Datasets:

HaimingW

/

PutnamBench-lean4

like 1

putnam_1987_a2

5bce53b3-b67e-5252-a4e8-df7427e65568

train

abbrev putnam_1987_a2_solution : ℕ := sorry -- 1984 /-- The sequence of digits $123456789101112131415161718192021 \dots$ is obtained by writing the positive integers in order. If the $10^n$-th digit in this sequence occurs in the part of the sequence in which the $m$-digit numbers are placed, define $f(n)$ to be $m$. For example, $f(2)=2$ because the $100$th digit enters the sequence in the placement of the two-digit integer $55$. Find, with proof, $f(1987)$. -/ theorem putnam_1987_a2 (seqind seqsize f : ℕ → ℕ) (hseqind : seqind 1 = 1 ∧ ∀ i ≥ 2, seqind i = seqind (i - 1) + (Nat.digits 10 (i - 1)).length) (hseqsize : ∀ i ≥ 1, ∀ j : Fin ((Nat.digits 10 i).length), seqsize (seqind i + j) = (Nat.digits 10 i).length) (hf : ∀ n : ℕ, f n = seqsize (10 ^ n)) : f 1987 = putnam_1987_a2_solution := sorry

import Mathlib -- 1984 /-- The sequence of digits $123456789101112131415161718192021 \dots$ is obtained by writing the positive integers in order. If the $10^n$-th digit in this sequence occurs in the part of the sequence in which the $m$-digit numbers are placed, define $f(n)$ to be $m$. For example, $f(2)=2$ because the $100$th digit enters the sequence in the placement of the two-digit integer $55$. Find, with proof, $f(1987)$. -/ theorem putnam_1987_a2 (seqind seqsize f : ℕ → ℕ) (hseqind : seqind 1 = 1 ∧ ∀ i ≥ 2, seqind i = seqind (i - 1) + (Nat.digits 10 (i - 1)).length) (hseqsize : ∀ i ≥ 1, ∀ j : Fin ((Nat.digits 10 i).length), seqsize (seqind i + j) = (Nat.digits 10 i).length) (hf : ∀ n : ℕ, f n = seqsize (10 ^ n)) : f 1987 = putnam_1987_a2_solution := by

import Mathlib abbrev putnam_1987_a2_solution : ℕ := sorry -- 1984 /-- The sequence of digits $123456789101112131415161718192021 \dots$ is obtained by writing the positive integers in order. If the $10^n$-th digit in this sequence occurs in the part of the sequence in which the $m$-digit numbers are placed, define $f(n)$ to be $m$. For example, $f(2)=2$ because the $100$th digit enters the sequence in the placement of the two-digit integer $55$. Find, with proof, $f(1987)$. -/ theorem putnam_1987_a2 (seqind seqsize f : ℕ → ℕ) (hseqind : seqind 1 = 1 ∧ ∀ i ≥ 2, seqind i = seqind (i - 1) + (Nat.digits 10 (i - 1)).length) (hseqsize : ∀ i ≥ 1, ∀ j : Fin ((Nat.digits 10 i).length), seqsize (seqind i + j) = (Nat.digits 10 i).length) (hf : ∀ n : ℕ, f n = seqsize (10 ^ n)) : f 1987 = putnam_1987_a2_solution := sorry

The sequence of digits $123456789101112131415161718192021 \dots$ is obtained by writing the positive integers in order. If the $10^n$-th digit in this sequence occurs in the part of the sequence in which the $m$-digit numbers are placed, define $f(n)$ to be $m$. For example, $f(2)=2$ because the $100$th digit enters the sequence in the placement of the two-digit integer $55$. Find, with proof, $f(1987)$.

Show that the value of $f(1987)$ is $1984$.

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putnam_2002_b5

5e4c3ab8-2311-5d37-be85-ddb673ef2085

train

theorem putnam_2002_b5 : ∃ n : ℕ, {b : ℕ | b ≥ 1 ∧ (Nat.digits b n).length = 3 ∧ List.Palindrome (Nat.digits b n)}.ncard ≥ 2002 := sorry

import Mathlib open Nat Set Topology Filter /-- A palindrome in base $b$ is a positive integer whose base-$b$ digits read the same backwards and forwards; for example, $2002$ is a 4-digit palindrome in base 10. Note that 200 is not a palindrome in base 10, but it is the 3-digit palindrome 242 in base 9, and 404 in base 7. Prove that there is an integer which is a 3-digit palindrome in base $b$ for at least 2002 different values of $b$. -/ theorem putnam_2002_b5 : ∃ n : ℕ, {b : ℕ | b ≥ 1 ∧ (Nat.digits b n).length = 3 ∧ List.Palindrome (Nat.digits b n)}.ncard ≥ 2002 := by

