Dataset Viewer
Auto-converted to Parquet Duplicate
name
stringlengths
14
14
lean4_statement
stringlengths
75
1.48k
βŒ€
informal_statement
stringlengths
47
898
βŒ€
informal_solution
stringlengths
16
303
βŒ€
tags
stringclasses
36 values
coq_statement
stringlengths
91
1.51k
βŒ€
isabelle_statement
stringlengths
160
2.65k
βŒ€
lean4_proof
float64
putnam_1981_a1
abbrev putnam_1981_a1_solution : ℝ := sorry -- 1/8 theorem putnam_1981_a1 (P : β„• β†’ β„• β†’ Prop := fun n k : β„• => 5^k ∣ ∏ m in Finset.Icc 1 n, (m^m : β„€)) (E : β„• β†’ β„•) (hE : βˆ€ n ∈ Ici 1, P n (E n) ∧ βˆ€ k : β„•, P n k β†’ k ≀ E n) : Tendsto (fun n : β„• => ((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' => ...
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...
null
putnam_1981_a3
abbrev putnam_1981_a3_solution : Prop := sorry -- False theorem putnam_1981_a3 (f : ℝ β†’ 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)) : (βˆƒ 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_integ...
null
putnam_1981_a5
abbrev putnam_1981_a5_solution : Prop := sorry -- True theorem putnam_1981_a5 (Q : Polynomial ℝ β†’ Polynomial ℝ := fun P : Polynomial ℝ => (X^2 + 1)*P*(derivative P) + X*(P^2 + (derivative P)^2)) (n : Polynomial ℝ β†’ ℝ := fun P : Polynomial ℝ => {x ∈ Ioi 1 | P.eval x = 0}.ncard) : (βˆ€ P : Polynomial ℝ, {x : ℝ | (Q P).eval...
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']
null
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 ...
null
putnam_1981_b1
abbrev putnam_1981_b1_solution : ℝ := sorry -- -1 theorem putnam_1981_b1 (f : β„• β†’ ℝ := fun n : β„• => (1/n^5) * βˆ‘ h in Finset.Icc 1 n, βˆ‘ 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_solutio...
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 ...
null
putnam_1981_b2
abbrev putnam_1981_b2_solution : ℝ := sorry -- 12 - 8 * Real.sqrt 2 theorem putnam_1981_b2 (P : ℝ Γ— ℝ Γ— ℝ β†’ Prop := fun (r, s, t) => 1 ≀ r ∧ r ≀ s ∧ s ≀ t ∧ t ≀ 4) (f : ℝ Γ— ℝ Γ— ℝ β†’ ℝ := fun (r, s, t) => (r - 1)^2 + (s/r - 1)^2 + (t/s - 1)^2 + (4/t - 1)^2) : (βˆƒ r : ℝ, βˆƒ s : ℝ, βˆƒ t : ℝ, P (r, s, t) ∧ f (r, s, t) = putnam...
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_so...
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...
null
putnam_1981_b3
theorem putnam_1981_b3 (P : β„• β†’ Prop := fun n : β„• => βˆ€ p : β„•, (Nat.Prime p ∧ p ∣ n^2 + 3) β†’ βˆƒ k : β„•, (p : β„€) ∣ (k : β„€)^2 + 3 ∧ k^2 < n) : βˆ€ n : β„•, βˆƒ m : β„•, (m : β„€) > 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 $k^2 < n$.
null
['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) -> ...
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...
null
putnam_1981_b4
abbrev putnam_1981_b4_solution : Prop := sorry -- False theorem putnam_1981_b4 (VAB : Set (Matrix (Fin 5) (Fin 7) ℝ) β†’ Prop) (Vrank : Set (Matrix (Fin 5) (Fin 7) ℝ) β†’ β„• β†’ Prop) (hVAB : βˆ€ V : Set (Matrix (Fin 5) (Fin 7) ℝ), VAB V = (βˆ€ A ∈ V, βˆ€ B ∈ V, βˆ€ r s : ℝ, r β€’ A + s β€’ B ∈ V)) (hVrank : βˆ€ (V : Set (Matrix (Fin 5) (F...
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 ra...
Show that the assertion is false.
['linear_algebra']
null
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 :: "(r...
null
putnam_1981_b5
abbrev putnam_1981_b5_solution : Prop := sorry -- True theorem putnam_1981_b5 (sumbits : List β„• β†’ β„€) (B : β„• β†’ β„€) (hsumbits : βˆ€ bits : List β„•, sumbits bits = βˆ‘ i : Fin bits.length, (bits[i] : β„€)) (hB : βˆ€ n > 0, B n = sumbits (Nat.digits 2 n)) : (βˆƒ q : β„š, Real.exp (βˆ‘' n : Set.Ici 1, B n / ((n : ℝ) * ((n : ℝ) + 1))) = q) ...
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' ...
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. e...
null
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 β†’ ℝ) := ContDiff ℝ 1 f ∧ ContDiff ℝ 1 g ∧ (βˆƒ x y : ℝ, f x = y ∧ g y = x ∧ (deriv f) x = 1 / (deriv g) y) theorem putnam_2007_a1 : βˆ€ a : ℝ, reflect_tangent (fun x => a * x^2 + a * ...
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']
null
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, ...
null
putnam_2007_a2
abbrev putnam_2007_a2_solution : ENNReal := sorry -- 4 theorem putnam_2007_a2 (Sinterpos : Set (Fin 2 β†’ ℝ) β†’ Prop) (Sinterneg : Set (Fin 2 β†’ ℝ) β†’ Prop) (Sconv : Set (Fin 2 β†’ ℝ) β†’ Prop) (hSinterpos : βˆ€ S : Set (Fin 2 β†’ ℝ), Sinterpos S = ((βˆƒ p ∈ S, p 0 > 0 ∧ p 1 > 0 ∧ p 0 * p 1 = 1) ∧ (βˆƒ p ∈ S, p 0 < 0 ∧ p 1 < 0 ∧ p 0 * ...
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']
null
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> boo...
null
putnam_2007_a4
abbrev putnam_2007_a4_solution : Set (Polynomial ℝ) := sorry -- {f : Polynomial ℝ | βˆƒ d : β„•, βˆƒ c β‰₯ (1 : β„€) - d, βˆ€ n : ℝ, f.eval n = (1 / 9) * ((10 ^ c) * (9 * n + 1) ^ d - 1)} theorem putnam_2007_a4 (S : Set (Polynomial ℝ)) (repunit : ℝ β†’ Prop := fun x ↦ x > 0 ∧ x = floor x ∧ βˆ€ d ∈ (digits 10 (floor x)), d = 1) (hS : βˆ€...
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) ...
