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PutnamBench

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putnam_1993_b1

abbrev putnam_1993_b1_solution : ℕ := sorry -- 3987 theorem putnam_1993_b1 (nallmexk : ℕ → Prop) (hnallmexk : ∀ n : ℕ, nallmexk n = (n > 0 ∧ ∀ m ∈ Set.Ioo 0 1993, ∃ k : ℤ, (m / 1993 < (k : ℝ) / n) ∧ ((k : ℝ) / n < (m + 1) / 1994))) : nallmexk putnam_1993_b1_solution ∧ (∀ n : ℕ, nallmexk n → n ≥ putnam_1993_b1_solution) := sorry

Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $\frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$.

Show that the smallest positive integer $n$ satisfying the condition is $n=3987$.

['algebra']

Section putnam_1993_b1. Require Import Reals. Open Scope R. Definition putnam_1993_b1_solution : nat := 3987. Theorem putnam_1993_b1: let cond (n: nat) := forall (m: nat), and (lt 0 m) (lt m 1993) -> exists (k: nat), INR m / 1993 < INR k / INR n < INR (m + 1) / 1994 in exists (mn: nat), cond mn /\ forall (n: nat), cond n -> ge n mn <-> mn = putnam_1993_b1_solution. Proof. Admitted. End putnam_1993_b1.

theory putnam_1993_b1 imports Complex_Main begin definition putnam_1993_b1_solution::nat where "putnam_1993_b1_solution \<equiv> undefined" (* 3987 *) theorem putnam_1993_b1: fixes nallmexk::"nat\<Rightarrow>bool" defines "nallmexk \<equiv> \<lambda>n::nat. (n > 0 \<and> (\<forall>m \<in> {0<..<1993}. (\<exists>k::int. (m / 1993 < k / n) \<and> (k / n < (m+1) / 1994))))" shows "putnam_1993_b1_solution = (LEAST n. nallmexk n)" sorry end

putnam_1993_b4

theorem putnam_1993_b4 (K : ℝ × ℝ → ℝ) (f g : ℝ → ℝ) (Kpos : ∀ x y : Set.Icc (0 : ℝ) 1, K (x, y) > 0) (Kcont : ContinuousOn K {(x, y) : ℝ × ℝ | x ∈ Set.Icc 0 1 ∧ y ∈ Set.Icc 0 1}) (fgpos : ∀ x : Set.Icc (0 : ℝ) 1, f x > 0 ∧ g x > 0) (fgcont : ContinuousOn f (Set.Icc 0 1) ∧ ContinuousOn g (Set.Icc 0 1)) (fgint : ∀ x : Set.Icc (0 : ℝ) 1, (∫ y in Set.Ioo 0 1, f y * K (x, y)) = g x ∧ (∫ y in Set.Ioo 0 1, g y * K (x, y)) = f x) : ∀ x : Set.Icc (0 : ℝ) 1, f x = g x := sorry

The function $K(x,y)$ is positive and continuous for $0 \leq x \leq 1,0 \leq y \leq 1$, and the functions $f(x)$ and $g(x)$ are positive and continuous for $0 \leq x \leq 1$. Suppose that for all $x$, $0 \leq x \leq 1$, $\int_0^1 f(y)K(x,y)\,dy=g(x)$ and $\int_0^1 g(y)K(x,y)\,dy=f(x)$. Show that $f(x)=g(x)$ for $0 \leq x \leq 1$.

['analysis']

Section putnam_1993_b4. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Theorem putnam_1993_b4: forall (K: (R * R) -> R) (f g: R -> R), (forall (xy : R * R), let (x, y) := xy in 0 <= x <= 1 /\ 0 <= y <= 1 /\ K xy > 0 /\ continuous K xy) /\ (forall (x: R), 0 <= x <= 1 /\ f x > 0 /\ g x > 0 /\ continuity f /\ continuity g) -> forall (x: R), 0 <= x <= 1 -> (RInt (fun y => f y * K (x, y)) 0 1) = g x /\ RInt (fun y => g y * K (x, y)) 0 1 = f x -> f = g. Proof. Admitted. End putnam_1993_b4.

theory putnam_1993_b4 imports Complex_Main "HOL-Analysis.Interval_Integral" begin theorem putnam_1993_b4: fixes K::"(real\<times>real) \<Rightarrow> real" and f g::"real\<Rightarrow>real" assumes Kpos : "\<forall>x::real \<in> {0..1}. \<forall>y::real \<in> {0..1}. K (x, y) > 0" and Kcont : "continuous_on {(x::real, y::real). x \<in> {0..1} \<and> y \<in> {0..1}} K" and fgpos : "\<forall>x::real \<in> {0..1}. f x > 0 \<and> g x > 0" and fgcont : "continuous_on {0..1::real} f \<and> continuous_on {0..1::real} g" and fgint : "\<forall>x::real \<in> {0..1}. interval_lebesgue_integral lebesgue 0 1 (\<lambda>y. f y * K (x, y)) = g x \<and> interval_lebesgue_integral lebesgue 0 1 (\<lambda>y. g y * K (x, y)) = f x" shows "\<forall>x::real \<in> {0..1}. f x = g x" sorry end

putnam_1993_b5

theorem putnam_1993_b5 (pdists : (Fin 4 → (Fin 2 → ℝ)) → Prop) (hpdists: ∀ p : Fin 4 → (Fin 2 → ℝ), pdists p = ∀ i j : Fin 4, i ≠ j → (Euclidean.dist (p i) (p j) = round (Euclidean.dist (p i) (p j)) ∧ Odd (round (Euclidean.dist (p i) (p j))))) : ¬∃ p : Fin 4 → (Fin 2 → ℝ), pdists p := sorry

Show there do not exist four points in the Euclidean plane such that the pairwise distances between the points are all odd integers.

