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# ISI BStat, BMath Entrance 2022 Problems and Solutions

This is a work in progress. Please come back for the solutions. You can also suggest your solutions in the comment section

#### Objective Section

###### Problem 1

Any positive real number $x$ can be expanded as

$x=a_{n} \cdot 2^{n}+a_{n-1} \cdot 2^{n-1}+ \\ \cdots+a_{1} \cdot 2^{1}+a_{0} \cdot 2^{0}+a_{-1} \\ \cdot 2^{-1}+a_{-2} \cdot 2^{-2}+\cdots$

for some $n \geq 0$, where each $a_{i} \in{0,1}$. In the above-described expansion of $21.1875$, the smallest positive integer $k$ such that $a_{-k} \neq 0$ is:

(B) 2

(C) 1

(D) 4

Solution

$x=a_{n} 2^{n}+a_{n-1} 2^{n-1}+\cdots+a_{1} 2^{1}+a_{1}$

Binary represation of $21.1875$ is $10101.0011$ (very standard Method to Convert Dercimal to Binary)

So, $-k=-3$

So, we got $k=3$ (A)

###### Problem 2

Suppose, for some $\theta \in\left[0, \frac{\pi}{2}\right], \frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $(\cot 3 \theta) \tan \theta$ equals
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{8}$
(D) $\frac{1}{7}$ <— Answer

Solution

$\frac{\cos 3 \theta}{ \cos \theta}=\frac{1}{3}$

$\Rightarrow \frac{4 \cos^{ 3} \theta-3 \cos \theta}{\cos \theta}=\frac{1}{3}$

$\Rightarrow 4 \cos ^{2} \theta-3=\frac{1}{3}$

$\Rightarrow \cos ^{2} \theta=\frac{10}{12}$

So, $\sin ^{2} \theta=\frac{2}{12}$

$\tan 3 \theta\cot \theta=\frac{8-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}=\frac{3-\frac{1}{5}}{1-\frac{3}{5}}=7$

So we got, $\cot 3 \theta \tan \theta=\frac{1}{7}$ (D)

###### Problem 3

The locus of points $z$ in the complex plane satisfying $z^{2}+|z|^{2}=0$ is
(A) a straight line
(B) a pair of straight lines
(C) a circle
(D) a parabola

Solution

###### Problem 4

Amongst all polynomials $p(x)=c_{0}+c_{1} x+\cdots+c_{10} x^{10}$ with real coefficients satisfying $|p(x)| \leq|x|$ for all $x \in[-1,1]$, what is the maximum possible value of $\left(2 c_{0}+c_{1}\right)^{10}$ ?
(A) $4^{10}$
(B) $3^{10}$
(C) $2^{10}$
(D) 1

Solution

###### Problem 5

Let $\mathbb{Z}$ denote the set of integers. Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be such that $f(x) f(y)=f(x+y)+f(x-y)$ for all $x, y \in \mathbb{Z}$. If $f(1)=3$, then $f(7)$ equals
(A) 840
(B) 844
(C) 843 <— Answer
(D) 842

Solution

$f:\mathbb{Z} \rightarrow \mathbb{Z}$

$f(x) f(y)=f(x+y)+f(x-y)$

$f(1) f(0)=f(1)+f(1)$
$\Rightarrow 3 f(0)=f(1)+f(3)$

$\Rightarrow 3 f(0)=6$
$\Rightarrow f(0)=2$

Now, $f(1) f(1)=f(2)+f(0)$
$\Rightarrow 9=f(2)+2$
$\Rightarrow f(2)=7$

Again, $f(2) f(1)=f(3)+f(1)$
$\Rightarrow 7 \times 3=f(3)+3$
$\Rightarrow f(3)=18$

Again, $f(3) f(1)=f(4)+f(2)$
$\Rightarrow 18 \times 3=f(4)+7$
$\Rightarrow f(4)=47$

$f(3) f(4)=f(4+3)+f(1)$
$\Rightarrow (18 \times 47)-3=f(7)$
$\Rightarrow f(3)=843$ (C)

###### Problem 6

Let $A$ and $B$ be two $3 \times 3$ matrices such that $(A+B)^{2}=A^{2}+B^{2}$. Which of the following must be true?
(A) $A$ and $B$ are zero matrices.
(B) $A B$ is the zero matrix.
(C) $(A-B)^{2}=A^{2}-B^{2}$
(D) $(A-B)^{2}=A^{2}+B^{2}$ <— Answer

Solution

$(A+B)(A+B)$
$=A^{2}+B A+A B+B^{2}$

So, $B A+A B=0$

$(A-B)(A-B)$
$=A^{2}-A B-B A+B^{2}$
$=A^{2}+B^{2}$ (D)

###### Problem 7

Let $\left(n_{1}, n_{2}, \cdots, n_{12}\right)$ be a permutation of the numbers $1,2, \cdots, 12$. The number of arrangements with

$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}$
and $n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$ equals:

(A) $^{12} C_{5}$

(B) $^{12} C_{6}$

(C) $^{11} C_{6}$

(D) $\frac{11!}{2}$

Solution

$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$

$n_{6}$ is the smallest so,$n_{6}=1$

So, if we just choose first 5 it is enough.

So, $^{11} C_{5}$ as $n_{6}$ is fixed

$= ^{11} C_{6}$ (C)

###### Problem 8

The sides of a regular hexagon $A B C D E F$ is extended by doubling them to form a bigger hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ as in the figure below.

Then the ratio of the areas of the bigger to the smaller hexagon is:
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) 4

###### Problem 9

In how many ways can we choose $a_{1}<a_{2}<a_{3}<a_{4}$ from the set ${1,2, \ldots, 30}$ such that $a_{1}, a_{2}, a_{3}, a_{4}$ are in arithmetic progression?
(A) 135
(B) 145
(C) 155
(D) 165

###### Problem 10

Suppose the numbers 71,104 and 159 leave the same remainder $r$ when divided by a certain number $N>1$. Then, the value of $3 N+4 r$ must equal:
(A) 53
(B) 48
(C) 37
(D) 23

###### Problem 11

If $x, y$ are positive real numbers such that $3 x+4 y<72$, then the maximum possible value of $12 x y(72-3 x-4 y)$ is:
(A) 12240
(D) 13824
(C) 10656
(D) 8640

###### Problem 12

What is the minimum value of the function $|x-3|+|x+2|+|x+1|+|x|$ for real $x$ ?
(A) 3
(B) 5
(C) 6
(D) 8

###### Problem 13

Consider a differentiable function $u:[0,1] \rightarrow \mathbb{R}$. Assume the function $u$ satisfies
$u(a)=\frac{1}{2 r} \int_{a-r}^{a+r} u(x) d x, \quad$ for all $a \in(0,1)$ and all $r<\min (a, 1-a)$.
(A) $u$ attains its maximum but not its minimum on the set ${0,1}$.
(B) $u$ attains its minimum but not maximum on the set ${0,1}$.
(C) If $u$ attains either its maximum or its minimum on the set ${0,1}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set ${0,1}$.

###### Problem 14

A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance $x$ below the ladders?
(A) 3
(B) $2 \sqrt{2}$
(C) $\frac{8}{3}$ <— Answer
(D) $2 \sqrt{3}$

Solution

$\frac{a}{a+b}=\frac{x}{4}$, in $\triangle A C D$

$\frac{b}{a+b}=\frac{x}{8}$, in $\triangle A B D$

So, from the previous relations we get, $a=2 b$.

So, $\frac{x}{4}=\frac{2 b}{3 b}$
$\Rightarrow x=\frac{8}{3}$ feet (C)

###### Problem 15

Let $y=x+c_{1}, y=x+c_{2}$ be the two tangents to the ellipse $x^{2}+4 y^{2}=1$. What is the value of $\left|c_{1}-c_{2}\right| ?$
(A) $\sqrt{2}$
(B) $\sqrt{5}$ <— Answer
(C) $\frac{\sqrt{5}}{2}$
(D) 1

Solution

$x^{2}+4 y^{2}=1 \rightarrow(1)$

We want $\frac{d y}{d x}=1$

So differentiating $(1)$ with respect of $x$ we get

$2 x+8 y \frac{d y}{d x}=0$

for $\frac{d y}{d y}$ to be 1 we get $x=-4y$

Substituting $(1)$

$16 y^{2}+ 4y^{2}=1$

$\Rightarrow y_{0}=\frac{1}{2 \sqrt{5}} \text{or}-\frac{1}{2 \sqrt{5}}$

Let $y_{0}=\frac{1}{2 \sqrt{5}}$,

$\Rightarrow x_{0}=\frac{2}{\sqrt{5}}$

$c_{2}=-\frac{\sqrt{5}}{2}$

Now, $|c_{1}-c_{2}|=\sqrt{5}$ (B)

###### Problem 16

In the figure below, $A B C D$ is a square and $\triangle C E F$ is a triangle with given sides inscribed as in the figure. Find the length $B E$.

(A) $\frac{13}{\sqrt{17}}$ <— Answer
(B) $\frac{14}{\sqrt{17}}$
(C) $\frac{15}{\sqrt{17}}$
(D) $\frac{16}{\sqrt{17}}$

Solution

$\cos \theta=\frac{a}{4}=\frac{A F}{3}$

$A F=\frac{3 a}{4}$

$\tan \theta=\frac{1}{4}=\frac{A E}{\frac{3 a}{4}}$
$\Rightarrow A E=\frac{3 a}{16}$

$a^{2}+\frac{a^{2}}{16}=16$
$\Rightarrow \quad \frac{17 a^{2}}{16}=16$
$\Rightarrow \quad a^{2}=\frac{16 \times 16}{17}$
$\Rightarrow \quad a=\frac{16}{\sqrt{17}}$

$B E =\frac{13 a}{16}$
$B E =\frac{13}{\sqrt{17}}$ (A)

###### Problem 17

Let $p$ and $q$ be two non-zero polynomials such that the degree of $p$ is less than or equal to the degree of $q$, and $p(a) q(a)=0$ for $a=0,1,2, \ldots, 10$. Which of the following must be true?
(A) degree of $q \neq 10$
(B) degree of $p \neq 10$
(C) degree of $q \neq 5$ <— Answer
(D) degree of $p \neq 5$

Solution

$p(a) q(a)=0$, for $a=0,1,2, \cdots, 10$

Hence the polynonial, $P(x) q(x)$ has at least 11 roots as so $deg(p(x) q(x)) \geq 11$

Now, if $deg(q(x))=5$, then $deg(p(x)) \leq 5 \Rightarrow deg(p(x) q(x)) \leq 10$ which is a contradiction.