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) * ...
null
putnam_2007_a5
theorem putnam_2007_a5 (G : Type*) [Group G] [Fintype G] (p : β„•) (n : β„•) (hp : Nat.Prime p) (hn : n = {g : G | orderOf g = p}.encard) : n = 0 ∨ p ∣ (n + 1) := sorry
null
null
[]
null
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 \...
null
putnam_2007_b1
theorem putnam_2007_b1 (f : Polynomial β„€) (hf : βˆ€ n : β„•, f.coeff n β‰₯ 0) (hfnconst : βˆƒ n : β„•, n > 0 ∧ f.coeff n > 0) (n : β„€) (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 $n = 1$
null
['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.t...
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'...
null
putnam_2007_b2
theorem putnam_2007_b2 (f : ℝ β†’ ℝ) (hf : ContDiffOn ℝ 1 f (Icc 0 1)) (hfint : ∫ x in (0)..1, f x = 0) (max : ℝ) (heqmax : βˆƒ x ∈ Icc (0 : ℝ) 1, |deriv f x| = max) (hmaxub : βˆ€ x ∈ Icc (0 : ℝ) 1, |deriv f x| ≀ max) : (βˆ€ Ξ± ∈ (Ioo (0 : ℝ) 1), |∫ x in (0)..Ξ±, 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 $\alpha \in (0,1)$, \[ \left| \int_0^\alpha f(x)\,dx \right| \leq \frac{1}{8} \max_{0 \leq x \leq 1} |f'(x)|. \]
null
['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...
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 "\<foral...
null
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 : β„€)) theorem putnam_2007_b3 (x : β„• β†’ ℝ) (hx0 : x 0 = 1) (hx : βˆ€ n : β„•, x (n + 1) = 3 * (x n) + ⌊(x n) * Real.sqrt 5βŒ‹) : (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'...
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...
null
putnam_2007_b4
abbrev putnam_2007_b4_solution : β„• β†’ β„• := sorry -- fun n ↦ 2 ^ (n + 1) theorem putnam_2007_b4 (n : β„•) (npos : n > 0) : ({(P, Q) : (Polynomial ℝ) Γ— (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']
null
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 :: r...
null
putnam_2007_b5
theorem putnam_2007_b5 (k : β„•) (kpos : k > 0) : (βˆƒ P : Finset.range k β†’ Polynomial ℝ, βˆ€ n : β„€, ⌊(n : ℝ) / kβŒ‹ ^ k = βˆ‘ i : Finset.range k, (P i).eval (n : ℝ) * ⌊(n : ℝ) / kβŒ‹ ^ (i : β„•)) := 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...
null
['algebra']
null
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\<...
null
putnam_1980_a2
abbrev putnam_1980_a2_solution : β„• β†’ β„• β†’ β„• := sorry -- (fun r s : β„• => (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2)) theorem putnam_1980_a2 (r s : β„•) (abcdlcm : β„• β†’ β„• β†’ β„• β†’ β„• β†’ Prop) (rspos : r > 0 ∧ s > 0) (habcdlcm : βˆ€ a b c d : β„•, abcdlcm a b c d = (a > 0 ∧ b > 0 ∧ c > 0 ∧ d > 0 ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat....
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...
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) <-> ...
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 :: "...
null
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 ...
null
putnam_1980_a4
theorem putnam_1980_a4 (abcvals : β„€ β†’ β„€ β†’ β„€ β†’ Prop) (habcvals : βˆ€ a b c : β„€, abcvals a b c = (Β¬(a = 0 ∧ b = 0 ∧ c = 0) ∧ |a| < 1000000 ∧ |b| < 1000000 ∧ |c| < 1000000)) : (βˆƒ a b c : β„€, abcvals a b c ∧ |a + b * Real.sqrt 2 + c * Real.sqrt 3| < 10 ^ (-(11 : ℝ))) ∧ (βˆ€ a b c : β„€, abcvals a b c β†’ |a + b * Real.sqrt 2 + c * ...
\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^{-2...
null
['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)) ) /\ ...
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...
null
putnam_1980_a5
theorem putnam_1980_a5 (P : Polynomial ℝ) (xeqs : ℝ β†’ Prop) (Pnonconst : P.degree > 0) (hxeqs : βˆ€ 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 $x$.
null
['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)...
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 ...
null
putnam_1980_a6
abbrev putnam_1980_a6_solution : ℝ := sorry -- 1 / Real.exp 1 theorem putnam_1980_a6 (C : Set (ℝ β†’ ℝ)) (uleint : ℝ β†’ Prop) (hC : C = {f : ℝ β†’ ℝ | ContDiffOn ℝ 1 f (Set.Icc 0 1) ∧ f 0 = 0 ∧ f 1 = 1}) (huleint : βˆ€ u : ℝ, uleint u = βˆ€ f ∈ C, u ≀ (∫ x in Set.Ioo 0 1, |deriv f x - f x|)) : uleint putnam_1980_a6_solution ∧ (...
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']
null
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 \<e...
null
putnam_1980_b1
abbrev putnam_1980_b1_solution : Set ℝ := sorry -- {c : ℝ | c β‰₯ 1 / 2} theorem putnam_1980_b1 (c : ℝ) : (βˆ€ 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)) \<longleftrightarr...
null
putnam_1980_b3
abbrev putnam_1980_b3_solution : Set ℝ := sorry -- {a : ℝ | a β‰₯ 3} theorem putnam_1980_b3 (a : ℝ) (u : β„• β†’ ℝ) (hu : u 0 = a ∧ (βˆ€ n : β„•, u (n + 1) = 2 * u n - n ^ 2)) : (βˆ€ n : β„•, 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 >...
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) = ...
null
putnam_1980_b4
theorem putnam_1980_b4 {T : Type} (X : Finset T) (A : Fin 1066 β†’ Finset T) (hX : X.card β‰₯ 10) (hA : βˆ€ i : Fin 1066, A i βŠ† X ∧ (A i).card > ((1 : β„š)/2) * X.card) : βˆƒ Y : Finset T, Y βŠ† X ∧ Y.card = 10 ∧ βˆ€ i : Fin 1066, βˆƒ 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 $x_1, x_2, \dots, x_{10}$.
null
['set_theory', 'combinatorics']
null
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...
null
putnam_1980_b5
abbrev putnam_1980_b5_solution : ℝ β†’ Prop := sorry -- fun t : ℝ => 1 β‰₯ t theorem putnam_1980_b5 (T : Set ℝ := Icc 0 1) (P : ℝ β†’ (ℝ β†’ ℝ) β†’ Prop := fun t : ℝ => fun f : ℝ β†’ ℝ => f 1 - 2*f (2/3) + f (1/3) β‰₯ t*(f (2/3) - 2*f (1/3) + f 0)) (Convex : (ℝ β†’ ℝ) β†’ Prop := fun f : ℝ β†’ ℝ => βˆ€ u ∈ T, βˆ€ v ∈ T, βˆ€ s ∈ T, f (s*u + (1 -...
