality that\n\n$$\nx_{1} \\leqslant x_{2} \\leqslant \\cdots \\leqslant x_{k} \\leqslant 0 \\leqslant x_{k+1} \\leqslant \\cdots \\leqslant x_{n}\n\\tag{2}\n$$\n\nCase 1: $n$ is odd.\n\nWithout loss of generality, assume that $k$ (from (2)) is even, because we may negate all the $x_{i}$ if $k$ is odd. We then have $x_{1} x_{2}, x_{3} x_{4}, \\ldots, x_{n-2} x_{n-1} \\geqslant 0$ because the factors are of the same sign. Let $L=x_{1} x_{2}+x_{3} x_{4}+\\cdots+x_{n-2} x_{n-1} \\geqslant 0$. We choose our $z_{i}$ so that this definition of $L$ agrees with the sum of the leftover terms in (1). Relabel the $x_{i}$ as $z_{i}$ such that\n\n$$\n\\left\\{z_{1}, z_{n-1}\\right\\},\\left\\{z_{2}, z_{n-2}\\right\\}, \\ldots,\\left\\{z_{(n-1) / 2}, z_{(n+1) / 2}\\right\\}\n$$\n\nare some permutation of\n\n$$\n\\left\\{x_{1}, x_{2}\\right\\},\\left\\{x_{3}, x_{4}\\right\\}, \\ldots,\\left\\{x_{n-2}, x_{n-1}\\right\\}\n$$\n\nand $z_{n}=x_{n}$. Then, we have $L=z_{1} z_{n-1}+\\cdots+z_{(n-1) / 2} z_{(n+1) / 2}$, as desired.\n\nCase 2: $n$ is even.\n\nLet $L=x_{1} x_{2}+x_{2} x_{3}+\\cdots+x_{n-1} x_{n}$. Assume without loss of generality $k \\neq 1$. Now, we have\n\n$$\n\\begin{gathered}\n2 L=\\left(x_{1} x_{2}+\\cdots+x_{n-1} x_{n}\\right)+\\left(x_{1} x_{2}+\\cdots+x_{n-1} x_{n}\\right) \\geqslant\\left(x_{2} x_{3}+\\cdots+x_{n-1} x_{n}\\right)+x_{k} x_{k+1} \\\\\n\\geqslant x_{2} x_{3}+\\cdots+x_{n-1} x_{n}+x_{n} x_{1} \\geqslant-1\n\\end{gathered}\n$$\n\nwhere the first inequality holds because the only negative term in $L$ is $x_{k} x_{k+1}$, the second inequality holds because $x_{1} \\leqslant x_{k} \\leqslant 0 \\leqslant x_{k+1} \\leqslant x_{n}$, and the third inequality holds because the $x_{i}$ are assumed to be Shiny. We thus have that $L \\geqslant-1 / 2$. We now choose a suitable $z_{i}$ such that the definition of $L$ matches the leftover terms in (1).\n\n\n\nRelabel the $x_{i}$ with $z_{i}$ in the following manner: $x_{2 i-1}=z_{-i}, x_{2 i}=z_{i}$ (again taking indices modulo $n$ ). We have that\n\n$$\nL=\\sum_{\\substack{i+j \\equiv 0,-1(\\bmod n) \\\\ i \\neq j j}} z_{i} z_{j}\n$$\n\nas desired.",
"We present another proof that $\\sum_{i<j} x_{i} x_{j} \\geqslant-(n-1) / 2$ for any Shiny $n$-tuple $\\left(x_{1}, \\ldots, x_{n}\\right)$. Assume an ordering of the $x_{i}$ as in (2), and let $\\ell=n-k$. Assume without loss of generality that $k \\geqslant \\ell$. Also assume $k \\neq n$, (as otherwise, all of the $x_{i}$ are nonpositive, and so the inequality is trivial). Define the sets of indices $S=\\{1,2, \\ldots, k\\}$ and $T=\\{k+1, \\ldots, n\\}$. Define the following sums:\n\n$$\nK=\\sum_{\\substack{i<j \\\\ i, j \\in S}} x_{i} x_{j}, \\quad M=\\sum_{\\substack{i \\in S \\\\ j \\in T}} x_{i} x_{j}, \\quad \\text { and } \\quad L=\\sum_{\\substack{i<j \\\\ i, j \\in T}} x_{i} x_{j}\n$$\n\nBy definition, $K, L \\geqslant 0$ and $M \\leqslant 0$. We aim to show that $K+L+M \\geqslant-(n-1) / 2$.\n\nWe split into cases based on whether $k=\\ell$ or $k>\\ell$.\n\nCase 1: $k>\\ell$.\n\nConsider all permutations $\\phi:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ such that $\\phi^{-1}(T)=\\{2,4, \\ldots, 2 \\ell\\}$. Note that there are $k ! \\ell$ ! such permutations $\\phi$. Define\n\n$$\nf(\\phi)=\\sum_{i=1}^{n-1} x_{\\phi(i)} x_{\\phi(i+1)}\n$$\n\nWe know that $f(\\phi) \\geqslant-1$ for every permutation $\\phi$ with the above property. Averaging $f(\\phi)$ over all $\\phi$ gives\n\n$$\n-1 \\leqslant \\frac{1}{k ! \\ell !} \\sum_{\\phi} f(\\phi)=\\frac{2 \\ell}{k \\ell} M+\\frac{2(k-\\ell-1)}{k(k-1)} K\n$$\n\nwhere the equality holds because there are $k \\ell$ products in $M$, of which $2 \\ell$ are selected for each $\\phi$, and there are $k(k-1) / 2$ products in $K$, of which $k-\\ell-1$ are selected for each $\\phi$. We now have\n\n$$\nK+L+M \\geqslant K+L+\\left(-\\frac{k}{2}-\\frac{k-\\ell-1}{k-1} K\\right)=-\\frac{k}{2}+\\frac{\\ell}{k-1} K+L .