import Mathlib open Nat Set Topology Filter /-- A palindrome in base $b$ is a positive integer whose base-$b$ digits read the same backwards and forwards; for example, $2002$ is a 4-digit palindrome in base 10. Note that 200 is not a palindrome in base 10, but it is the 3-digit palindrome 242 in base 9, and 404 in base 7. Prove that there is an integer which is a 3-digit palindrome in base $b$ for at least 2002 different values of $b$. -/ theorem putnam_2002_b5 : ∃ n : ℕ, {b : ℕ | b ≥ 1 ∧ (Nat.digits b n).length = 3 ∧ List.Palindrome (Nat.digits b n)}.ncard ≥ 2002 := sorry

A palindrome in base $b$ is a positive integer whose base-$b$ digits read the same backwards and forwards; for example, $2002$ is a 4-digit palindrome in base 10. Note that 200 is not a palindrome in base 10, but it is the 3-digit palindrome 242 in base 9, and 404 in base 7. Prove that there is an integer which is a 3-digit palindrome in base $b$ for at least 2002 different values of $b$.

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putnam_1981_b1

cf7adaca-307b-5dda-aa5c-523634e57376

train

abbrev putnam_1981_b1_solution : ℝ := sorry -- -1 /-- Find the value of $$\lim_{n \rightarrow \infty} \frac{1}{n^5}\sum_{h=1}^{n}\sum_{k=1}^{n}(5h^4 - 18h^2k^2 + 5k^4).$$ -/ theorem putnam_1981_b1 (f : ℕ → ℝ) (hf : f = fun n : ℕ => ((1 : ℝ)/n^5) * ∑ h in Finset.Icc 1 n, ∑ k in Finset.Icc 1 n, (5*(h : ℝ)^4 - 18*h^2*k^2 + 5*k^4)) : Tendsto f atTop (𝓝 putnam_1981_b1_solution) := sorry

import Mathlib open Topology Filter Set Polynomial Function -- -1 /-- Find the value of $$\lim_{n \rightarrow \infty} \frac{1}{n^5}\sum_{h=1}^{n}\sum_{k=1}^{n}(5h^4 - 18h^2k^2 + 5k^4).$$ -/ theorem putnam_1981_b1 (f : ℕ → ℝ) (hf : f = fun n : ℕ => ((1 : ℝ)/n^5) * ∑ h in Finset.Icc 1 n, ∑ k in Finset.Icc 1 n, (5*(h : ℝ)^4 - 18*h^2*k^2 + 5*k^4)) : Tendsto f atTop (𝓝 putnam_1981_b1_solution) := by

import Mathlib open Topology Filter Set Polynomial Function abbrev putnam_1981_b1_solution : ℝ := sorry -- -1 /-- Find the value of $$\lim_{n \rightarrow \infty} \frac{1}{n^5}\sum_{h=1}^{n}\sum_{k=1}^{n}(5h^4 - 18h^2k^2 + 5k^4).$$ -/ theorem putnam_1981_b1 (f : ℕ → ℝ) (hf : f = fun n : ℕ => ((1 : ℝ)/n^5) * ∑ h in Finset.Icc 1 n, ∑ k in Finset.Icc 1 n, (5*(h : ℝ)^4 - 18*h^2*k^2 + 5*k^4)) : Tendsto f atTop (𝓝 putnam_1981_b1_solution) := sorry

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

The limit equals $-1$.

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putnam_2002_a3

cbc068cd-3ee7-5a29-b478-c09312695265

train

theorem putnam_2002_a3 (n Tn : ℤ) (hn : n ≥ 2) (hTn : Tn = Set.ncard {S : Set ℤ | S ⊆ Set.Icc 1 n ∧ Nonempty S ∧ ∃ k : ℤ, k = ((1 : ℝ) / S.ncard) * (∑' s : S, s.1)}) : Even (Tn - n) := sorry

