Datasets:

XiaoHLim

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PutnamBench

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putnam_1981_a1

abbrev putnam_1981_a1_solution : ℝ := sorry -- 1/8 theorem putnam_1981_a1 (P : ℕ → ℕ → Prop := fun n k : ℕ => 5^k ∣ ∏ m in Finset.Icc 1 n, (m^m : ℤ)) (E : ℕ → ℕ) (hE : ∀ n ∈ Ici 1, P n (E n) ∧ ∀ k : ℕ, P n k → k ≤ E n) : Tendsto (fun n : ℕ => ((E n) : ℝ)/n^2) atTop (𝓝 putnam_1981_a1_solution) := sorry

Let $E(n)$ be the greatest integer $k$ such that $5^k$ divides $1^1 2^2 3^3 \cdots n^n$. Find $\lim_{n \rightarrow \infty} \frac{E(n)}{n^2}$.

The limit equals $\frac{1}{8}$.

['analysis', 'number_theory']

Section putnam_1981_a1. Require Import Nat Reals Coquelicot.Coquelicot. From mathcomp Require Import div. Definition putnam_1981_a1_solution := 1 / 8. Theorem putnam_1981_a1 (prod_n : (nat -> nat) -> nat -> nat := fix prod_n (m: nat -> nat) (n : nat) := match n with | O => m 0%nat | S n' => mul (m n') (prod_n m n') end) (P : nat -> nat -> Prop := fun n k => 5 ^ k %| prod_n (fun m => Nat.pow m m) (S n) = true) (f : nat -> nat) (hf : forall (n: nat), gt n 1 -> P n (f n) /\ forall (k: nat), P n k -> le k (f n)) : Lim_seq (fun n => INR (f n) / INR n ^ 2) = putnam_1981_a1_solution. Proof. Admitted. End putnam_1981_a1.

theory putnam_1981_a1 imports Complex_Main begin definition putnam_1981_a1_solution::real where "putnam_1981_a1_solution \<equiv> undefined" (* 1/8 *) theorem putnam_1981_a1: fixes P::"nat\<Rightarrow>nat\<Rightarrow>bool" and E::"nat\<Rightarrow>nat" defines "P \<equiv> \<lambda>n. \<lambda>k. 5^k dvd (\<Prod>m=1..n. m^m)" and "E \<equiv> \<lambda>n. (GREATEST k. P n k)" shows "(\<lambda>n. (E n) / n^2) \<longlonglongrightarrow> putnam_1981_a1_solution" sorry end

putnam_1981_a3

abbrev putnam_1981_a3_solution : Prop := sorry -- False theorem putnam_1981_a3 (f : ℝ → ℝ := fun t : ℝ => Real.exp (-t) * ∫ y in (Ico 0 t), ∫ x in (Ico 0 t), (Real.exp x - Real.exp y) / (x - y)) : (∃ L : ℝ, Tendsto f atTop (𝓝 L)) ↔ putnam_1981_a3_solution := sorry

Does the limit $$lim_{t \rightarrow \infty}e^{-t}\int_{0}^{t}\int_{0}^{t}\frac{e^x - e^y}{x - y} dx dy$$exist?

The limit does not exist.

['analysis']

Section putnam_1981_a3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_1981_a3_solution := 14. Theorem putnam_1981_a3: Lim_seq (fun k => exp (-1*INR k) * (RInt (fun x => (RInt (fun y => (exp x - exp y) / (x - y)) 0 (INR k))) 0 (INR k))) = putnam_1981_a3_solution. Proof. Admitted. End putnam_1981_a3.

theory putnam_1981_a3 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1981_a3_solution::bool where "putnam_1981_a3_solution \<equiv> undefined" (* False *) theorem putnam_1981_a3: fixes f::"real\<Rightarrow>real" defines "f \<equiv> \<lambda>t::real. exp (-t) * interval_lebesgue_integral lebesgue 0 t (\<lambda>y::real. interval_lebesgue_integral lebesgue 0 t (\<lambda>x::real. (exp x - exp y) / (x - y)))" shows "(\<exists>L::real. (f \<longlonglongrightarrow> L)) \<longleftrightarrow> putnam_1981_a3_solution" sorry end

putnam_1981_a5

abbrev putnam_1981_a5_solution : Prop := sorry -- True theorem putnam_1981_a5 (Q : Polynomial ℝ → Polynomial ℝ := fun P : Polynomial ℝ => (X^2 + 1)*P*(derivative P) + X*(P^2 + (derivative P)^2)) (n : Polynomial ℝ → ℝ := fun P : Polynomial ℝ => {x ∈ Ioi 1 | P.eval x = 0}.ncard) : (∀ P : Polynomial ℝ, {x : ℝ | (Q P).eval x = 0}.ncard ≥ 2*(n P) - 1) ↔ putnam_1981_a5_solution := sorry

Let $P(x)$ be a polynomial with real coefficients; let $$Q(x) = (x^2 + 1)P(x)P'(x) + x((P(x))^2 + (P'(x))^2).$$ Given that $P$ has $n$ distinct real roots all greater than $1$, prove or disprove that $Q$ must have at least $2n - 1$ distinct real roots.

$Q(x)$ must have at least $2n - 1$ distinct real roots.

['algebra']

theory putnam_1981_a5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1981_a5_solution::bool where "putnam_1981_a5_solution \<equiv> undefined" (* True *) theorem putnam_1981_a5: fixes Q::"(real poly) \<Rightarrow> (real poly)" and n::"(real poly) \<Rightarrow> real" defines "Q \<equiv> \<lambda>P. [: 1, 0, 1 :] * P * (pderiv P) + [: 0, 1 :] * (P^2 + (pderiv P)^2)" and "n \<equiv> \<lambda>P. card {x::real. x > 1 \<and> (poly P x) = 0}" shows "(\<forall>P. card {x::real. poly (Q P) x = 0} \<ge> 2 * (n P) - 1) \<longleftrightarrow> putnam_1981_a5_solution" sorry end

putnam_1981_b1

abbrev putnam_1981_b1_solution : ℝ := sorry -- -1 theorem putnam_1981_b1 (f : ℕ → ℝ := fun n : ℕ => (1/n^5) * ∑ h in Finset.Icc 1 n, ∑ k in Finset.Icc 1 n, (5*(h : ℝ)^4 - 18*h^2*k^2 + 5*k^4)) : Tendsto f atTop (𝓝 putnam_1981_b1_solution) := sorry

Find the value of $$\lim_{n \rightarrow \infty} \frac{1}{n^5}\sum_{h=1}^{n}\sum_{k=1}^{n}(5h^4 - 18h^2k^2 + 5k^4).$$

The limit equals $-1$.

['analysis']

Section putnam_1981_b1. Require Import Reals Coquelicot.Coquelicot. Definition putnam_1981_b1_solution := -1. Theorem putnam_1981_b1: Lim_seq (fun n => 1/(pow (INR n) 5) * (sum_n (fun r => (sum_n (fun s => 5 * pow (INR r) 4 - 18 * pow (INR r) 2 * pow (INR s) 2 + 5 * pow (INR s) 4) n)) n)) = putnam_1981_b1_solution. Proof. Admitted. End putnam_1981_b1.

theory putnam_1981_b1 imports Complex_Main begin definition putnam_1981_b1_solution::real where "putnam_1981_b1_solution \<equiv> undefined" (* -1 *) theorem putnam_1981_b1: fixes f::"nat\<Rightarrow>real" defines "f \<equiv> \<lambda>n. 1/n^5 * (\<Sum>h=1..n. (\<Sum>k=1..n. (5*h^4 -18*h^2*k^2 + 5*k^4)))" shows "f \<longlonglongrightarrow> putnam_1981_b1_solution" sorry end

putnam_1981_b2

abbrev putnam_1981_b2_solution : ℝ := sorry -- 12 - 8 * Real.sqrt 2 theorem putnam_1981_b2 (P : ℝ × ℝ × ℝ → Prop := fun (r, s, t) => 1 ≤ r ∧ r ≤ s ∧ s ≤ t ∧ t ≤ 4) (f : ℝ × ℝ × ℝ → ℝ := fun (r, s, t) => (r - 1)^2 + (s/r - 1)^2 + (t/s - 1)^2 + (4/t - 1)^2) : (∃ r : ℝ, ∃ s : ℝ, ∃ t : ℝ, P (r, s, t) ∧ f (r, s, t) = putnam_1981_b2_solution) ∧ ∀ r : ℝ, ∀ s : ℝ, ∀ t : ℝ, P (r, s, t) → f (r, s, t) ≥ putnam_1981_b2_solution := sorry

Determine the minimum value attained by $$(r - 1)^2 + (\frac{s}{r} - 1)^2 + (\frac{t}{s} - 1)^2 + (\frac{4}{t} - 1)^2$$ across all choices of real $r$, $s$, and $t$ that satisfy $1 \le r \le s \le t \le 4$.

The minimum is $12 - 8\sqrt{2}$.

['algebra']

Section putnam_1981_b2. Require Import Reals. Open Scope R. Definition putnam_1981_b2_solution := 12 - 8 * sqrt 2. Theorem putnam_1981_b2: let f (a b c: R) := pow (a-1) 2 + pow (b / a - 1) 2 + pow (c / b - 1) 2 + pow (4 / c - 1) 2 in (forall (a b c: R), 1 <= a /\ a <= b /\ b <= c /\ c <= 4 -> putnam_1981_b2_solution <= f a b c) /\ (exists (a b c: R), 1 <= a /\ a <= b /\ b <= c /\ c <= 4 -> putnam_1981_b2_solution = f a b c). Proof. Admitted. End putnam_1981_b2.

theory putnam_1981_b2 imports Complex_Main begin definition putnam_1981_b2_solution::real where "putnam_1981_b2_solution \<equiv> undefined" (* 12 - 8 * sqrt 2 *) theorem putnam_1981_b2: fixes P::"real\<Rightarrow>real\<Rightarrow>real\<Rightarrow>bool" and f::"real\<Rightarrow>real\<Rightarrow>real\<Rightarrow>real" defines "P \<equiv> \<lambda>r. \<lambda>s. \<lambda>t. 1 \<le> r \<and> r \<le> s \<and> s \<le> t \<and> t \<le> 4" and "f \<equiv> \<lambda>r. \<lambda>s. \<lambda>t. (r-1)^2 + (s/r - 1)^2 + (t/s - 1)^2 + (4/t - 1)^2" shows "putnam_1981_b2_solution = (LEAST z::real. \<exists>r. \<exists>s. \<exists>t. P r s t \<and> z = f r s t)" sorry end

putnam_1981_b3

theorem putnam_1981_b3 (P : ℕ → Prop := fun n : ℕ => ∀ p : ℕ, (Nat.Prime p ∧ p ∣ n^2 + 3) → ∃ k : ℕ, (p : ℤ) ∣ (k : ℤ)^2 + 3 ∧ k^2 < n) : ∀ n : ℕ, ∃ m : ℕ, (m : ℤ) > n ∧ P m := sorry

Prove that, for infinitely many positive integers $n$, all primes $p$ that divide $n^2 + 3$ also divide $k^2 + 3$ for some integer $k$ such that $k^2 < n$.

['number_theory']

Section putnam_1981_b3. Require Import Nat ZArith Znumtheory. From mathcomp Require Import div. Open Scope nat_scope. Theorem putnam_1981_b3: ~ exists (N: nat), forall (n: nat), (forall (p: nat), prime (Z.of_nat p) /\ p %| pow n 2 + 3 = true -> exists (m: nat), p %| pow m 2 + 3 = true /\ pow m 2 < n) -> n < N. Proof. Admitted. End putnam_1981_b3.

theory putnam_1981_b3 imports Complex_Main "HOL-Computational_Algebra.Primes" begin theorem putnam_1981_b3: fixes P::"nat\<Rightarrow>bool" defines "P \<equiv> \<lambda>n. \<forall>p::nat. (prime p \<and> p dvd (n^2 + 3)) \<longrightarrow> (\<exists>k::nat. k^2 < n \<and> p dvd (k^2 + 3))" shows "\<forall>n::nat. \<exists>m::nat. m > n \<and> P m" sorry end

putnam_1981_b4

abbrev putnam_1981_b4_solution : Prop := sorry -- False theorem putnam_1981_b4 (VAB : Set (Matrix (Fin 5) (Fin 7) ℝ) → Prop) (Vrank : Set (Matrix (Fin 5) (Fin 7) ℝ) → ℕ → Prop) (hVAB : ∀ V : Set (Matrix (Fin 5) (Fin 7) ℝ), VAB V = (∀ A ∈ V, ∀ B ∈ V, ∀ r s : ℝ, r • A + s • B ∈ V)) (hVrank : ∀ (V : Set (Matrix (Fin 5) (Fin 7) ℝ)) (k : ℕ), Vrank V k = ∃ A ∈ V, A.rank = k) : (∀ V : Set (Matrix (Fin 5) (Fin 7) ℝ), (VAB V ∧ Vrank V 0 ∧ Vrank V 1 ∧ Vrank V 2 ∧ Vrank V 4 ∧ Vrank V 5) → Vrank V 3) ↔ putnam_1981_b4_solution := sorry

Let $V$ be a set of $5$ by $7$ matrices, with real entries and with the property that $rA+sB \in V$ whenever $A,B \in V$ and $r$ and $s$ are scalars (i.e., real numbers). \emph{Prove or disprove} the following assertion: If $V$ contains matrices of ranks $0$, $1$, $2$, $4$, and $5$, then it also contains a matrix of rank $3$. [The rank of a nonzero matrix $M$ is the largest $k$ such that the entries of some $k$ rows and some $k$ columns form a $k$ by $k$ matrix with a nonzero determinant.]

Show that the assertion is false.

['linear_algebra']

theory putnam_1981_b4 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Analysis.Cartesian_Space" begin definition putnam_1981_b4_solution :: bool where "putnam_1981_b4_solution \<equiv> undefined" (* False *) theorem putnam_1981_b4: fixes VAB :: "(real^7^5) set \<Rightarrow> bool" and Vrank :: "(real^7^5) set \<Rightarrow> nat \<Rightarrow> bool" defines "VAB \<equiv> \<lambda> V :: (real^7^5) set. \<forall> A \<in> V. \<forall> B \<in> V. \<forall> r s :: real. r *\<^sub>R A + s *\<^sub>R B \<in> V" and "Vrank \<equiv> \<lambda> (V :: (real^7^5) set) (k :: nat). \<exists> A \<in> V. rank A = k" shows "(\<forall> V :: (real^7^5) set. (VAB V \<and> Vrank V 0 \<and> Vrank V 1 \<and> Vrank V 2 \<and> Vrank V 4 \<and> Vrank V 5) \<longrightarrow> Vrank V 3) \<longleftrightarrow> putnam_1981_b4_solution" sorry end

putnam_1981_b5

abbrev putnam_1981_b5_solution : Prop := sorry -- True theorem putnam_1981_b5 (sumbits : List ℕ → ℤ) (B : ℕ → ℤ) (hsumbits : ∀ bits : List ℕ, sumbits bits = ∑ i : Fin bits.length, (bits[i] : ℤ)) (hB : ∀ n > 0, B n = sumbits (Nat.digits 2 n)) : (∃ q : ℚ, Real.exp (∑' n : Set.Ici 1, B n / ((n : ℝ) * ((n : ℝ) + 1))) = q) ↔ putnam_1981_b5_solution := sorry

Let $B(n)$ be the number of ones in the base two expression for the positive integer $n$. For example, $B(6)=B(110_2)=2$ and $B(15)=B(1111_2)=4$. Determine whether or not $\exp \left(\sum_{n=1}^\infty \frac{B(n)}{n(n+1)}\right)$ is a rational number. Here $\exp(x)$ denotes $e^x$.

Show that the expression is a rational number.

['analysis', 'algebra']

Section putnam_1981_b5. Require Import BinNums Nat NArith Coquelicot.Coquelicot. Definition putnam_1981_b5_solution := True. Theorem putnam_1981_b5: let f := fix count_ones (n : positive) : nat := match n with | xH => 1 | xO n' => count_ones n' | xI n' => 1 + count_ones n' end in let k := Series (fun n => Rdefinitions.Rdiv (Raxioms.INR (f (Pos.of_nat n))) (Raxioms.INR (n + pow n 2))) in exists (a b: nat), Rtrigo_def.exp k = Rdefinitions.Rdiv (Raxioms.INR a) (Raxioms.INR b) <-> putnam_1981_b5_solution. Proof. Admitted. End putnam_1981_b5.

theory putnam_1981_b5 imports Complex_Main begin definition putnam_1981_b5_solution :: bool where "putnam_1981_b5_solution \<equiv> undefined" (* True *) theorem putnam_1981_b5: fixes B :: "nat \<Rightarrow> nat" defines "B \<equiv> \<lambda> n. card {k :: nat. odd (n div (2 ^ k))}" shows "(\<exists> q :: rat. exp (\<Sum> n :: nat. B (n + 1) / ((n + 1) * (n + 2))) = q) \<longleftrightarrow> putnam_1981_b5_solution" sorry end

putnam_2007_a1

abbrev putnam_2007_a1_solution : Set ℝ := sorry -- {2/3, 3/2, (13 + Real.sqrt 601)/12, (13 - Real.sqrt 601)/12} def reflect_tangent (f g : ℝ → ℝ) := ContDiff ℝ 1 f ∧ ContDiff ℝ 1 g ∧ (∃ x y : ℝ, f x = y ∧ g y = x ∧ (deriv f) x = 1 / (deriv g) y) theorem putnam_2007_a1 : ∀ a : ℝ, reflect_tangent (fun x => a * x^2 + a * x + 1/24) (fun y => a * y^2 + a * y + 1/24) ↔ a ∈ putnam_2007_a1_solution := sorry

Find all values of $\alpha$ for which the curves $y = \alpha*x^2 + \alpha*x + 1/24$ and $x = \alpha*y^2 + \alpha*y + 1/24$ are tangent to each other.

Show that the solution is the set \{2/3, 3/2, (13 + \sqrt{601})/12, (13 - \sqrt{601})/12}.

['algebra', 'geometry']

theory putnam_2007_a1 imports Complex_Main "HOL-Analysis.Derivative" begin (* Note: Modified definition of tangent to handle this, but this is a bit of cheating - You would have to know that this works *) definition putnam_2007_a1_solution :: "real set" where "putnam_2007_a1_solution \<equiv> undefined" (* {2/3, 3/2, (13 + sqrt 601)/12, (13 - sqrt 601)/12} *) definition reflect_tangent :: "(real \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool" where "reflect_tangent \<equiv> (\<lambda>f g::real\<Rightarrow>real. f C1_differentiable_on UNIV \<and> g C1_differentiable_on UNIV \<and> (\<exists>x y::real. f x = y \<and> g y = x \<and> deriv f x = 1 / deriv g y))" theorem putnam_2007_a1: shows "\<forall>a::real. (reflect_tangent (\<lambda>x::real. a*x^2 + a*x + 1/24) (\<lambda>y::real. a*y^2 + a*y + 1/24) \<longleftrightarrow> a \<in> putnam_2007_a1_solution)" sorry end

putnam_2007_a2

abbrev putnam_2007_a2_solution : ENNReal := sorry -- 4 theorem putnam_2007_a2 (Sinterpos : Set (Fin 2 → ℝ) → Prop) (Sinterneg : Set (Fin 2 → ℝ) → Prop) (Sconv : Set (Fin 2 → ℝ) → Prop) (hSinterpos : ∀ S : Set (Fin 2 → ℝ), Sinterpos S = ((∃ p ∈ S, p 0 > 0 ∧ p 1 > 0 ∧ p 0 * p 1 = 1) ∧ (∃ p ∈ S, p 0 < 0 ∧ p 1 < 0 ∧ p 0 * p 1 = 1))) (hSinterneg : ∀ S : Set (Fin 2 → ℝ), Sinterneg S = ((∃ p ∈ S, p 0 < 0 ∧ p 1 > 0 ∧ p 0 * p 1 = -1) ∧ (∃ p ∈ S, p 0 > 0 ∧ p 1 < 0 ∧ p 0 * p 1 = -1))) (hSconv : ∀ S : Set (Fin 2 → ℝ), Sconv S = (Convex ℝ S ∧ Sinterpos S ∧ Sinterneg S)) : (∃ S : Set (Fin 2 → ℝ), Sconv S ∧ MeasureTheory.volume S = putnam_2007_a2_solution) ∧ (∀ S : Set (Fin 2 → ℝ), Sconv S → MeasureTheory.volume S ≥ putnam_2007_a2_solution) := sorry

Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $xy=1$ and both branches of the hyperbola $xy=-1$. (A set $S$ in the plane is called \emph{convex} if for any two points in $S$ the line segment connecting them is contained in $S$.)

Show that the minimum is $4$.

['geometry']

theory putnam_2007_a2 imports Complex_Main "HOL-Analysis.Lebesgue_Measure" begin definition putnam_2007_a2_solution :: ennreal where "putnam_2007_a2_solution \<equiv> undefined" (* 4 *) theorem putnam_2007_a2: fixes Sinterpos :: "((real^2) set) \<Rightarrow> bool" and Sinterneg :: "((real^2) set) \<Rightarrow> bool" and Sconv :: "((real^2) set) \<Rightarrow> bool" assumes hSinterpos: "\<forall>S::(real^2) set. Sinterpos S = ((\<exists>p\<in>S. p$1 > 0 \<and> p$2 > 0 \<and> p$1*p$2 = 1) \<and> (\<exists>p\<in>S. p$1 < 0 \<and> p$2 < 0 \<and> p$1*p$2 = 1))" assumes hSinterneg: "\<forall>S::(real^2) set. Sinterneg S = ((\<exists>p\<in>S. p$1 < 0 \<and> p$2 > 0 \<and> p$1*p$2 = -1) \<and> (\<exists>p\<in>S. p$1 > 0 \<and> p$2 < 0 \<and> p$1*p$2 = -1))" assumes hSconv: "\<forall>S::(real^2) set. Sconv S = (convex S \<and> Sinterpos S \<and> Sinterneg S)" shows "(LEAST area::ennreal. (\<exists>S::(real^2) set. Sconv S \<and> emeasure lebesgue S = area)) = putnam_2007_a2_solution" sorry end

putnam_2007_a4

abbrev putnam_2007_a4_solution : Set (Polynomial ℝ) := sorry -- {f : Polynomial ℝ | ∃ d : ℕ, ∃ c ≥ (1 : ℤ) - d, ∀ n : ℝ, f.eval n = (1 / 9) * ((10 ^ c) * (9 * n + 1) ^ d - 1)} theorem putnam_2007_a4 (S : Set (Polynomial ℝ)) (repunit : ℝ → Prop := fun x ↦ x > 0 ∧ x = floor x ∧ ∀ d ∈ (digits 10 (floor x)), d = 1) (hS : ∀ f : Polynomial ℝ, f ∈ S ↔ (∀ n : ℝ, repunit n → repunit (f.eval n))) : (S = putnam_2007_a4_solution) := sorry

A \emph{repunit} is a positive integer whose digits in base 10 are all ones. Find all polynomials $f$ with real coefficients such that if $n$ is a repunit, then so is $f(n)$.

Show that the desired polynomials $f$ are those of the form \[ f(n) = \frac{1}{9}(10^c (9n+1)^d - 1) \] for integers $d \geq 0$ and $c \geq 1-d$.