['geometry', 'number_theory', 'linear_algebra']

theory putnam_1993_b5 imports Complex_Main "HOL-Analysis.Linear_Algebra" begin theorem putnam_1993_b5: fixes pdist::"(real \<times> real) \<Rightarrow> (real \<times> real) \<Rightarrow> (real \<times> real) \<Rightarrow> (real \<times> real) \<Rightarrow> bool" defines "pdist \<equiv> \<lambda>A. \<lambda>B. \<lambda>C. \<lambda>D. (dist A B) \<in> \<int> \<and> (dist A C) \<in> \<int> \<and> (dist A D) \<in> \<int> \<and> (dist B C) \<in> \<int> \<and> (dist B D) \<in> \<int> \<and> (dist C D) \<in> \<int> \<and> odd (round (dist A B)) \<and> odd (round (dist A C)) \<and> odd (round (dist A D)) \<and> odd (round (dist B C)) \<and> odd (round (dist B D)) \<and> odd (round (dist C D))" shows "\<not>(\<exists>A B C D. pdist A B C D)" sorry end

putnam_1993_b6

theorem putnam_1993_b6 (S : Fin 3 → ℕ) (f : Fin 3 → Fin 3 → (Fin 3 → ℕ) → (Fin 3 → ℕ)) (Spos : ∀ i : Fin 3, S i > 0) (hf : ∀ i j k : Fin 3, (i ≠ j ∧ i ≠ k ∧ j ≠ k) → ∀ S' : Fin 3 → ℕ, if S' i ≤ S' j then ((f i j S') i = 2 * S' i ∧ (f i j S') j = S' j - S' i ∧ (f i j S') k = S' k) else (f i j S' = S')) : ∃ (Ss : ℕ → (Fin 3 → ℕ)) (N : ℕ), Ss 0 = S ∧ (∃ i : Fin 3, Ss N i = 0) ∧ (∀ n : Fin N, ∃ i j : Fin 3, i ≠ j ∧ f i j (Ss n) = Ss ((n : ℕ) + 1)) := sorry

Let $S$ be a set of three, not necessarily distinct, positive integers. Show that one can transform $S$ into a set containing $0$ by a finite number of applications of the following rule: Select two of the three integers, say $x$ and $y$, where $x \leq y$ and replace them with $2x$ and $y-x$.

['algebra']

theory putnam_1993_b6 imports Complex_Main begin (* Note: Boosted domain to infinite set *) theorem putnam_1993_b6: fixes S::"nat\<Rightarrow>nat" and f::"nat \<Rightarrow> nat \<Rightarrow> (nat\<Rightarrow>nat) \<Rightarrow> (nat\<Rightarrow>nat)" assumes Spos : "\<forall>i \<in> {0..<3}. S i > 0" and hf : "\<forall>i \<in> {0..<3}. \<forall>j \<in> {0..<3}. \<forall>k \<in> {0..<3}. (i \<noteq> j \<and> i \<noteq> k \<and> j \<noteq> k) \<longrightarrow> (\<forall>S'::nat\<Rightarrow>nat. if (S' i \<le> S' j) then ( (f i j S') i = 2 * S' i \<and> (f i j S') j = S' j - S' i \<and> (f i j S') k = S' k) else (f i j S' = S'))" shows "\<exists>N::nat. \<exists>Ss::nat\<Rightarrow>(nat\<Rightarrow>nat). Ss 0 = S \<and> (\<exists>i \<in> {0..<3}. Ss N i = 0) \<and> (\<forall>n \<in> {0..<N}. \<exists>i \<in> {0..<3}. \<exists>j \<in> {0..<3}. (i \<noteq> j \<and> f i j (Ss n) = Ss (n+1)))" sorry end

putnam_2015_a2

abbrev putnam_2015_a2_solution : ℕ := sorry -- 181 theorem putnam_2015_a2 (a : ℕ → ℤ) (abase : a 0 = 1 ∧ a 1 = 2) (arec : ∀ n ≥ 2, a n = 4 * a (n - 1) - a (n - 2)) : Odd putnam_2015_a2_solution ∧ putnam_2015_a2_solution.Prime ∧ ((putnam_2015_a2_solution : ℤ) ∣ a 2015) := sorry

Let $a_0=1$, $a_1=2$, and $a_n=4a_{n-1}-a_{n-2}$ for $n \geq 2$. Find an odd prime factor of $a_{2015}$.