import Mathlib open Nat /-- Let $n \geq 2$ be an integer and $T_n$ be the number of non-empty subsets $S$ of $\{1, 2, 3, \dots, n\}$ with the property that the average of the elements of $S$ is an integer. Prove that $T_n - n$ is always even. -/ theorem putnam_2002_a3 (n Tn : ℤ) (hn : n ≥ 2) (hTn : Tn = Set.ncard {S : Set ℤ | S ⊆ Set.Icc 1 n ∧ Nonempty S ∧ ∃ k : ℤ, k = ((1 : ℝ) / S.ncard) * (∑' s : S, s.1)}) : Even (Tn - n) := by

import Mathlib open Nat /-- Let $n \geq 2$ be an integer and $T_n$ be the number of non-empty subsets $S$ of $\{1, 2, 3, \dots, n\}$ with the property that the average of the elements of $S$ is an integer. Prove that $T_n - n$ is always even. -/ theorem putnam_2002_a3 (n Tn : ℤ) (hn : n ≥ 2) (hTn : Tn = Set.ncard {S : Set ℤ | S ⊆ Set.Icc 1 n ∧ Nonempty S ∧ ∃ k : ℤ, k = ((1 : ℝ) / S.ncard) * (∑' s : S, s.1)}) : Even (Tn - n) := sorry

Let $n \geq 2$ be an integer and $T_n$ be the number of non-empty subsets $S$ of $\{1, 2, 3, \dots, n\}$ with the property that the average of the elements of $S$ is an integer. Prove that $T_n - n$ is always even.

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putnam_1981_a5

2e2e58f3-e2a6-5efa-886f-6af9cd957438

train

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

import Mathlib open Topology Filter Set Polynomial Function -- True /-- 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. -/ theorem putnam_1981_a5 (Q : Polynomial ℝ → Polynomial ℝ) (hQ : Q = fun P : Polynomial ℝ => (X^2 + 1)*P*(derivative P) + X*(P^2 + (derivative P)^2)) (n : Polynomial ℝ → ℝ) (hn : n = fun P : Polynomial ℝ => ({x ∈ Ioi 1 | P.eval x = 0}.ncard : ℝ)) : (∀ P : Polynomial ℝ, {x : ℝ | (Q P).eval x = 0}.ncard ≥ 2*(n P) - 1) ↔ putnam_1981_a5_solution := by

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

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

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

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putnam_1995_a3

39ef9e8e-8b6f-5c7e-b129-78b21ff62131

train

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

import Mathlib open Filter Topology Real /-- The number $d_{1}d_{2}\dots d_{9}$ has nine (not necessarily distinct) decimal digits. The number $e_{1}e_{2}\dots e_{9}$ is such that each of the nine 9-digit numbers formed by replacing just one of the digits $d_{i}$ is $d_{1}d_{2}\dots d_{9}$ by the corresponding digit $e_{i}$ ($1 \leq i \leq 9$) is divisible by 7. The number $f_{1}f_{2}\dots f_{9}$ is related to $e_{1}e_{2}\dots e_{9}$ is the same way: that is, each of the nine numbers formed by replacing one of the $e_{i}$ by the corresponding $f_{i}$ is divisible by 7. Show that, for each $i$, $d_{i}-f_{i}$ is divisible by 7. [For example, if $d_{1}d_{2}\dots d_{9} = 199501996$, then $e_{6}$ may be 2 or 9, since $199502996$ and $199509996$ are multiples of 7.] -/ theorem putnam_1995_a3 (relation : (Fin 9 → ℤ) → (Fin 9 → ℤ) → Prop) (digits_to_num : (Fin 9 → ℤ) → ℤ) (hdigits_to_num : digits_to_num = fun dig => ∑ i : Fin 9, (dig i) * 10^i.1) (hrelation : ∀ d e : (Fin 9 → ℤ), relation d e ↔ (∀ i : Fin 9, d i < 10 ∧ d i ≥ 0 ∧ e i < 10 ∧ e i ≥ 0) ∧ (∀ i : Fin 9, 7 ∣ (digits_to_num (fun j : Fin 9 => if j = i then e j else d j)))) : ∀ d e f : (Fin 9 → ℤ), ((relation d e) ∧ (relation e f)) → (∀ i : Fin 9, 7 ∣ d i - f i) := by