\n$$\n\nSince $k \\leqslant n-1$ and $K, L \\geqslant 0$, we get the desired inequality.\n\nCase 2: $k=\\ell=n / 2$.\n\nWe do a similar approach, considering all $\\phi:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ such that $\\phi^{-1}(T)=$ $\\{2,4, \\ldots, 2 \\ell\\}$, and defining $f$ the same way. Analogously to Case 1 , we have\n\n$$\n-1 \\leqslant \\frac{1}{k ! \\ell !} \\sum_{\\phi} f(\\phi)=\\frac{2 \\ell-1}{k \\ell} M\n$$\n\nbecause there are $k \\ell$ products in $M$, of which $2 \\ell-1$ are selected for each $\\phi$. Now, we have that\n\n$$\nK+L+M \\geqslant M \\geqslant-\\frac{n^{2}}{4(n-1)} \\geqslant-\\frac{n-1}{2}\n$$\n\nwhere the last inequality holds because $n \\geqslant 4$."
] | $-(n-1) / 2$ |
1,818 | Combinatorics | null | $\boxed{\frac{n(n+1)(2 n+1)}{6}}$ | [
"$\\frac{n(n+1)(2 n+1)}{6}$"
] | false | null | Expression | null | Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed only once). This way, we get $3 n$ sets of colors, split into three groups according to the orientation. It happens that for every set in any group, the same set appears in both of the other groups. Determine, in terms of $n$, the maximal possible number of colors that are present. | [
"Call a $n \\times n \\times 1$ box an $x$-box, a $y$-box, or a $z$-box, according to the direction of its short side. Let $C$ be the number of colors in a valid configuration. We start with the upper bound for $C$.\n\nLet $\\mathcal{C}_{1}, \\mathcal{C}_{2}$, and $\\mathcal{C}_{3}$ be the sets of colors which appear in the big cube exactly once, exactly twice, and at least thrice, respectively. Let $M_{i}$ be the set of unit cubes whose colors are in $\\mathcal{C}_{i}$, and denote $n_{i}=\\left|M_{i}\\right|$.\n\nConsider any $x$-box $X$, and let $Y$ and $Z$ be a $y$ - and a $z$-box containing the same set of colors as $X$ does.\n\nClaim. $4\\left|X \\cap M_{1}\\right|+\\left|X \\cap M_{2}\\right| \\leqslant 3 n+1$.\n\nProof. We distinguish two cases.\n\nCase 1: $X \\cap M_{1} \\neq \\varnothing$.\n\nA cube from $X \\cap M_{1}$ should appear in all three boxes $X, Y$, and $Z$, so it should lie in $X \\cap Y \\cap Z$. Thus $X \\cap M_{1}=X \\cap Y \\cap Z$ and $\\left|X \\cap M_{1}\\right|=1$.\n\nConsider now the cubes in $X \\cap M_{2}$. There are at most $2(n-1)$ of them lying in $X \\cap Y$ or $X \\cap Z$ (because the cube from $X \\cap Y \\cap Z$ is in $M_{1}$ ). Let $a$ be some other cube from $X \\cap M_{2}$. Recall that there is just one other cube $a^{\\prime}$ sharing a color with $a$. But both $Y$ and $Z$ should contain such cube, so $a^{\\prime} \\in Y \\cap Z$ (but $a^{\\prime} \\notin X \\cap Y \\cap Z$ ). The map $a \\mapsto a^{\\prime}$ is clearly injective, so the number of cubes $a$ we are interested in does not exceed $|(Y \\cap Z) \\backslash X|=n-1$. Thus $\\left|X \\cap M_{2}\\right| \\leqslant 2(n-1)+(n-1)=3(n-1)$, and hence $4\\left|X \\cap M_{1}\\right|+\\left|X \\cap M_{2}\\right| \\leqslant 4+3(n-1)=3 n+1$.\n\nCase 2: $X \\cap M_{1}=\\varnothing$.\n\nIn this case, the same argument applies with several changes. Indeed, $X \\cap M_{2}$ contains at most $2 n-1$ cubes from $X \\cap Y$ or $X \\cap Z$. Any other cube $a$ in $X \\cap M_{2}$ corresponds to some $a^{\\prime} \\in Y \\cap Z$ (possibly with $a^{\\prime} \\in X$ ), so there are at most $n$ of them. All this results in $\\left|X \\cap M_{2}\\right| \\leqslant(2 n-1)+n=3 n-1$, which is even better than we need (by the assumptions of our case).\n\nSumming up the inequalities from the Claim over all $x$-boxes $X$, we obtain\n\n$$\n4 n_{1}+n_{2} \\leqslant n(3 n+1) .