Show that one possible answer is $181$.

['number_theory']

Section putnam_2015_a2. Require Import Nat Reals Znumtheory. From mathcomp Require Import div. Definition putnam_2015_a2_solution : nat := 181. Theorem putnam_2015_a2: let A := fix a (n: nat) : nat := match n with | O => 1%nat | S O => 2%nat | S ((S n'') as n') => sub (4*(a n')) (a n'') end in odd putnam_2015_a2_solution = true /\ prime (Z.of_nat putnam_2015_a2_solution) /\ (putnam_2015_a2_solution %| A 2015%nat) = true. Proof. Admitted. End putnam_2015_a2.

theory putnam_2015_a2 imports Complex_Main "HOL-Computational_Algebra.Primes" begin (* there are several possible correct solutions; this is the one shown on the solutions document *) definition putnam_2015_a2_solution :: nat where "putnam_2015_a2_solution \<equiv> undefined" (* 181 *) theorem putnam_2015_a2: fixes a :: "nat \<Rightarrow> nat" assumes abase: "a 0 = 1 \<and> a 1 = 2" assumes arec: "\<forall>n::nat\<ge>2. a n = 4 * a (n-1) - a (n-2)" shows "odd putnam_2015_a2_solution \<and> prime putnam_2015_a2_solution \<and> putnam_2015_a2_solution dvd (a 2015)" sorry end

putnam_2015_a3

abbrev putnam_2015_a3_solution : ℂ := sorry -- 13725 theorem putnam_2015_a3 : Complex.log (∏ a : Fin 2015, ∏ b : Fin 2015, (1 + Complex.exp (2 * Real.pi * Complex.I * (a.1 + 1) * (b.1 + 1) / 2015))) / Complex.log 2 = putnam_2015_a3_solution := sorry

Compute $\log_2 \left( \prod_{a=1}^{2015}\prod_{b=1}^{2015}(1+e^{2\pi iab/2015}) \right)$. Here $i$ is the imaginary unit (that is, $i^2=-1$).

Show that the answer is $13725$.

['number_theory', 'algebra']

Section putnam_2015_a3. Require Import Reals ROrderedType Coquelicot.Coquelicot. Local Open Scope C. Definition putnam_2015_a3_solution : C := RtoC 13725. Theorem putnam_2015_a3: let Carg (z : C) : R := if Reqb (Im z) 0 then (if Rlt_dec (Re z) 0 then PI else R0) else atan ((Im z)/(Re z)) in let Clog (n: nat) (z : C) : C := (Re (ln (Cmod z) / ln (INR n)), Carg z) in let HCprod2_aux := fix Cprod2_aux (f : nat -> nat -> C) (n m : nat) {struct m} : C := match m with | O => f n O | S m' => Cmult (f n m') (Cprod2_aux f n m') end in let HCprod2 := fix Cprod2 (f : nat -> nat -> C) (n m : nat) {struct n}: C := match n with | O => 1%R | S n' => Cmult (HCprod2_aux f n m) (Cprod2 f n' m) end in let f (a b: nat) : C := Clog 2%nat (Re (1 + cos (2*PI*INR(a+1)*INR(b+1)/2015)), sin (2*PI*INR(a+1)*INR(b+1)/2015)) in HCprod2 f 2015%nat 2015%nat = putnam_2015_a3_solution. Proof. Admitted. End putnam_2015_a3.

theory putnam_2015_a3 imports Complex_Main begin definition putnam_2015_a3_solution :: complex where "putnam_2015_a3_solution \<equiv> undefined" (* 13725 *) theorem putnam_2015_a3: shows "log 2 (Re (\<Prod>a::nat=1..2015. \<Prod>b::nat=1..2015. (1 + exp (2*pi*\<i>*a*b/2015)))) = putnam_2015_a3_solution" sorry end

putnam_2015_a4

abbrev putnam_2015_a4_solution : ℝ := sorry -- 4 / 7 theorem putnam_2015_a4 (S : ℝ → Set ℕ := fun x ↦ {n : ℕ | n > 0 ∧ Even ⌊n * x⌋}) (f : ℝ → ℝ := fun x ↦ ∑' n : S x, 1 / 2 ^ (n : ℕ)) (p : ℝ → Prop := fun l ↦ ∀ x ∈ Set.Ico 0 1, f x ≥ l) (L : ℝ) (hpL : p L) (hLub : ∀ l : ℝ, p l → l ≤ L) : (L = putnam_2015_a4_solution) := sorry

For each real number $x$, let \[ f(x) = \sum_{n\in S_x} \frac{1}{2^n}, \] where $S_x$ is the set of positive integers $n$ for which $\lfloor nx \rfloor$ is even. What is the largest real number $L$ such that $f(x) \geq L$ for all $x \in [0,1)$? (As usual, $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$.)

Prove that $L = \frac{4}{7}$.