Degree of $q \neq 5$

###### Problem 18

For $n \in \mathbb{N}$, let $a_{n}$ be defined

𝑎𝑛=𝑛011+𝑛𝑥2𝑑𝑥

Then $\lim {n \rightarrow \infty} a{n}$

(A) equals 0 <— Answer
(B) equals $\frac{\pi}{4}$
(C) equals $\frac{\pi}{2}$
(D) does not exist

Solution

$a_{n}=\int_{0}^{n} \frac{1}{1+n x^{2}} d x$

$=\frac{1}{\sqrt{n}} \int_{0}^{n} \frac{1}{1+(\sqrt{n} x)^{2}} d(\sqrt{n} x)$

$=\frac{1}{\sqrt{n}}\left[\tan ^{-1}(\sqrt{n} x)\right]_{0}^{n}$

$a_{n}=\frac{1}{\sqrt{n}} \tan ^{-1}\left(n^{3 / 2}\right)$

$=\frac{1}{\sqrt{n}}\left(\frac{\pi}{2}-\cot ^{-1}\left(n^{3 / 2}\right)\right)$

$\Rightarrow a_{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{\sqrt{n}} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)$

$\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)=\frac{1}{n^{3 / 2}}$, as $\frac{1}{n^{3 / 2}} \rightarrow 0$

$\lim {n \rightarrow \infty} a{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{n^{2}}$

$\Rightarrow \lim {n \rightarrow \infty} a{n}=0$

###### Problem 19

A $3 \times 3$ magic square is a $3 \times 3$ rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square
(A) 90
(B) 96 <— Answer
(C) 94
(D) 99

Solution

Asrume the sum is $s$ and $x$ as shown. Fill the rest of the square setting the diagonals equal to $s$ we get.

$s-2x=40$

$s-x=68$

So we get that $s=96$

###### Problem 20

The number of positive integers $n$ less than or equal to 22 such that 7 divides $n^{5}+4 n^{4}+3 n^{3}+2022$ is
(A) 7
(B) 8
(C) 9
(D) 10 <— Answer

Solution

$n^{5}+4 n^{4}+3 n^{3}+2022$

$\left.\equiv n^{3}(n+1)(n+3)+6 (\bmod 7\right)$

checking $n \equiv \pm 0, \pm 1, \pm 2, \pm 3 (\bmod 7)$

Now we see that only $n=1, \pm 2$ is permitted

Thus, $n \in{1,2,5,8,9,12,15,16,10,22}$

###### Problem 21

In a class of 45 students, three students can write well using either hand. The number of students who can write well only with the right hand is 24 more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:
(A) 33
(B) 36 <— Answer
(C) 39
(D) 41

Solution

$R =$ Set of right handers

$L$= Set of left handers

$|R \cap L|=3, | R \cup L|=45$

$24+\left|R^{C} \cap L|=\right| R \cap L^{c}|=|\left(R^{c}\right)^{c} \cap L^{c} \mid \left|\left(R^{c} \cup L\right)^{c}\right|$

$=45-\mid R^{c} \cup L|$

$45-\left(\left|R^{c}\right|+|L|-| R^{c} \cap L \mid\right)$

$=\left|R^{c} \cap L\right|+\left(45-\left|R^{c}\right|\right)-|L|$

$=\left|R^{c} \cap L\right|+|R|-L \mid$

Thenefore, $|R|-|L|=24$

Moreover, $|R|+|L| =|R \cup L|+\mid R \cap L| =45+3=48$

Hence, $|R|=\frac{24+48}{2}=\frac{72}{2}=36$

###### Problem 22

Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
(B) $3^{10}$
(C) $2^{10} \omega$ <— Answer
(D) $3^{10} \omega^{2}$

Solution

$2 \equiv-1$ mod 3

$\Rightarrow 2^{k} \equiv(-1)^{k}$ mod 3

$\Rightarrow \omega^{2 k}=\omega^{(-1)^{k}}$

$\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1-\omega^{2}+\omega^{2^{{2}}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left(1-\omega^{2^{2}}+ \omega^{2^{3}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1-\omega^{2^{3}}+\omega^{2^{4}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left(1- \omega^{2^{4}} + \omega^{2{^5}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1- \omega^{2^{5}}+\omega^{2^{6}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left.(1-\ \omega^{2^{6}}+\omega^{2^{7}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1-\omega^{2^{7}}+\omega^{2^{8}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left(1-\omega^{2^{8}}+\omega^{2^{9}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1- \omega^{2^{9}}+\omega^{2^{10}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

Product $(-1)^{10} 2^{10} \cdot \omega^{10+5}=2^{10}$

###### Problem 23

The function $x^{2} \log {e} x$ in the interval $(0,2)$

(A) exactly one point of local maximum and no points of local minimum.
(B) exactly one point of local minimum and no points of local maximum. <— Answer
(C) points of local maximum as well as local minimum.
(D) neither a point of local maximum nor a point of local minimum.

Solution

$f(x) =x^{2} \log {e} x$ $\Rightarrow f^{\prime}(x)=x+2 x \log {e} x=x(1+2 \log _{e}^{x})$

###### Problem 24

The number of triples $(a, b, c)$ of positive integers satisfying the equation $(2,3,4) \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{2}{a b c}$
and such that $a<b<c$, equals:
(A) 3
(B) 2
(C) 1 <— Answer
(D) 0

Solution

1𝑎+1𝑏+1𝑐=2+1𝑎𝑏𝑐;𝑎,𝑏,𝑐<𝑎<𝑏<𝑐. 𝑎𝑏+𝑎𝑐+𝑏𝑐=𝑎𝑏𝑐+2;𝑎,𝑏,𝑐<

$\Rightarrow a(b+c)+b c=a b c+2$
$\Rightarrow a(b+c-b c)=2-b c$
$\Rightarrow a(b c-b-c)=b c-2$

$b c-b-c \mid b c-2$
$\Rightarrow b c-b-c \mid b+c-2$

$b c-b-c \leq b+c-2$
$\Rightarrow b c-2 b-2 c \leq-2$
$\Rightarrow(b-2)(c-2) \leq 2$

$b-2=1$, $c-2=2$

So, $b=3, c=4$

So, $a=2$ (C)

###### Problem 25

An urn contains 30 balls out of which one is special. If 6 of these balls are taken out at random, what is the probability that the special ball is chosen?
(A) $\frac{1}{30}$
(B) $\frac{1}{6}$
(C) $\frac{1}{5}$ <— Answer
(D) $\frac{1}{15}$

Solution

###### Problem 26

A triangle has sides of lengths $\sqrt{5}, 2 \sqrt{2}, \sqrt{3}$ units. Then, the radius of its inscribed circle is :
(A) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{2}$
(B) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{3}$
(C) $\sqrt{5}+\sqrt{3}+2 \sqrt{2}$
(D) $\frac{\sqrt{5}+\sqrt{3}-2 \sqrt{2}}{2}$ <— Answer

Solution

$r_{\Delta}=\frac{2 \times {\text {Area in }} \Delta \triangle A B C}{\text { perimeter of } \triangle A B C}$
$\Rightarrow r_{\Delta}=2 \times \frac{\frac{1}{2} \times \sqrt{3} \times \sqrt{5}}{\sqrt{3}+\sqrt{5}+2 \sqrt{2}}$
$\Rightarrow r_{\Delta}=\frac{\sqrt{3} \times \sqrt{5}(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{(\sqrt{3}+\sqrt{5})^{2}-(2 \sqrt{2})^{2}}$
$\frac{\sqrt{3} \times \sqrt{5} \times(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{2 \times \sqrt{3} \times \sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}-2 \sqrt{2}}{2}(D)$

###### Problem 27

Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled $80 \mathrm{~km}$, the triangle became right-angled.

When the first ship reaches the port, the second ship was still $120 \mathrm{~km}$ from the port. Find the initial distance of the ships from the port.
(A) $240 \mathrm{~km}$ <— Answer
(B) $300 \mathrm{~km}$
(C) $360 \mathrm{~km}$
(D) $180 \mathrm{~km}$

Solution

Let $P=Port$
$S_{1}=$ Ship 1
$S_{2}=$ Ship 2

When $S_{1}$ travelled $x$ km, $S_{2}$ travelled $80 km$ \& when $S_{1}$ travelled $d$ km, $S_{2}$ travelled $d-120 km$. Since they we moving at constant velocites.

$\frac{x}{80}=\frac{d}{d-120} \Rightarrow d-x=\frac{d(d-200)}{(d-120)}$

$d-x =(d-80) \cos 60^{\circ}$
$=\frac{1}{2}(d-80)$

So, $\frac{d(d-200)}{(d-120)}=\frac{1}{2}(d-80)$

$d^{2}-200 d-120 \times 80=0 \Rightarrow(d-240)(d+40)=0$

So, $d=240$ (A)

###### Problem 28

If $x_{1}>x_{2}>\cdots>x_{10}$ are real numbers, what is the least possible value of $\left(\frac{x_{1}-x_{10}}{x_{1}-x_{2}}\right)\left(\frac{x_{1}-x_{10}}{x_{2}-x_{3}}\right) \cdots\left(\frac{x_{1}-x_{10}}{x_{9}-x_{10}}\right)$

(A) $10^{10}$

(B) $10^{9}$

(C) $9^{9}$

(D) $9^{10}$

Solution

$x_{1}>x_{2}>\ldots>x_{10} \in \mathbb{R}$
$x_{i}-x_{j}>0 \quad \forall 1 \leq j < i \leq 10$

$(x_{1}-x_{2})+(x_{2}-x_{3})+\ldots+(x_{9}-x_{10})=(x_{1}-x_{10})$

By AM-GM Inequality we get, $[(x_{1}-x_{2})+\cdots+(x_{4}-x_{0})]^{9} \geq 9^{9}(x_{1}-x_{2})(x_{2}-x_{3}) \cdots(x_{9}-x_{10})$

$(x_{1}-x_{10})^{9} \geq 9^{9}(x_{1}-x_{2}) \ldots(x_{1}-x_{10})$

$\frac{(x_{1}-x_{10})}{(x_{1}-x_{2})} \times \frac{(x_{1}-x_{10})}{(x_{2}-x_{3})} \cdots \frac{(x_{1}-x_{6})}{(x_{1}-x_{10})} \geq 9^{9}$

Equality is achieved when $x_{i}$ are in AP eg $x_{i}=i$(C)

###### Problem 29

The range of values that the function

𝑓(𝑥)=𝑥2+2𝑥+42𝑥2+4𝑥+9=12

takes as $x$ varies over all real numbers in the domain of $f$ is:
(A) $\frac{3}{7}<f(x) \leq \frac{1}{2}$
(B) $\frac{3}{7} \leq f(x)<\frac{1}{2}$
(C) $\frac{3}{7}<f(x) \leq \frac{4}{9}$
(D) $\frac{3}{7} \leq f(x) \leq \frac{1}{2}$

Solution

$f(x)=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$

$=\frac{1}{2}-\frac{1}{2\left(2 x^{2}+4 x+9\right)}$

$=\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]}$

$2(x+1)^{2}+7 \geq 7$

$-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq-\frac{1}{14}$

$\frac{1}{2}>\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq \frac{1}{2}-\frac{1}{14}=\frac{3}{7}$

###### Problem 30

In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area $x$ of the remaining triangle.