['analysis', 'algebra', 'number_theory']

Section putnam_2007_a4. Require Import Reals Zpower Coquelicot.Coquelicot. Definition putnam_2007_a4_solution (f: R -> R) := exists (c d: Z), Z.ge d 0 /\ Z.ge c (1 - d) /\ f = (fun n => (IZR (Zpower 10 c) * IZR (Zpower (9 * floor n + 1) d) - 1) / 9). Theorem putnam_2007_a4: let repunit (n: R) := n = IZR (floor n) /\ n > 0 /\ exists (m: nat), n = sum_n (fun i => 10 ^ i) m in forall (c: nat -> R) (n: nat), let f (x: R) := sum_n (fun i => c i * x ^ i) (n + 1) in repunit (INR n) -> repunit (f (INR n)) <-> putnam_2007_a4_solution f. Proof. Admitted. End putnam_2007_a4.

theory putnam_2007_a4 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_2007_a4_solution :: "(real poly) set" where "putnam_2007_a4_solution \<equiv> undefined" (* {f :: real poly. \<exists> d :: nat. \<exists> c :: int. c \<ge> 1 - d \<and> (\<forall> n :: real. poly f n = (1 / 9) * ((10 powr c) * (9 * n + 1) ^ d - 1))} *) theorem putnam_2007_a4: fixes repunit :: "real \<Rightarrow> bool" defines "repunit \<equiv> \<lambda> x :: real. x > 0 \<and> x = real_of_int (floor x) \<and> (\<forall> i :: nat. (10 ^ i \<le> floor x) \<longrightarrow> ((floor x) div (10 ^ i)) mod 10 = 1)" shows "{f :: real poly. \<forall> n :: real. repunit n \<longrightarrow> repunit (poly f n)} = putnam_2007_a4_solution" sorry end

putnam_2007_a5

theorem putnam_2007_a5 (G : Type*) [Group G] [Fintype G] (p : ℕ) (n : ℕ) (hp : Nat.Prime p) (hn : n = {g : G | orderOf g = p}.encard) : n = 0 ∨ p ∣ (n + 1) := sorry

[]

theory putnam_2007_a5 imports Complex_Main "HOL-Algebra.Multiplicative_Group" "HOL-Computational_Algebra.Primes" begin theorem putnam_2007_a5: fixes G (structure) and p :: nat and n :: nat assumes hG: "group G \<and> finite (carrier G)" and hp: "prime p" and hn: "n \<equiv> card {g::'a. g \<in> carrier G \<and> (group.ord G) g = p}" shows "n = 0 \<or> p dvd (n+1)" sorry end

putnam_2007_b1

theorem putnam_2007_b1 (f : Polynomial ℤ) (hf : ∀ n : ℕ, f.coeff n ≥ 0) (hfnconst : ∃ n : ℕ, n > 0 ∧ f.coeff n > 0) (n : ℤ) (hn : n > 0) : f.eval n ∣ f.eval (f.eval n + 1) ↔ n = 1 := sorry

Let $f$ be a nonconstant polynomial with positive integer coefficients. Prove that if $n$ is a positive integer, then $f(n)$ divides $f(f(n) + 1)$ if and only if $n = 1$

['algebra']

Section putnam_2007_b1. Require Import Nat Reals Coquelicot.Coquelicot. From mathcomp Require Import div. Theorem putnam_2007_b1: forall (c: nat -> nat) (n: nat), gt n 0 /\ forall (x: nat), gt (c x) 0 -> let f (x: nat) := sum_n (fun i => INR (mul (c i) (x ^ i))) (n + 1) in Z.to_nat (floor (f n)) %| Z.to_nat (floor (f (Z.to_nat (floor (f n + 1))))) = true <-> n = 1%nat. Proof. Admitted. End putnam_2007_b1.

theory putnam_2007_b1 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin (* Note: Since the coefficient is defined for all natural numbers, the condition here is modified to nonnegative which is what makes sense *) theorem putnam_2007_b1: fixes f :: "int poly" and n :: int assumes hf: "\<forall>n'::nat. poly.coeff f n' \<ge> 0" and hfnconst: "\<exists>n'::nat. n' > 0 \<and> poly.coeff f n' > 0" and hn: "n > 0" shows "((poly f n) dvd (poly f (poly f n + 1))) \<longleftrightarrow> n = 1" sorry end

putnam_2007_b2

theorem putnam_2007_b2 (f : ℝ → ℝ) (hf : ContDiffOn ℝ 1 f (Icc 0 1)) (hfint : ∫ x in (0)..1, f x = 0) (max : ℝ) (heqmax : ∃ x ∈ Icc (0 : ℝ) 1, |deriv f x| = max) (hmaxub : ∀ x ∈ Icc (0 : ℝ) 1, |deriv f x| ≤ max) : (∀ α ∈ (Ioo (0 : ℝ) 1), |∫ x in (0)..α, f x| ≤ (1 / 8) * max) := sorry

Suppose that $f: [0,1] \to \mathbb{R}$ has a continuous derivative and that $\int_0^1 f(x)\,dx = 0$. Prove that for every $\alpha \in (0,1)$, \[ \left| \int_0^\alpha f(x)\,dx \right| \leq \frac{1}{8} \max_{0 \leq x \leq 1} |f'(x)|. \]

['analysis']

Section putnam_2007_b2. Require Import Reals Coquelicot.Coquelicot. Theorem putnam_2007_b2: forall (f: R -> R), (forall (x: R), 0 <= x <= 1 /\ continuity_pt f x /\ ex_derive f x /\ RInt f 0 1 = 0) -> exists (max_f_prime_abs: R), (forall (x: R), 0 <= x <= 1 -> max_f_prime_abs >= Rabs (Derive f x)) /\ (exists (x: R), 0 <= x <= 1 -> max_f_prime_abs = Rabs (Derive f x)) /\ forall (a: R), 0 < a < 1 -> Rabs (RInt f 0 a) = max_f_prime_abs / 8. Proof. Admitted. End putnam_2007_b2.

theory putnam_2007_b2 imports Complex_Main "HOL-Analysis.Derivative" "HOL-Analysis.Set_Integral" "HOL-Analysis.Lebesgue_Measure" begin theorem putnam_2007_b2: fixes f :: "real \<Rightarrow> real" assumes hf: "f C1_differentiable_on {0..1}" and hfint: "set_lebesgue_integral lebesgue {0..1} f = 0" shows "\<forall> \<alpha> \<in> {0<..<1}. \<bar>set_lebesgue_integral lebesgue {0..\<alpha>} f\<bar> \<le> (1 / 8) * (GREATEST y. \<exists> x \<in> {0..1}. \<bar>deriv f x\<bar> = y)" sorry end

putnam_2007_b3

abbrev putnam_2007_b3_solution : ℝ := sorry -- (2 ^ 2006 / Real.sqrt 5) * (((1 + Real.sqrt 5) / 2) ^ 3997 - ((1 + Real.sqrt 5) / 2) ^ (-3997 : ℤ)) theorem putnam_2007_b3 (x : ℕ → ℝ) (hx0 : x 0 = 1) (hx : ∀ n : ℕ, x (n + 1) = 3 * (x n) + ⌊(x n) * Real.sqrt 5⌋) : (x 2007 = putnam_2007_b3_solution) := sorry

Let $x_0 = 1$ and for $n \geq 0$, let $x_{n+1} = 3x_n + \lfloor x_n \sqrt{5} \rfloor$. In particular, $x_1 = 5$, $x_2 = 26$, $x_3 = 136$, $x_4 = 712$. Find a closed-form expression for $x_{2007}$. ($\lfloor a \rfloor$ means the largest integer $\leq a$.)

Prove that $x_{2007} = \frac{2^{2006}}{\sqrt{5}}(\alpha^{3997}-\alpha^{-3997})$, where $\alpha = \frac{1+\sqrt{5}}{2}$.

['analysis']

Section putnam_2007_b3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2007_b3_solution := let a := (1 + sqrt 5) / 2 in 2 ^ 2006 / sqrt 5 * (a ^ 3997 - Rpower a (-3997)). Theorem putnam_2007_b3: let fix x (n: nat) := match n with | O => 1 | S n' => 3 * x n' + IZR (floor (x n' * sqrt (INR n))) end in x 2007%nat = putnam_2007_b3_solution. Proof. Admitted. End putnam_2007_b3.

theory putnam_2007_b3 imports Complex_Main begin definition putnam_2007_b3_solution :: real where "putnam_2007_b3_solution \<equiv> undefined" (* (2 ^ 2006 / sqrt 5) * ((((1 + sqrt 5) / 2) powr 3997) - (((1 + sqrt 5) / 2) powr -3997)) *) theorem putnam_2007_b3: fixes x :: "nat \<Rightarrow> real" assumes hx0: "x 0 = 1" and hx: "\<forall> n :: nat. x (n + 1) = 3 * (x n) + floor (x n + sqrt 5)" shows "x 2007 = putnam_2007_b3_solution" sorry end

putnam_2007_b4

abbrev putnam_2007_b4_solution : ℕ → ℕ := sorry -- fun n ↦ 2 ^ (n + 1) theorem putnam_2007_b4 (n : ℕ) (npos : n > 0) : ({(P, Q) : (Polynomial ℝ) × (Polynomial ℝ) | P ^ 2 + Q ^ 2 = Polynomial.X ^ (2 * n) + 1 ∧ P.degree > Q.degree}.ncard = putnam_2007_b4_solution n) := sorry

Let $n$ be a positive integer. Find the number of pairs $P, Q$ of polynomials with real coefficients such that \[ (P(X))^2 + (Q(X))^2 = X^{2n} + 1 \] and $\deg P > \deg Q$.

Show that the number of pairs is $2^{n+1}$.

['algebra']

theory putnam_2007_b4 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_2007_b4_solution :: "nat \<Rightarrow> nat" where "putnam_2007_b4_solution \<equiv> undefined" (* \<lambda> n. 2 ^ (n + 1) *) theorem putnam_2007_b4: fixes n :: nat assumes npos: "n > 0" shows "card {(P :: real poly, Q :: real poly). P ^ 2 + Q ^ 2 = (monom 1 (2 * n)) + 1 \<and> degree P > degree Q} = putnam_2007_b4_solution n" sorry end

putnam_2007_b5

theorem putnam_2007_b5 (k : ℕ) (kpos : k > 0) : (∃ P : Finset.range k → Polynomial ℝ, ∀ n : ℤ, ⌊(n : ℝ) / k⌋ ^ k = ∑ i : Finset.range k, (P i).eval (n : ℝ) * ⌊(n : ℝ) / k⌋ ^ (i : ℕ)) := sorry

Let $k$ be a positive integer. Prove that there exist polynomials $P_0(n), P_1(n), \dots, P_{k-1}(n)$ (which may depend on $k$) such that for any integer $n$, \[ \left\lfloor \frac{n}{k} \right\rfloor^k = P_0(n) + P_1(n) \left\lfloor \frac{n}{k} \right\rfloor + \cdots + P_{k-1}(n) \left\lfloor \frac{n}{k} \right\rfloor^{k-1}. \] ($\lfloor a \rfloor$ means the largest integer $\leq a$.)

['algebra']

theory putnam_2007_b5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2007_b5: fixes k :: nat assumes kpos: "k > 0" shows "\<exists> P :: real poly list. length P = k \<and> (\<forall> n :: int. \<lfloor>n / k\<rfloor> ^ k = (\<Sum> i = 0..(k - 1). poly (P!i) n * \<lfloor>n / k\<rfloor> ^ i))" sorry end

putnam_1980_a2

abbrev putnam_1980_a2_solution : ℕ → ℕ → ℕ := sorry -- (fun r s : ℕ => (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2)) theorem putnam_1980_a2 (r s : ℕ) (abcdlcm : ℕ → ℕ → ℕ → ℕ → Prop) (rspos : r > 0 ∧ s > 0) (habcdlcm : ∀ a b c d : ℕ, abcdlcm a b c d = (a > 0 ∧ b > 0 ∧ c > 0 ∧ d > 0 ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) c) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a c) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm b c) d))) : {(a, b, c, d) : ℕ × ℕ × ℕ × ℕ | abcdlcm a b c d}.encard = putnam_1980_a2_solution r s := sorry

Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a,b,c,d)$ of positive integers such that $3^r \cdot 7^s=\text{lcm}[a,b,c]=\text{lcm}[a,b,d]=\text{lcm}[a,c,d]=\text{lcm}[b,c,d]$. The answer should be a function of $r$ and $s$. (Note that $\text{lcm}[x,y,z]$ denotes the least common multiple of $x,y,z$.)

Show that the number is $(1+4r+6r^2)(1+4s+6s^2)$.

['number_theory']

Section putnam_1980_a2. Require Import Nat Ensembles Finite_sets. Definition putnam_1980_a2_solution (m n: nat) := (6*m*m + 3*m + 1) * (6*n*n + 3*n + 1). Theorem putnam_1980_a2: let gcd3 (a b c: nat) := gcd (gcd a b) c in exists (f: Ensemble (nat*nat)), forall (m n: nat) (a b c d: nat), (f (m, n) <-> gcd3 a b c = 3 ^ m * 7 ^ n \/ gcd3 a b d = 3 ^ m * 7 ^ n \/ gcd3 a c d = 3 ^ m * 7 ^ n \/ gcd3 b c d = 3 ^ m * 7 ^ n) -> cardinal (nat*nat) f (putnam_1980_a2_solution m n). Proof. Admitted. End putnam_1980_a2.

theory putnam_1980_a2 imports Complex_Main begin definition putnam_1980_a2_solution :: "nat \<Rightarrow> nat \<Rightarrow> nat" where "putnam_1980_a2_solution \<equiv> undefined" (* \<lambda> r s :: nat. (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2) *) theorem putnam_1980_a2: fixes r s :: nat and abcdlcm :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool" defines "abcdlcm \<equiv> \<lambda> a b c d :: nat. a > 0 \<and> b > 0 \<and> c > 0 \<and> d > 0 \<and> 3 ^ r * 7 ^ s = lcm (lcm a b) c \<and> 3 ^ r * 7 ^ s = lcm (lcm a b) d \<and> 3 ^ r * 7 ^ s = lcm (lcm a c) d \<and> 3 ^ r * 7 ^ s = lcm (lcm b c) d" assumes rspos: "r > 0 \<and> s > 0" shows "card {(a, b, c, d). abcdlcm a b c d} = putnam_1980_a2_solution r s" sorry end

putnam_1980_a3

abbrev putnam_1980_a3_solution : ℝ := sorry -- Real.pi / 4 theorem putnam_1980_a3 : ∫ x in Set.Ioo 0 (Real.pi / 2), 1 / (1 + (Real.tan x) ^ (Real.sqrt 2)) = putnam_1980_a3_solution := sorry

Evaluate $\int_0^{\pi/2}\frac{dx}{1+(\tan x)^{\sqrt{2}}}$.

Show that the integral is $\pi/4$.

['analysis']

Section putnam_1980_a3. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1980_a3_solution := PI/4. Theorem putnam_1980_a3: let f (x: R) := 1/(1 + Rpower (tan x) (sqrt 2)) in RInt f 0 PI/2 = putnam_1980_a3_solution. Proof. Admitted. End putnam_1980_a3.

theory putnam_1980_a3 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1980_a3_solution :: real where "putnam_1980_a3_solution \<equiv> undefined" (* pi / 4 *) theorem putnam_1980_a3: shows "interval_lebesgue_integral lebesgue 0 (pi / 2) (\<lambda> x :: real. 1 / (1 + (tan x) powr (sqrt 2))) = putnam_1980_a3_solution" sorry end

putnam_1980_a4

theorem putnam_1980_a4 (abcvals : ℤ → ℤ → ℤ → Prop) (habcvals : ∀ a b c : ℤ, abcvals a b c = (¬(a = 0 ∧ b = 0 ∧ c = 0) ∧ |a| < 1000000 ∧ |b| < 1000000 ∧ |c| < 1000000)) : (∃ a b c : ℤ, abcvals a b c ∧ |a + b * Real.sqrt 2 + c * Real.sqrt 3| < 10 ^ (-(11 : ℝ))) ∧ (∀ a b c : ℤ, abcvals a b c → |a + b * Real.sqrt 2 + c * Real.sqrt 3| > 10 ^ (-(21 : ℝ))) := sorry

\begin{enumerate} \item[(a)] Prove that there exist integers $a,b,c$, not all zero and each of absolute value less than one million, such that $|a+b\sqrt{2}+c\sqrt{3}|<10^{-11}$. \item[(b)] Let $a,b,c$ be integers, not all zero and each of absolute value less than one million. Prove that $|a+b\sqrt{2}+c\sqrt{3}|>10^{-21}$. \end{enumerate}

['algebra']

Section putnam_1980_a4. Require Import Reals BinInt. Open Scope Z. Theorem putnam_1980_a4: (forall (a b c: Z), (~ (a = 0 \/ b = 0 /\ c = 0) /\ Z.abs a < 10^6 /\ Z.abs b < 10^6 /\ Z.abs c < 10^6) -> Rgt (Rabs (Rplus (Rplus (IZR a) (Rmult (IZR b) (sqrt 2))) (Rmult (IZR c) (sqrt 3)))) (Rpower 10 (-21)) ) /\ (exists (a b c: Z), Rlt (Rabs (Rplus (IZR a) (Rplus (Rmult (IZR b) (sqrt 2)) (Rmult (IZR c) (sqrt 3))))) (Rpower 10 (-11)) ). Proof. Admitted. End putnam_1980_a4.

theory putnam_1980_a4 imports Complex_Main begin theorem putnam_1980_a4: fixes abcvals :: "int \<Rightarrow> int \<Rightarrow> int \<Rightarrow> bool" defines "abcvals \<equiv> \<lambda> a b c :: int. \<not>(a = 0 \<and> b = 0 \<and> c = 0) \<and> \<bar>a\<bar> < 1000000 \<and> \<bar>b\<bar> < 1000000 \<and> \<bar>c\<bar> < 1000000" shows "(\<exists> a b c :: int. abcvals a b c \<and> \<bar>a + b * sqrt 2 + c * sqrt 3\<bar> < 10 powi -11) \<and> (\<forall> a b c :: int. abcvals a b c \<longrightarrow> \<bar>a + b * sqrt 2 + c * sqrt 3\<bar> > 10 powi -21)" sorry end

putnam_1980_a5

theorem putnam_1980_a5 (P : Polynomial ℝ) (xeqs : ℝ → Prop) (Pnonconst : P.degree > 0) (hxeqs : ∀ x : ℝ, xeqs x = (0 = (∫ t in (0)..x, P.eval t * Real.sin t) ∧ 0 = (∫ t in (0)..x, P.eval t * Real.cos t))) : {x : ℝ | xeqs x}.Finite := sorry

Let $P(t)$ be a nonconstant polynomial with real coefficients. Prove that the system of simultaneous equations $0=\int_0^xP(t)\sin t\,dt=\int_0^xP(t)\cos t\,dt$ has only finitely many real solutions $x$.

['analysis']

Section putnam_1980_a5. Require Import Reals Coquelicot.Coquelicot. Theorem putnam_1980_a5 (n : nat) (npos : gt n 0) (coeff : nat -> R) (hcoeff : coeff n <> 0) (p : R -> R := fun x => sum_n (fun i => coeff i * x ^ i) (S n)) (h1 : nat -> Prop := fun a => RInt (fun x => p x * sin x) 0 (INR a) = 0) (h2 : nat -> Prop := fun a => RInt (fun x => p x * cos x) 0 (INR a) = 0) : exists (m: nat), forall (b: nat), h1 b /\ h2 b -> lt b m. Proof. Admitted. End putnam_1980_a5.

theory putnam_1980_a5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Analysis.Interval_Integral" begin theorem putnam_1980_a5: fixes P :: "real poly" and xeqs :: "real \<Rightarrow> bool" defines "xeqs \<equiv> \<lambda> x :: real. 0 = interval_lebesgue_integral lebesgue 0 x (\<lambda> t. poly P t * sin t) \<and> 0 = interval_lebesgue_integral lebesgue 0 x (\<lambda> t. poly P t * cos t)" assumes Pnonconst: "degree P > 0" shows "finite {x :: real. xeqs x}" sorry end

putnam_1980_a6

abbrev putnam_1980_a6_solution : ℝ := sorry -- 1 / Real.exp 1 theorem putnam_1980_a6 (C : Set (ℝ → ℝ)) (uleint : ℝ → Prop) (hC : C = {f : ℝ → ℝ | ContDiffOn ℝ 1 f (Set.Icc 0 1) ∧ f 0 = 0 ∧ f 1 = 1}) (huleint : ∀ u : ℝ, uleint u = ∀ f ∈ C, u ≤ (∫ x in Set.Ioo 0 1, |deriv f x - f x|)) : uleint putnam_1980_a6_solution ∧ (∀ u : ℝ, uleint u → u ≤ putnam_1980_a6_solution) := sorry

Let $C$ be the class of all real valued continuously differentiable functions $f$ on the interval $0 \leq x \leq 1$ with $f(0)=0$ and $f(1)=1$. Determine the largest real number $u$ such that $u \leq \int_0^1|f'(x)-f(x)|\,dx$ for all $f$ in $C$.

Show that $u=1/e$.

['analysis']

theory putnam_1980_a6 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1980_a6_solution :: real where "putnam_1980_a6_solution \<equiv> undefined" (* 1 / exp 1 *) theorem putnam_1980_a6: fixes C :: "(real \<Rightarrow> real) set" and uleint :: "real \<Rightarrow> bool" defines "C \<equiv> {f :: real \<Rightarrow> real. f C1_differentiable_on {0..1} \<and> f 0 = 0 \<and> f 1 = 1}" and "uleint \<equiv> \<lambda> u :: real. \<forall> f \<in> C. u \<le> interval_lebesgue_integral lebesgue 0 1 (\<lambda> x. \<bar>deriv f x - f x\<bar>)" shows "(GREATEST u. uleint u) = putnam_1980_a6_solution" sorry end

putnam_1980_b1

abbrev putnam_1980_b1_solution : Set ℝ := sorry -- {c : ℝ | c ≥ 1 / 2} theorem putnam_1980_b1 (c : ℝ) : (∀ x : ℝ, (Real.exp x + Real.exp (-x)) / 2 ≤ Real.exp (c * x ^ 2)) ↔ c ∈ putnam_1980_b1_solution := sorry

For which real numbers $c$ is $(e^x+e^{-x})/2 \leq e^{cx^2}$ for all real $x$?

Show that the inequality holds if and only if $c \geq 1/2$.

['analysis']

Section putnam_1980_b1. Require Import Reals Rtrigo_def. Open Scope R. Definition putnam_1980_b1_solution (k: R) := k >= 1/2. Theorem putnam_1980_b1: forall (k: R), forall (x: R), cosh x <= exp (k*x*x) <-> putnam_1980_b1_solution k. Proof. Admitted. End putnam_1980_b1.

theory putnam_1980_b1 imports Complex_Main begin definition putnam_1980_b1_solution :: "real set" where "putnam_1980_b1_solution \<equiv> undefined" (* {c :: real. c \<ge> 1 / 2} *) theorem putnam_1980_b1: fixes c :: real shows "(\<forall> x :: real. (exp x + exp (-x)) / 2 \<le> exp (c * x ^ 2)) \<longleftrightarrow> c \<in> putnam_1980_b1_solution" sorry end

putnam_1980_b3

abbrev putnam_1980_b3_solution : Set ℝ := sorry -- {a : ℝ | a ≥ 3} theorem putnam_1980_b3 (a : ℝ) (u : ℕ → ℝ) (hu : u 0 = a ∧ (∀ n : ℕ, u (n + 1) = 2 * u n - n ^ 2)) : (∀ n : ℕ, u n > 0) ↔ a ∈ putnam_1980_b3_solution := sorry

For which real numbers $a$ does the sequence defined by the initial condition $u_0=a$ and the recursion $u_{n+1}=2u_n-n^2$ have $u_n>0$ for all $n \geq 0$? (Express the answer in the simplest form.)

Show that $u_n>0$ for all $n \geq 0$ if and only if $a \geq 3$.

['algebra']

Section putnam_1980_b3. Require Import Reals. Open Scope R. Definition putnam_1980_b3_solution (b: R) := b >= 3. Theorem putnam_1980_b3: forall (b: R), let A := fix a (n: nat) : R := match n with | O => b | S n' => 2 * a n' - INR (n' * n') end in forall (n: nat), A n > 0 <-> putnam_1980_b3_solution b. Proof. Admitted. End putnam_1980_b3.

theory putnam_1980_b3 imports Complex_Main begin definition putnam_1980_b3_solution :: "real set" where "putnam_1980_b3_solution \<equiv> undefined" (* {a :: real. a \<ge> 3} *) theorem putnam_1980_a3: fixes a :: real and u :: "nat \<Rightarrow> real" assumes hu: "u 0 = a \<and> (\<forall> n :: nat. u (n + 1) = 2 * (u n) - n ^ 2)" shows "(\<forall> n :: nat. u n > 0) \<longleftrightarrow> a \<in> putnam_1980_b3_solution" sorry end

putnam_1980_b4

theorem putnam_1980_b4 {T : Type} (X : Finset T) (A : Fin 1066 → Finset T) (hX : X.card ≥ 10) (hA : ∀ i : Fin 1066, A i ⊆ X ∧ (A i).card > ((1 : ℚ)/2) * X.card) : ∃ Y : Finset T, Y ⊆ X ∧ Y.card = 10 ∧ ∀ i : Fin 1066, ∃ y ∈ Y, y ∈ A i := sorry

Let $X$ be a finite set with at least $10$ elements; for each $i \in \{0, 1, ..., 1065\}$, let $A_i \subseteq X$ satisfy $|A_i| > \frac{1}{2}|X|$. Prove that there exist $10$ elements $x_1, x_2, \dots, x_{10} \in X$ such that each $A_i$ contains at least one of $x_1, x_2, \dots, x_{10}$.

['set_theory', 'combinatorics']

theory putnam_1980_b4 imports Complex_Main begin theorem putnam_1980_b4: fixes X :: "'t set" and A :: "nat \<Rightarrow> 't set" assumes Xfin: "finite X" and hX: "card X \<ge> 10" and hA: "\<forall> i \<in> {1 .. 1066}. A i \<subseteq> X \<and> card (A i) > (1 / 2) * card X" shows "\<exists> Y :: 't set. Y \<subseteq> X \<and> card Y = 10 \<and> (\<forall> i \<in> {1 .. 1066}. \<exists> y \<in> Y. y \<in> A i)" sorry end

putnam_1980_b5

abbrev putnam_1980_b5_solution : ℝ → Prop := sorry -- fun t : ℝ => 1 ≥ t theorem putnam_1980_b5 (T : Set ℝ := Icc 0 1) (P : ℝ → (ℝ → ℝ) → Prop := fun t : ℝ => fun f : ℝ → ℝ => f 1 - 2*f (2/3) + f (1/3) ≥ t*(f (2/3) - 2*f (1/3) + f 0)) (Convex : (ℝ → ℝ) → Prop := fun f : ℝ → ℝ => ∀ u ∈ T, ∀ v ∈ T, ∀ s ∈ T, f (s*u + (1 - s)*v) ≤ s*(f u) + (1 - s)*(f v)) (S : ℝ → Set (ℝ → ℝ) := fun t : ℝ => {f : ℝ → ℝ | (∀ x ∈ T, f x ≥ 0) ∧ StrictMonoOn f T ∧ Convex f ∧ ContinuousOn f T ∧ P t f}) : ∀ t : ℝ, t ≥ 0 → ((∀ f ∈ S t, ∀ g ∈ S t, f * g ∈ S t) ↔ putnam_1980_b5_solution t) := sorry

A function $f$ is convex on $[0, 1]$ if and only if $$f(su + (1-s)v) \le sf(u) + (1 - s)f(v)$$ for all $s \in [0, 1]$. Let $S_t$ denote the set of all nonnegative increasing convex continuous functions $f : [0, 1] \rightarrow \mathbb{R}$ such that $$f(1) - 2f\left(\frac{2}{3}\right) + f\left(\frac{1}{3}\right) \ge t\left(f\left(\frac{2}{3}\right) - 2f\left(\frac{1}{3}\right) + f(0)\right).$$ For which real numbers $t \ge 0$ is $S_t$ closed under multiplication?