['analysis']

Section putnam_2015_a4. Require Import Nat List Reals. From Coquelicot Require Import Coquelicot Hierarchy Rcomplements. Definition putnam_2015_a4_solution : R := 4/7. Theorem putnam_2015_a4: let f (x: R) := Series (fun n => if even (Z.to_nat (floor ((INR n)*x))) then 1/(2^n) else R0) in exists (L: R), forall (x: R), 0 <= x < 1 -> ((f x >= L) /\ forall (M: R), f x >= M -> L >= M) <-> L = putnam_2015_a4_solution. Proof. Admitted. End putnam_2015_a4.

theory putnam_2015_a4 imports Complex_Main "HOL-Analysis.Infinite_Sum" begin definition putnam_2015_a4_solution::real where "putnam_2015_a4_solution \<equiv> undefined" (* 4/7 *) theorem putnam_2015_a4: fixes S::"real\<Rightarrow>(nat set)" and f::"real\<Rightarrow>real" and p::"real\<Rightarrow>bool" defines "S \<equiv> \<lambda>x::real. {n::nat. n > 0 \<and> even (\<lfloor>n * x\<rfloor>)}" and "f \<equiv> \<lambda>x::real. \<Sum>\<^sub>\<infinity>n \<in> S x. 1 / (2 ^ n)" shows "putnam_2015_a4_solution = (GREATEST l::real. \<forall>x \<in> {0..<1}. f x \<ge> l)" sorry end

putnam_2015_a5

theorem putnam_2015_a5 (q : ℕ) (Nq : ℕ) (qodd : Odd q) (qpos : q > 0) (hNq : Nq = {a : ℕ | 0 < a ∧ a < (q : ℝ) / 4 ∧ Nat.gcd a q = 1}.encard) : Odd Nq ↔ ∃ p k : ℕ, q = p ^ k ∧ k > 0 ∧ p.Prime ∧ (p % 8 = 5 ∨ p % 8 = 7) := sorry

Let $q$ be an odd positive integer, and let $N_q$ denote the number of integers $a$ such that $0<a<q/4$ and $\gcd(a,q)=1$. Show that $N_q$ is odd if and only if $q$ is of the form $p^k$ with $k$ a positive integer and $p$ a prime congruent to $5$ or $7$ modulo $8$.

['number_theory']

Section putnam_2015_a5. Require Import Nat Reals Arith Znumtheory Ensembles Finite_sets. Local Open Scope nat_scope. Theorem putnam_2015_a5: forall (q: nat), odd q = true -> forall (Nq: Ensemble nat), forall (a: nat), ( and (Rle R0 (INR a)) (Rle (INR a) (Rdiv (INR q) (INR 4))) /\ gcd a q = 1 <-> Nq a ) -> exists (Nq_size: nat), cardinal nat Nq Nq_size /\ odd Nq_size = true <-> exists (p k: nat), prime (Z.of_nat p) /\ (p mod 8 = 5 \/ p mod 8 = 7) /\ q = p^k. Proof. Admitted. End putnam_2015_a5.

theory putnam_2015_a5 imports Complex_Main "HOL-Computational_Algebra.Primes" begin theorem putnam_2015_a5: fixes q :: nat and Nq :: nat assumes qodd: "odd q" and qpos: "q > 0" and hNq: "Nq = card {a::nat. 0 < a \<and> a < (real q)/4 \<and> gcd a q = 1}" shows "odd Nq \<longleftrightarrow> (\<exists>p k::nat. q = p^k \<and> k > 0 \<and> prime p \<and> (p mod 8 = 5 \<or> p mod 8 = 7))" sorry end

putnam_2015_a6

theorem putnam_2015_a6 (n : ℕ) (A B M : Matrix (Fin n) (Fin n) ℝ) (npos : n > 0) (hmul : A * M = M * B) (hpoly : Matrix.charpoly A = Matrix.charpoly B) : ∀ X : Matrix (Fin n) (Fin n) ℝ, (A - M * X).det = (B - X * M).det := sorry

Let $n$ be a positive integer. Suppose that $A$, $B$, and $M$ are $n \times n$ matrices with real entries such that $AM = MB$, and such that $A$ and $B$ have the same characteristic polynomial. Prove that $\det(A-MX)=\det(B-XM)$ for every $n \times n$ matrix $X$ with real entries.

['linear_algebra']

Section putnam_2015_a6. Theorem putnam_2015_a6: True. Proof. Admitted. End putnam_2015_a6.

theory putnam_2015_a6 imports Complex_Main "HOL-Analysis.Determinants" "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2015_a6: fixes n :: nat and A B M :: "real^'n^'n" and charpoly :: "real^'n^'n \<Rightarrow> (real poly)" assumes npos: "n > 0" and pncard: "CARD('n) = n" and hmul: "A ** M = M ** B" and hcharpoly: "\<forall>A'::real^'n^'n. charpoly A' = det (mat (monom 1 1) - map_matrix (\<lambda>a::real. monom a 0) A')" and hpoly: "charpoly A = charpoly B" shows "\<forall>X::real^'n^'n. det (A - M**X) = det (B - X**M)" sorry end

putnam_2015_b1

theorem putnam_2015_b1 (f : ℝ → ℝ) (nzeros : (ℝ → ℝ) → ℕ → Prop) (fdiff : ContDiff ℝ 2 f ∧ Differentiable ℝ (iteratedDeriv 2 f)) (hnzeros : ∀ f' : ℝ → ℝ, ∀ n : ℕ, nzeros f' n = ({x : ℝ | f' x = 0}.encard ≥ n)) (fzeros : nzeros f 5) : nzeros (f + 6 * deriv f + 12 * iteratedDeriv 2 f + 8 * iteratedDeriv 3 f) 2 := sorry

Let $f$ be a three times differentiable function (defined on $\mathbb{R}$ and real-valued) such that $f$ has at least five distinct real zeros. Prove that $f+6f'+12f''+8f'''$ has at least two distinct real zeros.