import Mathlib open Filter Topology Real /-- The number $d_{1}d_{2}\dots d_{9}$ has nine (not necessarily distinct) decimal digits. The number $e_{1}e_{2}\dots e_{9}$ is such that each of the nine 9-digit numbers formed by replacing just one of the digits $d_{i}$ is $d_{1}d_{2}\dots d_{9}$ by the corresponding digit $e_{i}$ ($1 \leq i \leq 9$) is divisible by 7. The number $f_{1}f_{2}\dots f_{9}$ is related to $e_{1}e_{2}\dots e_{9}$ is the same way: that is, each of the nine numbers formed by replacing one of the $e_{i}$ by the corresponding $f_{i}$ is divisible by 7. Show that, for each $i$, $d_{i}-f_{i}$ is divisible by 7. [For example, if $d_{1}d_{2}\dots d_{9} = 199501996$, then $e_{6}$ may be 2 or 9, since $199502996$ and $199509996$ are multiples of 7.] -/ theorem putnam_1995_a3 (relation : (Fin 9 → ℤ) → (Fin 9 → ℤ) → Prop) (digits_to_num : (Fin 9 → ℤ) → ℤ) (hdigits_to_num : digits_to_num = fun dig => ∑ i : Fin 9, (dig i) * 10^i.1) (hrelation : ∀ d e : (Fin 9 → ℤ), relation d e ↔ (∀ i : Fin 9, d i < 10 ∧ d i ≥ 0 ∧ e i < 10 ∧ e i ≥ 0) ∧ (∀ i : Fin 9, 7 ∣ (digits_to_num (fun j : Fin 9 => if j = i then e j else d j)))) : ∀ d e f : (Fin 9 → ℤ), ((relation d e) ∧ (relation e f)) → (∀ i : Fin 9, 7 ∣ d i - f i) := sorry

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

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putnam_2010_b5

b0b878ef-5fad-57f0-a05d-790669ed41a0

train

abbrev putnam_2010_b5_solution : Prop := sorry -- False /-- Is there a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) = f(f(x))$ for all $x$? -/ theorem putnam_2010_b5 : (∃ f : ℝ → ℝ, StrictMono f ∧ Differentiable ℝ f ∧ (∀ x : ℝ, deriv f x = f (f x))) ↔ putnam_2010_b5_solution := sorry

import Mathlib open Filter Topology Set -- False /-- Is there a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) = f(f(x))$ for all $x$? -/ theorem putnam_2010_b5 : (∃ f : ℝ → ℝ, StrictMono f ∧ Differentiable ℝ f ∧ (∀ x : ℝ, deriv f x = f (f x))) ↔ putnam_2010_b5_solution := by

import Mathlib open Filter Topology Set abbrev putnam_2010_b5_solution : Prop := sorry -- False /-- Is there a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) = f(f(x))$ for all $x$? -/ theorem putnam_2010_b5 : (∃ f : ℝ → ℝ, StrictMono f ∧ Differentiable ℝ f ∧ (∀ x : ℝ, deriv f x = f (f x))) ↔ putnam_2010_b5_solution := sorry

Is there a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) = f(f(x))$ for all $x$?

Show that the solution is no such function exists.

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putnam_1984_b5

5e815945-72f3-54f1-a936-85c5995f5ec7

train

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

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

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

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

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

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putnam_2017_b3

5ddca2c4-9951-5b34-906c-e1a77f325ee9

train

theorem putnam_2017_b3 (f : ℝ → ℝ) (c : ℕ → ℝ) (hc : ∀ n, c n = 0 ∨ c n = 1) (hf : ∀ x, f x = ∑' n : ℕ, (c n) * x^n) : f (2/3) = 3/2 → Irrational (f 1/2) := sorry

import Mathlib open Topology Filter /-- Suppose that $f(x) = \sum_{i=0}^\infty c_i x^i$ is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational. -/ theorem putnam_2017_b3 (f : ℝ → ℝ) (c : ℕ → ℝ) (hc : ∀ n, c n = 0 ∨ c n = 1) (hf : ∀ x, f x = ∑' n : ℕ, (c n) * x^n) : f (2/3) = 3/2 → Irrational (f 1/2) := by

import Mathlib open Topology Filter /-- Suppose that $f(x) = \sum_{i=0}^\infty c_i x^i$ is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational. -/ theorem putnam_2017_b3 (f : ℝ → ℝ) (c : ℕ → ℝ) (hc : ∀ n, c n = 0 ∨ c n = 1) (hf : ∀ x, f x = ∑' n : ℕ, (c n) * x^n) : f (2/3) = 3/2 → Irrational (f 1/2) := sorry

Suppose that $f(x) = \sum_{i=0}^\infty c_i x^i$ is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational.