$S_t$ is closed under multiplication if and only if $1 \ge t$.

['analysis', 'algebra']

theory putnam_1980_b5 imports Complex_Main begin definition putnam_1980_b5_solution :: "real \<Rightarrow> bool" where "putnam_1980_b5_solution \<equiv> undefined" (* \<lambda> t :: real. 1 \<ge> t *) theorem putnam_1980_b5: fixes T :: "real set" and P :: "real \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool" and Convex :: "(real \<Rightarrow> real) \<Rightarrow> bool" and S :: "real \<Rightarrow> (real \<Rightarrow> real) set" defines "T \<equiv> {0..1}" and "P \<equiv> \<lambda> (t :: real) (f :: real \<Rightarrow> real). f 1 - 2 * f (2 / 3) + f (1 / 3) \<ge> t * (f (2 / 3) - 2 * f (1 / 3) + f 0)" and "Convex \<equiv> \<lambda> f :: real \<Rightarrow> real. \<forall> u \<in> T. \<forall> v \<in> T. \<forall> s \<in> T. f (s * u + (1 - s) * v) \<le> s * (f u) + (1 - s) * (f v)" and "S \<equiv> \<lambda> t :: real. {f :: real \<Rightarrow> real. (\<forall> x \<in> T. f x \<ge> 0) \<and> strict_mono_on T f \<and> Convex f \<and> continuous_on T f \<and> P t f}" shows "\<forall> t :: real. t \<ge> 0 \<longrightarrow> ((\<forall> f \<in> S t. \<forall> g \<in> S t. (\<lambda> x. f x * g x) \<in> S t) \<longleftrightarrow> putnam_1980_b5_solution t)" sorry end

putnam_1980_b6

theorem putnam_1980_b6 (G : ℤ × ℤ → ℚ) (hG : ∀ d n : ℕ, d ≤ n → (d = 1 → G (d, n) = 1/(n : ℚ)) ∧ (d > 1 → G (d, n) = (d/(n : ℚ))*∑ i in Finset.Icc d n, G ((d : ℤ) - 1, (i : ℤ) - 1))) : ∀ d p : ℕ, 1 < d ∧ d ≤ p ∧ Prime p → ¬p ∣ (G (d, p)).den := sorry

For integers $d, n$ with $1 \le d \le n$, let $G(1, n) = \frac{1}{n}$ and $G(d, n) = \frac{d}{n}\sum_{i=d}^{n}G(d - 1, i - 1)$ for all $d > 1$. If $1 < d \le p$ for some prime $p$, prove that the reduced denominator of $G(d, p)$ is not divisible by $p$.

['number_theory', 'algebra']

Section putnam_1980_b6. Require Import Reals Nat Znumtheory QArith Coquelicot.Coquelicot. From mathcomp Require Import div. Theorem putnam_1980_b6: let A := fix f (n i: nat) := match (n,i) with | (O,i') => 1/INR i' | (S n', i') => (INR n' + 1)/(INR i') * sum_n (fun x => f n' (Nat.add n' x)) (Nat.sub i' n') end in forall (n p: nat), and (gt p n) (gt n 1) /\ prime (Z.of_nat p) -> exists (a b: nat), A n p = INR a/INR b /\ p %| b/(gcd a b) = false. Proof. Admitted. End putnam_1980_b6.

theory putnam_1980_b6 imports Complex_Main "HOL-Computational_Algebra.Primes" begin theorem putnam_1980_b6: fixes G :: "nat \<times> nat \<Rightarrow> rat" assumes hG: "\<forall> d n :: nat. d \<le> n \<longrightarrow> ((d = 1 \<longrightarrow> G (d, n) = 1 / n) \<and> (d > 1 \<longrightarrow> G (d, n) = (d / n) * (\<Sum> i = d..n. G (d - 1, i - 1))))" shows "\<forall> d p :: nat. (1 < d \<and> d \<le> p \<and> prime p) \<longrightarrow> \<not>(p dvd (snd (quotient_of (G (d, p)))))" sorry end

putnam_1984_a2

abbrev putnam_1984_a2_solution : ℚ := sorry -- 2 theorem putnam_1984_a2 : ∑' k : Set.Ici 1, (6 ^ (k : ℕ) / ((3 ^ ((k : ℕ) + 1) - 2 ^ ((k : ℕ) + 1)) * (3 ^ (k : ℕ) - 2 ^ (k : ℕ)))) = putnam_1984_a2_solution := sorry

Express $\sum_{k=1}^\infty (6^k/(3^{k+1}-2^{k+1})(3^k-2^k))$ as a rational number.

Show that the sum converges to $2$.

['analysis']

Section putnam_1984_a2. Require Import Reals Coquelicot.Coquelicot. Open Scope nat_scope. Definition putnam_1984_a2_solution := 2%R. Theorem putnam_1984_a2: let f (n: nat) := Rdiv (pow 6 n) ((pow 3 (n+1) - pow 2 (n+1)) * pow 3 n - pow 2 n) in Series f = putnam_1984_a2_solution. Proof. Admitted. End putnam_1984_a2.

theory putnam_1984_a2 imports Complex_Main begin definition putnam_1984_a2_solution :: "rat" where "putnam_1984_a2_solution \<equiv> undefined" (* 2 *) theorem putnam_1984_a2: shows "(\<Sum> (k :: nat) \<in> {1..}. 6^k/((3^(k+1) - 2^(k+1))*(3^k - 2^k))) = putnam_1984_a2_solution" sorry end

putnam_1984_a3

abbrev putnam_1984_a3_solution : MvPolynomial (Fin 3) ℝ := sorry -- (MvPolynomial.X 2) ^ 2 * ((MvPolynomial.X 0) ^ 2 - (MvPolynomial.X 1) ^ 2) theorem putnam_1984_a3 (n : ℕ) (a b : ℝ) (Mn : ℝ → Matrix (Fin (2 * n)) (Fin (2 * n)) ℝ) (polyabn : Fin 3 → ℝ) (npos : n > 0) (aneb : a ≠ b) (hMnx : ∀ x : ℝ, ∀ i : Fin (2 * n), (Mn x) i i = x) (hMna : ∀ x : ℝ, ∀ i j : Fin (2 * n), (i ≠ j ∧ Even ((i : ℕ) + j)) → (Mn x) i j = a) (hMnb : ∀ x : ℝ, ∀ i j : Fin (2 * n), (i ≠ j ∧ Odd ((i : ℕ) + j)) → (Mn x) i j = b) (hpolyabn : polyabn 0 = a ∧ polyabn 1 = b ∧ polyabn 2 = n) : Tendsto (fun x : ℝ => (Mn x).det / (x - a) ^ (2 * n - 2)) (𝓝[≠] a) (𝓝 (MvPolynomial.eval polyabn putnam_1984_a3_solution)) := sorry

Let $n$ be a positive integer. Let $a,b,x$ be real numbers, with $a \neq b$, and let $M_n$ denote the $2n \times 2n$ matrix whose $(i,j)$ entry $m_{ij}$ is given by \[ m_{ij}=\begin{cases} x & \text{if }i=j, \\ a & \text{if }i \neq j\text{ and }i+j\text{ is even}, \\ b & \text{if }i \neq j\text{ and }i+j\text{ is odd}. \end{cases} \] Thus, for example, $M_2=\begin{pmatrix} x & b & a & b \\ b & x & b & a \\ a & b & x & b \\ b & a & b & x \end{pmatrix}$. Express $\lim_{x \to a} \det M_n/(x-a)^{2n-2}$ as a polynomial in $a$, $b$, and $n$, where $\det M_n$ denotes the determinant of $M_n$.

Show that $\lim_{x \to a} \frac{\det M_n}{(x-a)^{2n-2}}=n^2(a^2-b^2)$.

['linear_algebra', 'analysis']

theory putnam_1984_a3 imports Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Analysis.Finite_Cartesian_Product" "HOL-Analysis.Determinants" begin definition putnam_1984_a3_solution :: "real poly poly poly" where "putnam_1984_a3_solution \<equiv> undefined" (* (monom 1 2) * ([:[:monom 1 2:]:] - [:monom 1 2:]) *) theorem putnam_1984_a3: fixes n :: "nat" and a b :: "real" and Mn :: "real \<Rightarrow> (real^'i^'i)" and ni :: "'i \<Rightarrow> nat" assumes hij : "CARD('i) = 2*n" and hn : "bij_betw ni (UNIV :: 'i set) {1..2*n}" and npos : "n > 0" and aneb : "a \<noteq> b" and hMnx : "\<forall> x :: real. \<forall> i :: 'i. (Mn x)$i$i = x" and hMna : "\<forall> x :: real. \<forall> i j :: 'i. (ni i \<noteq> nj j \<and> even (ni i + nj j)) \<longrightarrow> (Mn x)$i$j = a" and hMnb : "\<forall> x :: real. \<forall> i j :: 'i. (ni i \<noteq> nj j \<and> odd (ni i + nj j)) \<longrightarrow> (Mn x)$i$j = b" shows "filterlim (\<lambda> x :: real. det (Mn x) / (x - a) ^ (2 * n - 2)) (nhds (poly (poly (poly putnam_1984_a3_solution [:[:(n::real):]:]) [:b:]) a)) (at a)" sorry end

putnam_1984_a5

abbrev putnam_1984_a5_solution : ℕ × ℕ × ℕ × ℕ × ℕ := sorry -- (1, 9, 8, 4, 25) theorem putnam_1984_a5 (R : Set (Fin 3 → ℝ)) (w : (Fin 3 → ℝ) → ℝ) (hR : R = {p : Fin 3 → ℝ | (∀ i : Fin 3, p i ≥ 0) ∧ p 0 + p 1 + p 2 ≤ 1}) (hw : ∀ p : Fin 3 → ℝ, w p = 1 - p 0 - p 1 - p 2) : let (a, b, c, d, n) := putnam_1984_a5_solution; a > 0 ∧ b > 0 ∧ c > 0 ∧ d > 0 ∧ n > 0 ∧ (∫ p in R, (p 0) ^ 1 * (p 1) ^ 9 * (p 2) ^ 8 * (w p) ^ 4 = ((a)! * (b)! * (c)! * (d)! : ℝ) / (n)!) := sorry

Let $R$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x+y+z \leq 1$. Let $w=1-x-y-z$. Express the value of the triple integral $\iiint_R x^1y^9z^8w^4\,dx\,dy\,dz$ in the form $a!b!c!d!/n!$, where $a$, $b$, $c$, $d$, and $n$ are positive integers.

Show that the integral we desire is $1!9!8!4!/25!$.

['analysis']

Section putnam_1984_a5. Require Import Reals Factorial Coquelicot.Coquelicot. Open Scope R. Definition putnam_1984_a5_solution := INR (fact 9 * fact 8 * fact 4 / fact 25). Theorem putnam_1984_a5: RInt (fun z => RInt (fun y => RInt (fun x => x * pow y 9 * pow z 8 * pow (1 - x - y - z) 4) 0 (1 - y - z)) 0 (1 - z)) 0 1 = putnam_1984_a5_solution. Proof. Admitted. End putnam_1984_a5.

theory putnam_1984_a5 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Analysis.Set_Integral" begin definition putnam_1984_a5_solution :: "nat \<times> nat \<times> nat \<times> nat \<times> nat" where "putnam_1984_a5_solution \<equiv> undefined" (* (1, 9, 8, 4, 25) *) theorem putnam_1984_a5: fixes R :: "(real \<times> real \<times> real) set" and w :: "(real \<times> real \<times> real) \<Rightarrow> real" defines "R \<equiv> {(x :: real, y :: real, z :: real). x \<ge> 0 \<and> y \<ge> 0 \<and> z \<ge> 0 \<and> x + y + z \<le> 1}" and "w \<equiv> \<lambda> (x :: real, y :: real, z :: real). 1 - x - y - z" shows "let (a, b, c, d, n) = putnam_1984_a5_solution in a > 0 \<and> b > 0 \<and> c > 0 \<and> d > 0 \<and> n > 0 \<and> set_lebesgue_integral lebesgue R (\<lambda> (x, y, z). x^1 * y^9 * z^8 * (w (x, y, z))^4) = (real_of_nat ((fact a) * (fact b) * (fact c) * (fact d)))/(fact n)" sorry end

putnam_1984_a6

abbrev putnam_1984_a6_solution : Prop × ℕ := sorry -- (True, 4) theorem putnam_1984_a6 (lastnzdig : List ℕ → ℕ) (f : ℕ → ℕ) (kadistinct : ℕ → (ℕ → ℕ) → Prop := fun k : ℕ => fun a : ℕ → ℕ => (k ≥ 1 ∧ ∀ i j : Fin k, i ≠ j → a i ≠ a j)) (gpeq : (ℕ → ℕ) → ℕ → Prop := fun g : ℕ → ℕ => fun p : ℕ => (p > 0 ∧ ∀ s ≥ 1, g s = g (s + p))) (hlastnzdig : ∀ digs : List ℕ, (∃ i : Fin digs.length, digs[i] ≠ 0) → lastnzdig digs ≠ 0 ∧ (∃ i : Fin digs.length, digs[i] = lastnzdig digs ∧ ∀ j < i, digs[j] = 0)) (hf : ∀ n > 0, f n = lastnzdig (Nat.digits 10 (n)!)) : ∃ g : ℕ → ℕ, (∀ (k : ℕ) (a : ℕ → ℕ), kadistinct k a → g (∑ i : Fin k, a i) = f (∑ i : Fin k, 5 ^ (a i))) ∧ (if putnam_1984_a6_solution.1 = True then (gpeq g putnam_1984_a6_solution.2 ∧ ∀ p : ℕ, gpeq g p → p ≥ putnam_1984_a6_solution.2) else (¬∃ p : ℕ, gpeq g p)) := sorry

Let $n$ be a positive integer, and let $f(n)$ denote the last nonzero digit in the decimal expansion of $n!$. For instance, $f(5)=2$. \begin{enumerate} \item[(a)] Show that if $a_1,a_2,\dots,a_k$ are \emph{distinct} nonnegative integers, then $f(5^{a_1}+5^{a_2}+\dots+5^{a_k})$ depends only on the sum $a_1+a_2+\dots+a_k$. \item[(b)] Assuming part (a), we can define $g(s)=f(5^{a_1}+5^{a_2}+\dots+5^{a_k})$, where $s=a_1+a_2+\dots+a_k$. Find the least positive integer $p$ for which $g(s)=g(s + p)$, for all $s \geq 1$, or else show that no such $p$ exists. \end{enumerate}

Show that the least such $p$ is $p=4$.

['algebra', 'number_theory']

theory putnam_1984_a6 imports Complex_Main "HOL-Number_Theory.Cong" begin definition putnam_1984_a6_solution :: "bool \<times> nat" where "putnam_1984_a6_solution \<equiv> undefined" (* (True, 4) *) theorem putnam_1984_a6: fixes f :: "nat \<Rightarrow> nat" and kadistinct :: "nat \<Rightarrow> (nat \<Rightarrow> nat) \<Rightarrow> bool" and gpeq :: "(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> bool" defines "kadistinct \<equiv> \<lambda> (k :: nat) (a :: nat \<Rightarrow> nat). k \<ge> 1 \<and> (\<forall> i j :: nat. (i < k \<and> j < k \<and> i \<noteq> j) \<longrightarrow> a i \<noteq> a j)" and "gpeq \<equiv> \<lambda> (g :: nat \<Rightarrow> nat) (p :: nat). p > 0 \<and> (\<forall> (s :: nat) \<ge> 1. g s = g (s + p))" assumes hf : "\<forall> n > 0. f n = (if [n \<noteq> 0] (mod 10) then (n mod 10) else f (n div 10))" shows "let (b, n) = putnam_1984_a6_solution in \<exists> g :: nat \<Rightarrow> nat. (\<forall> (k :: nat) (a :: nat \<Rightarrow> nat). kadistinct k a \<longrightarrow> g (\<Sum> i=0..(k-1). a i) = f (\<Sum> i=0..(k-1). 5^(a i))) \<and> (if b then gpeq g n \<and> (\<forall> p :: nat. gpeq g p \<longrightarrow> p \<ge> n) else \<not>(\<exists> p :: nat. gpeq g p))" sorry end

putnam_1984_b1

abbrev putnam_1984_b1_solution : Polynomial ℝ × Polynomial ℝ := sorry -- (Polynomial.X + 3, -Polynomial.X - 2) theorem putnam_1984_b1 (f : ℕ → ℤ) (hf : ∀ n > 0, f n = ∑ i : Set.Icc 1 n, ((i)! : ℤ)) : let (P, Q) := putnam_1984_b1_solution; ∀ n ≥ 1, f (n + 2) = P.eval (n : ℝ) * f (n + 1) + Q.eval (n : ℝ) * f n := sorry

Let $n$ be a positive integer, and define $f(n)=1!+2!+\dots+n!$. Find polynomials $P(x)$ and $Q(x)$ such that $f(n+2)=P(n)f(n+1)+Q(n)f(n)$ for all $n \geq 1$.

Show that we can take $P(x)=x+3$ and $Q(x)=-x-2$.

['algebra']

Section putnam_1984_b1. Require Import Factorial ZArith Reals Coquelicot.Coquelicot. Open Scope Z. Definition putnam_1984_b1_solution (coeff1 coeff2 : nat -> Z) (n1 n2 : Z) := (coeff1 = fun x => match x with | O => 3 | S O => 1 | _ => 0 end) /\ (coeff2 = fun x => match x with | O => -2 | S O => -1 | _ => 0 end) /\ n1 = 1 /\ n2 = 1. Theorem putnam_1984_b1 (f : nat -> Z := fun n => (floor (sum_n (fun i => INR (fact (i + 1))) n))) (p: (nat -> Z) -> Z -> (nat -> Z) := fun coeff n => (fun x => (floor (sum_n (fun i => IZR ((coeff i) * (Z.of_nat (x ^ i)))) (Z.to_nat n + 1))))) : exists (coeff1 coeff2 : nat -> Z) (n1 n2 : Z), let fix F (n: nat) := match n with | O => f 0%nat | S O => f 1%nat | S ((S n'') as n') => (p coeff1 n1) n' * F n' + (p coeff2 n2) n'' * F n'' end in f = F -> putnam_1984_b1_solution coeff1 coeff2 n1 n2. Proof. Admitted. End putnam_1984_b1.

theory putnam_1984_b1 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1984_b1_solution :: "(real poly) \<times> (real poly)" where "putnam_1984_b1_solution \<equiv> undefined" (* (monom 1 1 + monom 3 0, monom (-1) 1 + monom (-2) 0) *) theorem putnam_1984_b1: fixes f :: "nat \<Rightarrow> nat" assumes hf : "\<forall> n > 0. f n = (\<Sum> i=1..n. (fact i))" shows "let (P, Q) = putnam_1984_b1_solution in \<forall> n \<ge> 1. f (n + 2) = (poly P n) * (f (n + 1)) + (poly Q n) * (f n)" sorry end

putnam_1984_b2

abbrev putnam_1984_b2_solution : ℝ := sorry -- 8 theorem putnam_1984_b2 (f : ℝ → ℝ → ℝ) (hf : ∀ u v : ℝ, f u v = (u - v) ^ 2 + (Real.sqrt (2 - u ^ 2) - 9 / v) ^ 2) : (∃ u : Set.Ioo 0 (Real.sqrt 2), ∃ v > 0, f u v = putnam_1984_b2_solution) ∧ (∀ u : Set.Ioo 0 (Real.sqrt 2), ∀ v > 0, f u v ≥ putnam_1984_b2_solution) := sorry

Find the minimum value of $(u-v)^2+(\sqrt{2-u^2}-\frac{9}{v})^2$ for $0<u<\sqrt{2}$ and $v>0$.

Show that the minimum value is $8$.

['geometry', 'analysis']

Section putnam_1984_b2. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1984_b2_solution := 8. Theorem putnam_1984_b2: let f (x y: R) := pow (x - y) 2 + pow (sqrt (2 - pow x 2) - 9 / y) 2 in exists (m: R), (forall (x y: R), 0 < x < sqrt 2 /\ y > 0 -> f x y >= m) /\ (exists (x y: R), 0 < x < sqrt 2 /\ y > 0 -> f x y = m) -> m = putnam_1984_b2_solution. Proof. Admitted. End putnam_1984_b2.

theory putnam_1984_b2 imports Complex_Main begin definition putnam_1984_b2_solution :: "real" where "putnam_1984_b2_solution \<equiv> undefined" (* 8 *) theorem putnam_1984_b2: fixes f :: "real \<Rightarrow> real \<Rightarrow> real" defines "f \<equiv> \<lambda> u v :: real. (u - v)^2 + (sqrt (2 - u^2) - 9/v)^2" shows "(LEAST (r :: real) . (\<exists> u \<in> {0<..<sqrt 2}. \<exists> v > 0. f u v = r)) = putnam_1984_b2_solution" sorry end

putnam_1984_b3

abbrev putnam_1984_b3_solution : Prop := sorry -- True theorem putnam_1984_b3 : (∀ (F : Type*) (_ : Fintype F), Fintype.card F ≥ 2 → (∃ mul : F → F → F, ∀ x y z : F, (mul x z = mul y z → x = y) ∧ (mul x (mul y z) ≠ mul (mul x y) z))) ↔ putnam_1984_b3_solution := sorry

Prove or disprove the following statement: If $F$ is a finite set with two or more elements, then there exists a binary operation $*$ on F such that for all $x,y,z$ in $F$, \begin{enumerate} \item[(i)] $x*z=y*z$ implies $x=y$ (right cancellation holds), and \item[(ii)] $x*(y*z) \neq (x*y)*z$ (\emph{no} case of associativity holds). \end{enumerate}

Show that the statement is true.

['abstract_algebra']

theory putnam_1984_b3 imports Complex_Main "HOL-Library.Cardinality" begin definition putnam_1984_b3_solution :: "bool" where "putnam_1984_b3_solution \<equiv> undefined" (* True *) theorem putnam_1984_b3: shows "((CARD('a) \<ge> 2 \<and> (\<exists> n :: nat. CARD('a) = n)) \<longrightarrow> (\<exists> mul :: 'a \<Rightarrow> 'a \<Rightarrow> 'a. \<forall> x y z :: 'a. (mul x z = mul y z \<longrightarrow> x = y) \<and> (mul x (mul y z) \<noteq> mul (mul x y) z))) \<longleftrightarrow> putnam_1984_b3_solution" sorry end

putnam_1984_b5

abbrev putnam_1984_b5_solution : ℤ × Polynomial ℝ × Polynomial ℕ := sorry -- (2, (Polynomial.X * (Polynomial.X - 1)) / 2, Polynomial.X) theorem putnam_1984_b5 (sumbits : List ℕ → ℕ) (d : ℕ → ℕ) (m : ℕ) (hsumbits : ∀ bits : List ℕ, sumbits bits = ∑ i : Fin bits.length, bits[i]) (hd : ∀ k : ℕ, d k = sumbits (Nat.digits 2 k)) (mpos : m > 0) : let (a, f, g) := putnam_1984_b5_solution; ∑ k : Set.Icc 0 (2 ^ m - 1), (-(1 : ℤ)) ^ (d k) * (k : ℕ) ^ m = (-1) ^ m * (a : ℝ) ^ (f.eval (m : ℝ)) * (g.eval m)! := sorry

For each nonnegative integer $k$, let $d(k)$ denote the number of $1$'s in the binary expansion of $k$ (for example, $d(0)=0$ and $d(5)=2$). Let $m$ be a positive integer. Express $\sum_{k=0}^{2^m-1} (-1)^{d(k)}k^m$ in the form $(-1)^ma^{f(m)}(g(m))!$, where $a$ is an integer and $f$ and $g$ are polynomials.

Show that $\sum_{k=0}^{2^m-1} (-1)^{d(k)}k^m=(-1)^m2^{m(m-1)/2}m!$.