['analysis']

Section putnam_2015_b1. Require Import Reals List Coquelicot.Derive. Local Open Scope R_scope. Theorem putnam_2015_b1 : forall (f: R -> R), continuity (Derive_n f 3) -> exists (l: list R), length l = 5%nat /\ NoDup l /\ forall x, In x l -> f x = 0 -> let g (x: R):= f x + 6 * (Derive_n f 1) x + 12 * (Derive_n f 2) x + 8 * (Derive_n f 3) x in exists (l': list R), length l = 2%nat /\ NoDup l' /\ forall x, In x l' -> g x = 0. Proof. Admitted. End putnam_2015_b1.

theory putnam_2015_b1 imports Complex_Main "HOL-Analysis.Derivative" begin theorem putnam_2015_b1: fixes f :: "real \<Rightarrow> real" and nzeros :: "(real \<Rightarrow> real) \<Rightarrow> nat \<Rightarrow> bool" assumes fdiff: "f C1_differentiable_on UNIV \<and> (deriv f) C1_differentiable_on UNIV \<and> ((deriv^^2) f) differentiable_on UNIV" and hnzeros: "\<forall>(f'::real\<Rightarrow>real)(n::nat). nzeros f' n = (card {x::real. f' x = 0} \<ge> n)" and fzeros: "nzeros f 5" shows "nzeros (\<lambda>x::real. f x + (6*deriv f x) + (12*(deriv^^2) f x) + (8*(deriv^^3) f x)) 2" sorry end

putnam_2015_b3

abbrev putnam_2015_b3_solution : Set (Matrix (Fin 2) (Fin 2) ℝ) := sorry -- {A : Matrix (Fin 2) (Fin 2) ℝ | (∃ α : ℝ, ∀ i j : Fin 2, A i j = α * 1) ∨ (∃ β : ℝ, A 0 0 = β * -3 ∧ A 0 1 = β * -1 ∧ A 1 0 = β * 1 ∧ A 1 1 = β * 3)} theorem putnam_2015_b3 (S : Set (Matrix (Fin 2) (Fin 2) ℝ)) (M : Matrix (Fin 2) (Fin 2) ℝ) (MinS : Prop) (hS : S = {M' : Matrix (Fin 2) (Fin 2) ℝ | (M' 0 1 - M' 0 0 = M' 1 0 - M' 0 1) ∧ (M' 0 1 - M' 0 0 = M' 1 1 - M' 1 0)}) (hMinS : MinS = (M ∈ S ∧ (∃ k > 1, M ^ k ∈ S))) : MinS ↔ M ∈ putnam_2015_b3_solution := sorry

Let $S$ be the set of all $2 \times 2$ real matrices $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ whose entries $a,b,c,d$ (in that order) form an arithmetic progression. Find all matrices $M$ in $S$ for which there is some integer $k>1$ such that $M^k$ is also in $S$.

Show that matrices of the form $\alpha A$ or $\beta B$, where $A=\left(\begin{smallmatrix} 1 & 1 \\ 1 & 1 \end{smallmatrix}\right)$, $B=\left(\begin{smallmatrix} -3 & -1 \\ 1 & 3 \end{smallmatrix}\right)$, and $\alpha,\beta \in \mathbb{R}$, are the only matrices in $S$ that satisfy the given condition.

['linear_algebra']

theory putnam_2015_b3 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin definition putnam_2015_b3_solution :: "(real^2^2) set" where "putnam_2015_b3_solution \<equiv> undefined" (* {M::real^2^2. (\<exists>\<alpha> \<beta>::real. M$1$1 = \<alpha>*1 + \<beta>*-3 \<and> M$1$2 = \<alpha>*1 + \<beta>*-1 \<and> M$2$1 = \<alpha>*1 + \<beta>*1 \<and> M$2$2 = \<alpha>*1 + \<beta>*3)} *) theorem putnam_2015_b3: fixes S :: "(real^2^2) set" and M :: "real^2^2" and Mpow :: "real^2^2 \<Rightarrow> nat \<Rightarrow> real^2^2" and MinS :: bool assumes hS: "S \<equiv> {M'::real^2^2. (M'$1$2 - M'$1$1 = M'$2$1 - M'$1$2) \<and> (M'$1$2 - M'$1$1 = M'$2$2 - M'$2$1)}" and hMpow: "\<forall>A::real^2^2. Mpow A 0 = mat 1 \<and> (\<forall>k::nat>0. Mpow A k = Mpow A (k-1) ** A)" and hMinS: "MinS \<equiv> (M \<in> S \<and> (\<exists>k::nat>1. Mpow M k \<in> S))" shows "MinS \<longleftrightarrow> M \<in> putnam_2015_b3_solution" sorry end