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putnam_1989_b6

2ad96435-dc80-5259-bc39-7cbcdcfc58bc

train

theorem putnam_1989_b6 (n : ℕ) (Sx : Set (Fin n → ℝ)) (fprop : (ℝ → ℝ) → Prop) (xext : (Fin n → ℝ) → (ℕ → ℝ)) (fxsum : (ℝ → ℝ) → (Fin n → ℝ) → ℝ) (fEV : (ℝ → ℝ) → ℝ) (hSx : Sx = {x : Fin n → ℝ | 0 < x ∧ StrictMono x ∧ x < 1}) (hfprop : fprop = (fun f : ℝ → ℝ => ContinuousOn f (Set.Icc 0 1) ∧ f 1 = 0)) (hfxsum : fxsum = (fun (f : ℝ → ℝ) (x : Fin n → ℝ) => ∑ i in Finset.Icc 0 n, ((xext x) (i + 1) - (xext x) i) * f ((xext x) (i + 1)))) (hfEV : fEV = (fun f : ℝ → ℝ => (∫ x in Sx, fxsum f x) / (∫ x in Sx, 1))) (npos : n ≥ 1) (hxext : ∀ x : Fin n → ℝ, (xext x) 0 = 0 ∧ (xext x) (n + 1) = 1 ∧ (∀ i : Fin n, (xext x) (i + 1) = x i)) : ∃ P : Polynomial ℝ, P.degree = n ∧ (∀ t ∈ Set.Icc 0 1, 0 ≤ P.eval t ∧ P.eval t ≤ 1) ∧ (∀ f : ℝ → ℝ, fprop f → fEV f = (∫ t in Set.Ioo 0 1, f t * P.eval t)) := sorry

import Mathlib open Nat Filter Topology Set -- Note: uses (ℝ → ℝ) instead of (Set.Icc 0 1 → ℝ) /-- Let $(x_1,x_2,\dots,x_n)$ be a point chosen at random from the $n$-dimensional region defined by $0<x_1<x_2<\dots<x_n<1$. Let $f$ be a continuous function on $[0,1]$ with $f(1)=0$. Set $x_0=0$ and $x_{n+1}=1$. Show that the expected value of the Riemann sum $\sum_{i=0}^n (x_{i+1}-x_i)f(x_{i+1})$ is $\int_0^1 f(t)P(t)\,dt$, where $P$ is a polynomial of degree $n$, independent of $f$, with $0 \leq P(t) \leq 1$ for $0 \leq t \leq 1$. -/ theorem putnam_1989_b6 (n : ℕ) (Sx : Set (Fin n → ℝ)) (fprop : (ℝ → ℝ) → Prop) (xext : (Fin n → ℝ) → (ℕ → ℝ)) (fxsum : (ℝ → ℝ) → (Fin n → ℝ) → ℝ) (fEV : (ℝ → ℝ) → ℝ) (hSx : Sx = {x : Fin n → ℝ | 0 < x ∧ StrictMono x ∧ x < 1}) (hfprop : fprop = (fun f : ℝ → ℝ => ContinuousOn f (Set.Icc 0 1) ∧ f 1 = 0)) (hfxsum : fxsum = (fun (f : ℝ → ℝ) (x : Fin n → ℝ) => ∑ i in Finset.Icc 0 n, ((xext x) (i + 1) - (xext x) i) * f ((xext x) (i + 1)))) (hfEV : fEV = (fun f : ℝ → ℝ => (∫ x in Sx, fxsum f x) / (∫ x in Sx, 1))) (npos : n ≥ 1) (hxext : ∀ x : Fin n → ℝ, (xext x) 0 = 0 ∧ (xext x) (n + 1) = 1 ∧ (∀ i : Fin n, (xext x) (i + 1) = x i)) : ∃ P : Polynomial ℝ, P.degree = n ∧ (∀ t ∈ Set.Icc 0 1, 0 ≤ P.eval t ∧ P.eval t ≤ 1) ∧ (∀ f : ℝ → ℝ, fprop f → fEV f = (∫ t in Set.Ioo 0 1, f t * P.eval t)) := by