['algebra', 'analysis']

Section putnam_1984_b5. Require Import Nat Reals Coquelicot.Coquelicot. Definition putnam_1984_b5_solution (coeff1 coeff2 : nat -> R) (n1 n2: nat) (a: Z) := a = 2%Z /\ (coeff1 = fun x => match x with | O => 0 | S O => -1 / 2 | S (S O) => 1 / 2 | _ => 0 end) /\ (coeff2 = fun x => match x with | O => 0 | S O => INR x | _ => 0 end) /\ n1 = 2%nat /\ n2 = 1%nat. Theorem putnam_1984_b5 (f : positive -> nat := fix f (n : positive) : nat := match n with | xH => 1%nat | xO n' => f n' | xI n' => add 1 (f n') end) (g : nat -> R := fun m => sum_n (fun k => (-1) ^ (f (Pos.of_nat k)) * INR k ^ m) (2 ^ m - 1)) (p: (nat -> R) -> nat -> (R -> R) := fun coeff n => (fun x => sum_n (fun i => coeff i * x ^ i) (n + 1))) : forall (m : nat), exists (a : Z) (coeff1 coeff2 : nat -> R) (n1 n2: nat), g m = (-1) ^ m * Rpower (IZR a) ((p coeff1 n1) (INR m)) * INR (fact (Z.to_nat (floor ((p coeff2 n2) (INR m))))) <-> putnam_1984_b5_solution coeff1 coeff2 n1 n2 a. Proof. Admitted. End putnam_1984_b5.

theory putnam_1984_b5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1984_b5_solution :: "int \<times> (real poly) \<times> (nat poly)" where "putnam_1984_b5_solution \<equiv> undefined" (* (2, ((monom 1 1) * ((monom 1 1) - 1)) * (monom (1/2) 0), monom 1 1) *) theorem putnam_1984_b5: fixes d :: "nat \<Rightarrow> nat" and m :: "nat" assumes mpos : "m > 0" and hd : "\<forall> n::nat. d n = (if n < 2 then n else ((n mod 2::nat) + d (n div 2::nat)))" shows "let (a, f, g) = putnam_1984_b5_solution in (\<Sum> k=0..(2^m-1). (-1::int)^(d k) * k^m) = (-1::int)^m * (a powr (poly f m)) * (fact (poly g m))" sorry end

putnam_2009_a1

abbrev putnam_2009_a1_solution : Prop := sorry -- True theorem putnam_2009_a1 : ((∀ f : (ℝ × ℝ) → ℝ, (∀ O v : ℝ × ℝ, v ≠ (0, 0) → f (O.1, O.2) + f (O.1 + v.1, O.2 + v.2) + f (O.1 + v.1 - v.2, O.2 + v.2 + v.1) + f (O.1 - v.2, O.2 + v.1) = 0) → ∀ P : ℝ × ℝ, f P = 0) ↔ putnam_2009_a1_solution) := sorry

Let $f$ be a real-valued function on the plane such that for every square $ABCD$ in the plane, $f(A)+f(B)+f(C)+f(D)=0$. Does it follow that $f(P)=0$ for all points $P$ in the plane?

Prove that $f$ is identically $0$.

['geometry', 'algebra']

Section putnam_2009_a1. Require Import Reals GeoCoq.Main.Tarski_dev.Ch16_coordinates_with_functions. Context `{T2D:Tarski_2D} `{TE:@Tarski_euclidean Tn TnEQD}. Open Scope R. Definition putnam_2009_a1_solution := True. Theorem putnam_2009_a1 (f: Tpoint -> R) : ((forall (A B C D: Tpoint), Square A B C D -> f A + f B + f C + f D = 0) -> forall (P : Tpoint), f P = 0) <-> putnam_2009_a1_solution. Proof. Admitted. End putnam_2009_a1.

theory putnam_2009_a1 imports Complex_Main begin definition putnam_2009_a1_solution :: bool where "putnam_2009_a1_solution \<equiv> undefined" (* True *) theorem putnam_2009_a1: shows "(\<forall>f::(real\<times>real)\<Rightarrow>real. (\<forall>A v::real\<times>real. v \<noteq> (0, 0) \<longrightarrow> f (fst A, snd A) + f (fst A+fst v, snd A+snd v) + f (fst A+fst v-snd v, snd A+snd v+fst v) + f (fst A-snd v, snd A+fst v) = 0) \<longrightarrow> (\<forall>P::real\<times>real. f P = 0)) \<longleftrightarrow> putnam_2009_a1_solution" sorry end

putnam_2009_a3

abbrev putnam_2009_a3_solution : ℝ := sorry -- 0 theorem putnam_2009_a3 (cos_matrix : (n : ℕ) → Matrix (Fin n) (Fin n) ℝ) (hM : ∀ n : ℕ, ∀ i j : Fin n, (cos_matrix n) i j = Real.cos (1 + n * i + j)) : Tendsto (fun n => (cos_matrix n).det) atTop (𝓝 putnam_2009_a3_solution) := sorry

Let $d_n$ be the determinant of the $n \times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos 1, \cos 2, \dots, \cos n^2$. (For example,\[ d_3 = \left|\begin{matrix} \cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{matrix} \right|. \]The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty} d_n$.

Show that the limit is 0.

['linear_algebra', 'analysis']

theory putnam_2009_a3 imports Complex_Main "HOL-Combinatorics.Permutations" begin definition putnam_2009_a3_solution :: real where "putnam_2009_a3_solution \<equiv> undefined" (* 0 *) theorem putnam_2009_a3: fixes cos_matrix :: "nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real)" and ndet :: "nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> real" assumes hM: "\<forall>n::nat. \<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. (cos_matrix n) i j = cos (1 + n * i + j)" defines "ndet \<equiv> (\<lambda>(n::nat)(A::nat\<Rightarrow>nat\<Rightarrow>real). (\<Sum>p\<in>{p'::nat\<Rightarrow>nat. p' permutes {0..(n-1)} \<and> (\<forall>i::nat\<ge>n. p' i = i)}. (sign p * (\<Prod>i::nat=0..(n-1). A i (p i)))))" shows "filterlim (\<lambda>n::nat. ndet n (cos_matrix n)) (nhds putnam_2009_a3_solution) at_top" sorry end

putnam_2009_a4

abbrev putnam_2009_a4_solution : Prop := sorry -- False theorem putnam_2009_a4 : ((∀ S : Set ℚ, 0 ∈ S → (∀ x ∈ S, x + 1 ∈ S ∧ x - 1 ∈ S) → (∀ x ∈ S, x ∉ ({0, 1} : Set ℚ) → 1 / (x * (x - 1)) ∈ S) → ∀ r : ℚ, r ∈ S) ↔ putnam_2009_a4_solution) := sorry

Let $S$ be a set of rational numbers such that \begin{enumerate} \item[(a)] $0 \in S$; \item[(b)] If $x \in S$ then $x+1\in S$ and $x-1\in S$; and \item[(c)] If $x\in S$ and $x\not\in\{0,1\}$, then $\frac{1}{x(x-1)}\in S$. \end{enumerate} Must $S$ contain all rational numbers?

Prove that $S$ need not contain all rationals.

['number_theory']

Section putnam_2009_a4. Require Import Ensembles QArith. Definition putnam_2009_a4_solution := False. Theorem putnam_2009_a4: forall (E: Ensemble Q), (forall (q: Q), E q <-> q = 0 /\ E q -> E (q + 1) /\ E (q - 1) /\ E q /\ q <> 0 /\ q <> 1 -> E (1 / (q * (q - 1)))) -> forall (q: Q), E q <-> putnam_2009_a4_solution. Proof. Admitted. End putnam_2009_a4.

theory putnam_2009_a4 imports Complex_Main begin definition putnam_2009_a4_solution :: bool where "putnam_2009_a4_solution \<equiv> undefined" (* False *) theorem putnam_2009_a4: shows "(\<forall>S::rat set. 0 \<in> S \<longrightarrow> (\<forall>x\<in>S. x+1 \<in> S \<and> x-1 \<in> S) \<longrightarrow> (\<forall>x\<in>S. x \<notin> {0, 1} \<longrightarrow> 1 / (x*(x-1)) \<in> S) \<longrightarrow> (\<forall>r::rat. r \<in> S)) \<longleftrightarrow> putnam_2009_a4_solution" sorry end

putnam_2009_a5

abbrev putnam_2009_a5_solution : Prop := sorry -- False theorem putnam_2009_a5 : (∃ (G : Type*) (_ : CommGroup G) (_ : Fintype G), ∏ g : G, orderOf g = 2^2009) ↔ putnam_2009_a5_solution := sorry

Is there a finite abelian group $G$ such that the product of the orders of all its elements is 2^{2009}?

Show that the answer is no such finite abelian group exists.

['abstract_algebra']

theory putnam_2009_a5 imports Complex_Main "HOL-Algebra.Multiplicative_Group" begin definition putnam_2009_a5_solution :: bool where "putnam_2009_a5_solution \<equiv> undefined" (* False *) theorem putnam_2009_a5: assumes pacount: "\<exists>pamap::'a\<Rightarrow>nat. surj pamap" shows "(\<exists>G::'a monoid. finite (carrier G) \<and> comm_group G \<and> (\<Prod>g\<in>(carrier G). (group.ord G) g) = 2^2009) \<longleftrightarrow> putnam_2009_a5_solution" sorry end

putnam_2009_b1

theorem putnam_2009_b1 (isquotprodprimefact : ℚ → Prop := fun q => (∃ (k m : ℕ) (a : Fin k → ℕ) (b : Fin m → ℕ), (∀ i : Fin k, Nat.Prime (a i)) ∧ (∀ j : Fin m, Nat.Prime (b j)) ∧ (q = (∏ i : Fin k, Nat.factorial (a i))/(∏ j : Fin m, Nat.factorial (b j))) )) : ∀ q : ℚ, q > 0 → isquotprodprimefact q := sorry

Show that every positive rational number can be written as a quotient of products of factorails of (not necessarily distinct) primes. For example, 10/9 = (2! * 5!)/(3! * 3! * 3!).

['number_theory']

Section putnam_2009_b1. Require Import List QArith Znumtheory Reals. Open Scope Q. Theorem putnam_2009_b1: let fix factl (l : list nat) : list nat := match l with | nil => nil | h :: t => fact h :: t end in forall (q: Q), q > 0 -> exists (n d: list nat), (forall x, (In x n \/ In x d)-> prime (Z.of_nat x)) /\ inject_Z (Z.of_nat (fold_left Nat.mul (factl n) 1%nat)) / inject_Z (Z.of_nat (fold_left Nat.mul (factl d) 1%nat)) = q. Proof. Admitted. End putnam_2009_b1.

theory putnam_2009_b1 imports Complex_Main "HOL-Computational_Algebra.Primes" begin (* uses (nat \<Rightarrow> nat) instead of (Fin k \<Rightarrow> nat) and (Fin m \<Rightarrow> nat) *) theorem putnam_2009_b1: fixes isquotprodprimefact :: "rat \<Rightarrow> bool" defines "isquotprodprimefact \<equiv> (\<lambda>q::rat. (\<exists>(k::nat)(m::nat)(a::nat\<Rightarrow>nat)(b::nat\<Rightarrow>nat). (\<forall>i::nat\<in>{0..(k-1)}. prime (a i)) \<and> (\<forall>j::nat\<in>{0..(m-1)}. prime (b j)) \<and> q = (\<Prod>i::nat=0..(k-1). fact (a i)) / (\<Prod>j::nat=0..(m-1). fact (b j))))" shows "\<forall>q::rat. (q > 0 \<longrightarrow> isquotprodprimefact q)" sorry end

putnam_2009_b3

abbrev putnam_2009_b3_solution : Set ℤ := sorry -- {n : ℤ | ∃ k ≥ 1, n = 2 ^ k - 1} theorem putnam_2009_b3 (mediocre : ℤ → Set ℤ → Prop := fun n S ↦ (S ⊆ Icc 1 n) ∧ ∀ a ∈ S, ∀ b ∈ S, 2 ∣ a + b → (a + b) / 2 ∈ S) (A : ℤ → ℤ := fun n ↦ {S : Set ℤ | mediocre n S}.ncard) : ({n : ℤ | n > 0 ∧ A (n + 2) - 2 * A (n + 1) + A n = 1} = putnam_2009_b3_solution) := sorry

Call a subset $S$ of $\{1, 2, \dots, n\}$ \emph{mediocre} if it has the following property: Whenever $a$ and $b$ are elements of $S$ whose average is an integer, that average is also an element of $S$. Let $A(n)$ be the number of mediocre subsets of $\{1,2,\dots,n\}$. [For instance, every subset of $\{1,2,3\}$ except $\{1,3\}$ is mediocre, so $A(3) = 7$.] Find all positive integers $n$ such that $A(n+2) - 2A(n+1) + A(n) = 1$.

Show that the answer is $n = 2^k - 1$ for some integer $k$.

['number_theory']

theory putnam_2009_b3 imports Complex_Main begin definition putnam_2009_b3_solution :: "nat set" where "putnam_2009_b3_solution \<equiv> undefined" (* {n::nat. (\<exists>k::nat\<ge>1. n = 2^k - 1)} *) theorem putnam_2009_b3: fixes mediocre :: "nat \<Rightarrow> nat set \<Rightarrow> bool" and A :: "nat \<Rightarrow> int" defines "mediocre \<equiv> (\<lambda>(n::nat)(S::nat set). S \<subseteq> {1..n} \<and> (\<forall>a\<in>S. \<forall>b\<in>S. 2 dvd (a + b) \<longrightarrow> (nat (round ((a + b)/2))) \<in> S))" and "A \<equiv> (\<lambda>n::nat. card {S::nat set. mediocre n S})" shows "{n::nat. n > 0 \<and> A (n+2) - 2*A (n+1) + A n = 1} = putnam_2009_b3_solution" sorry end

putnam_2009_b4

abbrev putnam_2009_b4_solution : ℕ := sorry -- 2020050 theorem putnam_2009_b4 (balanced : MvPolynomial (Fin 2) ℝ → Prop := fun P ↦ ∀ r > 0, (∫ x in Metric.sphere 0 r, MvPolynomial.eval x P) / (2 * Real.pi * r) = 0) (V : Set (MvPolynomial (Fin 2) ℝ)) [AddCommGroup V] [Module ℝ V] (hV : ∀ P : MvPolynomial (Fin 2) ℝ, P ∈ V ↔ balanced P ∧ P.totalDegree ≤ 2009) : (Module.rank V = putnam_2009_b4_solution) := sorry

Say that a polynomial with real coefficients in two variables, $x,y$, is \emph{balanced} if the average value of the polynomial on each circle centered at the origin is $0$. The balanced polynomials of degree at most $2009$ form a vector space $V$ over $\mathbb{R}$. Find the dimension of $V$.

Prove that the dimension of $V$ is $2020050$.

['algebra', 'linear_algebra']

theory putnam_2009_b4 imports Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Analysis.Set_Integral" "HOL-Analysis.Lebesgue_Measure" begin definition putnam_2009_b4_solution :: nat where "putnam_2009_b4_solution \<equiv> undefined" (* 2020050 *) definition mvpolyscale :: "real \<Rightarrow> (real poly poly) \<Rightarrow> (real poly poly)" where "mvpolyscale \<equiv> (\<lambda>(c::real)(P::real poly poly). smult (monom c 0) P)" interpretation mvpolyspace: vector_space "mvpolyscale" sorry theorem putnam_2009_b4: fixes balanced :: "(real poly poly) \<Rightarrow> bool" and V :: "(real poly poly) set" defines "balanced \<equiv> (\<lambda>P::real poly poly. (\<forall>r::real>0. (set_lebesgue_integral lebesgue (sphere 0 r) (\<lambda>x::real^2. poly (poly P (monom (x$2) 0)) (x$1))) / (2*pi*r) = 0))" assumes hV: "\<forall>P::real poly poly. (P \<in> V \<longleftrightarrow> (balanced P \<and> degree P \<le> 2009 \<and> (\<forall>i::nat. degree (coeff P i) \<le> 2009)))" shows "mvpolyspace.dim V = putnam_2009_b4_solution" sorry end

putnam_2009_b5

theorem putnam_2009_b5 (f : ℝ → ℝ) (hfdiff : DifferentiableOn ℝ f (Ioi 1)) (hf : ∀ x > 1, deriv f x = (x ^ 2 - (f x) ^ 2) / ((x ^ 2) * ((f x) ^ 2 + 1))) : (Tendsto f ⊤ ⊤) := sorry

Let $f: (1, \infty) \to \mathbb{R}$ be a differentiable function such that \[ f'(x) = \frac{x^2 - f(x)^2}{x^2 (f(x)^2 + 1)} \qquad \mbox{for all $x>1$.} \] Prove that $\lim_{x \to \infty} f(x) = \infty$.

['analysis']

Section putnam_2009_b5. Require Import Reals Coquelicot.Coquelicot. Theorem putnam_2009_b5: forall (f: R -> R) (x: R), (1 < x /\ ex_derive f x /\ Derive f x = (x ^ 2 - (f x) ^ 2) / (x ^ 2 * ((f x) ^ 2 + 1))) -> ~ ex_lim_seq (fun n => f (INR n)). Proof. Admitted. End putnam_2009_b5.

theory putnam_2009_b5 imports Complex_Main "HOL-Analysis.Derivative" begin (* uses (real \<Rightarrow> real) instead of ({1<..} \<Rightarrow> real) *) theorem putnam_2009_b5: fixes f :: "real \<Rightarrow> real" assumes hfdiff: "f differentiable_on {1<..}" and f: "\<forall>x::real>1. deriv f x = (x^2 - (f x)^2) / (x^2 * ((f x)^2 + 1))" shows "filterlim f at_top at_top" sorry end

putnam_2009_b6

theorem putnam_2009_b6 (n : ℕ) (npos : n > 0) : (∃ a : ℕ → ℤ, a 0 = 0 ∧ a 2009 = n ∧ ∀ i : Icc 1 2009, ((∃ j k : ℕ, j < i ∧ a i = a j + 2 ^ k) ∨ ∃ b c : ℕ, b < i ∧ c < i ∧ a b > 0 ∧ a c > 0 ∧ a i = (a b) % (a c))) := sorry

Prove that for every positive integer $n$, there is a sequence of integers $a_0, a_1, \dots, a_{2009}$ with $a_0 = 0$ and $a_{2009} = n$ such that each term after $a_0$ is either an earlier term plus $2^k$ for some nonnegative integer $k$, or of the form $b\,\mathrm{mod}\,c$ for some earlier positive terms $b$ and $c$. [Here $b\,\mathrm{mod}\,c$ denotes the remainder when $b$ is divided by $c$, so $0 \leq (b\,\mathrm{mod}\,c) < c$.]

['number_theory']

Section putnam_2009_b6. Require Import List ZArith Coquelicot.Coquelicot. Open Scope Z. Theorem putnam_2009_b6: forall (n: Z), n > 0 -> exists (a: list Z), length a = 2009%nat /\ nth 0 a 0 = 0 /\ nth 2008 a 0 = n /\ forall (i: nat), and (le 1 i) (lt i 2009) -> exists (j: nat), and (le 0 j) (lt j i) /\ ((exists (k: Z), k > 0 /\ nth i a 0 = nth j a 0 + 2 ^ k) \/ exists (b c: Z), b > 0 /\ c > 0 /\ nth i a 0 = b mod c). Proof. Admitted. End putnam_2009_b6.

theory putnam_2009_b6 imports Complex_Main begin (* uses (nat \<Rightarrow> int) instead of ({0..2009} \<Rightarrow> int) *) theorem putnam_2009_b6: fixes n :: nat assumes npos: "n > 0" shows "\<exists>a::nat\<Rightarrow>int. a 0 = 0 \<and> a 2009 = n \<and> (\<forall>i::nat\<in>{1..2009}. (\<exists>j::nat<i. \<exists>k::nat. a i = a j + 2^k) \<or> (\<exists>b::nat<i. \<exists>c::nat<i. a b > 0 \<and> a c > 0 \<and> a i = (a b) mod (a c)))" sorry end

putnam_2020_a1

abbrev putnam_2020_a1_solution : ℕ := sorry -- 508536 theorem putnam_2020_a1 : Set.ncard {x : ℕ | (2020 ∣ x) ∧ (Nat.log 10 x) + 1 ≤ 2020 ∧ (∃ k l, k ≥ l ∧ x = ∑ i in Finset.range (k-l+1), 10 ^ (i+l))} = putnam_2020_a1_solution := sorry

Find the number of positive integers $N$ satisfying: (i) $N$ is divisible by $2020$, (ii) $N$ has at most $2020$ decimal digits, (iii) The decimal digits of $N$ are a string of consecutive ones followed by a string of consecutive zeros.

Show that the solution is $508536$.

['number_theory', 'algebra']

Section putnam_2020_a1. Require Import Ensembles Finite_sets Rdefinitions Reals Rpower. From mathcomp Require Import bigop div fintype ssralg ssrnat ssrnum. Definition putnam_2020_a1_solution := 508536. Theorem putnam_2020_a1: exists (A: Ensemble nat), forall (n: nat), ( (2020 %| n = true /\ (Rle (Rlog (INR 10) (INR n) + R1) (INR 2020)) /\ exists (k l: nat), k >= l = true /\ n = \sum_(i < k-l+1) 10^(i+l)) <-> A n ) -> cardinal nat A putnam_2020_a1_solution. Proof. Admitted. End putnam_2020_a1.

theory putnam_2020_a1 imports Complex_Main begin definition putnam_2020_a1_solution::nat where "putnam_2020_a1_solution \<equiv> undefined" (* 508536 *) theorem putnam_2020_a1: fixes S::"nat set" defines "S \<equiv> {x. (2020 dvd x) \<and> \<lfloor>log 10 x\<rfloor> + 1 \<le> 2020 \<and> (\<exists>k l::nat. k \<ge> l \<and> x = (\<Sum>i=0..(k-l). 10^(i+l)))}" shows "card S = putnam_2020_a1_solution" sorry end

putnam_2020_a2

abbrev putnam_2020_a2_solution : ℕ → ℕ := sorry -- fun k ↦ 4 ^ k theorem putnam_2020_a2 (k : ℕ) : (∑ j in Finset.Icc 0 k, 2 ^ (k - j) * Nat.choose (k + j) j = putnam_2020_a2_solution k) := sorry

Let $k$ be a nonnegative integer. Evaluate \[ \sum_{j=0}^k 2^{k-j} \binom{k+j}{j}. \]

Show that the answer is $4^k$.

['algebra']

Section putnam_2020_a2. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2020_a2_solution := fun k => 4 ^ k. Theorem putnam_2020_a2 : (fun k => sum_n (fun j => 2 ^ (k - j) * Binomial.C (k + j) j) (k + 1)) = putnam_2020_a2_solution. Proof. Admitted. End putnam_2020_a2.

theory putnam_2020_a2 imports Complex_Main begin definition putnam_2020_a2_solution :: "nat \<Rightarrow> nat" where "putnam_2020_a2_solution \<equiv> undefined" (* \<lambda> k. 4 ^ k *) theorem putnam_2020_a2: fixes k :: nat shows "(\<Sum> j = 0..k. 2 ^ (k - j) * (k + j choose j)) = putnam_2020_a2_solution k" sorry end

putnam_2020_a3

abbrev putnam_2020_a3_solution : Prop := sorry -- False theorem putnam_2020_a3 (a : ℕ → ℝ) (ha0 : a 0 = Real.pi / 2) (ha : ∀ n : ℕ, n ≥ 1 → a n = Real.sin (a (n - 1))) : (∃ L : ℝ, Tendsto (fun m : ℕ => ∑ n : Icc 1 m, (a n)^2) atTop (𝓝 L)) ↔ putnam_2020_a3_solution := sorry

Let $a_0 = \pi/2$, and let $a_n = \sin(a_{n-1})$ for $n \geq 1$. Determine whether \[ \sum_{n=1}^\infty a_n^2 \] converges.

The series diverges.

['analysis']

Section putnam_2020_a3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2020_a3_solution := False. Theorem putnam_2020_a3 (a : nat -> R := fix a (n: nat) := match n with | O => PI / 2 | S n' => sin (a n') end) : ex_lim_seq (fun n => (a n) ^ 2) <-> putnam_2020_a3_solution. Proof. Admitted. End putnam_2020_a3.

theory putnam_2020_a3 imports Complex_Main begin definition putnam_2020_a3_solution :: bool where "putnam_2020_a3_solution \<equiv> undefined" (* False *) theorem putnam_2020_a3: fixes a :: "nat \<Rightarrow> real" assumes ha0: "a 0 = pi/2" and ha: "\<forall>n::nat\<ge>1. a n = sin (a (n-1))" shows "(\<exists>L::real. filterlim (\<lambda>m::nat. (\<Sum>n::nat=1..m. (a n)^2)) (nhds L) at_top) \<longleftrightarrow> putnam_2020_a3_solution" sorry end

putnam_2020_a5

abbrev putnam_2020_a5_solution : ℤ := sorry -- (Nat.fib 4040) - 1 theorem putnam_2020_a5 (a : ℤ → ℕ := fun n : ℤ => {S : Finset ℕ | (∀ k ∈ S, k > 0) ∧ ∑ k : S, Nat.fib k = n}.ncard) : a putnam_2020_a5_solution = 2020 ∧ ∀ n : ℤ, a n = 2020 → n ≤ putnam_2020_a5_solution := sorry

Let $a_n$ be the number of sets $S$ of positive integers for which \[ \sum_{k \in S} F_k = n, \] where the Fibonacci sequence $(F_k)_{k \geq 1}$ satisfies $F_{k+2} = F_{k+1} + F_k$ and begins $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$. Find the largest integer $n$ such that $a_n = 2020$.