\n$$\n\nObviously, we also have $n_{1}+n_{2}+n_{3}=n^{3}$.\n\nNow we are prepared to estimate $C$. Due to the definition of the $M_{i}$, we have $n_{i} \\geqslant i\\left|\\mathcal{C}_{i}\\right|$, so\n\n$$\nC \\leqslant n_{1}+\\frac{n_{2}}{2}+\\frac{n_{3}}{3}=\\frac{n_{1}+n_{2}+n_{3}}{3}+\\frac{4 n_{1}+n_{2}}{6} \\leqslant \\frac{n^{3}}{3}+\\frac{3 n^{2}+n}{6}=\\frac{n(n+1)(2 n+1)}{6} .\n$$\n\nIt remains to present an example of an appropriate coloring in the above-mentioned number of colors. For each color, we present the set of all cubes of this color. These sets are:\n\n1. $n$ singletons of the form $S_{i}=\\{(i, i, i)\\}$ (with $1 \\leqslant i \\leqslant n$ );\n2. $3\\left(\\begin{array}{c}n \\\\ 2\\end{array}\\right)$ doubletons of the forms $D_{i, j}^{1}=\\{(i, j, j),(j, i, i)\\}, D_{i, j}^{2}=\\{(j, i, j),(i, j, i)\\}$, and $D_{i, j}^{3}=$ $\\{(j, j, i),(i, i, j)\\}$ (with $1 \\leqslant i<j \\leqslant n)$;\n\n\n3. $2\\left(\\begin{array}{l}n \\\\ 3\\end{array}\\right)$ triplets of the form $T_{i, j, k}=\\{(i, j, k),(j, k, i),(k, i, j)\\}$ (with $1 \\leqslant i<j<k \\leqslant n$ or $1 \\leqslant i<k<j \\leqslant n)$.\n\nOne may easily see that the $i^{\\text {th }}$ boxes of each orientation contain the same set of colors, and that\n\n$$\nn+\\frac{3 n(n-1)}{2}+\\frac{n(n-1)(n-2)}{3}=\\frac{n(n+1)(2 n+1)}{6}\n$$\n\ncolors are used, as required.",
"We will approach a new version of the original problem. In this new version, each cube may have a color, or be invisible (not both). Now we make sets of colors for each $n \\times n \\times 1$ box as before (where \"invisible\" is not considered a color) and group them by orientation, also as before. Finally, we require that, for every non-empty set in any group, the same set must appear in the other 2 groups. What is the maximum number of colors present with these new requirements?\n\nLet us call strange a big $n \\times n \\times n$ cube whose painting scheme satisfies the new requirements, and let $D$ be the number of colors in a strange cube. Note that any cube that satisfies the original requirements is also strange, so $\\max (D)$ is an upper bound for the original answer.\n\nClaim. $D \\leqslant \\frac{n(n+1)(2 n+1)}{6}$.\n\nProof. The proof is by induction on $n$. If $n=1$, we must paint the cube with at most 1 color.\n\nNow, pick a $n \\times n \\times n$ strange cube $A$, where $n \\geqslant 2$. If $A$ is completely invisible, $D=0$ and we are done. Otherwise, pick a non-empty set of colors $\\mathcal{S}$ which corresponds to, say, the boxes $X, Y$ and $Z$ of different orientations.\n\nNow find all cubes in $A$ whose colors are in $\\mathcal{S}$ and make them invisible. Since $X, Y$ and $Z$ are now completely invisible, we can throw them away and focus on the remaining $(n-1) \\times(n-1) \\times(n-1)$ cube $B$. The sets of colors in all the groups for $B$ are the same as the sets for $A$, removing exactly the colors in $\\mathcal{S}$, and no others! Therefore, every nonempty set that appears in one group for $B$ still shows up in all possible orientations (it is possible that an empty set of colors in $B$ only matched $X, Y$ or $Z$ before these were thrown away, but remember we do not require empty sets to match anyway). In summary, $B$ is also strange.\n\nBy the induction hypothesis, we may assume that $B$ has at most $\\frac{(n-1) n(2 n-1)}{6}$ colors. Since there were at most $n^{2}$ different colors in $\\mathcal{S}$, we have that $A$ has at most $\\frac{(n-1) n(2 n-1)}{6}+n^{2}=$ $\\frac{n(n+1)(2 n+1)}{6}$ colors.\n\nFinally, the construction in the previous solution shows a painting scheme (with no invisible cubes) that reaches this maximum, so we are done."
] | $\frac{n(n+1)(2 n+1)}{6}$ |
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