import Mathlib open Nat Filter Topology Set -- Note: uses (ℝ → ℝ) instead of (Set.Icc 0 1 → ℝ) /-- Let $(x_1,x_2,\dots,x_n)$ be a point chosen at random from the $n$-dimensional region defined by $0<x_1<x_2<\dots<x_n<1$. Let $f$ be a continuous function on $[0,1]$ with $f(1)=0$. Set $x_0=0$ and $x_{n+1}=1$. Show that the expected value of the Riemann sum $\sum_{i=0}^n (x_{i+1}-x_i)f(x_{i+1})$ is $\int_0^1 f(t)P(t)\,dt$, where $P$ is a polynomial of degree $n$, independent of $f$, with $0 \leq P(t) \leq 1$ for $0 \leq t \leq 1$. -/ theorem putnam_1989_b6 (n : ℕ) (Sx : Set (Fin n → ℝ)) (fprop : (ℝ → ℝ) → Prop) (xext : (Fin n → ℝ) → (ℕ → ℝ)) (fxsum : (ℝ → ℝ) → (Fin n → ℝ) → ℝ) (fEV : (ℝ → ℝ) → ℝ) (hSx : Sx = {x : Fin n → ℝ | 0 < x ∧ StrictMono x ∧ x < 1}) (hfprop : fprop = (fun f : ℝ → ℝ => ContinuousOn f (Set.Icc 0 1) ∧ f 1 = 0)) (hfxsum : fxsum = (fun (f : ℝ → ℝ) (x : Fin n → ℝ) => ∑ i in Finset.Icc 0 n, ((xext x) (i + 1) - (xext x) i) * f ((xext x) (i + 1)))) (hfEV : fEV = (fun f : ℝ → ℝ => (∫ x in Sx, fxsum f x) / (∫ x in Sx, 1))) (npos : n ≥ 1) (hxext : ∀ x : Fin n → ℝ, (xext x) 0 = 0 ∧ (xext x) (n + 1) = 1 ∧ (∀ i : Fin n, (xext x) (i + 1) = x i)) : ∃ P : Polynomial ℝ, P.degree = n ∧ (∀ t ∈ Set.Icc 0 1, 0 ≤ P.eval t ∧ P.eval t ≤ 1) ∧ (∀ f : ℝ → ℝ, fprop f → fEV f = (∫ t in Set.Ioo 0 1, f t * P.eval t)) := sorry

Let $(x_1,x_2,\dots,x_n)$ be a point chosen at random from the $n$-dimensional region defined by $0<x_1<x_2<\dots<x_n<1$. Let $f$ be a continuous function on $[0,1]$ with $f(1)=0$. Set $x_0=0$ and $x_{n+1}=1$. Show that the expected value of the Riemann sum $\sum_{i=0}^n (x_{i+1}-x_i)f(x_{i+1})$ is $\int_0^1 f(t)P(t)\,dt$, where $P$ is a polynomial of degree $n$, independent of $f$, with $0 \leq P(t) \leq 1$ for $0 \leq t \leq 1$.

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putnam_1971_b2

0f152c45-007e-5b3f-8024-dbaa1a1410b6

train

abbrev putnam_1971_b2_solution : Set (ℝ → ℝ) := sorry -- {fun x : ℝ => (x^3 - x^2 - 1)/(2 * x * (x - 1))} /-- Find all functions $F : \mathbb{R} \setminus \{0, 1\} \to \mathbb{R}$ that satisfy $F(x) + F\left(\frac{x - 1}{x}\right) = 1 + x$ for all $x \in \mathbb{R} \setminus \{0, 1\}$. -/ theorem putnam_1971_b2 (S : Set ℝ) (hS : S = univ \ {0, 1}) (P : (ℝ → ℝ) → Prop) (hP : P = fun (F : ℝ → ℝ) => ∀ x ∈ S, F x + F ((x - 1)/x) = 1 + x) : (∀ F ∈ putnam_1971_b2_solution, P F) ∧ ∀ f : ℝ → ℝ, P f → ∃ F ∈ putnam_1971_b2_solution, (∀ x ∈ S, f x = F x) := sorry

import Mathlib open Set MvPolynomial -- {fun x : ℝ => (x^3 - x^2 - 1)/(2 * x * (x - 1))} /-- Find all functions $F : \mathbb{R} \setminus \{0, 1\} \to \mathbb{R}$ that satisfy $F(x) + F\left(\frac{x - 1}{x}\right) = 1 + x$ for all $x \in \mathbb{R} \setminus \{0, 1\}$. -/ theorem putnam_1971_b2 (S : Set ℝ) (hS : S = univ \ {0, 1}) (P : (ℝ → ℝ) → Prop) (hP : P = fun (F : ℝ → ℝ) => ∀ x ∈ S, F x + F ((x - 1)/x) = 1 + x) : (∀ F ∈ putnam_1971_b2_solution, P F) ∧ ∀ f : ℝ → ℝ, P f → ∃ F ∈ putnam_1971_b2_solution, (∀ x ∈ S, f x = F x) := by