The answer is $n=F_{4040}-1$.

['number_theory', 'combinatorics']

theory putnam_2020_a5 imports Complex_Main "HOL-Number_Theory.Fib" begin definition putnam_2020_a5_solution :: int where "putnam_2020_a5_solution \<equiv> undefined" (* (fib 4040) - 1 *) theorem putnam_2020_a5: fixes a :: "int \<Rightarrow> nat" defines "a \<equiv> (\<lambda>n::int. card {S::nat set. finite S \<and> (\<forall>k\<in>S. k > 0) \<and> (\<Sum>k\<in>S. fib k) = n})" shows "(GREATEST n::int. a n = 2020) = putnam_2020_a5_solution" sorry end

putnam_2020_a6

abbrev putnam_2020_a6_solution : ℝ := sorry -- Real.pi / 4 theorem putnam_2020_a6 (f : ℤ → (ℝ → ℝ) := fun N : ℤ => fun x : ℝ => ∑ n in Finset.Icc 0 N, (N + 1/2 - n)/((N + 1)*(2*n + 1)) * Real.sin ((2*n + 1)*x)) : (∀ N > 0, ∀ x : ℝ, f N x ≤ putnam_2020_a6_solution) ∧ ∀ M : ℝ, (∀ N > 0, ∀ x : ℝ, f N x ≤ M) → M ≥ putnam_2020_a6_solution := sorry

For a positive integer $N$, let $f_N$ be the function defined by \[ f_N(x) = \sum_{n=0}^N \frac{N+1/2-n}{(N+1)(2n+1)} \sin((2n+1)x). \] Determine the smallest constant $M$ such that $f_N(x) \leq M$ for all $N$ and all real $x$.

The smallest constant $M$ is $\pi/4$.

['algebra']

Section putnam_2020_a6. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2020_a6_solution := PI / 4. Theorem putnam_2020_a6 (f : Z -> (R -> R) := fun Nz : Z => fun x : R => sum_n (fun n => let N := IZR Nz in let n := INR n in (N + 1 / 2 - n) / ((N + 1) * (2 * n + 1)) * sin ((2 * n + 1) * x)) (Z.to_nat Nz + 1)) (M : R) (hM : forall (N: Z), Z.gt N 0 -> forall (x: R), f N x <= M) (hMlb : forall (n: R), (forall (N: Z), Z.gt N 0 -> forall (x: R), f N x <= n) -> n >= M) : M = putnam_2020_a6_solution. Proof. Admitted. End putnam_2020_a6.

theory putnam_2020_a6 imports Complex_Main begin definition putnam_2020_a6_solution :: real where "putnam_2020_a6_solution \<equiv> undefined" (* pi/4 *) theorem putnam_2020_a6: fixes f :: "int \<Rightarrow> (real \<Rightarrow> real)" assumes "f \<equiv> (\<lambda>N::int. (\<lambda>x::real. (\<Sum>n::int=0..N. (N + 1/2 - n) / ((N + 1)*(2*n + 1)) * sin ((2*n + 1)*x))))" shows "(LEAST M::real. (\<forall>N::int>0. \<forall>x::real. f N x \<le> M)) = putnam_2020_a6_solution" sorry end

putnam_2020_b1

abbrev putnam_2020_b1_solution : ℕ := sorry -- 1990 theorem putnam_2020_b1 (d : ℕ → ℕ := fun n : ℕ => ∑ i : Fin (Nat.digits 2 n).length, (Nat.digits 2 n)[i]!) (S : ℤ := ∑ k : Icc 1 2020, ((-1)^(d k))*(k : ℕ)^3) : S % 2020 = putnam_2020_b1_solution := sorry

For a positive integer $n$, define $d(n)$ to be the sum of the digits of $n$ when written in binary (for example, $d(13) = 1+1+0+1=3)$. Let \[ S = \sum_{k=1}^{2020} (-1)^{d(k)} k^3. \] Determine $S$ modulo 2020.

The answer is $1990$.

['algebra']

Section putnam_2020_b1. Require Import ZArith Reals Coquelicot.Coquelicot. Open Scope Z. Definition putnam_2020_b1_solution := 1990. Theorem putnam_2020_b1 (d : positive -> Z := fix d (n : positive) := match n with | xH => 1 | xO n' => d n'%positive | xI n' => 1 + d n'%positive end) (A := sum_n (fun k => IZR ((-1) ^ (d (Pos.of_nat (S k))) * (Z.of_nat k) ^ 3)) 2020) : (floor A) mod 2020 = putnam_2020_b1_solution. Proof. Admitted. End putnam_2020_b1.

theory putnam_2020_b1 imports Complex_Main begin definition putnam_2020_b1_solution :: nat where "putnam_2020_b1_solution \<equiv> undefined" (* 1990 *) theorem putnam_2020_b1: fixes d :: "int \<Rightarrow> nat" and S :: "int" assumes "d \<equiv> (\<lambda>n::int. if n = 0 then 0 else ((n mod 2) + d \<lfloor>n / 2\<rfloor>))" and "S \<equiv> \<Sum>k::int=1..2020. (-1)^(d k) * k^3" shows "S mod 2020 = putnam_2020_b1_solution" sorry end

putnam_2020_b4

abbrev putnam_2020_b4_solution : ℝ := sorry -- 1 / 4040 theorem putnam_2020_b4 (V : ℕ → Set (ℕ → ℤ) := fun n ↦ ({s : ℕ → ℤ | s 0 = 0 ∧ (∀ j ≥ 2 * n, s j = 0) ∧ (∀ j ∈ Icc 1 (2 * n), |s j - s (j - 1)| = 1)})) (q : ℕ → (ℕ → ℤ) → ℝ := fun n s ↦ 1 + ∑ j in Finset.Icc 1 (2 * n - 1), 3 ^ (s j)) (M : ℕ → ℝ := fun n ↦ (∑' v : V n, 1 / (q n v)) / (V n).ncard) : (M 2020 = putnam_2020_b4_solution) := sorry

Let $n$ be a positive integer, and let $V_n$ be the set of integer $(2n+1)$-tuples $\mathbf{v} = (s_0, s_1, \cdots, s_{2n-1}, s_{2n})$ for which $s_0 = s_{2n} = 0$ and $|s_j - s_{j-1}| = 1$ for $j=1,2,\cdots,2n$. Define \[ q(\mathbf{v}) = 1 + \sum_{j=1}^{2n-1} 3^{s_j}, \] and let $M(n)$ be the average of $\frac{1}{q(\mathbf{v})}$ over all $\mathbf{v} \in V_n$. Evaluate $M(2020)$.

Show that the answer is $\frac{1}{4040}$.

['algebra']

theory putnam_2020_b4 imports Complex_Main begin definition putnam_2020_b4_solution :: real where "putnam_2020_b4_solution \<equiv> undefined" (* 1 / 4040 *) theorem putnam_2020_b4: fixes V :: "nat \<Rightarrow> (nat \<Rightarrow> int) set" and q :: "nat \<Rightarrow> (nat \<Rightarrow> int) \<Rightarrow> real" and M :: "nat \<Rightarrow> real" defines "V \<equiv> \<lambda> n. {s. s 0 = 0 \<and> (\<forall> j \<ge> 2 * n. s j = 0) \<and> (\<forall> j \<in> {1 .. (2 * n)}. \<bar>s j - s (j - 1)\<bar> = 1)}" and "q \<equiv> \<lambda> n s. 1 + (\<Sum> j = 1 .. 2 * n - 1. 3 powr (s j))" and "M \<equiv> \<lambda> n. (\<Sum> v \<in> V n. 1 / (q n v)) / card (V n)" shows "M 2020 = putnam_2020_b4_solution" sorry end

putnam_2020_b5

theorem putnam_2020_b5 (z : Fin 4 → ℂ) (hzle1 : ∀ n, ‖z n‖ < 1) (hzne1 : ∀ n, z n ≠ 1) : 3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) ≠ 0:= sorry

For $j \in \{1, 2, 3, 4\}$, let $z_j$ be a complex number with $|z_j| = 1$ and $z_j \neq 1$. Prove that \[ 3 - z_1 - z_2 - z_3 - z_4 + z_1 z_2 z_3 z_4 \neq 0. \]

['algebra']

theory putnam_2020_b5 imports Complex_Main begin (* Note: Boosted domain to infinite set *) theorem putnam_2020_b5: fixes z :: "nat \<Rightarrow> complex" assumes hz: "\<forall>n \<in> {1..4}. norm (z n) = 1" and hzne1: "\<forall>n \<in> {1..4}. z n \<noteq> 1" shows "3 - z 1 - z 2 - z 3 - z 4 + z 1 * z 2 * z 3 * z 4 \<noteq> 0" sorry end

putnam_2020_b6

theorem putnam_2020_b6 (n : ℕ) (npos : n > 0) : ∑ k : Fin n, ((-1) ^ Int.floor ((k.1 + 1) * (Real.sqrt 2 - 1)) : ℝ) ≥ 0 := sorry

Let $n$ be a positive integer. Prove that $\sum_{k=1}^n(-1)^{\lfloor k(\sqrt{2}-1) \rfloor} \geq 0$.

['algebra']

Section putnam_2020_b6. Require Import Reals. From Coquelicot Require Import Coquelicot Hierarchy Rcomplements. Local Open Scope R. Theorem putnam_2020_b6: let A (k: nat) := (-1)^(Z.to_nat (floor (INR k * (sqrt 2 - 1)))) in let B (n: nat) := sum_n A n in forall (n: nat), B n >= 0. Proof. Admitted. End putnam_2020_b6.

theory putnam_2020_b6 imports Complex_Main begin theorem putnam_2020_b6: fixes n :: nat assumes npos: "n > 0" shows "(\<Sum>k::nat=1..n. (-1) ^ (nat \<lfloor>k * (sqrt 2 - 1)\<rfloor>)) \<ge> 0" sorry end

putnam_1970_a1

theorem putnam_1970_a1 (a b : ℝ) (ha : a > 0) (hb : b > 0) (f : ℝ → ℝ := fun x : ℝ => Real.exp (a*x) * Real.cos (b*x)) (p : ℕ → ℝ) (hp : ∃ c : ℝ, c > 0 ∧ ∀ x ∈ ball 0 c, ∑' n : ℕ, (p n)*x^n = f x) (S : Set ℕ := {n : ℕ | p n = 0}) : S = ∅ ∨ ¬Finite S := sorry

Prove that, for all $a > 0$ and $b > 0$, the power series of $e^{ax} \cos (bx)$ with respect to $x$ has either zero or infinitely many zero coefficients.

['analysis']

theory putnam_1970_a1 imports Complex_Main "HOL-Analysis.Elementary_Metric_Spaces" begin theorem putnam_1970_a1: fixes a b :: real and f :: "real \<Rightarrow> real" and p :: "nat \<Rightarrow> real" and S :: "nat set" defines "f \<equiv> (\<lambda>x::real. exp (a*x) * cos (b*x))" assumes hp: "\<exists>a::real>0. (\<forall>x::real\<in>(ball 0 a). (\<Sum>n::nat. (p n)*x^n) = f x)" defines "S \<equiv> {n::nat. p n = 0}" shows "S = {} \<or> infinite S" sorry end

putnam_1970_a3

abbrev putnam_1970_a3_solution : ℕ × ℕ := sorry -- (3, 1444) theorem putnam_1970_a3 (L : ℕ → ℕ) (hL : ∀ n : ℕ, L n ≤ (Nat.digits 10 n).length ∧ (∀ k : ℕ, k < L n → (Nat.digits 10 n)[k]! = (Nat.digits 10 n)[0]!) ∧ (L n ≠ (Nat.digits 10 n).length → (Nat.digits 10 n)[L n]! ≠ (Nat.digits 10 n)[0]!)) : (∃ n : ℕ, (Nat.digits 10 (n^2))[0]! ≠ 0 ∧ L (n^2) = putnam_1970_a3_solution.1) ∧ (∀ n : ℕ, (Nat.digits 10 (n^2))[0]! ≠ 0 → L (n^2) ≤ putnam_1970_a3_solution.1) ∧ (∃ m : ℕ, m^2 = putnam_1970_a3_solution.2) ∧ L (putnam_1970_a3_solution.2) = putnam_1970_a3_solution.1 ∧ (Nat.digits 10 putnam_1970_a3_solution.2)[0]! ≠ 0 ∧ ∀ n : ℕ, (Nat.digits 10 (n^2))[0]! ≠ 0 ∧ L (n^2) = putnam_1970_a3_solution.1 → n^2 ≥ putnam_1970_a3_solution.2 := sorry

Find the length of the longest possible sequence of equal nonzero digits (in base 10) in which a perfect square can terminate. Also, find the smallest square that attains this length.

The maximum attainable length is $3$; the smallest such square is $38^2 = 1444$.

['number_theory']

theory putnam_1970_a3 imports Complex_Main begin fun digits :: "nat \<Rightarrow> (nat list)" where "digits n = (if n < 10 then [n] else ([n mod 10::nat] @ digits (n div 10::nat)))" definition putnam_1970_a3_solution :: "nat \<times> nat" where "putnam_1970_a3_solution \<equiv> undefined" (* (3, 1444) *) theorem putnam_1970_a3: fixes L :: "nat \<Rightarrow> nat" assumes hL: "\<forall>n::nat. L n \<le> length (digits n) \<and> (\<forall>k::nat<(L n). (digits n)!k = (digits n)!0) \<and> (L n \<noteq> length (digits n) \<longrightarrow> (digits n)!(L n) \<noteq> (digits n)!0)" shows "(GREATEST d::nat. (\<exists>n::nat. (digits (n^2))!0 \<noteq> 0 \<and> d = L (n^2))) = fst putnam_1970_a3_solution \<and> (LEAST m::nat. (\<exists>n::nat. n^2 = m) \<and> (digits m)!0 \<noteq> 0 \<and> L m = fst putnam_1970_a3_solution) = snd putnam_1970_a3_solution" sorry end

putnam_1970_a4

theorem putnam_1970_a4 (x : ℕ → ℝ) (hxlim : Tendsto (fun n => x n - x (n-2)) atTop (𝓝 0)) : Tendsto (fun n => (x n - x (n-1))/n) atTop (𝓝 0) := sorry

Suppose $(x_n)$ is a sequence such that $\lim_{n \to \infty} (x_n - x_{n-2} = 0$. Prove that $\lim_{n \to \infty} \frac{x_n - x_{n-1}}{n} = 0$.

['analysis']

theory putnam_1970_a4 imports Complex_Main begin theorem putnam_1970_a4: fixes x :: "nat \<Rightarrow> real" assumes hxlim: "filterlim (\<lambda>n::nat. x n - x (n-2)) (nhds 0) at_top" shows "filterlim (\<lambda>n::nat. (x n - x (n-1)) / n) (nhds 0) at_top" sorry end

putnam_1970_b1

abbrev putnam_1970_b1_solution : ℝ := sorry -- Real.exp (2 * Real.log 5 - 4 + 2 * Real.arctan 2) theorem putnam_1970_b1 : Tendsto (fun n => 1/(n^4) * ∏ i in Finset.Icc (1 : ℤ) (2*n), ((n^2 + i^2) : ℝ)^((1 : ℝ)/n)) atTop (𝓝 putnam_1970_b1_solution) := sorry

Evaluate the infinite product $\lim_{n \to \infty} \frac{1}{n^4} \prod_{i = 1}^{2n} (n^2 + i^2)^{1/n}$.

Show that the solution is $e^{2 \log(5) - 4 + 2 arctan(2)}$.

['analysis']

theory putnam_1970_b1 imports Complex_Main begin definition putnam_1970_b1_solution :: real where "putnam_1970_b1_solution \<equiv> undefined" (* exp (2*ln 5 - 4 + 2*arctan 2) *) theorem putnam_1970_b1: shows "filterlim (\<lambda>n::nat. 1/(n^4) * (\<Prod>i::nat=1..(2*n). (n^2 + i^2) powr (1/n))) (nhds putnam_1970_b1_solution) at_top" sorry end

putnam_1970_b3

theorem putnam_1970_b3 (S : Set (ℝ × ℝ)) (a b : ℝ) (hab : a < b) (hS : ∀ s ∈ S, s.1 ∈ Ioo a b) (hSclosed : IsClosed S) : IsClosed {y | ∃ x : ℝ, ⟨x,y⟩ ∈ S} := sorry

A closed subset $S$ of $\mathbb{R}^2$ lies in $a < x < b$. Show that its projection on the y-axis is closed.

['analysis']

theory putnam_1970_b3 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin theorem putnam_1970_b3: fixes S :: "(real^2) set" and a b :: real assumes hab: "a < b" and hS: "\<forall>s\<in>S. s$1 \<in> {a<..<b}" and hSclosed: "closed S" shows "closed {y::real. (\<exists>s\<in>S. y = s$2)}" sorry end

putnam_1970_b5

theorem putnam_1970_b5 (ramp : ℤ → (ℝ → ℝ) := fun n => (fun x => if x ≤ -n then -n else (if -n < x ∧ x ≤ n then x else n))) (F : ℝ → ℝ) : Continuous F ↔ (∀ n : ℕ, Continuous ((ramp n) ∘ F)) := sorry

Let $u_n$ denote the function $u_n(x) = -n$ if $x \leq -n$, $x$ if $-n < x \leq n$, and $n$ otherwise. Let $F$ be a function on the reals. Show that $F$ is continuous if and only if $u_n \circ F$ is continuous for all natural numbers $n$.

['analysis']

theory putnam_1970_b5 imports Complex_Main begin theorem putnam_1970_b5: fixes ramp :: "int \<Rightarrow> (real \<Rightarrow> real)" and F :: "real \<Rightarrow> real" defines "ramp \<equiv> (\<lambda>n::int. (\<lambda>x::real. if x \<le> -n then -n else if (-n < x \<and> x \<le> n) then x else n))" shows "continuous_on UNIV F \<longleftrightarrow> (\<forall>n::nat. continuous_on UNIV ((ramp n) \<circ> F))" sorry end

putnam_1986_a1

abbrev putnam_1986_a1_solution : ℝ := sorry -- 18 theorem putnam_1986_a1 (S : Set ℝ := {x : ℝ | x ^ 4 + 36 ≤ 13 * x ^ 2}) (f : ℝ → ℝ := fun x ↦ x ^ 3 - 3 * x) : (∀ x ∈ S, f x ≤ putnam_1986_a1_solution ∧ ∃ x ∈ S, f x = putnam_1986_a1_solution) := sorry

Find, with explanation, the maximum value of $f(x)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36 \leq 13x^2$.

Show that the maximum value is $18$.

['algebra', 'analysis']

Section putnam_1986_a1. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1986_a1_solution := 18. Theorem putnam_1986_a1: let f (x: R) := pow x 3 in let on_S (x: R) := pow x 4 - 13 * pow x 2 + 36 <= 0 in exists (m: R), (forall (x: R), on_S x -> m >= f x) /\ (exists (x: R), on_S x -> m = f x) <-> m = putnam_1986_a1_solution. Proof. Admitted. End putnam_1986_a1.

theory putnam_1986_a1 imports Complex_Main begin definition putnam_1986_a1_solution::real where "putnam_1986_a1_solution \<equiv> undefined" (* 18 *) theorem putnam_1986_a1: fixes f::"real\<Rightarrow>real" defines "f \<equiv> \<lambda>x. x^3 - 3 * x" shows "putnam_1986_a1_solution = (GREATEST y. (\<exists>x. y = f x \<and> x^4 + 36 \<le> 13 * x^2))" sorry end

putnam_1986_a2

abbrev putnam_1986_a2_solution : ℕ := sorry -- 3 theorem putnam_1986_a2 : (Nat.floor ((10 ^ 20000 : ℝ) / (10 ^ 100 + 3)) % 10 = putnam_1986_a2_solution) := sorry

What is the units (i.e., rightmost) digit of \[ \left\lfloor \frac{10^{20000}}{10^{100}+3}\right\rfloor ? \]

Show that the answer is $3$.

['algebra']

Section putnam_1986_a2. Require Import Nat. Definition putnam_1986_a2_solution := 3. Theorem putnam_1986_a2: 10 ^ (20000) / (10 ^ (100) + 3) mod 10 = putnam_1986_a2_solution. Proof. Admitted. End putnam_1986_a2.

theory putnam_1986_a2 imports Complex_Main begin definition putnam_1986_a2_solution::nat where "putnam_1986_a2_solution \<equiv> undefined" (* 3 *) theorem putnam_1986_a2: shows "putnam_1986_a2_solution = \<lfloor>(10^20000) / (10^100 + 3)\<rfloor> mod 10" sorry end

putnam_1986_a3

abbrev putnam_1986_a3_solution : ℝ := sorry -- Real.pi / 2 theorem putnam_1986_a3 (cot : ℝ → ℝ := fun θ ↦ cos θ / sin θ) (arccot : ℝ → ℝ) (harccot : ∀ t : ℝ, t ≥ 0 → arccot t ∈ Set.Ioc 0 (Real.pi / 2) ∧ cot (arccot t) = t) : (∑' n : ℕ, arccot (n ^ 2 + n + 1) = putnam_1986_a3_solution) := sorry

Evaluate $\sum_{n=0}^\infty \mathrm{Arccot}(n^2+n+1)$, where $\mathrm{Arccot}\,t$ for $t \geq 0$ denotes the number $\theta$ in the interval $0 < \theta \leq \pi/2$ with $\cot \theta = t$.

Show that the sum equals $\pi/2$.

['analysis']

Section putnam_1986_a3. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1986_a3_solution := PI / 2. Theorem putnam_1986_a3: Series (fun n => 1/ atan (pow (INR n) 2 + INR n + 1)) = putnam_1986_a3_solution. Proof. Admitted. End putnam_1986_a3.

theory putnam_1986_a3 imports Complex_Main begin definition putnam_1986_a3_solution::real where "putnam_1986_a3_solution \<equiv> undefined" (* pi / 2 *) theorem putnam_1986_a3: fixes arccot::"real\<Rightarrow>real" defines "arccot \<equiv> \<lambda>y. (THE x. 0 < x \<and> x \<le> pi/2 \<and> cot x = y)" shows "(\<Sum>n::nat. arccot (n^2 + n + 1)) = putnam_1986_a3_solution" sorry end

putnam_1986_a4

abbrev putnam_1986_a4_solution : ℚ × ℚ × ℚ × ℚ × ℚ × ℚ × ℚ := sorry -- (1, 4, 2, 3, -4, 2, 1) theorem putnam_1986_a4 (f : ℕ → ℕ := fun n ↦ {A : Matrix (Fin n) (Fin n) ℤ | (∀ i j : Fin n, A i j ∈ ({-1, 0, 1} : Set ℤ)) ∧ ∃ S : ℤ, ∀ ϕ : Perm (Fin n), ∑ i : Fin n, A i (ϕ i) = S}.ncard) : let (a1, b1, a2, b2, a3, b3, a4) := putnam_1986_a4_solution; (∀ n > 0, f n = a1 * b1 ^ n + a2 * b2 ^ n + a3 * b3 ^ n + a4) := sorry

A \emph{transversal} of an $n\times n$ matrix $A$ consists of $n$ entries of $A$, no two in the same row or column. Let $f(n)$ be the number of $n \times n$ matrices $A$ satisfying the following two conditions: \begin{enumerate} \item[(a)] Each entry $\alpha_{i,j}$ of $A$ is in the set $\{-1,0,1\}$. \item[(b)] The sum of the $n$ entries of a transversal is the same for all transversals of $A$. \end{enumerate} An example of such a matrix $A$ is \[ A = \left( \begin{array}{ccc} -1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{array} \right). \] Determine with proof a formula for $f(n)$ of the form \[ f(n) = a_1 b_1^n + a_2 b_2^n + a_3 b_3^n + a_4, \] where the $a_i$'s and $b_i$'s are rational numbers.

Prove that $f(n) = 4^n + 2 \cdot 3^n - 4 \cdot 2^n + 1$.

['linear_algebra']

theory putnam_1986_a4 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Combinatorics.Permutations" begin definition putnam_1986_a4_solution::"rat\<times>rat\<times>rat\<times>rat\<times>rat\<times>rat\<times>rat" where "putnam_1986_a4_solution \<equiv> undefined" (* (1, 4, 2, 3, -4, 2, 1) *) theorem putnam_1986_a4: fixes n::nat and f::nat defines "f \<equiv> card {A::int^'a^'a. CARD('a) = n \<and> (\<forall>i::'a. \<forall>j::'a. (A$i$j) \<in> {-1, 0, 1}) \<and> (\<exists>S::int. \<forall>f::'a\<Rightarrow>'a. f permutes UNIV \<longrightarrow> S = (\<Sum>i::'a \<in> UNIV. A$i$(f i)))}" assumes npos : "n > 0" shows "let (a1, b1, a2, b2, a3, b3, a4) = putnam_1986_a4_solution in (f = a1 * b1^n + a2 * b2^n + a3 * b3^n + a4)" sorry end

putnam_1986_a6

abbrev putnam_1986_a6_solution : (ℕ → ℕ) → ℕ → ℝ := sorry -- fun b n ↦ (∏ i : Finset.Icc 1 n, b i) / Nat.factorial n theorem putnam_1986_a6 (n : ℕ) (npos : n > 0) (a : ℕ → ℝ) (b : ℕ → ℕ) (bpos : ∀ i ∈ Finset.Icc 1 n, b i > 0) (binj : ∀ i ∈ Finset.Icc 1 n, ∀ j ∈ Finset.Icc 1 n, b i = b j → i = j) (f : Polynomial ℝ) (hf : ∀ x : ℝ, (1 - x) ^ n * f.eval x = 1 + ∑ i : Finset.Icc 1 n, (a i) * x ^ (b i)) : (f.eval 1 = putnam_1986_a6_solution b n) := sorry

Let $a_1, a_2, \dots, a_n$ be real numbers, and let $b_1, b_2, \dots, b_n$ be distinct positive integers. Suppose that there is a polynomial $f(x)$ satisfying the identity \[ (1-x)^n f(x) = 1 + \sum_{i=1}^n a_i x^{b_i}. \] Find a simple expression (not involving any sums) for $f(1)$ in terms of $b_1, b_2, \dots, b_n$ and $n$ (but independent of $a_1, a_2, \dots, a_n$).