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\le...
$S_t$ is closed under multiplication if and only if $1 \ge t$.
['analysis', 'algebra']
null
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>...
null
putnam_1980_b6
theorem putnam_1980_b6 (G : β„€ Γ— β„€ β†’ β„š) (hG : βˆ€ d n : β„•, d ≀ n β†’ (d = 1 β†’ G (d, n) = 1/(n : β„š)) ∧ (d > 1 β†’ G (d, n) = (d/(n : β„š))*βˆ‘ i in Finset.Icc d n, G ((d : β„€) - 1, (i : β„€) - 1))) : βˆ€ d p : β„•, 1 < d ∧ d ≀ p ∧ Prime p β†’ Β¬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 $p$.
null
['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 => ...
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) *...
null
putnam_1984_a2
abbrev putnam_1984_a2_solution : β„š := sorry -- 2 theorem putnam_1984_a2 : βˆ‘' k : Set.Ici 1, (6 ^ (k : β„•) / ((3 ^ ((k : β„•) + 1) - 2 ^ ((k : β„•) + 1)) * (3 ^ (k : β„•) - 2 ^ (k : β„•)))) = 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_...
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
null
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 : β„•) (a b : ℝ) (Mn : ℝ β†’ Matrix (Fin (2 * n)) (Fin (2 * n)) ℝ) (polyabn : Fin 3 β†’ ℝ) (npos : n > 0) (aneb : a β‰  b) (hMnx : βˆ€ x : ℝ, βˆ€ i : Fin (2 * n), ...
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}....
Show that $\lim_{x \to a} \frac{\det M_n}{(x-a)^{2n-2}}=n^2(a^2-b^2)$.
['linear_algebra', 'analysis']
null
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:])...
null
putnam_1984_a5
abbrev putnam_1984_a5_solution : β„• Γ— β„• Γ— β„• Γ— β„• Γ— β„• := sorry -- (1, 9, 8, 4, 25) theorem putnam_1984_a5 (R : Set (Fin 3 β†’ ℝ)) (w : (Fin 3 β†’ ℝ) β†’ ℝ) (hR : R = {p : Fin 3 β†’ ℝ | (βˆ€ i : Fin 3, p i β‰₯ 0) ∧ p 0 + p 1 + p 2 ≀ 1}) (hw : βˆ€ p : Fin 3 β†’ ℝ, w p = 1 - p 0 - p 1 - p 2) : let (a, b, c, d, n) := putnam_1984_a5_solution;...
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 ...
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 ...
null
putnam_1984_a6
abbrev putnam_1984_a6_solution : Prop Γ— β„• := sorry -- (True, 4) theorem putnam_1984_a6 (lastnzdig : List β„• β†’ β„•) (f : β„• β†’ β„•) (kadistinct : β„• β†’ (β„• β†’ β„•) β†’ Prop := fun k : β„• => fun a : β„• β†’ β„• => (k β‰₯ 1 ∧ βˆ€ i j : Fin k, i β‰  j β†’ a i β‰  a j)) (gpeq : (β„• β†’ β„•) β†’ β„• β†’ Prop := fun g : β„• β†’ β„• => fun p : β„• => (p > 0 ∧ βˆ€ s β‰₯ 1, g s = g ...
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...
Show that the least such $p$ is $p=4$.
['algebra', 'number_theory']
null
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>...
null
putnam_1984_b1
abbrev putnam_1984_b1_solution : Polynomial ℝ Γ— Polynomial ℝ := sorry -- (Polynomial.X + 3, -Polynomial.X - 2) theorem putnam_1984_b1 (f : β„• β†’ β„€) (hf : βˆ€ n > 0, f n = βˆ‘ i : Set.Icc 1 n, ((i)! : β„€)) : let (P, Q) := putnam_1984_b1_solution; βˆ€ n β‰₯ 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 =...
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 \<Righ...
null
putnam_1984_b2
abbrev putnam_1984_b2_solution : ℝ := sorry -- 8 theorem putnam_1984_b2 (f : ℝ β†’ ℝ β†’ ℝ) (hf : βˆ€ u v : ℝ, f u v = (u - v) ^ 2 + (Real.sqrt (2 - u ^ 2) - 9 / v) ^ 2) : (βˆƒ u : Set.Ioo 0 (Real.sqrt 2), βˆƒ v > 0, f u v = putnam_1984_b2_solution) ∧ (βˆ€ u : Set.Ioo 0 (Real.sqrt 2), βˆ€ v > 0, f u v β‰₯ putnam_1984_b2_solution) := s...
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 ...
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" ...
null
putnam_1984_b3
abbrev putnam_1984_b3_solution : Prop := sorry -- True theorem putnam_1984_b3 : (βˆ€ (F : Type*) (_ : Fintype F), Fintype.card F β‰₯ 2 β†’ (βˆƒ mul : F β†’ F β†’ F, βˆ€ x y z : F, (mul x z = mul y z β†’ x = y) ∧ (mul x (mul y z) β‰  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 associat...
Show that the statement is true.
['abstract_algebra']
null
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 \<R...
null
putnam_1984_b5
abbrev putnam_1984_b5_solution : β„€ Γ— Polynomial ℝ Γ— Polynomial β„• := sorry -- (2, (Polynomial.X * (Polynomial.X - 1)) / 2, Polynomial.X) theorem putnam_1984_b5 (sumbits : List β„• β†’ β„•) (d : β„• β†’ β„•) (m : β„•) (hsumbits : βˆ€ bits : List β„•, sumbits bits = βˆ‘ i : Fin bits.length, bits[i]) (hd : βˆ€ k : β„•, d k = sumbits (Nat.digits 2...
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 | _...
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_b...
null
putnam_2009_a1
abbrev putnam_2009_a1_solution : Prop := sorry -- True theorem putnam_2009_a1 : ((βˆ€ f : (ℝ Γ— ℝ) β†’ ℝ, (βˆ€ O v : ℝ Γ— ℝ, v β‰  (0, 0) β†’ 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) β†’ βˆ€ P : ℝ Γ— 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...
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...
null
putnam_2009_a3
abbrev putnam_2009_a3_solution : ℝ := sorry -- 0 theorem putnam_2009_a3 (cos_matrix : (n : β„•) β†’ Matrix (Fin n) (Fin n) ℝ) (hM : βˆ€ n : β„•, βˆ€ 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 ...
Show that the limit is 0.