(B) 13
(C) 14
orms a squole of
(A) 12
(D) 15

#### Subjective Section

Note. In this question-paper, $\mathbb{R}$ denotes the set of real numbers.

###### Problem 1

Consider a board having 2 rows and $n$ columns. Thus there are $2 n$ cells in the board. Each cell is to be filled in by 0 or 1 .
(a) In how many ways can this be done such that each row sum and each column sum is even?
(b) In how many ways can this be done such that each row sum and each column sum is odd?

Solution

###### Problem 2

Consider the function

𝑓(𝑥)=𝑘=1𝑚(𝑥𝑘)4,𝑥,

where $m>1$ is an integer. Show that $f$ has a unique minimum and find the point where the minimum is attained.

Solution

###### Problem 3

Consider the parabola $C: y^{2}=4 x$ and the straight line $L$ : $y=x+2$. Let $P$ be a variable point on $L$. Draw the two tangents from $P$ to $C$ and let $Q_{1}$ and $Q_{2}$ denote the two points of contact on $C$. Let $Q$ be the mid-point of the line segment joining $Q_{1}$ and $Q_{2}$. Find the locus of $Q$ as $P$ moves along $L$.

Solution

###### Problem 4

Let $P(x)$ be an odd degree polynomial in $x$ with real coefficients. Show that the equation $P(P(x))=0$ has at least as many distinct real roots as the equation $P(x)=0$.

Solution

Since $P(x)$ and $P(P(x))$ both are odd degree polynomial it must have at least one real root.

Suppose, the distinct real roots of $P(x)$ be $\alpha_{1}, \alpha_{2,}, \ldots, \alpha_{k}$

Roots of $P(P(x))$ happens for all $x$ such that $p(x)=\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$

$P(P(x))=\left(P(x)-\alpha_{1}\right)\left(P(x)-\alpha_{2}\right)\left(P(x)-\alpha_{3}\right) \cdots\left(P(x)-\alpha_{k}\right)Q(x)$

Now, $P(x)=\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are $k$ different odd degree polynomials therefore there shoul we at least $K$ distinct real roots of $P(P(x))$. They should be distinct because $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are themselves distinct.

###### Problem 5

For any positive integer $n$, and $i=1,2$, let $f_{i}(n)$ denote the number of divisors of $n$ of the form $3 k+i$ (including 1 and $n$ ). Define, for any positive integer $n$.

𝑓(𝑛)=𝑓1(𝑛)𝑓2(𝑛).

Find the values of $f\left(5^{2022}\right)$ and $f\left(21^{2022}\right)$.

Solution

(i) Factors of $5^{2022}$ :
$1,5^{1}, 5^{2}, 5^{3}, \ldots, 5^{2021}, 5^{2022}$

$5^{\text {odd }}\equiv (-1)^{\text {odd }}(\bmod 3)$

$\quad\quad\equiv 2(\bmod 3)$

$5^{\text {even }} \equiv(-1)^{\text {even }} \equiv 1 (\bmod 3)$

$f_{1}\left(5^{2022}\right)=1012$
$f_{2}\left(5^{2022}\right)=1011$
$f\left(5^{2022}\right)=1$

(ii) $21^{2022}=3^{2022} \times 7^{2022}$

factors without 3 as a factor are only considered:
$1,7^{1}, 7^{2}, 7^{3}, \cdots, 7^{2021}, 7^{2022}$

$7^{k} \equiv 1^{k} \equiv 1(\bmod 3) \forall k \in N_{0}$

$f_{1}\left(21^{2022}\right)=2023$
$f_{2}\left(21^{2022}\right)=0$
$f\left(2^{2022}\right)=2023$

###### Problem 6

Consider a sequence $P_{1}, P_{2}, \ldots$ of points in the plane such that $P_{1}, P_{2}, P_{3}$ are non-collinear and for every $n \geq 4, P_{n}$ is the midpoint of the line segment joining $P_{n-2}$ and $P_{n-3}$. Let $L$ denote the line segment joining $P_{1}$ and $P_{5}$. Prove the following:
(a) The area of the triangle formed by the points $P_{n}, P_{n-1,} P_{n-2}$ converges to zero as $n$ goes to infinity.
(b) The point $P_{9}$ lies on $L$.

Solution

(a) Let us denole the complex no. connespida to $P_{n}$ by $p_{n}$.
Then

satisfies the recurrence relation
$2 p_{n}=p_{n-2}+p_{n-3}, n \geq 4$

It suffices to find three independent solutions to $2 z_{n}=z_{n-2}+z_{n-3}, n \geq 4$ to find the genenal solution. Let's try to find solutions of the form $z_{n}=\lambda^{n}$.

$2 \lambda^{n}=\lambda^{n-2}+\lambda^{n-3}$
$\Rightarrow 2 \lambda^{3}=\lambda+1$
$\Rightarrow 2 \lambda^{3}-2 \lambda^{2}+2 \lambda^{2}-2 \lambda+\lambda-1=0$
$\Rightarrow (\lambda-1)\left(2 \lambda^{2}+2 \lambda+1\right)=0$
$\Rightarrow \lambda=1, \frac{-2 \pm \sqrt{4-8}}{4}$
$=1,-\frac{1}{2} \pm 1 / 2$ Let $\varepsilon=e^\frac{i2 \pi}{8}$
So the roots are $1,\frac{\varepsilon^{3}}{\sqrt{2}} \frac{\varepsilon^{5}}{\sqrt{2}}$

Thenefone, the general sol is
$$Z_{n}=A+B \frac{\varepsilon^{3 n}}{(\sqrt{2})^{n}}+C \frac{\varepsilon^{5 n}}{(\sqrt{2})^{n}}$$
The (complex) constants can be adjusted to make
$$z_{1}=p_{1}, z_{2}=p_{2}, z_{3}=p_{3} .$$
As $n \rightarrow \infty$
$$\left|\left(\frac{\varepsilon^{3}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0, \left|\left(\frac{\varepsilon^{5}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0$$
Thus $Z n \rightarrow A$
Thenefore, eventually all points are contained in a ball of radius $\varepsilon$ about $A$, no matter how small $\varepsilon>0$ might be. Thus, the anea of $z_{n} z_{n-1} z_{n-2}$ must go to zero.

(b) $p_{9}=\frac{p_{7}+p_{6}}{2}$
$=\frac{\frac{1}{2}\left(p_{5}+p_{4}\right)+\frac{1}{2}\left(p_{4}+p_{3}\right)}{2}$
$=\frac{2 p_{4}+p_{5}+p_{3}}{4}$
$=\frac{p_{2}+p_{1}+p_{5}+p_{3}}{4}$
$=\frac{p_{1}+p_{5}+\left(p_{2}+p_{3}\right)}{4}$
$=\frac{p_{1}+p_{5}+2 p_{5}}{4}$
$=\frac{p_{1}+3 p_{5}}{4}$

7. Let

𝑃(𝑥)=1+2𝑥+7𝑥2+13𝑥3,𝑥

Calculate for all $x \in \mathbb{R}$.
lim𝑛→∞(𝑃(𝑥𝑛))𝑛.

Find the minimum value of

Find the smallest positive real number $k$ such that the following inequality holds

This is a work in progress. Please come back for the solutions. You can also suggest your solutions in the comment section

#### Objective Section

###### Problem 1

Any positive real number $x$ can be expanded as

$x=a_{n} \cdot 2^{n}+a_{n-1} \cdot 2^{n-1}+ \\ \cdots+a_{1} \cdot 2^{1}+a_{0} \cdot 2^{0}+a_{-1} \\ \cdot 2^{-1}+a_{-2} \cdot 2^{-2}+\cdots$

for some $n \geq 0$, where each $a_{i} \in{0,1}$. In the above-described expansion of $21.1875$, the smallest positive integer $k$ such that $a_{-k} \neq 0$ is:

(B) 2

(C) 1

(D) 4

Solution

$x=a_{n} 2^{n}+a_{n-1} 2^{n-1}+\cdots+a_{1} 2^{1}+a_{1}$

Binary represation of $21.1875$ is $10101.0011$ (very standard Method to Convert Dercimal to Binary)

So, $-k=-3$

So, we got $k=3$ (A)

###### Problem 2

Suppose, for some $\theta \in\left[0, \frac{\pi}{2}\right], \frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $(\cot 3 \theta) \tan \theta$ equals
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{8}$
(D) $\frac{1}{7}$ <— Answer

Solution

$\frac{\cos 3 \theta}{ \cos \theta}=\frac{1}{3}$

$\Rightarrow \frac{4 \cos^{ 3} \theta-3 \cos \theta}{\cos \theta}=\frac{1}{3}$

$\Rightarrow 4 \cos ^{2} \theta-3=\frac{1}{3}$

$\Rightarrow \cos ^{2} \theta=\frac{10}{12}$

So, $\sin ^{2} \theta=\frac{2}{12}$

$\tan 3 \theta\cot \theta=\frac{8-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}=\frac{3-\frac{1}{5}}{1-\frac{3}{5}}=7$

So we got, $\cot 3 \theta \tan \theta=\frac{1}{7}$ (D)

###### Problem 3

The locus of points $z$ in the complex plane satisfying $z^{2}+|z|^{2}=0$ is
(A) a straight line
(B) a pair of straight lines
(C) a circle
(D) a parabola

Solution

###### Problem 4

Amongst all polynomials $p(x)=c_{0}+c_{1} x+\cdots+c_{10} x^{10}$ with real coefficients satisfying $|p(x)| \leq|x|$ for all $x \in[-1,1]$, what is the maximum possible value of $\left(2 c_{0}+c_{1}\right)^{10}$ ?
(A) $4^{10}$
(B) $3^{10}$
(C) $2^{10}$
(D) 1

Solution

###### Problem 5

Let $\mathbb{Z}$ denote the set of integers. Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be such that $f(x) f(y)=f(x+y)+f(x-y)$ for all $x, y \in \mathbb{Z}$. If $f(1)=3$, then $f(7)$ equals
(A) 840
(B) 844
(C) 843 <— Answer
(D) 842

Solution

$f:\mathbb{Z} \rightarrow \mathbb{Z}$

$f(x) f(y)=f(x+y)+f(x-y)$

$f(1) f(0)=f(1)+f(1)$
$\Rightarrow 3 f(0)=f(1)+f(3)$

$\Rightarrow 3 f(0)=6$
$\Rightarrow f(0)=2$

Now, $f(1) f(1)=f(2)+f(0)$
$\Rightarrow 9=f(2)+2$
$\Rightarrow f(2)=7$

Again, $f(2) f(1)=f(3)+f(1)$
$\Rightarrow 7 \times 3=f(3)+3$
$\Rightarrow f(3)=18$

Again, $f(3) f(1)=f(4)+f(2)$
$\Rightarrow 18 \times 3=f(4)+7$
$\Rightarrow f(4)=47$

$f(3) f(4)=f(4+3)+f(1)$
$\Rightarrow (18 \times 47)-3=f(7)$
$\Rightarrow f(3)=843$ (C)

###### Problem 6

Let $A$ and $B$ be two $3 \times 3$ matrices such that $(A+B)^{2}=A^{2}+B^{2}$. Which of the following must be true?
(A) $A$ and $B$ are zero matrices.
(B) $A B$ is the zero matrix.
(C) $(A-B)^{2}=A^{2}-B^{2}$
(D) $(A-B)^{2}=A^{2}+B^{2}$ <— Answer

Solution

$(A+B)(A+B)$
$=A^{2}+B A+A B+B^{2}$

So, $B A+A B=0$

$(A-B)(A-B)$
$=A^{2}-A B-B A+B^{2}$
$=A^{2}+B^{2}$ (D)

###### Problem 7

Let $\left(n_{1}, n_{2}, \cdots, n_{12}\right)$ be a permutation of the numbers $1,2, \cdots, 12$. The number of arrangements with

$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}$
and $n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$ equals:

(A) $^{12} C_{5}$

(B) $^{12} C_{6}$

(C) $^{11} C_{6}$

(D) $\frac{11!}{2}$

Solution

$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$

$n_{6}$ is the smallest so,$n_{6}=1$

So, if we just choose first 5 it is enough.