Show that $f(1) = b_1 b_2 \dots b_n / n!$.

['algebra']

Section putnam_1986_a6. Require Import Reals Factorial Coquelicot.Coquelicot. Definition putnam_1986_a6_solution (m: nat -> R) (n: nat) := let fix prod_n (m: nat -> R) (n : nat) : R := match n with | O => m 0%nat | S n' => m n' * prod_n m n' end in prod_n m n / INR (fact n). Theorem putnam_1986_a6: forall (n: nat), forall (a m: nat -> R) (i j: nat), Nat.lt i j -> 0 < m i < m j -> let p (x: R) := sum_n (fun n => a n * Rpower x (m n)) n in exists (q: R -> R), forall (x: R), p x = (1 - x) ^ n * (q x) -> q 1 = putnam_1986_a6_solution m n. Proof. Admitted. End putnam_1986_a6.

theory putnam_1986_a6 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1986_a6_solution::"(nat\<Rightarrow>nat) \<Rightarrow> nat \<Rightarrow> real" where "putnam_1986_a6_solution \<equiv> undefined" (* \<lambda>b. \<lambda>n. (\<Prod>i=1..n. b i) / fact n *) theorem putnam_1986_a6: fixes n::nat and a::"nat\<Rightarrow>real" and b::"nat\<Rightarrow>nat" and f::"real poly" assumes npos : "n > 0" and bpos : "\<forall>i \<in> {1..n::nat}. b i > 0" and binj : "\<forall>i \<in> {1..n::nat}. \<forall>j \<in> {1..n::nat}. b i = b j \<longrightarrow> i = j" and hf : "\<forall>x::real. (1 - x) ^ n * (poly f x) = 1 + (\<Sum>i=1..n. (a i) * x ^ (b i))" shows "poly f 1 = putnam_1986_a6_solution b n" sorry end

putnam_1986_b2

abbrev putnam_1986_b2_solution : Finset (ℂ × ℂ × ℂ) := sorry -- {(0, 0, 0), (0, -1, 1), (1, 0, -1), (-1, 1, 0)} theorem putnam_1986_b2 : ({T : ℂ × ℂ × ℂ | ∃ x y z : ℂ, T = (x - y, y - z, z - x) ∧ x * (x - 1) + 2 * y * z = y * (y - 1) + 2 * z * x ∧ y * (y - 1) + 2 * z * x = z * (z - 1) + 2 * x * y} = putnam_1986_b2_solution) := sorry

Prove that there are only a finite number of possibilities for the ordered triple $T=(x-y,y-z,z-x)$, where $x,y,z$ are complex numbers satisfying the simultaneous equations \[ x(x-1)+2yz = y(y-1)+2zx = z(z-1)+2xy, \] and list all such triples $T$.

Show that the possibilities for $T$ are $(0, 0, 0), \, (0, -1, 1), \, (1, 0, -1), \, (-1, 1, 0)$.

['algebra']

Section putnam_1986_b2. Require Import Reals Ensembles Finite_sets Coquelicot.Coquelicot. Open Scope C. Definition putnam_1986_b2_solution (xyz : C*C*C) := xyz = (RtoC 0, RtoC 0, RtoC 0) \/ xyz = (RtoC 1, RtoC 0, RtoC (-1)) \/ xyz = (RtoC (-1), RtoC 1, RtoC 0) \/ xyz = (RtoC 0, RtoC (-1), RtoC 1). Theorem putnam_1986_b2: exists (n: nat), forall (E: Ensemble (C*C*C)) (xyz: C*C*C), let x := fst (fst xyz) in let y := snd (fst xyz) in let z := snd xyz in (E (x-y,y-z,z-x) <-> x * (x - 1) * 2 * y * z = y * (y - 1) * 2 * z * x /\ y * (y - 1) * 2 * z * x = z * (z - 1) + 2 * x * y) -> exists (n: nat), cardinal (C*C*C) E n /\ putnam_1986_b2_solution xyz. Proof. Admitted. End putnam_1986_b2.

theory putnam_1986_b2 imports Complex_Main begin definition putnam_1986_b2_solution::"(complex\<times>complex\<times>complex) set" where "putnam_1986_b2_solution \<equiv> undefined" (* {(0, 0, 0), (0, -1, 1), (1, 0, -1), (-1, 1, 0)} *) theorem putnam_1986_b2: shows "putnam_1986_b2_solution = {T. \<exists>x y z::complex. T = (x - y, y - z, z - x) \<and> x*(x-1) + 2*y*z = y*(y-1) + 2*z*x \<and> y*(y-1) + 2*z*x = z*(z-1) + 2*x*y}" sorry end

putnam_1986_b3

theorem putnam_1986_b3 (cong : Polynomial ℤ → Polynomial ℤ → ℤ → Prop := fun f g m ↦ ∀ i : ℕ, m ∣ (f - g).coeff i) (n p : ℕ) (nppos : n > 0 ∧ p > 0) (pprime : Nat.Prime p) (f g h r s : Polynomial ℤ) (hcoprime : cong (r * f + s * g) 1 p) (hprod : cong (f * g) h p) : (∃ F G : Polynomial ℤ, cong F f p ∧ cong G g p ∧ cong (F * G) h (p ^ n)) := sorry

Let $\Gamma$ consist of all polynomials in $x$ with integer coefficients. For $f$ and $g$ in $\Gamma$ and $m$ a positive integer, let $f \equiv g \pmod{m}$ mean that every coefficient of $f-g$ is an integral multiple of $m$. Let $n$ and $p$ be positive integers with $p$ prime. Given that $f,g,h,r$ and $s$ are in $\Gamma$ with $rf+sg\equiv 1 \pmod{p}$ and $fg \equiv h \pmod{p}$, prove that there exist $F$ and $G$ in $\Gamma$ with $F \equiv f \pmod{p}$, $G \equiv g \pmod{p}$, and $FG \equiv h \pmod{p^n}$.

['number_theory', 'algebra']

theory putnam_1986_b3 imports Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Computational_Algebra.Primes" begin theorem putnam_1986_b3: fixes cong::"(int poly) \<Rightarrow> (int poly) \<Rightarrow> int \<Rightarrow> bool" and n p::nat and f g h r s::"int poly" defines "cong \<equiv> \<lambda>f. \<lambda>g. \<lambda>m. \<forall>i::nat. m dvd (coeff (f - g) i)" assumes nppos : "n > 0 \<and> p > 0" and pprime : "prime p" and hcoprime : "cong (r * f + s * g) [:1:] p" and hprod : "cong (f * g) h p" shows "\<exists>F G:: int poly. cong F f p \<and> cong G g p \<and> cong (F * G) h (p ^ n)" sorry end

putnam_1986_b4

abbrev putnam_1986_b4_solution : Prop := sorry -- True theorem putnam_1986_b4 (G : ℝ → ℝ) (hGeq : ∀ r : ℝ, ∃ m n : ℤ, G r = |r - sqrt (m ^ 2 + 2 * n ^ 2)|) (hGlb : ∀ r : ℝ, ∀ m n : ℤ, G r ≤ |r - sqrt (m ^ 2 + 2 * n ^ 2)|) : (Tendsto G ⊤ (𝓝 0) ↔ putnam_1986_b4_solution) := sorry

For a positive real number $r$, let $G(r)$ be the minimum value of $|r - \sqrt{m^2+2n^2}|$ for all integers $m$ and $n$. Prove or disprove the assertion that $\lim_{r\to \infty}G(r)$ exists and equals $0$.

Show that the limit exists and equals $0$.

['analysis']

Section putnam_1986_b4. Require Import Reals Coquelicot.Coquelicot. Definition putnam_1986_b4_solution := True. Theorem putnam_1986_b4 (G : R -> R) (hGeq : forall (r: R), exists (m n: Z), G r = Rabs (r - sqrt (IZR (m ^ 2 + 2 * n ^ 2)))) (hGlb : forall (r: R), forall (m n: Z), G r <= Rabs (r - sqrt (IZR (m ^ 2 + 2 * n ^ 2)))) : Lim_seq (fun n => G (INR n)) = 0 <-> putnam_1986_b4_solution. Proof. Admitted. End putnam_1986_b4.

theory putnam_1986_b4 imports Complex_Main begin definition putnam_1986_b4_solution::bool where "putnam_1986_b4_solution \<equiv> undefined" (* True *) theorem putnam_1986_b4: fixes G::"real\<Rightarrow>real" defines "G \<equiv> \<lambda>r. (LEAST y. \<exists> m n::int. y = \<bar>r - sqrt (m^2 + 2*n^2)\<bar>)" shows "(G \<longlonglongrightarrow> 0) \<longleftrightarrow> putnam_1986_b4_solution" sorry end

putnam_1986_b5

abbrev putnam_1986_b5_solution : Prop := sorry -- False theorem putnam_1986_b5 (f : MvPolynomial (Fin 3) ℝ := (X 0) ^ 2 + (X 1) ^ 2 + (X 2) ^ 2 + (X 0) * (X 1) * (X 2)) (perms : Set (Set (MvPolynomial (Fin 3) ℝ)) := {{X 0, X 1, X 2}, {X 0, -X 1, -X 2}, {-X 0, X 1, -X 2}, {-X 0, -X 1, X 2}}) : ((∀ pqr : Fin 3 → MvPolynomial (Fin 3) ℝ, (∀ xyz : Fin 3 → ℝ, MvPolynomial.eval (fun i ↦ MvPolynomial.eval xyz (pqr i)) f = MvPolynomial.eval xyz f) → ({pqr 0, pqr 1, pqr 2} ∈ perms)) ↔ putnam_1986_b5_solution) := sorry

Let $f(x,y,z) = x^2+y^2+z^2+xyz$. Let $p(x,y,z), q(x,y,z)$, $r(x,y,z)$ be polynomials with real coefficients satisfying \[ f(p(x,y,z), q(x,y,z), r(x,y,z)) = f(x,y,z). \] Prove or disprove the assertion that the sequence $p,q,r$ consists of some permutation of $\pm x, \pm y, \pm z$, where the number of minus signs is $0$ or $2$.

Prove that the assertion is false.

['algebra']

theory putnam_1986_b5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1986_b5_solution::bool where "putnam_1986_b5_solution \<equiv> undefined" (* False *) theorem putnam_1986_b5: fixes f::"real poly poly poly" and perms::"((real poly poly poly) set) set" and eval::"(real poly poly poly) \<Rightarrow>real\<Rightarrow>real\<Rightarrow>real\<Rightarrow>real" defines "f \<equiv> [:[:[: 0, 0, 1:], 0, 1:], [:0, [:0, 1:]:], 1:]" (* x^2 + y^2 + z^2 + xyz = ((0 + 0z + z^2) + 0y + y^2) + (0 + (0 + z)y) x + x^2 *) and "perms \<equiv> { {[:0, 1:], [:[:0, 1:]:], [:[:[:0, 1:]:]:]}, {[:0, 1:], [:[:0, -1:]:], [:[:[:0, -1:]:]:]}, {[:0, -1:], [:[:0, 1:]:], [:[:[:0, -1:]:]:]}, {[:0, -1:], [:[:0, -1:]:], [:[:[:0, 1:]:]:]} }" and "eval \<equiv> \<lambda>P. \<lambda>x. \<lambda>y. \<lambda>z. poly (poly (poly P [:[:z:]:]) [:y:]) x" shows "(\<forall>p q r::real poly poly poly. (\<forall>x y z::real. eval f x y z = eval f (eval p x y z) (eval q x y z) (eval r x y z)) \<longrightarrow> ({p, q, r} \<in> perms)) \<longleftrightarrow> putnam_1986_b5_solution" sorry end

putnam_1986_b6

theorem putnam_1986_b6 (n : ℕ) (npos : n > 0) (F : Type*) [Field F] (A B C D : Matrix (Fin n) (Fin n) F) (hsymm : IsSymm (A * Bᵀ) ∧ IsSymm (C * Dᵀ)) (hid : A * Dᵀ - B * Cᵀ = 1) : (Aᵀ * D - Cᵀ * B = 1) := sorry

Suppose $A,B,C,D$ are $n \times n$ matrices with entries in a field $F$, satisfying the conditions that $AB^T$ and $CD^T$ are symmetric and $AD^T - BC^T = I$. Here $I$ is the $n \times n$ identity matrix, and if $M$ is an $n \times n$ matrix, $M^T$ is its transpose. Prove that $A^T D - C^T B = I$.

['linear_algebra']

theory putnam_1986_b6 imports Complex_Main "HOL-Algebra.Ring" "HOL-Analysis.Finite_Cartesian_Product" begin theorem putnam_1986_b6: fixes n::nat and F::"('a::{semiring_1, minus}, 'm) ring_scheme" (structure) and A B C D::"'a^'b^'b" assumes npos : "n > 0" and Ffield : "field F" and matdim : "CARD('b) = n" and hsymm1 : "A ** transpose B = transpose (A ** transpose B)" and hsymm2 : "C ** transpose D = transpose (C ** transpose D)" and hid : "A ** transpose D - B ** transpose C = 1" shows "transpose A ** D - transpose C ** B = 1" sorry end

putnam_2000_a1

abbrev putnam_2000_a1_solution : ℝ → Set ℝ := sorry -- (fun A : ℝ => Set.Ioo 0 (A ^ 2)) theorem putnam_2000_a1 (A : ℝ) (Apos : A > 0) : ({S : ℝ | ∃ x : ℕ → ℝ, (∀ j : ℕ, x j > 0) ∧ (∑' j : ℕ, x j) = A ∧ (∑' j : ℕ, (x j) ^ 2) = S} = putnam_2000_a1_solution A) := sorry

Let $A$ be a positive real number. What are the possible values of $\sum_{j=0}^\infty x_j^2$, given that $x_0,x_1,\ldots$ are positive numbers for which $\sum_{j=0}^\infty x_j=A$?

Show that the possible values comprise the interval $(0,A^2)$.

['analysis']

Section putnam_2000_a1. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2000_a1_solution (x A: R) := 0 < x < A ^ 2. Theorem putnam_2000_a1: forall (A: R), A > 0 -> forall (x: nat -> R), Series x = A -> putnam_2000_a1_solution (Series (fun j => x j ^ 2)) A. Proof. Admitted. End putnam_2000_a1.

theory putnam_2000_a1 imports Complex_Main begin definition putnam_2000_a1_solution :: "real \<Rightarrow> real set" where "putnam_2000_a1_solution \<equiv> undefined" (* \<lambda> A. {0 <..< A ^ 2} *) theorem putnam_2000_a1: fixes A :: real assumes Apos: "A > 0" shows "{S :: real. \<exists> x :: nat \<Rightarrow> real. (\<forall> j :: nat. x j > 0) \<and> (\<Sum> j :: nat. x j) = A \<and> (\<Sum> j :: nat. (x j) ^ 2) = S} = putnam_2000_a1_solution A" sorry end

putnam_2000_a2

theorem putnam_2000_a2 : ∀ n : ℕ, ∃ N : ℤ, ∃ i : Fin 6 → ℕ, N > n ∧ N = (i 0)^2 + (i 1)^2 ∧ N + 1 = (i 2)^2 + (i 3)^2 ∧ N + 2 = (i 4)^2 + (i 5)^2 := sorry

Prove that there exist infinitely many integers $n$ such that $n,n+1,n+2$ are each the sum of the squares of two integers.

['number_theory']

theory putnam_2000_a2 imports Complex_Main begin definition sum_of_squares:: "int \<Rightarrow> bool" where "sum_of_squares n \<equiv> \<exists>a b::int. n = a^2 + b^2" theorem putnam_2000_a2: shows "\<forall>n :: int. \<exists>m::int. m > n \<and> sum_of_squares m \<and> sum_of_squares (m+1) \<and> sum_of_squares (m+2)" sorry end

putnam_2000_a4

theorem putnam_2000_a4 : ∃ y : ℝ, Tendsto (fun B : ℝ => ∫ x in Set.Ioo 0 B, Real.sin x * Real.sin (x ^ 2)) atTop (𝓝 y) := sorry

Show that the improper integral $\lim_{B \to \infty} \int_0^B \sin(x)\sin(x^2)\,dx$ converges.

['analysis']

Section putnam_2000_a4. Require Import Reals Coquelicot.Coquelicot. Theorem putnam_2000_a4: ex_lim_seq (fun n => sum_n (fun x => sin (INR x) * sin ((INR x) ^ 2)) n). Proof. Admitted. End putnam_2000_a4.

theory putnam_2000_a4 imports Complex_Main "HOL-Analysis.Interval_Integral" begin theorem putnam_2000_a4: shows "\<exists> y :: real. ((\<lambda> B. interval_lebesgue_integral lebesgue 0 B (\<lambda> x. sin x * sin (x ^ 2))) \<longlongrightarrow> y) at_top" sorry end

putnam_2000_a5

theorem putnam_2000_a5 (r : ℝ) (z : Fin 2 → ℝ) (p : Fin 3 → (Fin 2 → ℝ)) (rpos : r > 0) (pdiff : ∀ n m : Fin 3, (n ≠ m) → (p n ≠ p m)) (pint : ∀ (n : Fin 3) (i : Fin 2), p n i = round (p n i)) (pcirc : ∀ n : Fin 3, p n ∈ Metric.sphere z r) : ∃ n m : Fin 3, (n ≠ m) ∧ (dist (p n) (p m) ≥ r ^ ((1 : ℝ) / 3)) := sorry

Three distinct points with integer coordinates lie in the plane on a circle of radius $r>0$. Show that two of these points are separated by a distance of at least $r^{1/3}$.

['algebra']

theory putnam_2000_a5 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin theorem putnam_2000_a5: fixes r :: real and z :: "real^2" and S :: "(real^2) set" assumes rpos: "r > 0" and Scard: "finite S \<and> card S = 3" and pint: "\<forall> p \<in> S. p$1 = round (p$1) \<and> p$2 = round (p$2)" and pcirc: "\<forall> p \<in> S. p \<in> sphere z r" shows "\<exists> p \<in> S. \<exists> q \<in> S. dist p q \<ge> r powr (1 / 3)" sorry end

putnam_2000_a6

theorem putnam_2000_a6 (f : Polynomial ℤ) (a : ℕ → ℤ) (ha0 : a 0 = 0) (ha : ∀ n : ℕ, a (n + 1) = f.eval (a n)) : ((∃ m > 0, a m = 0) → (a 1 = 0 ∨ a 2 = 0)) := sorry

Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0,a_1,\ldots$ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n\geq 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.

['algebra']

theory putnam_2000_a6 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2000_a6: fixes f :: "int poly" and a :: "nat \<Rightarrow> int" assumes ha0: "a 0 = 0" and ha: "\<forall>n::nat. a (n + 1) = poly f (a n)" shows "(\<exists>m::nat>0. a m = 0) \<longrightarrow> (a 1 = 0 \<or> a 2 = 0)" sorry end

putnam_2000_b1

theorem putnam_2000_b1 (N : ℕ) (a b c : Fin N → ℤ) (Nge1 : N ≥ 1) (hodd : ∀ j : Fin N, Odd (a j) ∨ Odd (b j) ∨ Odd (c j)) : (∃ r s t : ℤ, {j : Fin N | Odd (r * a j + s * b j + t * c j)}.ncard ≥ (4 * N : ℝ) / 7) := sorry

Let $a_j,b_j,c_j$ be integers for $1\leq j\leq N$. Assume for each $j$, at least one of $a_j,b_j,c_j$ is odd. Show that there exist integers $r$, $s$, $t$ such that $ra_j+sb_j+tc_j$ is odd for at least $4N/7$ values of $j$, $1\leq j\leq N$.

['algebra']

Section putnam_2000_b1. Require Import List Nat Reals ZArith. Open Scope Z. Theorem putnam_2000_b1: forall (a b c: nat -> Z) (n: nat), (forall (j: nat), and (le 1 j) (le j n) -> Z.odd (a j) =true \/ Z.odd (b j) = true \/ Z.odd (c j) = true) -> exists (l: list nat), ge (length l) (4 * n / 7) /\ forall (j: nat), In j l -> and (le 1 j) (le j n) /\ exists (r s t: Z), Z.odd (Z.add (Z.add (Z.mul r (a j)) (Z.mul s (b j))) (Z.mul t (c j))) = true. Proof. Admitted. End putnam_2000_b1.

theory putnam_2000_b1 imports Complex_Main begin (* uses (nat \<Rightarrow> int) instead of (Fin N \<Rightarrow> int) *) theorem putnam_2000_b1: fixes n :: nat and a b c :: "nat \<Rightarrow> int" assumes Nge1: "N \<ge> 1" and hodd: "\<forall>j::nat\<in>{0..(N-1)}. odd (a j) \<or> odd (b j) \<or> odd (c j)" shows "\<exists>r s t::int. card {j::nat\<in>{0..(N-1)}. odd (r * a j + s * b j + t * c j)} \<ge> 4*N/7" sorry end

putnam_2000_b2

theorem putnam_2000_b2 : (∀ m n : ℕ, m ≥ 1 → n ≥ m → n ∣ Nat.gcd m n * Nat.choose n m) := sorry

Prove that the expression \[ \frac{gcd(m,n)}{n}\binom{n}{m} \] is an integer for all pairs of integers $n\geq m\geq 1$.

['number_theory', 'algebra']

Section putnam_2000_b2. Require Import Nat Reals. Open Scope R. Theorem putnam_2000_b2: forall (n m: nat), and (ge n m) (ge m 1) -> exists (c: Z), INR (gcd m n) / INR n * Binomial.C n m = IZR c. Proof. Admitted. End putnam_2000_b2.

theory putnam_2000_b2 imports Complex_Main begin theorem putnam_2000_b2: shows "\<forall>m n::nat. (m \<ge> 1 \<longrightarrow> n \<ge> m \<longrightarrow> n dvd ((gcd m n) * (n choose m)))" sorry end

putnam_2000_b3

theorem putnam_2000_b3 (N : ℕ) (Npos : N > 0) (a : Fin (N + 1) → ℝ) (haN : a N ≠ 0) (f : ℝ → ℝ := fun t ↦ ∑ j : Icc 1 N, a j * Real.sin (2 * Real.pi * j * t)) (mult : (ℝ → ℝ) → ℝ → ℕ) (hmult : ∀ g : ℝ → ℝ, ∀ t : ℝ, (∃ c : ℕ, iteratedDeriv c g t ≠ 0) → (iteratedDeriv (mult g t) g t ≠ 0 ∧ ∀ k < (mult g t), iteratedDeriv k g t = 0)) (M : ℕ → ℕ := fun k ↦ ∑' t : Ico (0 : ℝ) 1, mult (iteratedDeriv k f) t) : ((∀ i j : ℕ, i ≤ j → M i ≤ M j) ∧ Tendsto M ⊤ (𝓝 (2 * N))) := sorry

Let $f(t)=\sum_{j=1}^N a_j \sin(2\pi jt)$, where each $a_j$ is real and $a_N$ is not equal to $0$. Let $N_k$ denote the number of zeroes (including multiplicities) of $\frac{d^k f}{dt^k}$. Prove that \[ N_0\leq N_1\leq N_2\leq \cdots \mbox{ and } \lim_{k\to\infty} N_k = 2N. \]

['analysis']

theory putnam_2000_b3 imports Complex_Main "HOL-Analysis.Derivative" begin (* uses (nat \<Rightarrow> real) instead of (Fin (N + 1) \<Rightarrow> real) *) theorem putnam_2000_b3: fixes N :: nat and a :: "nat \<Rightarrow> real" and f :: "real \<Rightarrow> real" and mult :: "(real \<Rightarrow> real) \<Rightarrow> real \<Rightarrow> nat" and M :: "nat \<Rightarrow> nat" assumes Npos: "N > 0" and haN: "a N \<noteq> 0" defines "f \<equiv> (\<lambda>t::real. (\<Sum>j::nat=1..N. a j * sin (2*pi*j*t)))" assumes hmult: "\<forall>(g::real\<Rightarrow>real)(t::real). ((\<exists>c::nat. (deriv^^c) g t \<noteq> 0) \<longrightarrow> mult g t = (LEAST k::nat. (deriv^^k) g t \<noteq> 0))" defines "M \<equiv> (\<lambda>k::nat. (\<Sum>\<^sub>\<infinity>t::real\<in>{0..<1}. mult ((deriv^^k) f) t))" shows "(\<forall>i j::nat. (i \<le> j \<longrightarrow> M i \<le> M j)) \<and> filterlim M (nhds (2*N)) at_top" sorry end

putnam_2000_b4

theorem putnam_2000_b4 (f : ℝ → ℝ) (hfcont : Continuous f) (hf : ∀ x : ℝ, f (2 * x ^ 2 - 1) = 2 * x * f x) : ∀ x : ℝ, x ∈ Icc (-1) 1 → f x = 0 := sorry

Let $f(x)$ be a continuous function such that $f(2x^2-1)=2xf(x)$ for all $x$. Show that $f(x)=0$ for $-1\leq x\leq 1$.