putnam_2015_b4

abbrev putnam_2015_b4_solution : ℤ × ℕ := sorry -- (17, 21) theorem putnam_2015_b4 (quotientof : ℚ → (ℤ × ℕ)) (hquotientof : ∀ q : ℚ, quotientof q = (q.num, q.den)) : quotientof (∑' t : (Fin 3 → ℤ), if (∀ n : Fin 3, t n > 0) ∧ t 0 < t 1 + t 2 ∧ t 1 < t 2 + t 0 ∧ t 2 < t 0 + t 1 then 2^(t 0)/(3^(t 1)*5^(t 2)) else 0) = putnam_2015_b4_solution := sorry

Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a,b,c$. Express \[ \sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c} \] as a rational number in lowest terms.

The answer is $17/21$.

['algebra']

Section putnam_2015_b4. Require Import Nat List Reals Coquelicot.Coquelicot. From mathcomp Require Import div. Open Scope nat_scope. Definition putnam_2015_b4_solution := (17, 21). Theorem putnam_2015_b4: let fix exprl2 (l : list nat) : R := match l with | a :: b :: c :: _ => Rdiv (2 ^ a) (3 ^ b * 5 ^ c) | _ => R0 end in let fix exprl (l : list (list nat)) : list R := match l with | nil => nil | h :: t => exprl2 h :: exprl t end in forall (E: list (list nat)) (l: list nat), (In l E <-> (length l = 3 /\ let a := nth 0 l 0 in let b := nth 1 l 0 in let c := nth 2 l 0 in a > 0 /\ b > 0 /\ c > 0 /\ a < c + b /\ b < a + c /\ c < a + b)) -> exists (p q: nat), coprime p q = true /\ fold_left Rplus (exprl E) R0 = Rdiv (INR p) (INR q) /\ (p, q) = putnam_2015_b4_solution. Proof. Admitted. End putnam_2015_b4.

theory putnam_2015_b4 imports Complex_Main "HOL-Analysis.Infinite_Sum" "HOL-Analysis.Finite_Cartesian_Product" begin definition putnam_2015_b4_solution :: "int \<times> int" where "putnam_2015_b4_solution \<equiv> undefined" (* (17,21) *) theorem putnam_2015_b4: shows "\<exists>q::rat. (\<Sum>\<^sub>\<infinity>t::nat^3. if ((t$1 > 0 \<and> t$2 > 0 \<and> t$3 > 0) \<and> t$1 < t$2 + t$3 \<and> t$2 < t$3 + t$1 \<and> t$3 < t$1 + t$2) then (2^(t$1) / (3^(t$2)*5^(t$3))) else 0) = q \<and> quotient_of q = putnam_2015_b4_solution" sorry end

putnam_2015_b5

abbrev putnam_2015_b5_solution : ℕ := sorry -- 4 theorem putnam_2015_b5 (P : ℕ → ℕ := fun n ↦ {pi : Finset.Icc 1 n → Finset.Icc 1 n | Bijective pi ∧ ∀ i j : Finset.Icc 1 n, Nat.dist i j = 1 → Nat.dist (pi i) (pi j) ≤ 2}.ncard) : (∀ n : ℕ, n ≥ 2 → (P (n + 5) : ℤ) - (P (n + 4) : ℤ) - (P (n + 3) : ℤ) + (P n : ℤ) = putnam_2015_b5_solution) := sorry

Let $P_n$ be the number of permutations $\pi$ of $\{1,2,\dots,n\}$ such that \[ |i-j| = 1 \mbox{ implies } |\pi(i) -\pi(j)| \leq 2 \] for all $i,j$ in $\{1,2,\dots,n\}$. Show that for $n \geq 2$, the quantity \[ P_{n+5} - P_{n+4} - P_{n+3} + P_n \] does not depend on $n$, and find its value.

Prove that answer is $4$.

['algebra']

Section putnam_2015_b5. Require Import Reals. From mathcomp Require Import fintype perm ssrbool. Open Scope nat_scope. Definition putnam_2015_b5_solution := 4. Theorem putnam_2015_b5: let cond (n: nat) (π: {perm 'I_n}) : bool := forallb (fun i => forallb (fun j => if Z.to_nat (Z.abs (Z.of_nat (nat_of_ord i) - Z.of_nat (nat_of_ord j))) =? 1 then Z.to_nat (Z.abs (Z.of_nat (nat_of_ord (π i)) - Z.of_nat (nat_of_ord (π j)))) <=? 2 else true ) (ord_enum n) ) (ord_enum n) in let P (n : nat) : nat := let perms := enum 'S_n in let valid_perms := filter (fun π => cond n π) perms in length valid_perms in forall (n: nat), n >= 2 -> P (n+5) - P (n+4) - P (n+3) + P n = putnam_2015_b5_solution. Proof. Admitted. End putnam_2015_b5.