So, $^{11} C_{5}$ as $n_{6}$ is fixed

$= ^{11} C_{6}$ (C)

###### Problem 8

The sides of a regular hexagon $A B C D E F$ is extended by doubling them to form a bigger hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ as in the figure below.

Then the ratio of the areas of the bigger to the smaller hexagon is:
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) 4

###### Problem 9

In how many ways can we choose $a_{1}<a_{2}<a_{3}<a_{4}$ from the set ${1,2, \ldots, 30}$ such that $a_{1}, a_{2}, a_{3}, a_{4}$ are in arithmetic progression?
(A) 135
(B) 145
(C) 155
(D) 165

###### Problem 10

Suppose the numbers 71,104 and 159 leave the same remainder $r$ when divided by a certain number $N>1$. Then, the value of $3 N+4 r$ must equal:
(A) 53
(B) 48
(C) 37
(D) 23

###### Problem 11

If $x, y$ are positive real numbers such that $3 x+4 y<72$, then the maximum possible value of $12 x y(72-3 x-4 y)$ is:
(A) 12240
(D) 13824
(C) 10656
(D) 8640

###### Problem 12

What is the minimum value of the function $|x-3|+|x+2|+|x+1|+|x|$ for real $x$ ?
(A) 3
(B) 5
(C) 6
(D) 8

###### Problem 13

Consider a differentiable function $u:[0,1] \rightarrow \mathbb{R}$. Assume the function $u$ satisfies
$u(a)=\frac{1}{2 r} \int_{a-r}^{a+r} u(x) d x, \quad$ for all $a \in(0,1)$ and all $r<\min (a, 1-a)$.
(A) $u$ attains its maximum but not its minimum on the set ${0,1}$.
(B) $u$ attains its minimum but not maximum on the set ${0,1}$.
(C) If $u$ attains either its maximum or its minimum on the set ${0,1}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set ${0,1}$.

###### Problem 14

A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance $x$ below the ladders?
(A) 3
(B) $2 \sqrt{2}$
(C) $\frac{8}{3}$ <— Answer
(D) $2 \sqrt{3}$

Solution

$\frac{a}{a+b}=\frac{x}{4}$, in $\triangle A C D$

$\frac{b}{a+b}=\frac{x}{8}$, in $\triangle A B D$

So, from the previous relations we get, $a=2 b$.

So, $\frac{x}{4}=\frac{2 b}{3 b}$
$\Rightarrow x=\frac{8}{3}$ feet (C)

###### Problem 15

Let $y=x+c_{1}, y=x+c_{2}$ be the two tangents to the ellipse $x^{2}+4 y^{2}=1$. What is the value of $\left|c_{1}-c_{2}\right| ?$
(A) $\sqrt{2}$
(B) $\sqrt{5}$ <— Answer
(C) $\frac{\sqrt{5}}{2}$
(D) 1

Solution

$x^{2}+4 y^{2}=1 \rightarrow(1)$

We want $\frac{d y}{d x}=1$

So differentiating $(1)$ with respect of $x$ we get

$2 x+8 y \frac{d y}{d x}=0$

for $\frac{d y}{d y}$ to be 1 we get $x=-4y$

Substituting $(1)$

$16 y^{2}+ 4y^{2}=1$

$\Rightarrow y_{0}=\frac{1}{2 \sqrt{5}} \text{or}-\frac{1}{2 \sqrt{5}}$

Let $y_{0}=\frac{1}{2 \sqrt{5}}$,

$\Rightarrow x_{0}=\frac{2}{\sqrt{5}}$

$c_{2}=-\frac{\sqrt{5}}{2}$

Now, $|c_{1}-c_{2}|=\sqrt{5}$ (B)

###### Problem 16

In the figure below, $A B C D$ is a square and $\triangle C E F$ is a triangle with given sides inscribed as in the figure. Find the length $B E$.

(A) $\frac{13}{\sqrt{17}}$ <— Answer
(B) $\frac{14}{\sqrt{17}}$
(C) $\frac{15}{\sqrt{17}}$
(D) $\frac{16}{\sqrt{17}}$

Solution

$\cos \theta=\frac{a}{4}=\frac{A F}{3}$

$A F=\frac{3 a}{4}$

$\tan \theta=\frac{1}{4}=\frac{A E}{\frac{3 a}{4}}$
$\Rightarrow A E=\frac{3 a}{16}$

$a^{2}+\frac{a^{2}}{16}=16$
$\Rightarrow \quad \frac{17 a^{2}}{16}=16$
$\Rightarrow \quad a^{2}=\frac{16 \times 16}{17}$
$\Rightarrow \quad a=\frac{16}{\sqrt{17}}$

$B E =\frac{13 a}{16}$
$B E =\frac{13}{\sqrt{17}}$ (A)

###### Problem 17

Let $p$ and $q$ be two non-zero polynomials such that the degree of $p$ is less than or equal to the degree of $q$, and $p(a) q(a)=0$ for $a=0,1,2, \ldots, 10$. Which of the following must be true?
(A) degree of $q \neq 10$
(B) degree of $p \neq 10$
(C) degree of $q \neq 5$ <— Answer
(D) degree of $p \neq 5$

Solution

$p(a) q(a)=0$, for $a=0,1,2, \cdots, 10$

Hence the polynonial, $P(x) q(x)$ has at least 11 roots as so $deg(p(x) q(x)) \geq 11$

Now, if $deg(q(x))=5$, then $deg(p(x)) \leq 5 \Rightarrow deg(p(x) q(x)) \leq 10$ which is a contradiction.

Degree of $q \neq 5$

###### Problem 18

For $n \in \mathbb{N}$, let $a_{n}$ be defined

𝑎𝑛=𝑛011+𝑛𝑥2𝑑𝑥

Then $\lim {n \rightarrow \infty} a{n}$

(A) equals 0 <— Answer
(B) equals $\frac{\pi}{4}$
(C) equals $\frac{\pi}{2}$
(D) does not exist

Solution

$a_{n}=\int_{0}^{n} \frac{1}{1+n x^{2}} d x$

$=\frac{1}{\sqrt{n}} \int_{0}^{n} \frac{1}{1+(\sqrt{n} x)^{2}} d(\sqrt{n} x)$

$=\frac{1}{\sqrt{n}}\left[\tan ^{-1}(\sqrt{n} x)\right]_{0}^{n}$

$a_{n}=\frac{1}{\sqrt{n}} \tan ^{-1}\left(n^{3 / 2}\right)$

$=\frac{1}{\sqrt{n}}\left(\frac{\pi}{2}-\cot ^{-1}\left(n^{3 / 2}\right)\right)$

$\Rightarrow a_{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{\sqrt{n}} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)$

$\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)=\frac{1}{n^{3 / 2}}$, as $\frac{1}{n^{3 / 2}} \rightarrow 0$

$\lim {n \rightarrow \infty} a{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{n^{2}}$

$\Rightarrow \lim {n \rightarrow \infty} a{n}=0$

###### Problem 19

A $3 \times 3$ magic square is a $3 \times 3$ rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square
(A) 90
(B) 96 <— Answer
(C) 94
(D) 99

Solution

Asrume the sum is $s$ and $x$ as shown. Fill the rest of the square setting the diagonals equal to $s$ we get.

$s-2x=40$

$s-x=68$

So we get that $s=96$

###### Problem 20

The number of positive integers $n$ less than or equal to 22 such that 7 divides $n^{5}+4 n^{4}+3 n^{3}+2022$ is
(A) 7
(B) 8
(C) 9
(D) 10 <— Answer

Solution

$n^{5}+4 n^{4}+3 n^{3}+2022$

$\left.\equiv n^{3}(n+1)(n+3)+6 (\bmod 7\right)$

checking $n \equiv \pm 0, \pm 1, \pm 2, \pm 3 (\bmod 7)$

Now we see that only $n=1, \pm 2$ is permitted

Thus, $n \in{1,2,5,8,9,12,15,16,10,22}$

###### Problem 21

In a class of 45 students, three students can write well using either hand. The number of students who can write well only with the right hand is 24 more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:
(A) 33
(B) 36 <— Answer
(C) 39
(D) 41

Solution

$R =$ Set of right handers

$L$= Set of left handers

$|R \cap L|=3, | R \cup L|=45$

$24+\left|R^{C} \cap L|=\right| R \cap L^{c}|=|\left(R^{c}\right)^{c} \cap L^{c} \mid \left|\left(R^{c} \cup L\right)^{c}\right|$

$=45-\mid R^{c} \cup L|$

$45-\left(\left|R^{c}\right|+|L|-| R^{c} \cap L \mid\right)$

$=\left|R^{c} \cap L\right|+\left(45-\left|R^{c}\right|\right)-|L|$

$=\left|R^{c} \cap L\right|+|R|-L \mid$

Thenefore, $|R|-|L|=24$

Moreover, $|R|+|L| =|R \cup L|+\mid R \cap L| =45+3=48$

Hence, $|R|=\frac{24+48}{2}=\frac{72}{2}=36$

###### Problem 22

Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
(B) $3^{10}$
(C) $2^{10} \omega$ <— Answer
(D) $3^{10} \omega^{2}$

Solution

$2 \equiv-1$ mod 3

$\Rightarrow 2^{k} \equiv(-1)^{k}$ mod 3

$\Rightarrow \omega^{2 k}=\omega^{(-1)^{k}}$

$\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1-\omega^{2}+\omega^{2^{{2}}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left(1-\omega^{2^{2}}+ \omega^{2^{3}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1-\omega^{2^{3}}+\omega^{2^{4}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left(1- \omega^{2^{4}} + \omega^{2{^5}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1- \omega^{2^{5}}+\omega^{2^{6}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left.(1-\ \omega^{2^{6}}+\omega^{2^{7}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1-\omega^{2^{7}}+\omega^{2^{8}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

$\left(1-\omega^{2^{8}}+\omega^{2^{9}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$

$\left(1- \omega^{2^{9}}+\omega^{2^{10}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$

Product $(-1)^{10} 2^{10} \cdot \omega^{10+5}=2^{10}$

###### Problem 23

The function $x^{2} \log {e} x$ in the interval $(0,2)$

(A) exactly one point of local maximum and no points of local minimum.
(B) exactly one point of local minimum and no points of local maximum. <— Answer
(C) points of local maximum as well as local minimum.
(D) neither a point of local maximum nor a point of local minimum.