['linear_algebra', 'analysis']
null
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 \<Right...
null
putnam_2009_a4
abbrev putnam_2009_a4_solution : Prop := sorry -- False theorem putnam_2009_a4 : ((βˆ€ S : Set β„š, 0 ∈ S β†’ (βˆ€ x ∈ S, x + 1 ∈ S ∧ x - 1 ∈ S) β†’ (βˆ€ x ∈ S, x βˆ‰ ({0, 1} : Set β„š) β†’ 1 / (x * (x - 1)) ∈ S) β†’ βˆ€ r : β„š, 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 <-> ...
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\...
null
putnam_2009_a5
abbrev putnam_2009_a5_solution : Prop := sorry -- False theorem putnam_2009_a5 : (βˆƒ (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']
null
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. fini...
null
putnam_2009_b1
theorem putnam_2009_b1 (isquotprodprimefact : β„š β†’ Prop := fun q => (βˆƒ (k m : β„•) (a : Fin k β†’ β„•) (b : Fin m β†’ β„•), (βˆ€ i : Fin k, Nat.Prime (a i)) ∧ (βˆ€ j : Fin m, Nat.Prime (b j)) ∧ (q = (∏ i : Fin k, Nat.factorial (a i))/(∏ j : Fin m, Nat.factorial (b j))) )) : βˆ€ q : β„š, q > 0 β†’ 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!).
null
['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...
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::...
null
putnam_2009_b3
abbrev putnam_2009_b3_solution : Set β„€ := sorry -- {n : β„€ | βˆƒ k β‰₯ 1, n = 2 ^ k - 1} theorem putnam_2009_b3 (mediocre : β„€ β†’ Set β„€ β†’ Prop := fun n S ↦ (S βŠ† Icc 1 n) ∧ βˆ€ a ∈ S, βˆ€ b ∈ S, 2 ∣ a + b β†’ (a + b) / 2 ∈ S) (A : β„€ β†’ β„€ := fun n ↦ {S : Set β„€ | mediocre n S}.ncard) : ({n : β„€ | n > 0 ∧ A (n + 2) - 2 * A (n + 1) + A n ...
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 $...
Show that the answer is $n = 2^k - 1$ for some integer $k$.
['number_theory']
null
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 \<Rightarro...
null
putnam_2009_b4
abbrev putnam_2009_b4_solution : β„• := sorry -- 2020050 theorem putnam_2009_b4 (balanced : MvPolynomial (Fin 2) ℝ β†’ Prop := fun P ↦ βˆ€ 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 : βˆ€ P : MvPolynomial (Fin 2) ℝ, P ∈ ...
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']
null
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) \...
null
putnam_2009_b5
theorem putnam_2009_b5 (f : ℝ β†’ ℝ) (hfdiff : DifferentiableOn ℝ f (Ioi 1)) (hf : βˆ€ 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 $\lim_{x \to \infty} f(x) = \infty$.
null
['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) ...
null
putnam_2009_b6
theorem putnam_2009_b6 (n : β„•) (npos : n > 0) : (βˆƒ a : β„• β†’ β„€, a 0 = 0 ∧ a 2009 = n ∧ βˆ€ i : Icc 1 2009, ((βˆƒ j k : β„•, j < i ∧ a i = a j + 2 ^ k) ∨ βˆƒ b c : β„•, 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 $c$....
null
['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) /\ ((ex...
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. \<...
null
putnam_2020_a1
abbrev putnam_2020_a1_solution : β„• := sorry -- 508536 theorem putnam_2020_a1 : Set.ncard {x : β„• | (2020 ∣ x) ∧ (Nat.log 10 x) + 1 ≀ 2020 ∧ (βˆƒ k l, k β‰₯ l ∧ x = βˆ‘ 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 ...
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...
null
putnam_2020_a2
abbrev putnam_2020_a2_solution : β„• β†’ β„• := sorry -- fun k ↦ 4 ^ k theorem putnam_2020_a2 (k : β„•) : (βˆ‘ 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" s...
null
putnam_2020_a3
abbrev putnam_2020_a3_solution : Prop := sorry -- False theorem putnam_2020_a3 (a : β„• β†’ ℝ) (ha0 : a 0 = Real.pi / 2) (ha : βˆ€ n : β„•, n β‰₯ 1 β†’ a n = Real.sin (a (n - 1))) : (βˆƒ L : ℝ, Tendsto (fun m : β„• => βˆ‘ 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_solut...
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>...
null
putnam_2020_a5
abbrev putnam_2020_a5_solution : β„€ := sorry -- (Nat.fib 4040) - 1 theorem putnam_2020_a5 (a : β„€ β†’ β„• := fun n : β„€ => {S : Finset β„• | (βˆ€ k ∈ S, k > 0) ∧ βˆ‘ k : S, Nat.fib k = n}.ncard) : a putnam_2020_a5_solution = 2020 ∧ βˆ€ n : β„€, a n = 2020 β†’ 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']
null
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 \<an...
null
putnam_2020_a6
abbrev putnam_2020_a6_solution : ℝ := sorry -- Real.pi / 4 theorem putnam_2020_a6 (f : β„€ β†’ (ℝ β†’ ℝ) := fun N : β„€ => fun x : ℝ => βˆ‘ n in Finset.Icc 0 N, (N + 1/2 - n)/((N + 1)*(2*n + 1)) * Real.sin ((2*n + 1)*x)) : (βˆ€ N > 0, βˆ€ x : ℝ, f N x ≀ putnam_2020_a6_solution) ∧ βˆ€ M : ℝ, (βˆ€ N > 0, βˆ€ x : ℝ, f N x ≀ M) β†’ M β‰₯ putnam_2...
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)) (...
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 ...
null
putnam_2020_b1
abbrev putnam_2020_b1_solution : β„• := sorry -- 1990 theorem putnam_2020_b1 (d : β„• β†’ β„• := fun n : β„• => βˆ‘ i : Fin (Nat.digits 2 n).length, (Nat.digits 2 n)[i]!) (S : β„€ := βˆ‘ k : Icc 1 2020, ((-1)^(d k))*(k : β„•)^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 ...
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 /...
null
putnam_2020_b4
abbrev putnam_2020_b4_solution : ℝ := sorry -- 1 / 4040 theorem putnam_2020_b4 (V : β„• β†’ Set (β„• β†’ β„€) := fun n ↦ ({s : β„• β†’ β„€ | s 0 = 0 ∧ (βˆ€ j β‰₯ 2 * n, s j = 0) ∧ (βˆ€ j ∈ Icc 1 (2 * n), |s j - s (j - 1)| = 1)})) (q : β„• β†’ (β„• β†’ β„€) β†’ ℝ := fun n s ↦ 1 + βˆ‘ j in Finset.Icc 1 (2 * n - 1), 3 ^ (s j)) (M : β„• β†’ ℝ := fun n ↦ (βˆ‘' v : ...