theory putnam_2015_b5 imports Complex_Main "HOL-Combinatorics.Permutations" begin definition putnam_2015_b5_solution::nat where "putnam_2015_b5_solution \<equiv> undefined" (* 4 *) theorem putnam_2015_b5: fixes P::"nat\<Rightarrow>nat" defines "P \<equiv> \<lambda>n. card ({pi::nat\<Rightarrow>nat. pi permutes {1..n} \<and> (\<forall>i\<in>{1..n}. \<forall>j\<in>{1..n}. abs(i - j) = 1 \<longrightarrow> abs(pi i - pi j) \<le> 2)})" shows "\<forall>n::nat \<ge> 2. P (n+5) - P (n+4) - P(n+3) + P n = putnam_2015_b5_solution" sorry end

putnam_2015_b6

abbrev putnam_2015_b6_solution : ℝ := sorry -- Real.pi ^ 2 / 16 theorem putnam_2015_b6 (A : ℕ → ℕ) (hA : ∀ k > 0, A k = {j : ℕ | Odd j ∧ j ∣ k ∧ j < Real.sqrt (2 * k)}.encard) : ∑' k : Set.Ici 1, (-1 : ℝ) ^ ((k : ℝ) - 1) * (A k / (k : ℝ)) = putnam_2015_b6_solution := sorry

For each positive integer $k$, let $A(k)$ be the number of odd divisors of $k$ in the interval $[1,\sqrt{2k})$. Evaluate $\sum_{k=1}^\infty (-1)^{k-1}\frac{A(k)}{k}$.

Show that the sum converges to $\pi^2/16$.

['analysis', 'number_theory']

theory putnam_2015_b6 imports Complex_Main "HOL-Analysis.Infinite_Sum" begin definition putnam_2015_b6_solution :: real where "putnam_2015_b6_solution \<equiv> undefined" (* pi^2 / 16 *) theorem putnam_2015_b6: fixes A :: "nat \<Rightarrow> nat" assumes hA: "\<forall>k::nat>0. A k = card {j::nat. odd j \<and> j dvd k \<and> j < sqrt (2*k)}" shows "(\<Sum>\<^sub>\<infinity>k::nat\<in>{1..}. (-1)^(k-1) * (A k / k)) = putnam_2015_b6_solution" sorry end

putnam_1996_a1

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Section putnam_1996_a1. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Theorem putnam_1996_a1: exists (minA: R), forall (A: R), let packable (n1 n2 a1 a2: R) := (n1 + n2) <= Rmax a1 a2 /\ Rmax n1 n2 <= Rmin a1 a2 in forall (n1 n2: R), pow n1 2 + pow n2 2 = 1 -> exists (a1 a2: R), a1 * a2 = minA /\ packable n1 n2 a1 a2 /\ exists (a1 a2: R), a1 * a2 = A /\ packable n1 n2 a1 a2 -> minA <= A. Proof. Admitted. End putnam_1996_a1.

putnam_2009_a2

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Section putnam_2009_a2. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2009_a2_solution : (R -> R) := fun x => Rpower 2 (-1 / 12) * (sin (6 * x + PI / 4) / (cos (6 * x + PI / 4)) ^ 2) ^ (1 / 6). Theorem putnam_2009_a2: forall (f g h: R -> R), exists (a b: R), a < b /\ forall (x: R), a < x < b -> (ex_derive f x /\ ex_derive g x /\ ex_derive h x /\ Derive f x = 2 * f (f (g (h x))) + 1 / g (h x) /\ f 0 = 1 /\ Derive g x = 2 * f (g (g (h x))) + 4 / f (h x) /\ g 0 = 1 /\ Derive h x = 3 * f (g (h (h x))) + 1 / f (g x) /\ h 0 = 1) -> f = putnam_2009_a2_solution. Proof. Admitted. End putnam_2009_a2.

putnam_1975_a5

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Section putnam_1975_a5. Require Import Basics Factorial Reals Coquelicot.Series. Open Scope R. Theorem putnam_1975_a5: let f0 (x: R) := Rpower (exp 1) x in let fix compose_n {A: Type} (f : A -> A) (n : nat) := match n with | O => fun x => x | S n' => compose f (compose_n f n') end in Series (fun n => ((compose_n f0 n) 1)/(INR (fact n))) = Rpower (exp 1) (exp 1). Proof. Admitted. End putnam_1975_a5.

putnam_1975_b6

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Section putnam_1975_b6. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Theorem putnam_1975_b6 (h : nat -> R := fun n => sum_n (fun r => 1 / (INR r + 1)) n) : forall (n: nat), gt n 2 -> INR n * INR (n + 1) ^ (1 / n) < h n < INR n - INR (n - 1) * Rpower (INR n) (-R1 / INR (n - 1)) . Proof. Admitted. End putnam_1975_b6.