import Mathlib open Set MvPolynomial abbrev putnam_1971_b2_solution : Set (ℝ → ℝ) := sorry -- {fun x : ℝ => (x^3 - x^2 - 1)/(2 * x * (x - 1))} /-- Find all functions $F : \mathbb{R} \setminus \{0, 1\} \to \mathbb{R}$ that satisfy $F(x) + F\left(\frac{x - 1}{x}\right) = 1 + x$ for all $x \in \mathbb{R} \setminus \{0, 1\}$. -/ theorem putnam_1971_b2 (S : Set ℝ) (hS : S = univ \ {0, 1}) (P : (ℝ → ℝ) → Prop) (hP : P = fun (F : ℝ → ℝ) => ∀ x ∈ S, F x + F ((x - 1)/x) = 1 + x) : (∀ F ∈ putnam_1971_b2_solution, P F) ∧ ∀ f : ℝ → ℝ, P f → ∃ F ∈ putnam_1971_b2_solution, (∀ x ∈ S, f x = F x) := sorry

Find all functions $F : \mathbb{R} \setminus \{0, 1\} \to \mathbb{R}$ that satisfy $F(x) + F\left(\frac{x - 1}{x}\right) = 1 + x$ for all $x \in \mathbb{R} \setminus \{0, 1\}$.

The only such function is $F(x) = \frac{x^3 - x^2 - 1}{2x(x - 1)}$.

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putnam_2019_a1

809c958b-410b-5e57-b3d3-79f29094facd

train

abbrev putnam_2019_a1_solution : Set ℤ := sorry -- {n : ℤ | n ≥ 0 ∧ ¬Int.ModEq 9 n 3 ∧ ¬Int.ModEq 9 n 6} /-- Determine all possible values of the expression \[ A^3+B^3+C^3-3ABC \] where $A, B$, and $C$ are nonnegative integers. -/ theorem putnam_2019_a1 : {n : ℤ | ∃ A B C : ℤ, A ≥ 0 ∧ B ≥ 0 ∧ C ≥ 0 ∧ A^3 + B^3 + C^3 - 3*A*B*C = n} = putnam_2019_a1_solution := sorry

import Mathlib open Topology Filter -- {n : ℤ | n ≥ 0 ∧ ¬Int.ModEq 9 n 3 ∧ ¬Int.ModEq 9 n 6} /-- Determine all possible values of the expression \[ A^3+B^3+C^3-3ABC \] where $A, B$, and $C$ are nonnegative integers. -/ theorem putnam_2019_a1 : {n : ℤ | ∃ A B C : ℤ, A ≥ 0 ∧ B ≥ 0 ∧ C ≥ 0 ∧ A^3 + B^3 + C^3 - 3*A*B*C = n} = putnam_2019_a1_solution := by

import Mathlib open Topology Filter abbrev putnam_2019_a1_solution : Set ℤ := sorry -- {n : ℤ | n ≥ 0 ∧ ¬Int.ModEq 9 n 3 ∧ ¬Int.ModEq 9 n 6} /-- Determine all possible values of the expression \[ A^3+B^3+C^3-3ABC \] where $A, B$, and $C$ are nonnegative integers. -/ theorem putnam_2019_a1 : {n : ℤ | ∃ A B C : ℤ, A ≥ 0 ∧ B ≥ 0 ∧ C ≥ 0 ∧ A^3 + B^3 + C^3 - 3*A*B*C = n} = putnam_2019_a1_solution := sorry

Determine all possible values of the expression \[ A^3+B^3+C^3-3ABC \] where $A, B$, and $C$ are nonnegative integers.

The answer is all nonnegative integers not congruent to $3$ or $6 \pmod{9}$.