Solution

$f(x) =x^{2} \log {e} x$ $\Rightarrow f^{\prime}(x)=x+2 x \log {e} x=x(1+2 \log _{e}^{x})$

###### Problem 24

The number of triples $(a, b, c)$ of positive integers satisfying the equation $(2,3,4) \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{2}{a b c}$
and such that $a<b<c$, equals:
(A) 3
(B) 2
(C) 1 <— Answer
(D) 0

Solution

1𝑎+1𝑏+1𝑐=2+1𝑎𝑏𝑐;𝑎,𝑏,𝑐<𝑎<𝑏<𝑐. 𝑎𝑏+𝑎𝑐+𝑏𝑐=𝑎𝑏𝑐+2;𝑎,𝑏,𝑐<

$\Rightarrow a(b+c)+b c=a b c+2$
$\Rightarrow a(b+c-b c)=2-b c$
$\Rightarrow a(b c-b-c)=b c-2$

$b c-b-c \mid b c-2$
$\Rightarrow b c-b-c \mid b+c-2$

$b c-b-c \leq b+c-2$
$\Rightarrow b c-2 b-2 c \leq-2$
$\Rightarrow(b-2)(c-2) \leq 2$

$b-2=1$, $c-2=2$

So, $b=3, c=4$

So, $a=2$ (C)

###### Problem 25

An urn contains 30 balls out of which one is special. If 6 of these balls are taken out at random, what is the probability that the special ball is chosen?
(A) $\frac{1}{30}$
(B) $\frac{1}{6}$
(C) $\frac{1}{5}$ <— Answer
(D) $\frac{1}{15}$

Solution

###### Problem 26

A triangle has sides of lengths $\sqrt{5}, 2 \sqrt{2}, \sqrt{3}$ units. Then, the radius of its inscribed circle is :
(A) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{2}$
(B) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{3}$
(C) $\sqrt{5}+\sqrt{3}+2 \sqrt{2}$
(D) $\frac{\sqrt{5}+\sqrt{3}-2 \sqrt{2}}{2}$ <— Answer

Solution

$r_{\Delta}=\frac{2 \times {\text {Area in }} \Delta \triangle A B C}{\text { perimeter of } \triangle A B C}$
$\Rightarrow r_{\Delta}=2 \times \frac{\frac{1}{2} \times \sqrt{3} \times \sqrt{5}}{\sqrt{3}+\sqrt{5}+2 \sqrt{2}}$
$\Rightarrow r_{\Delta}=\frac{\sqrt{3} \times \sqrt{5}(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{(\sqrt{3}+\sqrt{5})^{2}-(2 \sqrt{2})^{2}}$
$\frac{\sqrt{3} \times \sqrt{5} \times(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{2 \times \sqrt{3} \times \sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}-2 \sqrt{2}}{2}(D)$

###### Problem 27

Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled $80 \mathrm{~km}$, the triangle became right-angled.

When the first ship reaches the port, the second ship was still $120 \mathrm{~km}$ from the port. Find the initial distance of the ships from the port.
(A) $240 \mathrm{~km}$ <— Answer
(B) $300 \mathrm{~km}$
(C) $360 \mathrm{~km}$
(D) $180 \mathrm{~km}$

Solution

Let $P=Port$
$S_{1}=$ Ship 1
$S_{2}=$ Ship 2

When $S_{1}$ travelled $x$ km, $S_{2}$ travelled $80 km$ \& when $S_{1}$ travelled $d$ km, $S_{2}$ travelled $d-120 km$. Since they we moving at constant velocites.

$\frac{x}{80}=\frac{d}{d-120} \Rightarrow d-x=\frac{d(d-200)}{(d-120)}$

$d-x =(d-80) \cos 60^{\circ}$
$=\frac{1}{2}(d-80)$

So, $\frac{d(d-200)}{(d-120)}=\frac{1}{2}(d-80)$

$d^{2}-200 d-120 \times 80=0 \Rightarrow(d-240)(d+40)=0$

So, $d=240$ (A)

###### Problem 28

If $x_{1}>x_{2}>\cdots>x_{10}$ are real numbers, what is the least possible value of $\left(\frac{x_{1}-x_{10}}{x_{1}-x_{2}}\right)\left(\frac{x_{1}-x_{10}}{x_{2}-x_{3}}\right) \cdots\left(\frac{x_{1}-x_{10}}{x_{9}-x_{10}}\right)$

(A) $10^{10}$

(B) $10^{9}$

(C) $9^{9}$

(D) $9^{10}$

Solution

$x_{1}>x_{2}>\ldots>x_{10} \in \mathbb{R}$
$x_{i}-x_{j}>0 \quad \forall 1 \leq j < i \leq 10$

$(x_{1}-x_{2})+(x_{2}-x_{3})+\ldots+(x_{9}-x_{10})=(x_{1}-x_{10})$

By AM-GM Inequality we get, $[(x_{1}-x_{2})+\cdots+(x_{4}-x_{0})]^{9} \geq 9^{9}(x_{1}-x_{2})(x_{2}-x_{3}) \cdots(x_{9}-x_{10})$

$(x_{1}-x_{10})^{9} \geq 9^{9}(x_{1}-x_{2}) \ldots(x_{1}-x_{10})$

$\frac{(x_{1}-x_{10})}{(x_{1}-x_{2})} \times \frac{(x_{1}-x_{10})}{(x_{2}-x_{3})} \cdots \frac{(x_{1}-x_{6})}{(x_{1}-x_{10})} \geq 9^{9}$

Equality is achieved when $x_{i}$ are in AP eg $x_{i}=i$(C)

###### Problem 29

The range of values that the function

𝑓(𝑥)=𝑥2+2𝑥+42𝑥2+4𝑥+9=12

takes as $x$ varies over all real numbers in the domain of $f$ is:
(A) $\frac{3}{7}<f(x) \leq \frac{1}{2}$
(B) $\frac{3}{7} \leq f(x)<\frac{1}{2}$
(C) $\frac{3}{7}<f(x) \leq \frac{4}{9}$
(D) $\frac{3}{7} \leq f(x) \leq \frac{1}{2}$

Solution

$f(x)=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$

$=\frac{1}{2}-\frac{1}{2\left(2 x^{2}+4 x+9\right)}$

$=\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]}$

$2(x+1)^{2}+7 \geq 7$

$-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq-\frac{1}{14}$

$\frac{1}{2}>\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq \frac{1}{2}-\frac{1}{14}=\frac{3}{7}$

###### Problem 30

In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area $x$ of the remaining triangle.

(B) 13
(C) 14
orms a squole of
(A) 12
(D) 15

#### Subjective Section

Note. In this question-paper, $\mathbb{R}$ denotes the set of real numbers.

###### Problem 1

Consider a board having 2 rows and $n$ columns. Thus there are $2 n$ cells in the board. Each cell is to be filled in by 0 or 1 .
(a) In how many ways can this be done such that each row sum and each column sum is even?
(b) In how many ways can this be done such that each row sum and each column sum is odd?

Solution

###### Problem 2

Consider the function

𝑓(𝑥)=𝑘=1𝑚(𝑥𝑘)4,𝑥,

where $m>1$ is an integer. Show that $f$ has a unique minimum and find the point where the minimum is attained.

Solution

###### Problem 3

Consider the parabola $C: y^{2}=4 x$ and the straight line $L$ : $y=x+2$. Let $P$ be a variable point on $L$. Draw the two tangents from $P$ to $C$ and let $Q_{1}$ and $Q_{2}$ denote the two points of contact on $C$. Let $Q$ be the mid-point of the line segment joining $Q_{1}$ and $Q_{2}$. Find the locus of $Q$ as $P$ moves along $L$.

Solution

###### Problem 4

Let $P(x)$ be an odd degree polynomial in $x$ with real coefficients. Show that the equation $P(P(x))=0$ has at least as many distinct real roots as the equation $P(x)=0$.

Solution

Since $P(x)$ and $P(P(x))$ both are odd degree polynomial it must have at least one real root.

Suppose, the distinct real roots of $P(x)$ be $\alpha_{1}, \alpha_{2,}, \ldots, \alpha_{k}$

Roots of $P(P(x))$ happens for all $x$ such that $p(x)=\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$

$P(P(x))=\left(P(x)-\alpha_{1}\right)\left(P(x)-\alpha_{2}\right)\left(P(x)-\alpha_{3}\right) \cdots\left(P(x)-\alpha_{k}\right)Q(x)$

Now, $P(x)=\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are $k$ different odd degree polynomials therefore there shoul we at least $K$ distinct real roots of $P(P(x))$. They should be distinct because $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are themselves distinct.

###### Problem 5

For any positive integer $n$, and $i=1,2$, let $f_{i}(n)$ denote the number of divisors of $n$ of the form $3 k+i$ (including 1 and $n$ ). Define, for any positive integer $n$.

𝑓(𝑛)=𝑓1(𝑛)𝑓2(𝑛).

Find the values of $f\left(5^{2022}\right)$ and $f\left(21^{2022}\right)$.

Solution

(i) Factors of $5^{2022}$ :
$1,5^{1}, 5^{2}, 5^{3}, \ldots, 5^{2021}, 5^{2022}$

$5^{\text {odd }}\equiv (-1)^{\text {odd }}(\bmod 3)$

$\quad\quad\equiv 2(\bmod 3)$

$5^{\text {even }} \equiv(-1)^{\text {even }} \equiv 1 (\bmod 3)$

$f_{1}\left(5^{2022}\right)=1012$
$f_{2}\left(5^{2022}\right)=1011$
$f\left(5^{2022}\right)=1$

(ii) $21^{2022}=3^{2022} \times 7^{2022}$

factors without 3 as a factor are only considered:
$1,7^{1}, 7^{2}, 7^{3}, \cdots, 7^{2021}, 7^{2022}$

$7^{k} \equiv 1^{k} \equiv 1(\bmod 3) \forall k \in N_{0}$

$f_{1}\left(21^{2022}\right)=2023$
$f_{2}\left(21^{2022}\right)=0$
$f\left(2^{2022}\right)=2023$

###### Problem 6

Consider a sequence $P_{1}, P_{2}, \ldots$ of points in the plane such that $P_{1}, P_{2}, P_{3}$ are non-collinear and for every $n \geq 4, P_{n}$ is the midpoint of the line segment joining $P_{n-2}$ and $P_{n-3}$. Let $L$ denote the line segment joining $P_{1}$ and $P_{5}$. Prove the following:
(a) The area of the triangle formed by the points $P_{n}, P_{n-1,} P_{n-2}$ converges to zero as $n$ goes to infinity.
(b) The point $P_{9}$ lies on $L$.