putnam_1997_a1

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Section putnam_1997_a1. Require Import Reals Rgeom ZArith GeoCoq.Main.Tarski_dev.Ch16_coordinates_with_functions GeoCoq.Main.Annexes.midpoint_theorems GeoCoq.Main.Highschool.circumcenter. Context `{T2D:Tarski_2D} `{TE:@Tarski_euclidean Tn TnEQD}. Open Scope R. Definition putnam_1997_a1_solution := 28. Theorem putnam_1997_a1 (pt_to_R : Tpoint -> (R * R)) (dist : Tpoint -> Tpoint -> R := fun A B => let (a, b) := pt_to_R A in let (c, d) := pt_to_R B in dist_euc a b c d) (A B C : Tpoint) (Hp Op Mp Fp : Tpoint) (l1 : dist Hp Op = 11) (l2 : dist Op Mp = 5) (s : Rectangle Hp Op Mp Fp) (hHp : Bet A Fp Hp) (* H as the intersection of the altitudes *) (hOp : is_circumcenter Op A B C) (* O the center of the circumscribed circle *) (hMp : Midpoint B C Mp) (* M the midpoint of BC *) (hFp : Perp A C B Fp /\ Col A C Fp) (* foot of the altitude *) : dist B C = putnam_1997_a1_solution. Proof. Admitted. End putnam_1997_a1.

putnam_2003_b5

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Section putnam_2003_b5. Require Import Reals GeoCoq.Main.Tarski_dev.Ch16_coordinates_with_functions GeoCoq.Axioms.Definitions GeoCoq.Main.Highschool.triangles. Context `{T2D:Tarski_2D} `{TE:@Tarski_euclidean Tn TnEQD}. Open Scope R. Definition putnam_2003_b5_solution (pt_to_R : Tpoint -> (R * R)) (dist : Tpoint -> Tpoint -> R) (P Op : Tpoint) := sqrt 3 * (1 - (dist P Op) ^ 2 - 1). Theorem putnam_2003_b5 (pt_to_R : Tpoint -> (R * R)) (F_to_R : F -> R) (dist : Tpoint -> Tpoint -> R := fun A B => let (a, b) := pt_to_R A in let (c, d) := pt_to_R B in dist_euc a b c d) (Triangle : Tpoint -> Tpoint -> Tpoint -> Prop := fun x y z => ~ Col x y z) (* copied from GeoCoq.Axioms.euclidean_axioms *) (A B C Op Op' P: Tpoint) (fixpoint : dist Op Op' = R1) (hABC : OnCircle A Op Op' /\ OnCircle B Op Op' /\ OnCircle C Op Op') (hABC' : Main.Highschool.triangles.equilateral A B C) (hp : InCircle P Op Op') (a : R := dist P A) (b : R := dist P B) (c : R := dist P C) : exists (A' B' C' : Tpoint) (D: Cs O E A' B' C'), Triangle A' B' C' /\ dist A' B' = a /\ dist B' C' = b /\ dist C' A' = c /\ F_to_R (signed_area A' B' C' D A' B' C') = (putnam_2003_b5_solution pt_to_R dist P Op). Proof. Admitted. End putnam_2003_b5.

putnam_2014_a4

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Section putnam_2014_a4. From mathcomp.analysis Require Import probability. From mathcomp Require Import all_ssreflect. From mathcomp Require Import ssralg poly ssrnum ssrint interval finmap. From mathcomp Require Import mathcomp_extra boolp classical_sets functions. From mathcomp Require Import cardinality fsbigop. From HB Require Import structures. From mathcomp.analysis Require Import exp numfun lebesgue_measure lebesgue_integral. From mathcomp.analysis Require Import reals ereal signed topology normedtype sequences esum measure. From mathcomp.analysis Require Import exp numfun lebesgue_measure lebesgue_integral kernel. Context d (T : measurableType d) (R : realType) (P : probability T R). Definition putnam_2014_a4_solution : R := 1 / 3. Local Open Scope ring_scope. Theorem putnam_2014_a4 (X : {RV P >-> R}) (ed := @expectation _ _ _ P X = 1%:E) (ed2 := @expectation _ _ _ P (X * X) = 2%:E) (ed3 := @expectation _ _ _ P (X * X * X) = 5%:E) (minval : R) (de := distribution P X) : forall (P : probability T R), (minval <= (pmf X 0) /\ exists (P : probability T R), minval = (pmf X 0)) <-> minval = putnam_2014_a4_solution. Proof. Admitted. End putnam_2014_a4.

putnam_1987_a3

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Section putnam_1987_a3. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1987_a3_solution1 := False. Definition putnam_1987_a3_solution2 := True. Theorem putnam_1987_a3: exists (f g: R -> R) (x: R), ((Derive_n f 2) x - 2 * (Derive_n f 1) x + f x = 2 * exp x /\ forall (x: R), f x > 0) -> forall (x: R), Derive_n f 1 x > 0 <-> putnam_1987_a3_solution1 /\ ((Derive_n g 2) x - 2 * (Derive_n g 1) x + g x = 2 * exp x /\ forall (x: R), Derive_n g 1 x > 0) -> forall (x: R), g x > 0 <-> putnam_1987_a3_solution2. Proof. Admitted. End putnam_1987_a3.