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(\m...
Show that the answer is $\frac{1}{4040}$.
['algebra']
null
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" ...
null
putnam_2020_b5
theorem putnam_2020_b5 (z : Fin 4 β†’ β„‚) (hzle1 : βˆ€ n, β€–z nβ€– < 1) (hzne1 : βˆ€ n, z n β‰  1) : 3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) β‰  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 \[ 3 - z_1 - z_2 - z_3 - z_4 + z_1 z_2 z_3 z_4 \neq 0. \]
null
['algebra']
null
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 * ...
null
putnam_2020_b6
theorem putnam_2020_b6 (n : β„•) (npos : n > 0) : βˆ‘ k : Fin n, ((-1) ^ Int.floor ((k.1 + 1) * (Real.sqrt 2 - 1)) : ℝ) β‰₯ 0 := sorry
Let $n$ be a positive integer. Prove that $\sum_{k=1}^n(-1)^{\lfloor k(\sqrt{2}-1) \rfloor} \geq 0$.
null
['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 pu...
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
null
putnam_1970_a1
theorem putnam_1970_a1 (a b : ℝ) (ha : a > 0) (hb : b > 0) (f : ℝ β†’ ℝ := fun x : ℝ => Real.exp (a*x) * Real.cos (b*x)) (p : β„• β†’ ℝ) (hp : βˆƒ c : ℝ, c > 0 ∧ βˆ€ x ∈ ball 0 c, βˆ‘' n : β„•, (p n)*x^n = f x) (S : Set β„• := {n : β„• | p n = 0}) : S = βˆ… ∨ Β¬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.
null
['analysis']
null
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::...
null
putnam_1970_a3
abbrev putnam_1970_a3_solution : β„• Γ— β„• := sorry -- (3, 1444) theorem putnam_1970_a3 (L : β„• β†’ β„•) (hL : βˆ€ n : β„•, L n ≀ (Nat.digits 10 n).length ∧ (βˆ€ k : β„•, k < L n β†’ (Nat.digits 10 n)[k]! = (Nat.digits 10 n)[0]!) ∧ (L n β‰  (Nat.digits 10 n).length β†’ (Nat.digits 10 n)[L n]! β‰  (Nat.digits 10 n)[0]!)) : (βˆƒ n : β„•, (Nat.digits...
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']
null
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 put...
null
putnam_1970_a4
theorem putnam_1970_a4 (x : β„• β†’ 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 $\lim_{n \to \infty} \frac{x_n - x_{n-1}}{n} = 0$.
null
['analysis']
null
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
null
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 : β„€) (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']
null
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_b...
null
putnam_1970_b3
theorem putnam_1970_b3 (S : Set (ℝ Γ— ℝ)) (a b : ℝ) (hab : a < b) (hS : βˆ€ s ∈ S, s.1 ∈ Ioo a b) (hSclosed : IsClosed S) : IsClosed {y | βˆƒ 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.
null
['analysis']
null
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)}" sor...
null
putnam_1970_b5
theorem putnam_1970_b5 (ramp : β„€ β†’ (ℝ β†’ ℝ) := fun n => (fun x => if x ≀ -n then -n else (if -n < x ∧ x ≀ n then x else n))) (F : ℝ β†’ ℝ) : Continuous F ↔ (βˆ€ n : β„•, 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 $n$.
null
['analysis']
null
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 "con...
null
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 β†’ ℝ := fun x ↦ x ^ 3 - 3 * x) : (βˆ€ x ∈ S, f x ≀ putnam_1986_a1_solution ∧ βˆƒ 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),...
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 ...
null
putnam_1986_a2
abbrev putnam_1986_a2_solution : β„• := 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
null
putnam_1986_a3
abbrev putnam_1986_a3_solution : ℝ := sorry -- Real.pi / 2 theorem putnam_1986_a3 (cot : ℝ β†’ ℝ := fun ΞΈ ↦ cos ΞΈ / sin ΞΈ) (arccot : ℝ β†’ ℝ) (harccot : βˆ€ t : ℝ, t β‰₯ 0 β†’ arccot t ∈ Set.Ioc 0 (Real.pi / 2) ∧ cot (arccot t) = t) : (βˆ‘' n : β„•, 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 "...
null
putnam_1986_a4
abbrev putnam_1986_a4_solution : β„š Γ— β„š Γ— β„š Γ— β„š Γ— β„š Γ— β„š Γ— β„š := sorry -- (1, 4, 2, 3, -4, 2, 1) theorem putnam_1986_a4 (f : β„• β†’ β„• := fun n ↦ {A : Matrix (Fin n) (Fin n) β„€ | (βˆ€ i j : Fin n, A i j ∈ ({-1, 0, 1} : Set β„€)) ∧ βˆƒ S : β„€, βˆ€ Ο• : Perm (Fin n), βˆ‘ i : Fin n, A i (Ο• i) = S}.ncard) : let (a1, b1, a2, b2, a3, b3, a4) :=...
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 ...
Prove that $f(n) = 4^n + 2 \cdot 3^n - 4 \cdot 2^n + 1$.
['linear_algebra']
null
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 p...
null
putnam_1986_a6
abbrev putnam_1986_a6_solution : (β„• β†’ β„•) β†’ β„• β†’ ℝ := sorry -- fun b n ↦ (∏ i : Finset.Icc 1 n, b i) / Nat.factorial n theorem putnam_1986_a6 (n : β„•) (npos : n > 0) (a : β„• β†’ ℝ) (b : β„• β†’ β„•) (bpos : βˆ€ i ∈ Finset.Icc 1 n, b i > 0) (binj : βˆ€ i ∈ Finset.Icc 1 n, βˆ€ j ∈ Finset.Icc 1 n, b i = b j β†’ i = j) (f : Polynomial ℝ) (hf ...
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 $b_1, b_2, \dots, b_n$ a...
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...
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:...
null
putnam_1986_b2
abbrev putnam_1986_b2_solution : Finset (β„‚ Γ— β„‚ Γ— β„‚) := sorry -- {(0, 0, 0), (0, -1, 1), (1, 0, -1), (-1, 1, 0)} theorem putnam_1986_b2 : ({T : β„‚ Γ— β„‚ Γ— β„‚ | βˆƒ x y z : β„‚, 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_solu...
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_b...