['analysis']

theory putnam_2000_b4 imports Complex_Main begin theorem putnam_2000_b4: fixes f :: "real \<Rightarrow> real" assumes hf : "\<forall>x. f (2 * x^2 - 1) = 2 * x * f x" and f_cont : "continuous_on UNIV f" shows "\<forall>x. x \<ge> -1 \<and> x \<le> 1 \<longrightarrow> f x = 0" sorry end

putnam_2000_b5

theorem putnam_2000_b5 (S : ℕ → Set ℤ) (hSfin : ∀ n : ℕ, Set.Finite (S n)) (hSpos : ∀ n : ℕ, ∀ s ∈ S n, s > 0) (hSdef : ∀ n : ℕ, ∀ a : ℤ, a ∈ S (n + 1) ↔ Xor' (a - 1 ∈ S n) (a ∈ S n)) : (∀ n : ℕ, ∃ N ≥ n, S N = S 0 ∪ {M : ℤ | M - N ∈ S 0}) := sorry

Let $S_0$ be a finite set of positive integers. We define finite sets $S_1,S_2,\ldots$ of positive integers as follows: the integer $a$ is in $S_{n+1}$ if and only if exactly one of $a-1$ or $a$ is in $S_n$. Show that there exist infinitely many integers $N$ for which $S_N=S_0\cup\{N+a: a\in S_0\}$.

['algebra']

theory putnam_2000_b5 imports Complex_Main begin theorem putnam_2000_b5: fixes S :: "nat \<Rightarrow> int set" assumes hSfin: "\<forall>n::nat. finite (S n)" and hSpos: "\<forall>n::nat. (\<forall>s\<in>(S n). s > 0)" and hSdef: "\<forall>n::nat. (\<forall>a::int. (a \<in> S (n + 1) \<longleftrightarrow> (((a - 1) \<in> S n) \<noteq> (a \<in> S n))))" shows "\<forall>n::nat. \<exists>N::nat\<ge>n. S N = S 0 \<union> {M::int. M - N \<in> S 0}" sorry end

putnam_1995_a1

theorem putnam_1995_a1 (S : Set ℝ) (hS : ∀ a ∈ S, ∀ b ∈ S, a * b ∈ S) (T U : Set ℝ) (hsub : T ⊆ S ∧ U ⊆ S) (hunion : T ∪ U = S) (hdisj : T ∩ U = ∅) (hT3 : ∀ a ∈ T, ∀ b ∈ T, ∀ c ∈ T, a * b * c ∈ T) (hU3 : ∀ a ∈ U, ∀ b ∈ U, ∀ c ∈ U, a * b * c ∈ U) : (∀ a ∈ T, ∀ b ∈ T, a * b ∈ T) ∨ (∀ a ∈ U, ∀ b ∈ U, a * b ∈ U) := sorry

Let $S$ be a set of real numbers which is closed under multiplication (that is, if $a$ and $b$ are in $S$, then so is $ab$). Let $T$ and $U$ be disjoint subsets of $S$ whose union is $S$. Given that the product of any {\em three} (not necessarily distinct) elements of $T$ is in $T$ and that the product of any three elements of $U$ is in $U$, show that at least one of the two subsets $T,U$ is closed under multiplication.

['algebra']

theory putnam_1995_a1 imports Complex_Main begin theorem putnam_1995_a1: fixes S :: "real set" and T U :: "real set" assumes hS: "\<forall>a\<in>S. \<forall>b\<in>S. a * b \<in> S" and hsub: "T \<subseteq> S \<and> U \<subseteq> S" and hunion: "T \<union> U = S" and hdisj: "T \<inter> U = {}" and hT3: "\<forall>a\<in>T. \<forall>b\<in>T. \<forall>c\<in>T. a * b * c \<in> T" and hU3: "\<forall>a\<in>U. \<forall>b\<in>U. \<forall>c\<in>U. a * b * c \<in> U" shows "(\<forall>a\<in>T. \<forall>b\<in>T. a * b \<in> T) \<or> (\<forall>a\<in>U. \<forall>b\<in>U. a * b \<in> U)" sorry end

putnam_1995_a2

abbrev putnam_1995_a2_solution : Set (ℝ × ℝ) := sorry -- {x | let ⟨a,b⟩ := x; a = b} theorem putnam_1995_a2 (habconv : (ℝ × ℝ) → Prop := fun ⟨a,b⟩ => ∃ limit : ℝ, Tendsto (fun t : ℝ => ∫ x in (Set.Icc b t), (sqrt (sqrt (x + a) - sqrt x) - sqrt (sqrt x - sqrt (x - b)))) atTop (𝓝 limit)) : ∀ ab : ℝ × ℝ, habconv ab ↔ ab ∈ putnam_1995_a2_solution := sorry

For what pairs $(a,b)$ of positive real numbers does the improper integral \[ \int_{b}^{\infty} \left( \sqrt{\sqrt{x+a}-\sqrt{x}} - \sqrt{\sqrt{x}-\sqrt{x-b}} \right)\,dx \] converge?

Show that the solution is those pairs $(a,b)$ where $a = b$.

['analysis']

Section putnam_1995_a2. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1995_a2_solution (a b: R) := a = b. Theorem putnam_1995_a2: forall (a b: R), a > 0 /\ b > 0 /\ ex_lim_seq (fun n => RInt (fun x => sqrt (sqrt (x + a) - sqrt x) - sqrt (sqrt x - (x - b))) b (INR n)) <-> putnam_1995_a2_solution a b. Proof. Admitted. End putnam_1995_a2.

theory putnam_1995_a2 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1995_a2_solution :: "(real \<times> real) set" where "putnam_1995_a2_solution \<equiv> undefined" (* {(a::real,b::real). a = b} *) theorem putnam_1995_a2: fixes habconv :: "(real \<times> real) \<Rightarrow> bool" defines "habconv \<equiv> (\<lambda>(a::real,b::real). (\<exists>limit::real. filterlim (\<lambda>t::real. interval_lebesgue_integral lebesgue b t (\<lambda>x::real. sqrt (sqrt (x+a) - sqrt x) - sqrt (sqrt x - sqrt (x-b)))) (nhds limit) at_top))" shows "\<forall>ab::real\<times>real. habconv ab \<longleftrightarrow> ab \<in> putnam_1995_a2_solution" sorry end

putnam_1995_a3

theorem putnam_1995_a3 (relation : (Fin 9 → ℤ) → (Fin 9 → ℤ) → Prop) (digits_to_num : (Fin 9 → ℤ) → ℤ := fun dig => ∑ i : Fin 9, (dig i) * 10^i.1) (hrelation : ∀ d e : (Fin 9 → ℤ), relation d e ↔ (∀ i : Fin 9, d i < 10 ∧ d i ≥ 0 ∧ e i < 10 ∧ e i ≥ 0) ∧ (∀ i : Fin 9, 7 ∣ (digits_to_num (fun j : Fin 9 => if j = i then e j else d j)))) : ∀ d e f : (Fin 9 → ℤ), ((relation d e) ∧ (relation e f)) → (∀ i : Fin 9, 7 ∣ d i - f i) := sorry

The number $d_{1}d_{2}\dots d_{9}$ has nine (not necessarily distinct) decimal digits. The number $e_{1}e_{2}\dots e_{9}$ is such that each of the nine 9-digit numbers formed by replacing just one of the digits $d_{i}$ is $d_{1}d_{2}\dots d_{9}$ by the corresponding digit $e_{i}$ ($1 \leq i \leq 9$) is divisible by 7. The number $f_{1}f_{2}\dots f_{9}$ is related to $e_{1}e_{2}\dots e_{9}$ is the same way: that is, each of the nine numbers formed by replacing one of the $e_{i}$ by the corresponding $f_{i}$ is divisible by 7. Show that, for each $i$, $d_{i}-f_{i}$ is divisible by 7. [For example, if $d_{1}d_{2}\dots d_{9} = 199501996$, then $e_{6}$ may be 2 or 9, since $199502996$ and $199509996$ are multiples of 7.]

['number_theory']

theory putnam_1995_a3 imports Complex_Main begin (* uses (nat \<Rightarrow> int) instead of (Fin 9 \<Rightarrow> int) *) theorem putnam_1995_a3: fixes relation :: "(nat \<Rightarrow> int) \<Rightarrow> (nat \<Rightarrow> int) \<Rightarrow> bool" and digits_to_num :: "(nat \<Rightarrow> int) \<Rightarrow> int" defines "digits_to_num \<equiv> (\<lambda>dig::nat\<Rightarrow>int. (\<Sum>i::nat=0..8. (dig i) * 10^i))" assumes hrelation: "\<forall>d e::nat\<Rightarrow>int. (relation d e \<longleftrightarrow> ((\<forall>i::nat\<in>{0..8}. d i < 10 \<and> d i \<ge> 0 \<and> e i < 10 \<and> e i \<ge> 0) \<and> (\<forall>i::nat\<in>{0..8}. 7 dvd (digits_to_num (\<lambda>j::nat. if j = i then e j else d j)))))" shows "\<forall>d e f::nat\<Rightarrow>int. ((relation d e \<and> relation e f) \<longrightarrow> (\<forall>i::nat\<in>{0..8}. 7 dvd (d i - f i)))" sorry end

putnam_1995_a4

theorem putnam_1995_a4 (n : ℕ) (hn : n > 0) (necklace : Fin n → ℤ) (hnecklacesum : ∑ i : Fin n, necklace i = n - 1) : ∃ cut : Fin n, ∀ k : Fin n, ∑ i : {j : Fin n | j.1 ≤ k}, necklace (cut + i) ≤ k := sorry

Suppose we have a necklace of $n$ beads. Each bead is labeled with an integer and the sum of all these labels is $n-1$. Prove that we can cut the necklace to form a string whose consecutive labels $x_{1},x\_{2},\dots,x_{n}$ satisfy \[\sum_{i=1}^{k} x_{i} \leq k-1 \qquad \mbox{for} \quad k=1,2,\dots,n.\]

['combinatorics']

theory putnam_1995_a4 imports Complex_Main begin (* uses (nat \<Rightarrow> int) instead of (Fin n \<Rightarrow> int) *) theorem putnam_1995_a4: fixes n :: nat and necklace :: "nat \<Rightarrow> int" assumes hn: "n > 0" and hnecklacesum: "(\<Sum>i::nat=0..(n-1). necklace i) = n - 1" shows "\<exists>cut::nat\<in>{0..(n-1)}. \<forall>k::nat\<in>{0..(n-1)}. (\<Sum>i::nat\<le>k. necklace ((cut + i) mod n)) \<le> k" sorry end

putnam_1995_a5

abbrev putnam_1995_a5_solution : Prop := sorry -- True theorem putnam_1995_a5 (hdiffx : (n : ℕ) → (Fin n → (ℝ → ℝ)) → Prop := (fun (n : ℕ) (x : Fin n → (ℝ → ℝ)) => ∀ i : Fin n, Differentiable ℝ (x i))) (ha : (n : ℕ) → (Fin n → Fin n → ℝ) → Prop := (fun (n : ℕ) (a : Fin n → Fin n → ℝ) => ∀ i j : Fin n, a i j > 0)) (hcomb : (n : ℕ) → (Fin n → (ℝ → ℝ)) → (Fin n → Fin n → ℝ) → Prop := (fun (n : ℕ) (x : Fin n → (ℝ → ℝ)) (a : Fin n → Fin n → ℝ) => ∀ t : ℝ, ∀ i : Fin n, (deriv (x i)) t = ∑ j : Fin n, (a i j) * ((x j) t))) (hxlim : (n : ℕ) → (Fin n → (ℝ → ℝ)) → Prop := (fun (n : ℕ) (x : Fin n → (ℝ → ℝ)) => ∀ i : Fin n, Tendsto (x i) atTop (𝓝 0))) : putnam_1995_a5_solution ↔ (∀ (n : ℕ) (x : Fin n → (ℝ → ℝ)) (a : Fin n → Fin n → ℝ), (n > 0 ∧ hdiffx n x ∧ ha n a ∧ hcomb n x a ∧ hxlim n x) → ¬(∀ b : Fin n → ℝ, (∀ t : ℝ, ∑ i : Fin n, (b i) * ((x i) t) = 0) → (∀ i : Fin n, b i = 0))) := sorry

Let $x_{1},x_{2},\dots,x_{n}$ be differentiable (real-valued) functions of a single variable $f$ which satisfy \begin{align*} \frac{dx_{1}}{dt} &= a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} \ \frac{dx_{2}}{dt} &= a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} \ \vdots && \vdots \ \frac{dx_{n}}{dt} &= a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} \end{align*} for some constants $a_{ij}>0$. Suppose that for all $i$, $x_{i}(t) \to 0$ as $t \to \infty$. Are the functions $x_{1},x_{2},\dots,x_{n}$ necessarily linearly dependent?

Show that the answer is yes, the functions must be linearly dependent.

['linear_algebra', 'analysis']

theory putnam_1995_a5 imports Complex_Main "HOL-Analysis.Derivative" begin (* uses (nat \<Rightarrow> (real \<Rightarrow> real)) instead of (Fin n \<Rightarrow> (real \<Rightarrow> real)) and (nat \<Rightarrow> nat \<Rightarrow> real) instead of (Fin n \<Rightarrow> Fin n \<Rightarrow> real) *) definition putnam_1995_a5_solution :: bool where "putnam_1995_a5_solution \<equiv> undefined" (* True *) theorem putnam_1995_a5: fixes hdiffx :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<Rightarrow> real)) \<Rightarrow> bool" and ha :: "nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> bool" and hcomb :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<Rightarrow> real)) \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> bool" and hxlim :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<Rightarrow> real)) \<Rightarrow> bool" defines "hdiffx \<equiv> (\<lambda>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real)). (\<forall>i::nat\<in>{0..(n-1)}. (x i) differentiable_on UNIV))" and "ha \<equiv> (\<lambda>(n::nat)(a::nat\<Rightarrow>nat\<Rightarrow>real). (\<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. a i j > 0))" and "hcomb \<equiv> (\<lambda>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real))(a::nat\<Rightarrow>nat\<Rightarrow>real). (\<forall>t::real. \<forall>i::nat\<in>{0..(n-1)}. deriv (x i) t = (\<Sum>j::nat=0..(n-1). (a i j) * ((x j) t))))" and "hxlim \<equiv> (\<lambda>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real)). (\<forall>i::nat\<in>{0..(n-1)}. filterlim (x i) (nhds 0) at_top))" shows "putnam_1995_a5_solution \<longleftrightarrow> (\<forall>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real))(a::nat\<Rightarrow>nat\<Rightarrow>real). (n > 0 \<and> hdiffx n x \<and> ha n a \<and> hcomb n x a \<and> hxlim n x) \<longrightarrow> \<not>(\<forall>b::nat\<Rightarrow>real. (\<forall>t::real. (\<Sum>i::nat=0..(n-1). (b i) * ((x i) t)) = 0) \<longrightarrow> (\<forall>i::nat\<in>{0..(n-1)}. b i = 0)))" sorry end

putnam_1995_b1

theorem putnam_1995_b1 (part_ct : Finpartition (Finset.range 9) → (Finset.range 9) → ℕ) (hp : ∀ partition k, part_ct partition k = (Exists.choose (Finpartition.exists_mem partition k.2)).card) : ∀ Pt1 Pt2 : Finpartition (Finset.range 9), ∃ x y : Finset.range 9, x ≠ y ∧ part_ct Pt1 x = part_ct Pt1 y ∧ part_ct Pt2 x = part_ct Pt2 y := sorry

For a partition $\pi$ of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, let $\pi(x)$ be the number of elements in the part containing $x$. Prove that for any two partitions $\pi$ and $\pi'$, there are two distinct numbers $x$ and $y$ in $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that $\pi(x) = \pi(y)$ and $\pi'(x) = \pi'(y)$. [A {\em partition} of a set $S$ is a collection of disjoint subsets (parts) whose union is $S$.]

['combinatorics']

theory putnam_1995_b1 imports Complex_Main "HOL-Library.Disjoint_Sets" begin theorem putnam_1995_b1: fixes part_ct :: "(nat set set) \<Rightarrow> nat \<Rightarrow> nat" assumes hp: "\<forall>(partition::nat set set)(k::nat). (\<forall>part\<in>partition. (k \<in> part \<longrightarrow> part_ct partition k = card part))" shows "\<forall>Pt1 Pt2::nat set set. ((partition_on {1..9} Pt1 \<and> partition_on {1..9} Pt2) \<longrightarrow> (\<exists>x::nat\<in>{1..9}. \<exists>y::nat\<in>{1..9}. x \<noteq> y \<and> part_ct Pt1 x = part_ct Pt1 y \<and> part_ct Pt2 x = part_ct Pt2 y))" sorry end

putnam_1995_b3

abbrev putnam_1995_b3_solution : ℕ → ℤ := sorry -- fun n => if n = 1 then 45 else if n = 2 then 10 * 45^2 else 0 theorem putnam_1995_b3 (n : ℕ) (hn : n > 0) (digits_set := {f : ℕ → ℕ | f 0 ≠ 0 ∧ (∀ i : Fin (n ^ 2), f i ≤ 9) ∧ (∀ i ≥ n ^ 2, f i = 0)}) (digits_to_matrix : (ℕ → ℕ) → Matrix (Fin n) (Fin n) ℤ := fun f => (fun i j => f (i.1 * n + j.1))) : ∑' f : digits_set, (digits_to_matrix f).det = putnam_1995_b3_solution n := sorry

To each positive integer with $n^{2}$ decimal digits, we associate the determinant of the matrix obtained by writing the digits in order across the rows. For example, for $n=2$, to the integer 8617 we associate $\det \left( \begin{array}{cc} 8 & 6 \ 1 & 7 \end{array} \right) = 50$. Find, as a function of $n$, the sum of all the determinants associated with $n^{2}$-digit integers. (Leading digits are assumed to be nonzero; for example, for $n=2$, there are 9000 determinants.)

Show that the solution is $45$ if $n = 1$, $45^2*10$ if $n = 2$, and $0$ if $n$ is greater than 3.

['linear_algebra']

theory putnam_1995_b3 imports Complex_Main "HOL-Analysis.Determinants" begin (* Boosted the domain/range of digits_set to nat *) definition putnam_1995_b3_solution :: "nat \<Rightarrow> int" where "putnam_1995_b3_solution \<equiv> undefined" (* (\<lambda>n::nat. if n = 1 then 45 else if n = 2 then (10 * 45^2) else 0) *) theorem putnam_1995_b3: fixes n :: nat and digits_set :: "(nat \<Rightarrow> nat) set" and pnind :: "'n::finite \<Rightarrow> nat" and digits_to_matrix :: "(nat \<Rightarrow> nat) \<Rightarrow> (int^'n^'n)" assumes hn: "n > 0" defines "digits_set \<equiv> {f::nat\<Rightarrow>nat. f 0 \<noteq> 0 \<and> (\<forall>i::nat\<in>{0..(n^2-1)}. f i \<le> 9) \<and> (\<forall>i::nat\<ge>n^2. f i = 0)}" assumes pncard: "CARD('n) = n" and hpnind: "pnind ` UNIV = {0..(n-1)}" defines "digits_to_matrix \<equiv> (\<lambda>f::nat\<Rightarrow>nat. (\<chi> i j::'n. f ((pnind i) * n + (pnind j))))" shows "(\<Sum>f\<in>digits_set. det (digits_to_matrix f)) = putnam_1995_b3_solution n" sorry end

putnam_1995_b4

abbrev putnam_1995_b4_solution : ℤ × ℤ × ℤ × ℤ := sorry -- ⟨3,1,5,2⟩ theorem putnam_1995_b4 (contfrac : ℝ) (hcontfrac : contfrac = 2207 - 1/contfrac) : let ⟨a,b,c,d⟩ := putnam_1995_b4_solution; contfrac^((1 : ℝ)/8) = (a + b * sqrt c)/d := sorry

Evaluate \[ \sqrt[8]{2207 - \frac{1}{2207-\frac{1}{2207-\dots}}}. \] Express your answer in the form $\frac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are integers.

Show that the solution is $(3 + 1*\sqrt{5})/2.

['algebra']

Section putnam_1995_b4. Require Import Reals ZArith Coquelicot.Coquelicot. Open Scope R. Definition putnam_1995_b4_solution (a b c d: Z) := (a, b, c, d) = (3%Z,1%Z,5%Z,2%Z). Theorem putnam_1995_b4: exists (a b c d: Z), exists (contfrac: R), contfrac = 2207 - 1 / contfrac -> pow contfrac (1 / 8) = (IZR a + IZR b * sqrt (IZR c))/IZR d <-> putnam_1995_b4_solution a b c d. Proof. Admitted. End putnam_1995_b4.

theory putnam_1995_b4 imports Complex_Main begin definition putnam_1995_b4_solution :: "int \<times> int \<times> int \<times> int" where "putnam_1995_b4_solution \<equiv> undefined" (* (3,1,5,2) *) theorem putnam_1995_b4: fixes contfrac :: real assumes hcontfrac: "contfrac = 2207 - 1/contfrac" shows "let (a,b,c,d) = putnam_1995_b4_solution in (contfrac powr (1/8) = (a + b * sqrt c) / d)" sorry end

putnam_1995_b6

theorem putnam_1995_b6 (S : ℝ → Set ℕ := fun α => {x : ℕ | ∃ n : ℕ, n ≥ 1 ∧ x = floor (n * α)}) : ¬ ∃ α β γ : ℝ, α > 0 ∧ β > 0 ∧ γ > 0 ∧ (S α) ∩ (S β) = ∅ ∧ (S β) ∩ (S γ) = ∅ ∧ (S α) ∩ (S γ) = ∅ ∧ ℕ+ = (S α) ∪ (S β) ∪ (S γ) := sorry

For a positive real number $\alpha$, define \[ S(\alpha) = \{ \lfloor n\alpha \rfloor : n = 1,2,3,\dots \}. \] Prove that $\{1,2,3,\dots\}$ cannot be expressed as the disjoint union of three sets $S(\alpha), S(\beta)$ and $S(\gamma)$. [As usual, $\lfloor x \rfloor$ is the greatest integer $\leq x$.]

['algebra', 'number_theory']

theory putnam_1995_b6 imports Complex_Main begin theorem putnam_1995_b6: fixes S :: "real \<Rightarrow> (nat set)" defines "S \<equiv> (\<lambda>\<alpha>::real. {x::nat. (\<exists>n::nat. n \<ge> 1 \<and> x = \<lfloor>n*\<alpha>\<rfloor>)})" shows "\<not>(\<exists>\<alpha> \<beta> \<gamma>::real. \<alpha> > 0 \<and> \<beta> > 0 \<and> \<gamma> > 0 \<and> (S \<alpha>) \<inter> (S \<beta>) = {} \<and> (S \<beta>) \<inter> (S \<gamma>) = {} \<and> (S \<alpha>) \<inter> (S \<gamma>) = {} \<and> {1::nat..} = (S \<alpha>) \<union> (S \<beta>) \<union> (S \<gamma>))" sorry end

putnam_1989_a1

abbrev putnam_1989_a1_solution : ℕ := sorry -- 1 theorem putnam_1989_a1 (pdigalt : List ℕ → Prop) (hpdigalt : ∀ pdig : List ℕ, pdigalt pdig = Odd pdig.length ∧ (∀ i : Fin pdig.length, pdig.get i = if Even (i : ℕ) then 1 else 0)) : {p : ℕ | p > 0 ∧ p.Prime ∧ pdigalt (Nat.digits 10 p)}.encard = putnam_1989_a1_solution := sorry

How many primes among the positive integers, written as usual in base $10$, are alternating $1$'s and $0$'s, beginning and ending with $1$?

Show that there is only one such prime.