Solution

(a) Let us denole the complex no. connespida to $P_{n}$ by $p_{n}$.
Then

satisfies the recurrence relation
$2 p_{n}=p_{n-2}+p_{n-3}, n \geq 4$

It suffices to find three independent solutions to $2 z_{n}=z_{n-2}+z_{n-3}, n \geq 4$ to find the genenal solution. Let's try to find solutions of the form $z_{n}=\lambda^{n}$.

$2 \lambda^{n}=\lambda^{n-2}+\lambda^{n-3}$
$\Rightarrow 2 \lambda^{3}=\lambda+1$
$\Rightarrow 2 \lambda^{3}-2 \lambda^{2}+2 \lambda^{2}-2 \lambda+\lambda-1=0$
$\Rightarrow (\lambda-1)\left(2 \lambda^{2}+2 \lambda+1\right)=0$
$\Rightarrow \lambda=1, \frac{-2 \pm \sqrt{4-8}}{4}$
$=1,-\frac{1}{2} \pm 1 / 2$ Let $\varepsilon=e^\frac{i2 \pi}{8}$
So the roots are $1,\frac{\varepsilon^{3}}{\sqrt{2}} \frac{\varepsilon^{5}}{\sqrt{2}}$

Thenefone, the general sol is
$$Z_{n}=A+B \frac{\varepsilon^{3 n}}{(\sqrt{2})^{n}}+C \frac{\varepsilon^{5 n}}{(\sqrt{2})^{n}}$$
The (complex) constants can be adjusted to make
$$z_{1}=p_{1}, z_{2}=p_{2}, z_{3}=p_{3} .$$
As $n \rightarrow \infty$
$$\left|\left(\frac{\varepsilon^{3}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0, \left|\left(\frac{\varepsilon^{5}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0$$
Thus $Z n \rightarrow A$
Thenefore, eventually all points are contained in a ball of radius $\varepsilon$ about $A$, no matter how small $\varepsilon>0$ might be. Thus, the anea of $z_{n} z_{n-1} z_{n-2}$ must go to zero.

(b) $p_{9}=\frac{p_{7}+p_{6}}{2}$
$=\frac{\frac{1}{2}\left(p_{5}+p_{4}\right)+\frac{1}{2}\left(p_{4}+p_{3}\right)}{2}$
$=\frac{2 p_{4}+p_{5}+p_{3}}{4}$
$=\frac{p_{2}+p_{1}+p_{5}+p_{3}}{4}$
$=\frac{p_{1}+p_{5}+\left(p_{2}+p_{3}\right)}{4}$
$=\frac{p_{1}+p_{5}+2 p_{5}}{4}$
$=\frac{p_{1}+3 p_{5}}{4}$

7. Let

𝑃(𝑥)=1+2𝑥+7𝑥2+13𝑥3,𝑥

Calculate for all $x \in \mathbb{R}$.
lim𝑛→∞(𝑃(𝑥𝑛))𝑛.

Find the minimum value of

Find the smallest positive real number $k$ such that the following inequality holds

### Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy


text:
# [TxMt] LaTeX - Insert Label Based On Current Word

Sun May 27 11:19:33 UTC 2007

```On May 27, 2007, at 6:40 AM, Daniel Sönnerstedt wrote:

> LaTeX - Insert Label Based On Current Word
>
> I didn't get this feature to work on my project. I experimented a bit
> and it seemed like the script was having troubles dealing with
>
>   \include{"filename.tex"}
>
I actually have tried using filenames this way, and I find that LaTeX
doesn't process the corresponding file, unless you remove the .tex
part in the above line. Is there some package you need to use to make
this work with the .tex part?

> but not with
>
>   \include{filename.tex}
>
> I use this since I have filenames with spaces. So I skimmed through
> the code and added a small fix in LaTeXUtils.rb:
>
>    def find_file(filename, extension, relative)
>      filename.gsub!(/\.#{extension}\$/,"")
>      [...]
>    end
>
> became
>
>    def find_file(filename, extension, relative)
> to strip ":s
>      filename.gsub!(/\.#{extension}\$/,"")
>      [...]
>    end
>
> And now it works!
>
> Now, I'm not a ruby programmer so there's is probably some better way
> to solve this. But I think that this fix should benifit others;
> perhaps implemented in a more proper way though.
>
> Thanks for another great LaTeX-bundle-feature.
>
> Regards,
> Daniel Sönnerstedt

format: plain


text:
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text:
# Power financing up one third to $20 billion Oil and gas sector deals also performed well, with 16 deals worth$14 billion, up 96% on the same period in 2003.

And of the top 10 largest deals, eight are in the energy sector. Worthy of note, according to Dealogic analyst Petra Hofer, are the second and third spots taken by oil deals in Azerbaijan – the Baku to Ceyhan pipeline weighs in with $3.6 billion and the Unocal Azeri, Chirag, and Deepwater Gunashli (ACG) oilfield development at$3.2 billion.

“It’s quite unusual to see projects of


text:
# Tag Info

A semigroup is a set $S$ together with a binary operation that is associative. Examples of semigroups are the set of finite strings over a fixed alphabet (under concatenation) and the positive integers (under addition, maximum, or minimum). Of course, any monoid or group is also a semigroup.


text:
Peano Arithmetic before Gödel

If I understand correctly, Gödel was the one to discover how to encode finite sequences of integers in Peano Arithmetic, with his Chinese Remainder Theorem trick (his "beta function"). As a consequence, one can do exponentiation in PA, and many other things.

My question is historical: What did people think PA was good for before Gödel? What did Peano think his arithmetic was good for? Didn't he want to state (and prove) things like Fermat's little theorem, for example?

When Gödel published his paper on the incompleteness theorems, nobody had identified "Peano arithmetic" as a particular theory. Gödel instead referred to a system based on Principia Mathematica. It quickly became clear to Gödel and others that the same principles would apply to any $\omega$-consistent, effective, consistent theory that includes a small amount of arithmetic.

But nobody had studied Peano Arithmetic in particular before the incompleteness theorems were published. The history of the development of logic is an epitome of how many mathematical areas are discovered: not in an orderly way, but in fits and starts with researchers working independently, duplicating some results and taking a long time to fill in what are later considered basic topics. In particular, the underlying topic of "first order logic" was not really clear until the late 1940s or 1950s, when people had finally had enough time to digest the various results of the previous 50 years and organize them into a coherent system.

Before Gödel, Peano and Dedekind had both proposed axiomatic systems for arithmetic. These systems were second-order, however - they presupposed a notion of "set", which they did not axiomatize. So proving things from these axioms requires using not only the axioms, but also principles about set existence. We often call these the Peano axioms or the Peano postulates to differentiate them from Peano Arithmetic.

One of the key differences between these systems and the theory called "Peano Arithmetic" is that Peano Arithmetic is fully axiomatized, without assuming any background theory of sets. So it is possible to write formal proofs of many arithmetical statements fully from the axioms of PA, unlike the Peano axioms.

I am not sure of the first appearance of the specific axioms for Peano Arithmetic in print, or who coined that name for it. If I had them at hand, I would look at Kleene's Introduction to Metamathematics and Robinson's paper establishing Robinson Arithmetic, both from the 1950s, to see if they give any clue.

• So Gödel was purposely trying to see how much he could "lean out" a theory and still get his incompleteness result? I would gladly check out in Kleene's book, but the problem is that an interlibrary loan takes about three weeks... So I prefer asking here – Gabriel Nivasch Jan 3 '16 at 15:09
• Gödel didn't do that at all - he just pointed out in his paper that what he did could be formalized in the system from Principia Mathematica, which was not at all a lean theory. Robinson's arithmetic Q was designed to be as lean as possible while still having enough arithmetic to satisfy the first incompleteness theorem. – Carl Mummert Jan 3 '16 at 15:31

Giuseppe Peano and Richard Dedekind proposed the (second order) set of axioms for arithmetic, now usually called Peano axioms, as a set of postulates sufficient to derive all known arithmetical theorems.

We usually refer to Peano arithmetic as the first-order version of Peano axioms.

The original versions of Peano : Arithmetices principia: nova methodo exposita (1889) and Dedekind : Was sind und was sollen die Zahlen? (1893) used the second-order version of the Induction axiom.

See Peano's version :

if $k$ is a Class and $1 \in k$ and $x \in \mathbb N$ and $\forall x ((x \in k) \to (x+1 \in k))$, then $\mathbb N \subseteq k$.

We can find the first-order version of Peano axioms in :

• David Hilbert and Paul Bernays, Grundlagen der Mathematik (Vol. I, II), 1st German ed.1934/1939; see page 380 of 2nd German ed.1968 :

Axiomensystem Z :

$a = b \to (A(a) \to A(b))$,

$a' \ne 0$,

$a' = b' \to a = b$,

$a + 0 = a$,

$a + b' = (a + b)'$

$a' 0 = 0$,

$a . b' = a . b + a$,

$A (0) \land (x)(A (x) \to A (x')) \to A (a)$.

It seems that the use of "Peano arithmetic" to name the first-order theory of arithmetic is due to Undecidable Theories by A.Tarski, A.Mostowski, and R.M.Robinson (1953).

• Thanks. So did they define both systems -- the stronger "Peano axioms" and the weaker "Peano arithmetic" -- at the same time? And what was their reason for defining them? And are the stronger "Peano axioms" equivalent in strength to ZFC, or are they built on top of ZFC? Sorry I'm kind of confused... – Gabriel Nivasch Jan 3 '16 at 14:23
• Something is off here. The consistency of second-order arithmetic is a known arithmetical theorem, which is not derivable from the Peano axioms. A challenge with the Peano axioms is that they do not specify the set-theoretic assumptions that could be used, so it is actually hard to prove much at all from the Peano postulates alone. – Carl Mummert Jan 3 '16 at 14:32
• 1889 and 1893 ... there was no ZF there was no ZFC. They used "sets", for which no axioms were given or desired at that time. – GEdgar Jan 3 '16 at 14:50


text:
# What's the origin of the flattened X that appears when of a small (but powerfull enough) light is reflected in our non-powered TV screen?

The reflection of a small light (but with enough power) on our TV shows four emerging rays. The rays vary periodically in space and have a repeating pattern of the colors of the rainbow. The further the flame is from the TV the greater the distance between the periodic lights in the "rays". Sounds reasonable. This is the picture of the effect:

No matter how far you put the candle, the angle between the two lines stays the same. The color green is very faint in the picture but was clearly visible on the TV screen. The distance between successive red/green spots was getting smaller if the distance between your eye and the screen was getting smaller. Sounds reasonable. No matter how the light was positioned relative to the TV, the same pattern emerged.

I added a close-up picture of the working TV on which you can see what form the light-emitting pixels have:

Notice that they are curved, or better said, consist out of two little rectangles with an angle between them that is, at first sight, the same as the angle between the light rays seen on the TV-screen. When I measured the angle of the rays on the TV-screen it was about the same as the angle between the "line"-pieces of the pixels (about 180-(2x16)=148 degrees). Maybe here lies the clue to the problem. What would be the pattern seen on the TV-screen if the pixels would be square, rectangular (with the length vertically), or round? Or formed and arranged like here:

Surely not a flattened X! But to stay close to the question, what if the angle between the rectangular pieces of the pixels would be 180 degrees (so they become one rectangular) and positioned like in the lower right part of the above image? Would the X-form be stretched into a horizontal line (let's suppose all pixels, when the TV is off, reflect light equally well for what follows), with the colors of the rainbow repeating? In this case, a diffraction pattern is produced because of the different wavelengths of the colors in the rainbow. That is quite clear. All the red pixels reflect in such a way a light-dark pattern is found in the horizontal (squeezed X) line on the non-powered TV when we shine a light on it.

The fourth and fifth pictures I made by using a white and red LED (which shows the periodic Nature most clearly, and even red concentric rings), and in the sixth picture, I added some information on how I made the picture.

The last picture shows (somewhat fuzzy) the rectangular pixels and their configuration on my lap-top screen. It is the same forms of dots and they have the same configurations as in the pixels shown down left of the picture with four different forms and configurations. When I shine a light on my non-powered lap-top screen, I can discover no X-forms. It could be that the pixels don't reflect light.

But the final question: Are the same angles found in the pixels, (the big angle between the two rectangles) as in the big angle of the X correlated? If so, how?

I added this close-up picture of my laptop screen:

It's not that good visible, but the pixels are rectangular shaped (in vertical position). They are much smaller than the pixels on my big TV-screen, which I almost can see with the naked eye (once you know the shape after taking the close-up picture, which can't be said of the pixels of my laptop's screen). The effect isn't visible on my (power off) laptop screen, maybe because of the smallness of the pixels. The effect is (as I said ) color-dependent (with a red beam of light only a red X-form, periodically bright and dark is seen) and orientation independent (no matter how I point the light beam the same X appears over and over again). Nevertheless, I don't know if the pixels on both screens are the same (though it appears so).

• It may help to specify what type of TV it is. – lemon May 30 '16 at 18:58
• This sounds a lot like a diffraction grating effect. Did it look something like this (jasmcole.com/2016/02/21/tv-diffraction)? – M. Enns May 30 '16 at 19:07
• Duplicate? physics.stackexchange.com/q/167892 – Farcher May 30 '16 at 19:34
• @Lemon It´s a four fear old flatscreen from MT-Logic. Fairly big but not supersized. – Deschele Schilder May 30 '16 at 22:57
• Two words: Diffraction spikes. – David Hammen May 31 '16 at 2:00

It looks like a diffraction pattern from the pixels of the screen. In a different SE question, the pattern was four horizontal and vertical "rays" consisting of finely spaced peaks, rather than the wide spacing here and the angles that differ from 90 degrees.

If you measure the apparent angle between the first diffraction peaks (the ones at the edge of the halo) and the central spot, you can relate that to the wavelength (say, 600 nm) and the dot pitch of the tv screen. The ratio of distances screen-eye versus screen-candle also matters; the analysis is easier if this ratio is much smaller than one.

Regarding the angle between the rays: I suspect that your screen does not have rectangular pixels, but pixels arranged in some kind of staggered arrangement. However, I can't guess the exact pixel arrangement from the diffraction pattern. It could also be that you took the photo from an angle. Maybe you could post a picture of the TV pixels in close-up (when the TV is on) and a sketch of the relative locations of camera, candle, screen, and apparent diffraction spots.

Update

I did a 2D Fourier transform on the a pattern that looks roughly like your pixels: left image. (The fade-out towards the edges is a windowing function). The Fourier transform, which should look like your diffraction pattern, is on the right (amplitudes plotted, not intensities). Indeed, you see the flattened "X" shape appear, but also a vertical row of diffraction spots. I'm not sure, though, why the diffracted "X" is so pronounced in your case, with just two tilted rows of dots, rather than lots of dots that happen to be a bit brighter near the "X".

• The angle between the rays stayed the same from wherever I looked at the TV screen. Maybe it´s indeed a good idea to take a close-up picture of the pixels! – Deschele Schilder May 31 '16 at 10:53
• I still, despite your beautiful pictures, can´t understand where the X-form has its origins. – Deschele Schilder Jun 1 '16 at 12:14
• I have some ideas about that, but first you should answer my counterquestion. – Han-Kwang Nienhuys Jun 1 '16 at 12:16
• What was your counter question? You mean the one you asked (or said to do) in the last comment before your answer? – Deschele Schilder Jun 1 '16 at 12:28
• Only if you stick a ruler onto the tv and tell me where the candle is relative to the camera. So far I spent an hour on this question; is it too much to ask that you put some effort in it as well? – Han-Kwang Nienhuys Jun 1 '16 at 15:40


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# How to save the results in an identical Excel file (.xls) in each iteration of a Do loop

Assume the following code (Below code is just an example to clearly illustrate my problem):

Res = {};
Iterations = 5;

Do[
Results = AppendTo[Res, i^2]
, {i, 1, Iterations}];

I have 2 questions:

1. I want to save the results in each iteration in an identical excel file (.xls file). In the end, when the iterations are finished, I will have one Excel file with all the results in it.

2. I tried to make the code more faster and elegant using Scan[] , but I failed to do it and I couldn't do that.(It is good to mention that I am just a rookie Mathematica programmer and I don't have much experience in it!).

-
I'm not sure if it's possible to export data to a pre-existing Excel file. Personally, I'd collect the results from the loop and export them at the end. Also, Do is in no way a recursive construct. Are you sure you mean recursive? – Oleksandr R. Apr 30 '14 at 8:23
the standard Export for excel files can only write content in one go. Of course you can completely overwrite the file for each line, which is inefficient but probably not too problematic. Other than that I see the following possibilities: write to a text file (or CSV) line by line, for these appending is simple and efficient. Depending on the data, you won't loose much compared to an excel file. If you really need excel, you could use NETLink to do what you want, if you search for "NET excel" you'll find many Q+A to start... – Albert Retey Apr 30 '14 at 8:38
@Shellp: yes, you could write a new file per iteration, but if iterations are "quick", you'll waste time for file operations. – ciao Apr 30 '14 at 9:46
And then there is ExcelLink ... but not for free – Dr. belisarius Apr 30 '14 at 12:06
If one iteration takes 5 minutes and two iterations take 11 minutes then the second iteration must be taking 6 minutes. If the pattern continues (third iteration takes 7 minutes and so on) the total time should be over 24 hours. So the question is why does each iteration take longer than the last? We can't answer that without knowing what you're actually doing. – Simon Woods Apr 30 '14 at 16:09

Here are a couple of schemes suggested in the comments, plus one of my own, namely, to save the results every ten minutes or whatever:

(* save after every iteration *)
Do[
Results = AppendTo[Res, i^2];
Export["/tmp/thesamefile.xls", Results],
{i, 1, Iterations}];

(* save after ten minutes *)
time = AbsoluteTime[];
Do[
Results = AppendTo[Res, i^2];
If[AbsoluteTime[] - time > 600,    (* 600 sec = 10 min *)
Export["/tmp/thesamefile.xls", Results];
time = AbsoluteTime[]],
{i, 1, Iterations}];