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::comp...
null
putnam_1986_b3
theorem putnam_1986_b3 (cong : Polynomial β„€ β†’ Polynomial β„€ β†’ β„€ β†’ Prop := fun f g m ↦ βˆ€ i : β„•, m ∣ (f - g).coeff i) (n p : β„•) (nppos : n > 0 ∧ p > 0) (pprime : Nat.Prime p) (f g h r s : Polynomial β„€) (hcoprime : cong (r * f + s * g) 1 p) (hprod : cong (f * g) h p) : (βˆƒ F G : Polynomial β„€, cong F f p ∧ cong G g p ∧ cong ...
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 $\Gamm...
null
['number_theory', 'algebra']
null
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. \<lambd...
null
putnam_1986_b4
abbrev putnam_1986_b4_solution : Prop := sorry -- True theorem putnam_1986_b4 (G : ℝ β†’ ℝ) (hGeq : βˆ€ r : ℝ, βˆƒ m n : β„€, G r = |r - sqrt (m ^ 2 + 2 * n ^ 2)|) (hGlb : βˆ€ r : ℝ, βˆ€ m n : β„€, 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 ^...
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>)" s...
null
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}}) : ((βˆ€ pqr : Fin 3 β†’ MvPolyno...
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...
Prove that the assertion is false.
['algebra']
null
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 pol...
null
putnam_1986_b6
theorem putnam_1986_b6 (n : β„•) (npos : n > 0) (F : Type*) [Field F] (A B C D : Matrix (Fin n) (Fin n) F) (hsymm : IsSymm (A * Bα΅€) ∧ IsSymm (C * Dα΅€)) (hid : A * Dα΅€ - B * Cα΅€ = 1) : (Aα΅€ * D - Cα΅€ * 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 $A^T D - C^T B = I$.
null
['linear_algebra']
null
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" ...
null
putnam_2000_a1
abbrev putnam_2000_a1_solution : ℝ β†’ Set ℝ := sorry -- (fun A : ℝ => Set.Ioo 0 (A ^ 2)) theorem putnam_2000_a1 (A : ℝ) (Apos : A > 0) : ({S : ℝ | βˆƒ x : β„• β†’ ℝ, (βˆ€ j : β„•, x j > 0) ∧ (βˆ‘' j : β„•, x j) = A ∧ (βˆ‘' j : β„•, (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 \<Rightarro...
null
putnam_2000_a2
theorem putnam_2000_a2 : βˆ€ n : β„•, βˆƒ N : β„€, βˆƒ i : Fin 6 β†’ β„•, 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.
null
['number_theory']
null
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_s...
null
putnam_2000_a4
theorem putnam_2000_a4 : βˆƒ 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.
null
['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
null
putnam_2000_a5
theorem putnam_2000_a5 (r : ℝ) (z : Fin 2 β†’ ℝ) (p : Fin 3 β†’ (Fin 2 β†’ ℝ)) (rpos : r > 0) (pdiff : βˆ€ n m : Fin 3, (n β‰  m) β†’ (p n β‰  p m)) (pint : βˆ€ (n : Fin 3) (i : Fin 2), p n i = round (p n i)) (pcirc : βˆ€ n : Fin 3, p n ∈ Metric.sphere z r) : βˆƒ n m : Fin 3, (n β‰  m) ∧ (dist (p n) (p m) β‰₯ 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 $r^{1/3}$.
null
['algebra']
null
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)...
null
putnam_2000_a6
theorem putnam_2000_a6 (f : Polynomial β„€) (a : β„• β†’ β„€) (ha0 : a 0 = 0) (ha : βˆ€ n : β„•, a (n + 1) = f.eval (a n)) : ((βˆƒ m > 0, a m = 0) β†’ (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 $a_2=0$.
null
['algebra']
null
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...
null
putnam_2000_b1
theorem putnam_2000_b1 (N : β„•) (a b c : Fin N β†’ β„€) (Nge1 : N β‰₯ 1) (hodd : βˆ€ j : Fin N, Odd (a j) ∨ Odd (b j) ∨ Odd (c j)) : (βˆƒ r s t : β„€, {j : Fin N | Odd (r * a j + s * b j + t * c j)}.ncard β‰₯ (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 $j$, $1\leq j\leq N$.
null
['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 (...
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)" ...
null
putnam_2000_b2
theorem putnam_2000_b2 : (βˆ€ m n : β„•, m β‰₯ 1 β†’ n β‰₯ m β†’ 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 $n\geq m\geq 1$.
null
['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
null
putnam_2000_b3
theorem putnam_2000_b3 (N : β„•) (Npos : N > 0) (a : Fin (N + 1) β†’ ℝ) (haN : a N β‰  0) (f : ℝ β†’ ℝ := fun t ↦ βˆ‘ j : Icc 1 N, a j * Real.sin (2 * Real.pi * j * t)) (mult : (ℝ β†’ ℝ) β†’ ℝ β†’ β„•) (hmult : βˆ€ g : ℝ β†’ ℝ, βˆ€ t : ℝ, (βˆƒ c : β„•, iteratedDeriv c g t β‰  0) β†’ (iteratedDeriv (mult g t) g t β‰  0 ∧ βˆ€ k < (mult g t), iteratedDeriv ...
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 \[ N_0\leq N_1\leq N_2\leq \cdots \mbox{ and } \lim_{k\to\infty} N_k = 2N. \]
null
['analysis']
null
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) \<Rightar...
null
putnam_2000_b4
theorem putnam_2000_b4 (f : ℝ β†’ ℝ) (hfcont : Continuous f) (hf : βˆ€ x : ℝ, f (2 * x ^ 2 - 1) = 2 * x * f x) : βˆ€ x : ℝ, x ∈ Icc (-1) 1 β†’ 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 $-1\leq x\leq 1$.
null
['analysis']
null
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
null
putnam_2000_b5
theorem putnam_2000_b5 (S : β„• β†’ Set β„€) (hSfin : βˆ€ n : β„•, Set.Finite (S n)) (hSpos : βˆ€ n : β„•, βˆ€ s ∈ S n, s > 0) (hSdef : βˆ€ n : β„•, βˆ€ a : β„€, a ∈ S (n + 1) ↔ Xor' (a - 1 ∈ S n) (a ∈ S n)) : (βˆ€ n : β„•, βˆƒ N β‰₯ n, S N = S 0 βˆͺ {M : β„€ | 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 $S_N=S_0\cup\{N+a: a\in S_0\}$.
null
['algebra']
null
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...
null
putnam_1995_a1
theorem putnam_1995_a1 (S : Set ℝ) (hS : βˆ€ a ∈ S, βˆ€ b ∈ S, a * b ∈ S) (T U : Set ℝ) (hsub : T βŠ† S ∧ U βŠ† S) (hunion : T βˆͺ U = S) (hdisj : T ∩ U = βˆ…) (hT3 : βˆ€ a ∈ T, βˆ€ b ∈ T, βˆ€ c ∈ T, a * b * c ∈ T) (hU3 : βˆ€ a ∈ U, βˆ€ b ∈ U, βˆ€ c ∈ U, a * b * c ∈ U) : (βˆ€ a ∈ T, βˆ€ b ∈ T, a * b ∈ T) ∨ (βˆ€ a ∈ U, βˆ€ 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 ele...
null
['algebra']
null
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: "\...
null
putnam_1995_a2
abbrev putnam_1995_a2_solution : Set (ℝ Γ— ℝ) := sorry -- {x | let ⟨a,b⟩ := x; a = b} theorem putnam_1995_a2 (habconv : (ℝ Γ— ℝ) β†’ Prop := fun ⟨a,b⟩ => βˆƒ 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)) : βˆ€ ab : ℝ Γ— ℝ, habconv ab ↔ ab ...