['algebra', 'number_theory']

Section putnam_1989_a1. Require Import Nat Reals ZArith Znumtheory Coquelicot.Coquelicot. Open Scope R. Definition putnam_1989_a1_solution (x: R) := x = INR 101. Theorem putnam_1989_a1: let a (n: nat) : R := sum_n (fun n => if odd n then INR (10^(n-1)) else R0) (2*n+2) in forall (n: nat), prime (floor (a n)) -> putnam_1989_a1_solution (a n). Proof. Admitted. End putnam_1989_a1.

theory putnam_1989_a1 imports Complex_Main "HOL-Computational_Algebra.Primes" begin definition putnam_1989_a1_solution::nat where "putnam_1989_a1_solution \<equiv> undefined" (* 1 *) theorem putnam_1989_a1: fixes pdigalt::"(nat list) \<Rightarrow> bool" defines "pdigalt \<equiv> \<lambda>pdig. odd (length pdig) \<and> (\<forall>i \<in> {0..<(length pdig)}. pdig!i = (if (even i) then 1 else 0))" shows "putnam_1989_a1_solution = card {p::nat. p > 0 \<and> prime p \<and> (\<forall>dig. (foldr (\<lambda>a b. a + 10 * b) dig 0) = p \<longrightarrow> pdigalt dig)}" sorry end

putnam_1989_a2

abbrev putnam_1989_a2_solution : ℝ → ℝ → ℝ := sorry -- (fun a b : ℝ => (Real.exp (a ^ 2 * b ^ 2) - 1) / (a * b)) theorem putnam_1989_a2 (a b : ℝ) (abpos : a > 0 ∧ b > 0) : ∫ x in Set.Ioo 0 a, ∫ y in Set.Ioo 0 b, Real.exp (max (b ^ 2 * x ^ 2) (a ^ 2 * y ^ 2)) = putnam_1989_a2_solution a b := sorry

Evaluate $\int_0^a \int_0^b e^{\max\{b^2x^2,a^2y^2\}}\,dy\,dx$ where $a$ and $b$ are positive.

Show that the value of the integral is $(e^{a^2b^2}-1)/(ab)$.

['analysis']

Section putnam_1989_a2. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1989_a2_solution (a b: R) := (exp (pow (a*b) 2) - 1)/(a * b). Theorem putnam_1989_a2: forall (a b: R), let f (x y: R) := Rmax (pow (b*x) 2) (pow (a*y) 2) in RInt (fun x => (RInt (fun y => exp (f x y)) 0 b)) 0 a = putnam_1989_a2_solution a b. Proof. Admitted. End putnam_1989_a2.

theory putnam_1989_a2 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1989_a2_solution::"real\<Rightarrow>real\<Rightarrow>real" where "putnam_1989_a2_solution \<equiv> undefined" (* \<lambda>a b::real. (exp (a^2 * b^2) - 1) / (a * b) *) theorem putnam_1989_a2: fixes a b::real assumes abpos : "a > 0 \<and> b > 0" shows "set_lebesgue_integral lebesgue {(x::real, y::real). x \<in> {0..a} \<and> y \<in> {0..b}} (\<lambda> (x, y). exp (max (b^2 * x^2) (a^2 * y^2))) = putnam_1989_a2_solution a b" sorry end

putnam_1989_a3

theorem putnam_1989_a3 (z : ℂ) (hz : 11 * z ^ 10 + 10 * I * z ^ 9 + 10 * I * z - 11 = 0) : (‖z‖ = 1) := sorry

Prove that if \[ 11z^{10}+10iz^9+10iz-11=0, \] then $|z|=1.$ (Here $z$ is a complex number and $i^2=-1$.)

['algebra']

Section putnam_1989_a3. Require Import Reals Coquelicot.Coquelicot. From Coqtail Require Import Cpow. Open Scope C. Theorem putnam_1989_a3: let f (z: C) := 11 * Cpow z 10 + 10 * Ci * Cpow z 9 + 10 * Ci * z - 11 in forall (x: C), f x = 0 <-> Cmod x = R1. Proof. Admitted. End putnam_1989_a3.

theory putnam_1989_a3 imports Complex_Main begin theorem putnam_1989_a3: fixes z :: complex assumes hz: "11 * z ^ 10 + 10 * \<i> * z ^ 9 + 10 * \<i> * z - 11 = 0" shows "norm z = 1" sorry end

putnam_1989_a6

theorem putnam_1989_a6 (F : Type*) [Field F] [Fintype F] (hF : Fintype.card F = 2) (α : PowerSeries F) (hα : ∀ n : ℕ, let bin := [1] ++ (digits 2 n) ++ [1]; PowerSeries.coeff F n α = ite (∀ i j : Fin bin.length, i < j → bin.get i = 1 → bin.get j = 1 → (∀ k, i < k → k < j → bin.get k = 0) → Even ((j : ℕ) - (i : ℕ) - 1)) 1 0) : (α ^ 3 + PowerSeries.X * α + 1 = 0) := sorry

Let $\alpha=1+a_1x+a_2x^2+\cdots$ be a formal power series with coefficients in the field of two elements. Let \[ a_n = \begin{cases} 1 & \parbox{2in}{if every block of zeros in the binary expansion of $n$ has an even number of zeros in the block} \\[.3in] 0 & \text{otherwise.} \end{cases} \] (For example, $a_{36}=1$ because $36=100100_2$ and $a_{20}=0$ because $20=10100_2.$) Prove that $\alpha^3+x\alpha+1=0.$

['algebra', 'abstract_algebra']

theory putnam_1989_a6 imports Complex_Main "HOL-Algebra.Ring" "HOL-Computational_Algebra.Formal_Power_Series" "HOL-Library.Cardinality" begin fun digits :: "nat \<Rightarrow> nat list" where "digits n = (if n = 0 then [] else ([n mod 2] @ digits (n div 2)))" theorem putnam_1989_a6: fixes \<alpha> :: "('a::field) fps" and X :: "'a fps" defines "\<alpha> \<equiv> Abs_fps (\<lambda> n :: nat. let bin = [1] @ (digits n) @ [1] in (if (\<forall> i \<in> {0 ..< length bin}. \<forall> j \<in> {0 ..< length bin}. i < j \<longrightarrow> bin!i = 1 \<longrightarrow> bin!j = 1 \<longrightarrow> (\<forall> k :: nat. i < k \<longrightarrow> k < j \<longrightarrow> bin!k = 0) \<longrightarrow> even (j - i - 1)) then 1 else 0))" and "X \<equiv> Abs_fps (\<lambda> n :: nat. if n = 1 then 1 else 0)" assumes hF: "CARD ('a) = 2" shows "\<alpha> ^ 3 + X * \<alpha> + 1 = 0" sorry end

putnam_1989_b2

abbrev putnam_1989_b2_solution : Prop := sorry -- True theorem putnam_1989_b2 (pow : (S : Type) → ℕ → S → S) (hpow1 : ∀ (S : Type) (_ : Semigroup S), ∀ s : S, pow S 1 s = s) (hpown : ∀ (S : Type) (_ : Semigroup S), ∀ s : S, ∀ n > 0, pow S (n + 1) s = s * (pow S n s)) : ((∀ (S : Type) (_ : Nonempty S) (_ : Semigroup S) (_ : IsCancelMul S), (∀ a : S, Finite {x | ∃ n : ℕ, n > 0 ∧ pow S n a = x}) → ∃ (_ : Group S), True) ↔ putnam_1989_b2_solution) := sorry

Let $S$ be a non-empty set with an associative operation that is left and right cancellative ($xy=xz$ implies $y=z$, and $yx=zx$ implies $y=z$). Assume that for every $a$ in $S$ the set $\{a^n:\,n=1, 2, 3, \ldots\}$ is finite. Must $S$ be a group?

Prove that $S$ must be a group.

['abstract_algebra']

theory putnam_1989_b2 imports Complex_Main "HOL-Algebra.Complete_Lattice" "HOL-Library.FuncSet" begin record 'a semigroup = "'a partial_object" + mult :: "['a, 'a] \<Rightarrow> 'a" (infixl "\<bullet>\<index>" 70) locale semigroup = fixes G (structure) assumes m_closed [intro, simp]: "\<lbrakk>x \<in> carrier G; y \<in> carrier G\<rbrakk> \<Longrightarrow> x \<bullet> y \<in> carrier G" and m_assoc: "\<lbrakk>x \<in> carrier G; y \<in> carrier G; z \<in> carrier G\<rbrakk> \<Longrightarrow> (x \<bullet> y) \<bullet> z = x \<bullet> (y \<bullet> z)" (* Note: this problem includes a quantifier over algebraic structures of a certain type. We therefore assume that this type has cardinality at least that of the reals.*) definition putnam_1989_b2_solution :: bool where "putnam_1989_b2_solution \<equiv> undefined" (* True *) theorem putnam_1989_b2: fixes cancel :: "'a semigroup \<Rightarrow> bool" and group :: "'a semigroup \<Rightarrow> bool" defines "cancel \<equiv> \<lambda> S. \<forall> x \<in> carrier S. \<forall> y \<in> carrier S. \<forall> z \<in> carrier S. (x \<bullet>\<^bsub>S\<^esub> y = x \<bullet>\<^bsub>S\<^esub> z \<longrightarrow> y = z) \<and> (y \<bullet>\<^bsub>S\<^esub> x = z \<bullet>\<^bsub>S\<^esub> x \<longrightarrow> y = z)" and "group \<equiv> \<lambda> S. \<exists> e \<in> carrier S. \<forall> x \<in> carrier S. x \<bullet>\<^bsub>S\<^esub> e = x \<and> e \<bullet>\<^bsub>S\<^esub> x = x \<and> (\<exists> y \<in> carrier S. x \<bullet>\<^bsub>S\<^esub> y = e \<and> y \<bullet>\<^bsub>S\<^esub> x = e)" assumes pacard: "\<exists> pamap :: 'a \<Rightarrow> real. surj pamap" shows "(\<forall> S :: 'a semigroup. (carrier S \<noteq> {} \<and> cancel S \<and> (\<forall> a \<in> carrier S. finite {x \<in> carrier S. \<exists> n :: nat. x = foldr (\<bullet>\<^bsub>S\<^esub>) (replicate n a) a})) \<longrightarrow> group S) \<longleftrightarrow> putnam_1989_b2_solution" sorry end

putnam_1989_b3

abbrev putnam_1989_b3_solution : ℕ → ℝ → ℝ := sorry -- fun n c ↦ c * factorial n / (3 ^ n * ∏ m in Finset.Icc (1 : ℤ) n, (1 - 2 ^ (-m))) theorem putnam_1989_b3 (f : ℝ → ℝ) (hfdiff : Differentiable ℝ f) (hfderiv : ∀ x > 0, deriv f x = -3 * f x + 6 * f (2 * x)) (hdecay : ∀ x ≥ 0, |f x| ≤ Real.exp (-Real.sqrt x)) (μ : ℕ → ℝ := fun n ↦ ∫ x in Set.Ioi 0, x ^ n * (f x)) : ((∀ n : ℕ, μ n = putnam_1989_b3_solution n (μ 0)) ∧ (∃ L : ℝ, Tendsto (fun n ↦ (μ n) * 3 ^ n / factorial n) ⊤ (𝓝 L)) ∧ (Tendsto (fun n ↦ (μ n) * 3 ^ n / factorial n) ⊤ (𝓝 0) → μ 0 = 0)) := sorry

Let $f$ be a function on $[0,\infty)$, differentiable and satisfying \[ f'(x)=-3f(x)+6f(2x) \] for $x>0$. Assume that $|f(x)|\le e^{-\sqrt{x}}$ for $x\ge 0$ (so that $f(x)$ tends rapidly to $0$ as $x$ increases). For $n$ a non-negative integer, define \[ \mu_n=\int_0^\infty x^n f(x)\,dx \] (sometimes called the $n$th moment of $f$). \begin{enumerate} \item[a)] Express $\mu_n$ in terms of $\mu_0$. \item[b)] Prove that the sequence $\{\mu_n \frac{3^n}{n!}\}$ always converges, and that the limit is $0$ only if $\mu_0=0$. \end{enumerate}

Show that for each $n \geq 0$, $\mu_n = \frac{n!}{3^n} \left( \prod_{m=1}^{n}(1 - 2^{-m}) \right)^{-1} \mu_0$.

['analysis']

theory putnam_1989_b3 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1989_b3_solution :: "nat \<Rightarrow> real \<Rightarrow> real" where "putnam_1989_b3_solution \<equiv> undefined" (* \<lambda> n c. c * fact n / (3 ^ n * (\<Prod> m = 1..n. 1 - 2 powi (-m))) *) theorem putnam_1989_b3: fixes f :: "real \<Rightarrow> real" and \<mu> :: "nat \<Rightarrow> real" defines "\<mu> \<equiv> \<lambda> n. interval_lebesgue_integral lebesgue 0 \<infinity> (\<lambda> x. x ^ n * f x)" assumes hfdiff: "f differentiable_on {0..}" and hfderiv: "\<forall> x > 0. deriv f x = -3 * f x + 6 * f (2 * x)" and hdecay: "\<forall> x \<ge> 0. \<bar>f x\<bar> \<le> exp (-sqrt x)" shows "(\<forall> n :: nat. \<mu> n = putnam_1989_b3_solution n (\<mu> 0)) \<and> convergent (\<lambda> n :: nat. \<mu> n * 3 ^ n / fact n) \<and> (((\<lambda> n :: nat. \<mu> n * 3 ^ n / fact n) \<longlonglongrightarrow> 0) \<longrightarrow> (\<mu> 0 = 0))" sorry end

putnam_1989_b4

abbrev putnam_1989_b4_solution : Prop := sorry -- True theorem putnam_1989_b4 : ((∃ S : Type, Countable S ∧ Infinite S ∧ ∃ C : Set (Set S), ¬Countable C ∧ (∀ R ∈ C, R ≠ ∅) ∧ (∀ A ∈ C, ∀ B ∈ C, A ≠ B → (A ∩ B).Finite)) ↔ putnam_1989_b4_solution) := sorry

Can a countably infinite set have an uncountable collection of non-empty subsets such that the intersection of any two of them is finite?

Prove that such a collection exists.

['set_theory']

theory putnam_1989_b4 imports Complex_Main "HOL-Library.Countable_Set" begin definition putnam_1989_b4_solution :: bool where "putnam_1989_b4_solution \<equiv> undefined" (* True *) theorem putnam_1989_b4: shows "(\<exists> C :: nat set set. uncountable C \<and> (\<forall> R \<in> C. R \<noteq> {}) \<and> (\<forall> A \<in> C. \<forall> B \<in> C. A \<noteq> B \<longrightarrow> finite (A \<inter> B))) \<longleftrightarrow> putnam_1989_b4_solution" sorry end

putnam_1997_a3

abbrev putnam_1997_a3_solution : ℝ := sorry -- Real.sqrt (Real.exp 1) theorem putnam_1997_a3 (series1 : ℝ → ℝ := fun x => ∑' n : ℕ, (-1)^n * x^(2*n + 1)/(∏ i : Finset.range n, 2 * ((i : ℝ) + 1))) (series2 : ℝ → ℝ := fun x => ∑' n : ℕ, x^(2*n)/(∏ i : Finset.range n, (2 * ((i : ℝ) + 1))^2)) : Tendsto (fun t => ∫ x in Set.Icc 0 t, series1 x * series2 x) atTop (𝓝 (putnam_1997_a3_solution)) := sorry

Evaluate \begin{gather*} \int_0^\infty \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right) \\ \left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2 \cdot 6^2}+\cdots\right)\,dx. \end{gather*}

Show that the solution is $\sqrt{e}$.

['analysis']

Section putnam_1997_a3. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1997_a3_solution := sqrt (exp 1). Theorem putnam_1997_a3: let fix even_fact (n : nat) : R := match n with | O => 1 | S n' => (2 * INR n) * even_fact n' end in let fix even_fact_sqr (n : nat) : R := match n with | O => 1 | S n' => pow (2 * INR n) 2 * even_fact n' end in let f (x: R) := Series (fun n => pow (-1) n * pow x (2 * n + 1) / even_fact n) in let g (x: R) := Series (fun n => pow x (2 * n) / even_fact_sqr n) in Lim_seq (fun n => sum_n (fun m => RInt (fun x => f x * g x) 0 (INR m)) n) = putnam_1997_a3_solution. Proof. Admitted. End putnam_1997_a3.

theory putnam_1997_a3 imports Complex_Main "HOL-Analysis.Interval_Integral" begin definition putnam_1997_a3_solution::real where "putnam_1997_a3_solution \<equiv> undefined" (* sqrt (exp 1) *) theorem putnam_1997_a3: fixes series1 series2::"real\<Rightarrow>real" defines "series1 \<equiv> \<lambda>x. (\<Sum>n::nat. (-1)^n * x^(2*n+1) / (\<Prod>i=0..<n. 2 * (i + 1)))" and "series2 \<equiv> \<lambda>x. (\<Sum>n::nat. x^(2*n) / (\<Prod>i=0..<n. (2 * (i+1))^2))" shows "((\<lambda>t::real. interval_lebesgue_integral lebesgue 0 t (\<lambda>x. series1 x * series2 x)) \<longlongrightarrow> putnam_1997_a3_solution) at_top" sorry end

putnam_1997_a4

theorem putnam_1997_a4 (G : Type*) [Group G] (φ : G → G) (hφ : ∀ g1 g2 g3 h1 h2 h3 : G, (g1 * g2 * g3 = 1 ∧ h1 * h2 * h3 = 1) → φ g1 * φ g2 * φ g3 = φ h1 * φ h2 * φ h3) : ∃ a : G, let ψ := fun g => a * φ g; ∀ x y : G, ψ (x * y) = ψ x * ψ y := sorry

Let $G$ be a group with identity $e$ and $\phi:G\rightarrow G$ a function such that \[\phi(g_1)\phi(g_2)\phi(g_3)=\phi(h_1)\phi(h_2)\phi(h_3)\] whenever $g_1g_2g_3=e=h_1h_2h_3$. Prove that there exists an element $a\in G$ such that $\psi(x)=a\phi(x)$ is a homomorphism (i.e. $\psi(xy)=\psi(x)\psi(y)$ for all $x,y\in G$).

['abstract_algebra']

theory putnam_1997_a4 imports Complex_Main begin theorem putnam_1997_a4: fixes Gmul::"'g \<Rightarrow> 'g \<Rightarrow> 'g" (infixl "\<^bold>*" 70) and e::'g and inv::"'g\<Rightarrow>'g" and \<phi>::"'g\<Rightarrow>'g" assumes hgroup : "group (\<^bold>*) e inv" and hphi : "\<forall> g1 g2 g3 h1 h2 h3::'g. (g1 \<^bold>* g2 \<^bold>* g3 = e \<and> h1 \<^bold>* h2 \<^bold>* h3 = e) \<longrightarrow> \<phi> g1 \<^bold>* \<phi> g2 \<^bold>* \<phi> g3 = \<phi> h1 \<^bold>* \<phi> h2 \<^bold>* \<phi> h3" shows "\<exists>a::'g. \<forall>x y::'g. a \<^bold>* \<phi> (x \<^bold>* y) = (a \<^bold>* \<phi> x) \<^bold>* (a \<^bold>* \<phi> y)" sorry end

putnam_1997_a5

abbrev putnam_1997_a5_solution : Prop := sorry -- True theorem putnam_1997_a5 (N := fun (n : ℕ+) => {t : Fin n → ℕ+ | (∀ i j : Fin n, i < j → t i <= t j) ∧ (∑ i : Fin n, (1 : ℝ)/(t i) = 1) }) : Odd (N 10).ncard ↔ putnam_1997_a5_solution := sorry

Let $N_n$ denote the number of ordered $n$-tuples of positive integers $(a_1,a_2,\ldots,a_n)$ such that $1/a_1 + 1/a_2 +\ldots + 1/a_n=1$. Determine whether $N_{10}$ is even or odd.

Show that $N_{10}$ is odd.

['number_theory']

Section putnam_1997_a5. Require Import Nat Ensembles Finite_sets List Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1997_a5_solution := True. Theorem putnam_1997_a5: forall (E: Ensemble (list nat)) (l: list nat), length l = 10%nat /\ (E l <-> sum_n (fun i => 1/ INR (nth i l 0%nat)) 10 = 1) -> exists (m: nat), cardinal (list nat) E m /\ odd m = true <-> putnam_1997_a5_solution. Proof. Admitted. End putnam_1997_a5.

theory putnam_1997_a5 imports Complex_Main begin definition putnam_1997_a5_solution where "putnam_1997_a5_solution \<equiv> undefined" (* True *) theorem putnam_1997_a5: fixes N::"nat\<Rightarrow>nat" defines "N \<equiv> \<lambda>n. card {t::nat list. (size t = n) \<and> (\<forall>i \<in> {0..<n}. t!i > 0) \<and> (\<forall>i \<in> {0..<n}. \<forall>j \<in> {0..<n}. i < j \<longrightarrow> t!i \<le> t!j) \<and> (\<Sum>i=0..<n. 1 / (t!i)) = 1}" shows "odd (N 10) \<longleftrightarrow> putnam_1997_a5_solution" sorry end

putnam_1997_a6

abbrev putnam_1997_a6_solution : ℤ → ℤ → ℝ := sorry -- fun n k => Nat.choose (n.toNat-1) (k.toNat-1) theorem putnam_1997_a6 (n : ℤ) (hn : n > 0) (C : ℝ) (x : ℝ → (ℤ → ℝ)) (hx0 : ∀ c : ℝ, x c 0 = 0) (hx1 : ∀ c : ℝ, x c 1 = 1) (hxk : ∀ c : ℝ, ∀ k ≥ 0, x c (k + 2) = (c*(x c (k + 1)) - (n - k)*(x c k))/(k + 1)) (S : Set ℝ := {c : ℝ | x c (n + 1) = 0}) (hC : C = sSup S) : ∀ k : Set.Icc 1 n, x C k = putnam_1997_a6_solution n k := sorry

For a positive integer $n$ and any real number $c$, define $x_k$ recursively by $x_0=0$, $x_1=1$, and for $k\geq 0$, \[x_{k+2}=rac{cx_{k+1}-(n-k)x_k}{k+1}.\] Fix $n$ and then take $c$ to be the largest value for which $x_{n+1}=0$. Find $x_k$ in terms of $n$ and $k$, $1\leq k\leq n$.

Show that the solution is that $x_k = {n - 1 \choose k - 1}$.

['algebra']

Section putnam_1997_a6. Require Import Binomial Nat Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1997_a6_solution (n k: nat) := Binomial.C (n - 1) (k - 1). Theorem putnam_1997_a6: let fix x (n: nat) (c: R) (k: nat) : R := match k with | O => 0 | S O => 1 | S ((S k'') as k') => (c * x n c k' - INR (n - k) * x n c k'') / INR k' end in forall (n: nat), exists (maxc: R), forall (c: R), x n c (S n) = 0 /\ x n maxc (S n) = 0 -> c <= maxc -> forall (k: nat), and (le 1 k) (le k n) -> x n c k = putnam_1997_a6_solution n k. Proof. Admitted. End putnam_1997_a6.

theory putnam_1997_a6 imports Complex_Main begin definition putnam_1997_a6_solution :: "int \<Rightarrow> int \<Rightarrow> real" where "putnam_1997_a6_solution \<equiv> undefined" (* \<lambda> n k. (nat n-1) choose (nat k-1) *) theorem putnam_1997_a6: fixes n :: "int" and C :: "real" and x :: "real \<Rightarrow> (int \<Rightarrow> real)" and S :: "real set" defines "S \<equiv> {c :: real. x c (n + 1) = 0}" assumes hx0 : "\<forall> c :: real. x c 0 = 0" and hx1 : "\<forall> c :: real. x c 1 = 1" and hxk : "\<forall> (c :: real) (k :: nat). x c (k + 2) = (c * (x c (k+1)) - (n-k) * (x c k))/(k+1)" and hC : "C = (GREATEST s. s \<in> S)" and hn : "n > 0" shows "\<forall> k \<in> {1..n}. x C k = putnam_1997_a6_solution n k" sorry end

putnam_1997_b1

abbrev putnam_1997_b1_solution : ℕ → ℝ := sorry -- fun n => n noncomputable def dist_to_int : ℝ → ℝ := fun r => |r - round r| theorem putnam_1997_b1 (F : ℕ → ℝ := fun n => ∑ m in Finset.Icc 1 (6 * n - 1), min (dist_to_int (m/(6*n)) ) (dist_to_int (m/(3*n)))) : ∀ n, n > 0 → F n = putnam_1997_b1_solution n := sorry

Let $\{x\}$ denote the distance between the real number $x$ and the nearest integer. For each positive integer $n$, evaluate \[F_n=\sum_{m=1}^{6n-1} \min(\{\frac{m}{6n}\},\{\frac{m}{3n}\}).\] (Here $\min(a,b)$ denotes the minimum of $a$ and $b$.)

Show that the solution is $n$.

['algebra']

Section putnam_1997_b1. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1997_b1_solution (n: nat) := INR n. Theorem putnam_1997_b1: let rnd (x: R) := Rmin (Rabs (IZR (floor x) - x)) (Rabs (IZR (floor (x + 1)) - x)) in forall (n: nat), gt n 0 -> sum_n (fun m => Rmin (rnd ((INR m + 1) / (6 * INR n))) (rnd ((INR m + 1) / (3 * INR n)))) (6 * n - 1) = putnam_1997_b1_solution n. Proof. Admitted. End putnam_1997_b1.

theory putnam_1997_b1 imports Complex_Main begin definition putnam_1997_b1_solution :: "nat \<Rightarrow> real" where "putnam_1997_b1_solution \<equiv> undefined" (* \<lambda> n. n *) definition dist_to_int :: "real \<Rightarrow> real" where "dist_to_int r = \<bar>r - round r\<bar>" theorem putnam_1997_b1: fixes F :: "nat \<Rightarrow> real" defines "F \<equiv> \<lambda> n. (\<Sum> m \<in> {1..6*n-1}. min (dist_to_int (m/(6*n))) (dist_to_int (m/(3*n))))" shows "\<forall> n :: nat > 0. F n = putnam_1997_b1_solution n" sorry end