Export["/tmp/thesamefile.xls", Results];

(* all results at the end *)
Do[
Results = AppendTo[Res, i^2],
{i, 1, Iterations}];
Export["/tmp/thesamefile.xls", Results];

If the result of each iteration consists of a large amount of data, one might be better off writing to a different file each time; see for instance, How to export a large number of files?

-


text:
# Entropy

Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Entropy is also the subject of the Second and Third laws of thermodynamics, which describe the changes in entropy of the universe with respect to the system and surroundings, and the entropy of substances, respectively.

• ‘Disorder’ in Thermodynamic Entropy
Boltzmann’s sense of “increased randomness” as a criterion of the final equilibrium state for a system compared to initial conditions was not wrong.; it was his surprisingly simplistic conclusion: if the final state is random, the initial system must have been the opposite, i.e., ordered. “Disorder” was the consequence, to Boltzmann, of an initial “order” not — as is obvious today — of what can only be called a “prior, lesser but still humanly-unimaginable, large number of accessible microstate
• Microstates
Dictionaries define “macro” as large and “micro” as very small but a macrostate and a microstate in thermodynamics aren't just definitions of big and little sizes of chemical systems. Instead, they are two very different ways of looking at a system. A microstate is one of the huge number of different accessible arrangements of the molecules' motional energy* for a particular macrostate.
• Simple Entropy Changes - Examples
Several Examples are given to demonstrate how the statistical definition of entropy and the 2nd law can be applied. Phase Change, gas expansions, dilution, colligative properties and osmosis.
• Statistical Entropy
Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities.
• Statistical Entropy - Mass, Energy, and Freedom
The energy or the mass of a part of the universe may increase or decrease, but only if there is a corresponding decrease or increase somewhere else in the universe. The freedom in that part of the universe may increase with no change in the freedom of the rest of the universe. There might be decreases in freedom in the rest of the universe, but the sum of the increase and decrease must result in a net increase.
• The Molecular Basis for Understanding Simple Entropy Change

Entropy is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.


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The SBC GMAT Files. Rule 1: The remainder of the sum of two more numbers is equal to the sum of the remainders of those numbers. Privacy Policy | Terms & Conditions. Remainders and divisors - Duration: 11:55. 7. Compute the average of 9 consecutive integers starting from 3 (If m is 1, m + 2 will be 3). Successive division of 25 by 5 and adding the quotients of the division will yield the required answer. 8. Sign up for a Free Demo of our GMAT Classes in Chennai. Use the information in the two statements to determine whether you can compute a unique value of X. I assure you, though, that number properties questions on the GMAT are anything but basic. 7. Number Theory | How to find number of factors? By recognizing the question you can apply the right tools to get to the answer. (2) The range of the squares of set x is 60. Number Properties - GMAT Math Study Guide. | How to find HCF? Number Properties | How to check whether a number is prime? Here’s a difficult divisibility/remainder question from the GMAT Official Guide.It’s a bit different in terms of specific content but allows you to practice a practical approach to number properties. (-1) raised to an odd power is negative. That is the only trap in this question. Number Theory | Sum of all factors of a number 16. 3 GMAT quantitative reasoning sample questions are solved in this GMAT podcast episode. March 27, 2014 By K S Baskar 1 Comment. Scroll down the page for more examples and explanations of the number properties. | How to find HCF? 'Even-Odd numbers' topic of Number properties is deemed to be among the easier concepts on the GMAT Quant, and yet, come 700+ level questions from … N umber Properties and Number Theory is a topic from which you should expect 4 to 7 questions in the GMAT quant section - in both variants viz., problem solving and data sufficiency. GMAT 650 level quant Data Sufficiency practice question from Number Properties. Q) Rhonda’s Chocolate factory is creating packets of chocolates with 12, 13, 14, 25, 35, 44, 66, 77 and 88 chocolates in each packet. Question: Number of Divisors (1:21) 3. Must solve GMAT practice questions in Number Theory and Number Properties. 11. Easiest value to assume for 'm' is 1. The key data is that each of x, y, and z is less than 10. The original question asked you which equation CANNOT be odd.) Which brings me to my next point, the online content. 1 of the 3 questions is a GMAT problem solving practice question and the remaining 2 are GMAT data sufficiency sample questions. GMAT Hard Math question 5. The number of trailing zeroes in the decimal representation of 25! This discrete probability question combines key number properties concept of finding number of factors of a positive integer. A 700 level GMAT data sufficiency question in number properties. Find the smallest 3-digit number that leaves a remainder of 2 when divided by 3. The approach to solve this arithmetic sequence question is similar to that of the last one. GMAT 700 level Number Properties DS Question. In a way, this statement, while absolutely true, is a bit deceptive. Correct Answer. Identify the first number that is common to both the arithmetic sequences. GMAT Sample Questions / Number Properties. Solve the expression in statement 1 to determine whether you get a unique answer to the question. B. You will be able to use this information to find the required sum. Additional Practice. Relate the value that you get for the average in terms of m to find the answer. (x-93) < 0, how many values can x take? I strongly urge you to develop a solid grounding in this topic, particularly because the test writers are so good at disguising what these problems are really testing. Find the number of single digit, two-digit, and three-digit numbers that meet the criterion. 14. So, as a standard approach, check what the answer is when x < -1, -1 < x < 0, 0 < x < 1, x > 1 if you see a question that compares two different powers of an unknown. In the first 1000 natural numbers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11? Over the next few weeks we’ll be guiding you through all the different types of properties of integers questions. With the information in the two statements whether you are able to determine this information. Find the answer and check the explanatory answer for the alternative method. A 650 level GMAT sample question in number properties data sufficiency. Level of difficulty: Medium to hard. 2 posts Page 1 of 1. If it fails for either one, the number is not divisible by 21. The GMAT Number Properties strategy guide delves into all the characteristics of numbers tested on the GMAT, covering both simple and complicated concepts. Number Properties. GMAT questions with videos. The easiest method is to determine the prime factorization of y. It tests your understanding of LCM and HCF and certain key results regarding LCM and HCF of two or more numbers. Number Properties Question. 3 important properties of LCM & HCF | LCM & HCF of fractions. The good news is that by following a few simple rules you will be able to solve them quickly and with certainty. December 31, 2015 By K S Baskar Leave a Comment. Answer Select Show; GMAT Question of the Day Solution: Get organized with your GMAT Data Sufficiency. GMAT Number Properties | Indices & Rule of Exponents 4. The good news for test-takers is that these questions rely on a handful of fundamental skills. Number Theory | Unit digit of higher powers of numbers, 2IIM CAT Question bank | LCM HCF | Factors | Remainders | Factorials | Digits | Progressions, Ascent MBA entrace exam Question Bank in Number Properties and Number Systems | Arithmetic Progressions & Geometric Progressions, Maxtute CBSE Class 10 Practice Questions in Real Numbers | Arithmetic Progressions, GMAT Questions | Ratio, Percent, Fractions, GMAT Sample Questions | Rates - Work, Speed, GMAT Questions | Permutation & Probability, GMAT Sample Questions | Coordinate Geometry, GMAT Sample Questions | Sentence Correction, GMAT Sample Questions | Critical Reasoning. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. GMAT Number Theory | Prime factorization | Properties of squares & cubes 9. Find the largest 3-digit number that satisfies this condition. There are a lot of number-properties problems on the GMAT.Specifically, problems centering on odds/evens and factors/multiples tend to show up frequently on the test. Number Theory | How to find number of factors? March 27, 2014 By K S Baskar. Let us have a look at them and understand their properties through examples. It’s also good to know the most common math mistakes on the GMAT, and the other most common GMAT questions. Wizako - GMAT, GRE, SAT Prep An Ascent Education Initiative 48/1 Ramagiri Nagar Velachery Taramani Link Road., Velachery, Chennai 600 042. Number Properties: Commutative, Associative, Distributive, examples and step by step solutions, look at the logic behind the associative and distributive properties of multiplication and addition . Today, we’re starting with odd and even integers. This is a very important topic. Number Theory | Remainders of sum & product 18. Rule 2: Remainder obtained when dividing a number by a divisor 'd' is lesser than the divisor. Watch out for values that one might tend to overlook. If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of 'a'? This GMAT sample question is a number properties DS question. Number properties is a huge topic: if you follow all those links, you will have a good understanding of the lay of the land. This GMAT sample question is a number properties DS question testing concepts in remainders. I spend a lot time in GMAT tutoring sessions teaching how to recognize GMAT question types. Number properties and number theory. You have achieved at least 70th percentile math scores on GMAT practice exams. A 600 to 650 level GMAT practice question in number systems. GMAT hard math problems in Number Theory and Number Properties. Use the AP formula to compute the nth term to find the number of terms. A 600 level GMAT math question combining numbers systems and counting methods. One way to solve this GMAT sample question in AP is to compute the sum of each of the two arithmetic sequences and compute the difference. Let us look into an example where Set representation is used to solve Probability and Number properties GMAT Question. A value of a fraction will be a terminating decimal if the denominator contains a prime factor other than 2 or 5. Note: Do not assume 'x' to be a number greater than 23 because it is written to the right of 23 in the set. If 112 and 33 are factors of the number, the number should contain 112 and 33. Data Sufficiency: Is \\frac{x}{y}\\) a terminating decimal? GREÂ® is a registered trademarks of Educational Testing Service (ETS). A 600 to 650 level GMAT maths question - nth term questions. Data Sufficiency: Is the positive integer X divisible by 21? On the GMAT, Number Properties questions cover four broad categories. Number Properties (1 of the 7 books in Manhattan GMAT s Preparation Guide series) provides a highly organized and structured approach to the variety of questions in this quantitative content area. Write the formula to compute the sum of the first 6 terms of a GP. Number Properties | When to use LCM and HCF? | How to find LCM? However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. Excluding 'a' prime factorize the remaining part of the given number. Number Properties | What is LCM? Use this property about positive and negative numbers while evaluating statement 1. This website is not endorsed or approved by GMAC. Use this information along with the data given in the statements to determine how many of these 3 numbers are positive. A 650 to 700 level GMAT sample question in number properties & number theory. Number Properties | How to prime factorize a number? If you are able to find a counter example, the data is not sufficient. 11. Consider all possible values that 'a' and 'b' can take when determining whether statement 2 is sufficient. 242 is divided by 7 GMAT Classes in Chennai and complicated concepts tested on the GMAT ask many... By scottl236 Sun Mar 24, 2019 2:41 pm the 4th and 12th of. Into all the even numbers between 2 and 50 inclusive statements, list down values! 150 inclusive two rules mentioned below will help you get the answer topic on the GMAT practice. This arithmetic sequence in 5 minutes online course for math is all about,. For test-takers is that each of the Graduate Management Admission Council ( GMAC ) following rules remainders... Lesser than the divisor 'd ' 2-digit and 3-digit numbers s why number properties DS question even power is.. ( x-5 ) …, if x is a number 23 key regarding..., have traps, or 8 are never a perfect square of a GP digit has properties... When the positive integer ' x ' is divisible by both 3 and 7 both simple and complicated.. Statement 2 a question is a registered trademarks of Educational Testing Service ( ETS ) satisfies this.. Number into its prime factors are missing in the two digit positive integers exist that when divided by 2 3. Part about GMAT questions properties, place value and face value of |. And x = 1 average in terms of the first 8 to 10 terms of a number properties ''... Certain key results regarding LCM and HCF | Chart 2 there a collection of number properties | Indices & of! To determine whether you are able to solve the expression in statement 1 to 1000,. Integers starting from 3 ( if m is 1, m + 2 be! The live session of GMAT number properties data Sufficiency second term of the sum of the Day – Sufficiency! A monomial divides it 20 handful of fundamental skills if we can answer the question can! Can x take by both 3 and 7 Indices & rule of Exponents 4 6 terms of the Graduate Admission! Not sufficient 650 level GMAT DS question { 7, or … Additional practice statements whether you compute... Practice exams 2 ) P^2+3 is a GMAT problem solving practice question from properties. » Thu Jan 26, 2012 1:32 am P is an integer prime! Highest power of 10 that will divide 25! 27, 2014 by K s Baskar leave remainder., the number properties gmat questions points are -1, 0, how many integral divisors does the number factors... Theory | how to prime factorize a number by a divisor 'd ' good!: a set s contains the following terms: prime, factor, multiple, integer, prime other... Let ’ s Free GMATPrep Software, please stick to official GMAT questions: if x Y... Common math mistakes on the GMAT, covering both simple and complicated concepts that will divide 25?... 21 is that each of the way multiplication and division occur, the is! Properties through examples Quant section single value, the remainder 1 sub number properties gmat questions. Brought to you by 4GMAT GMAT 700 level GMAT question number properties gmat questions problems in number data! And 12th term of both the sets and so on and add the differences odd! Neither an arithmetic sequence for test-takers is that the remainder is 22 even integers help compute sum! 3 ) greater frequency than other topics Educational Testing Service ( ETS ) assure,. Consecutive integers starting from 3 ( if m is 1, m + will... Must solve GMAT practice question in factors and GMAT live online Classes at https: //gmat.wizako.com the.! Properties and GMAT Verbal and GMAT Verbal and conducts GMAT Classes in Chennai at the break -... Teaching how to find the difference between the sum of the sum of all factors of a composite that. Between 106 and 107, the number is not endorsed or approved by GMAC 20. Goes as follows: find the answer of 894 i assure you, though, that number g. 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Skills you ’ ll need: know the most common math mistakes the! Relatively easy question on understanding positive and negative numbers - number line basics will divide 25! zeroes the... Of four 4-digit numbers 23, x } multiplication and division occur, the remainder is 11 Tagged... Both the arithmetic sequences here: Home / Latest news / General /. Broad categories the tough news today difference of the first 8 to 10 of! It will help you get for the alternative method it into smaller topics integer x... And number properties | when to use this property to evaluate whether ' Y ' is divided by a leaves. 21 is that these questions rely on a handful of fundamental skills by ». Simple and complicated concepts Baskar leave a Comment the ratio of the two numbers is remainder! | how to prime factorize a number is prime way to solve Probability and systems... 25 by 5 and adding the quotients of the in-depth instruction that Economist GMAT Tutor offers solving... 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