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)) <-> putna...
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" d...
null
putnam_1995_a3
theorem putnam_1995_a3 (relation : (Fin 9 β†’ β„€) β†’ (Fin 9 β†’ β„€) β†’ Prop) (digits_to_num : (Fin 9 β†’ β„€) β†’ β„€ := fun dig => βˆ‘ i : Fin 9, (dig i) * 10^i.1) (hrelation : βˆ€ d e : (Fin 9 β†’ β„€), relation d e ↔ (βˆ€ i : Fin 9, d i < 10 ∧ d i β‰₯ 0 ∧ e i < 10 ∧ e i β‰₯ 0) ∧ (βˆ€ i : Fin 9, 7 ∣ (digits_to_num (fun j : Fin 9 => if j = i then e ...
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. ...
null
['number_theory']
null
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" de...
null
putnam_1995_a4
theorem putnam_1995_a4 (n : β„•) (hn : n > 0) (necklace : Fin n β†’ β„€) (hnecklacesum : βˆ‘ i : Fin n, necklace i = n - 1) : βˆƒ cut : Fin n, βˆ€ k : Fin n, βˆ‘ 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 \[\sum_{i=1}^{k} x_{i} \leq k-1 \qquad \mbox{for} \quad k=1,2,\dots,n.\]
null
['combinatorics']
null
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\...
null
putnam_1995_a5
abbrev putnam_1995_a5_solution : Prop := sorry -- True theorem putnam_1995_a5 (hdiffx : (n : β„•) β†’ (Fin n β†’ (ℝ β†’ ℝ)) β†’ Prop := (fun (n : β„•) (x : Fin n β†’ (ℝ β†’ ℝ)) => βˆ€ i : Fin n, Differentiable ℝ (x i))) (ha : (n : β„•) β†’ (Fin n β†’ Fin n β†’ ℝ) β†’ Prop := (fun (n : β„•) (a : Fin n β†’ Fin n β†’ ℝ) => βˆ€ i j : Fin n, a i j > 0)) (hcom...
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...
Show that the answer is yes, the functions must be linearly dependent.
['linear_algebra', 'analysis']
null
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_...
null
putnam_1995_b1
theorem putnam_1995_b1 (part_ct : Finpartition (Finset.range 9) β†’ (Finset.range 9) β†’ β„•) (hp : βˆ€ partition k, part_ct partition k = (Exists.choose (Finpartition.exists_mem partition k.2)).card) : βˆ€ Pt1 Pt2 : Finpartition (Finset.range 9), βˆƒ x y : Finset.range 9, x β‰  y ∧ part_ct Pt1 x = part_ct Pt1 y ∧ part_ct Pt2 x = pa...
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 part...
null
['combinatorics']
null
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 ...
null
putnam_1995_b3
abbrev putnam_1995_b3_solution : β„• β†’ β„€ := sorry -- fun n => if n = 1 then 45 else if n = 2 then 10 * 45^2 else 0 theorem putnam_1995_b3 (n : β„•) (hn : n > 0) (digits_set := {f : β„• β†’ β„• | f 0 β‰  0 ∧ (βˆ€ i : Fin (n ^ 2), f i ≀ 9) ∧ (βˆ€ i β‰₯ n ^ 2, f i = 0)}) (digits_to_matrix : (β„• β†’ β„•) β†’ Matrix (Fin n) (Fin n) β„€ := fun f => (f...
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 o...
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']
null
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) ...
null
putnam_1995_b4
abbrev putnam_1995_b4_solution : β„€ Γ— β„€ Γ— β„€ Γ— β„€ := 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 ...
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...
null
putnam_1995_b6
theorem putnam_1995_b6 (S : ℝ β†’ Set β„• := fun Ξ± => {x : β„• | βˆƒ n : β„•, n β‰₯ 1 ∧ x = floor (n * Ξ±)}) : Β¬ βˆƒ Ξ± Ξ² Ξ³ : ℝ, Ξ± > 0 ∧ Ξ² > 0 ∧ Ξ³ > 0 ∧ (S Ξ±) ∩ (S Ξ²) = βˆ… ∧ (S Ξ²) ∩ (S Ξ³) = βˆ… ∧ (S Ξ±) ∩ (S Ξ³) = βˆ… ∧ β„•+ = (S Ξ±) βˆͺ (S Ξ²) βˆͺ (S Ξ³) := 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 $\leq x$.]
null
['algebra', 'number_theory']
null
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 \<...
null
putnam_1989_a1
abbrev putnam_1989_a1_solution : β„• := sorry -- 1 theorem putnam_1989_a1 (pdigalt : List β„• β†’ Prop) (hpdigalt : βˆ€ pdig : List β„•, pdigalt pdig = Odd pdig.length ∧ (βˆ€ i : Fin pdig.length, pdig.get i = if Even (i : β„•) then 1 else 0)) : {p : β„• | p > 0 ∧ p.Prime ∧ pdigalt (Nat.digits 10 p)}.encard = putnam_1989_a1_solution :=...
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)) -...
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) \<...
null
putnam_1989_a2
abbrev putnam_1989_a2_solution : ℝ β†’ ℝ β†’ 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...
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...
null
putnam_1989_a3
theorem putnam_1989_a3 (z : β„‚) (hz : 11 * z ^ 10 + 10 * I * z ^ 9 + 10 * I * z - 11 = 0) : (β€–zβ€– = 1) := sorry
Prove that if \[ 11z^{10}+10iz^9+10iz-11=0, \] then $|z|=1.$ (Here $z$ is a complex number and $i^2=-1$.)
null
['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
null
putnam_1989_a6
theorem putnam_1989_a6 (F : Type*) [Field F] [Fintype F] (hF : Fintype.card F = 2) (Ξ± : PowerSeries F) (hΞ± : βˆ€ n : β„•, let bin := [1] ++ (digits 2 n) ++ [1]; PowerSeries.coeff F n Ξ± = ite (βˆ€ i j : Fin bin.length, i < j β†’ bin.get i = 1 β†’ bin.get j = 1 β†’ (βˆ€ k, i < k β†’ k < j β†’ bin.get k = 0) β†’ Even ((j : β„•) - (i : β„•) - 1))...
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, $a_{36}=1$ b...
null
['algebra', 'abstract_algebra']
null
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) f...
null
putnam_1989_b2
abbrev putnam_1989_b2_solution : Prop := sorry -- True theorem putnam_1989_b2 (pow : (S : Type) β†’ β„• β†’ S β†’ S) (hpow1 : βˆ€ (S : Type) (_ : Semigroup S), βˆ€ s : S, pow S 1 s = s) (hpown : βˆ€ (S : Type) (_ : Semigroup S), βˆ€ s : S, βˆ€ n > 0, pow S (n + 1) s = s * (pow S n s)) : ((βˆ€ (S : Type) (_ : Nonempty S) (_ : Semigroup S) ...
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']
null
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 \<i...
null
putnam_1989_b3
abbrev putnam_1989_b3_solution : β„• β†’ ℝ β†’ ℝ := sorry -- fun n c ↦ c * factorial n / (3 ^ n * ∏ m in Finset.Icc (1 : β„€) n, (1 - 2 ^ (-m))) theorem putnam_1989_b3 (f : ℝ β†’ ℝ) (hfdiff : Differentiable ℝ f) (hfderiv : βˆ€ x > 0, deriv f x = -3 * f x + 6 * f (2 * x)) (hdecay : βˆ€ x β‰₯ 0, |f x| ≀ Real.exp (-Real.sqrt x)) (ΞΌ : β„• β†’...
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 m...
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']
null
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: ...
null
putnam_1989_b4
abbrev putnam_1989_b4_solution : Prop := sorry -- True theorem putnam_1989_b4 : ((βˆƒ S : Type, Countable S ∧ Infinite S ∧ βˆƒ C : Set (Set S), Β¬Countable C ∧ (βˆ€ R ∈ C, R β‰  βˆ…) ∧ (βˆ€ A ∈ C, βˆ€ B ∈ C, A β‰  B β†’ (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']
null
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>...
null
putnam_1997_a3
abbrev putnam_1997_a3_solution : ℝ := sorry -- Real.sqrt (Real.exp 1) theorem putnam_1997_a3 (series1 : ℝ β†’ ℝ := fun x => βˆ‘' n : β„•, (-1)^n * x^(2*n + 1)/(∏ i : Finset.range n, 2 * ((i : ℝ) + 1))) (series2 : ℝ β†’ ℝ := fun x => βˆ‘' n : β„•, x^(2*n)/(∏ i : Finset.range n, (2 * ((i : ℝ) + 1))^2)) : Tendsto (fun t => ∫ x in Set...
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 (...
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...
null
putnam_1997_a4
theorem putnam_1997_a4 (G : Type*) [Group G] (Ο† : G β†’ G) (hΟ† : βˆ€ g1 g2 g3 h1 h2 h3 : G, (g1 * g2 * g3 = 1 ∧ h1 * h2 * h3 = 1) β†’ Ο† g1 * Ο† g2 * Ο† g3 = Ο† h1 * Ο† h2 * Ο† h3) : βˆƒ a : G, let ψ := fun g => a * Ο† g; βˆ€ 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 $x,y\in G$)...
null
['abstract_algebra']
null
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 ...
null
putnam_1997_a5
abbrev putnam_1997_a5_solution : Prop := sorry -- True theorem putnam_1997_a5 (N := fun (n : β„•+) => {t : Fin n β†’ β„•+ | (βˆ€ i j : Fin n, i < j β†’ t i <= t j) ∧ (βˆ‘ 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) -> ...
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> ...
null
putnam_1997_a6
abbrev putnam_1997_a6_solution : β„€ β†’ β„€ β†’ ℝ := sorry -- fun n k => Nat.choose (n.toNat-1) (k.toNat-1) theorem putnam_1997_a6 (n : β„€) (hn : n > 0) (C : ℝ) (x : ℝ β†’ (β„€ β†’ ℝ)) (hx0 : βˆ€ c : ℝ, x c 0 = 0) (hx1 : βˆ€ c : ℝ, x c 1 = 1) (hxk : βˆ€ c : ℝ, βˆ€ k β‰₯ 0, x c (k + 2) = (c*(x c (k + 1)) - (n - k)*(x c k))/(k + 1)) (S : Set ℝ ...
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'...
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 \<Right...
null
putnam_1997_b1
abbrev putnam_1997_b1_solution : β„• β†’ ℝ := sorry -- fun n => n noncomputable def dist_to_int : ℝ β†’ ℝ := fun r => |r - round r| theorem putnam_1997_b1 (F : β„• β†’ ℝ := fun n => βˆ‘ m in Finset.Icc 1 (6 * n - 1), min (dist_to_int (m/(6*n)) ) (dist_to_int (m/(3*n)))) : βˆ€ n, n > 0 β†’ 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...
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 ...
null
End of preview. Expand in Data Studio

YAML Metadata Warning:empty or missing yaml metadata in repo card

Check out the documentation for more information.

Link to the repository on GitHub: https://github.com/trishullab/PUTNAM

PutnamBench

PutnamBench is a benchmark for the evaluation of theorem-proving algorithms on competition mathematics problems sourced from the William Lowell Putnam Mathematical Competition years 1965 - 2023. Our formalizations currently support three formal languages: Lean 4 $\land$ Isabelle $\land$ Coq. PutnamBench comprises over 1300 manual formalizations, aggregated over all languages.

PutnamBench aims to support research in automated mathematical reasoning by providing a multilingual benchmark for evaluating theorem-proving algorithms. It is released under permissive licenses (Apache 2.0 for Lean 4 and Isabelle, MIT for Coq), and we encourage community contributions (TODO: After initial release)

Author List

George Tsoukalas (corresponding author email: george.tsoukalas@utexas.edu) Jasper Lee John Jennings Jimmy Xin Michelle Ding Michael Jennings Amitayush Thakur Swarat Chaudhuri

Statistics

Language Count
Lean 4 514
Isabelle 514
Coq 309

We also report the number of problems in a certain category. Note that some problems fall under multiple categories.

Category Total Quantity
Algebra 218
Analysis 176
Number Theory 97
Linear Algebra 43
Abstract Algebra 25
Geometry 22
Combinatorics 12
Set Theory 4
Probability 2

license: apache-2.0

Downloads last month
12