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putnam_1977_a2

3c04332a-fbcf-5760-bfbc-2a67a9db1c30

train

abbrev putnam_1977_a2_solution : ℝ → ℝ → ℝ → ℝ → Prop := sorry -- fun a b c d ↦ d = a ∧ b = -c ∨ d = b ∧ a = -c ∨ d = c ∧ a = -b /-- Find all real solutions $(a, b, c, d)$ to the equations $a + b + c = d$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{d}$. -/ theorem putnam_1977_a2 : ∀ a b c d : ℝ, putnam_1977_a2_solution a b c d ↔ a ≠ 0 → b ≠ 0 → c ≠ 0 → d ≠ 0 → (a + b + c = d ∧ 1 / a + 1 / b + 1 / c = 1 / d) := sorry

import Mathlib -- fun a b c d ↦ d = a ∧ b = -c ∨ d = b ∧ a = -c ∨ d = c ∧ a = -b /-- Find all real solutions $(a, b, c, d)$ to the equations $a + b + c = d$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{d}$. -/ theorem putnam_1977_a2 : ∀ a b c d : ℝ, putnam_1977_a2_solution a b c d ↔ a ≠ 0 → b ≠ 0 → c ≠ 0 → d ≠ 0 → (a + b + c = d ∧ 1 / a + 1 / b + 1 / c = 1 / d) := by

import Mathlib abbrev putnam_1977_a2_solution : ℝ → ℝ → ℝ → ℝ → Prop := sorry -- fun a b c d ↦ d = a ∧ b = -c ∨ d = b ∧ a = -c ∨ d = c ∧ a = -b /-- Find all real solutions $(a, b, c, d)$ to the equations $a + b + c = d$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{d}$. -/ theorem putnam_1977_a2 : ∀ a b c d : ℝ, putnam_1977_a2_solution a b c d ↔ a ≠ 0 → b ≠ 0 → c ≠ 0 → d ≠ 0 → (a + b + c = d ∧ 1 / a + 1 / b + 1 / c = 1 / d) := sorry

Find all real solutions $(a, b, c, d)$ to the equations $a + b + c = d$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{d}$.

Prove that the solutions are $d = a$ and $b = -c$, $d = b$ and $a = -c$, or $d = c$ and $a = -b$, with $a, b, c, d$ nonzero.

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putnam_1998_a5

17cb60f9-16e8-535a-877c-d7699d277d67

train

theorem putnam_1998_a5 (k : ℕ) (c : Fin k → (EuclideanSpace ℝ (Fin 2))) (r : Fin k → ℝ) (hr : ∀ i, r i > 0) (E : Set (EuclideanSpace ℝ (Fin 2))) (hE : E ⊆ ⋃ i, ball (c i) (r i)) : ∃ (n : ℕ) (t : Fin n → Fin k), (∀ i j, i ≠ j → (ball (c (t i)) (r (t i)) ∩ ball (c (t j)) (r (t j)) = ∅)) ∧ E ⊆ ⋃ i : Fin n, ball (c (t i)) (3 * (r (t i))) := sorry

import Mathlib open Set Function Metric /-- Let $\mathcal F$ be a finite collection of open discs in $\mathbb R^2$ whose union contains a set $E\subseteq \mathbb R^2$. Show that there is a pairwise disjoint subcollection $D_1,\ldots, D_n$ in $\mathcal F$ such that \[E\subseteq \cup_{j=1}^n 3D_j.\] Here, if $D$ is the disc of radius $r$ and center $P$, then $3D$ is the disc of radius $3r$ and center $P$. -/ theorem putnam_1998_a5 (k : ℕ) (c : Fin k → (EuclideanSpace ℝ (Fin 2))) (r : Fin k → ℝ) (hr : ∀ i, r i > 0) (E : Set (EuclideanSpace ℝ (Fin 2))) (hE : E ⊆ ⋃ i, ball (c i) (r i)) : ∃ (n : ℕ) (t : Fin n → Fin k), (∀ i j, i ≠ j → (ball (c (t i)) (r (t i)) ∩ ball (c (t j)) (r (t j)) = ∅)) ∧ E ⊆ ⋃ i : Fin n, ball (c (t i)) (3 * (r (t i))) := by

import Mathlib open Set Function Metric /-- Let $\mathcal F$ be a finite collection of open discs in $\mathbb R^2$ whose union contains a set $E\subseteq \mathbb R^2$. Show that there is a pairwise disjoint subcollection $D_1,\ldots, D_n$ in $\mathcal F$ such that \[E\subseteq \cup_{j=1}^n 3D_j.\] Here, if $D$ is the disc of radius $r$ and center $P$, then $3D$ is the disc of radius $3r$ and center $P$. -/ theorem putnam_1998_a5 (k : ℕ) (c : Fin k → (EuclideanSpace ℝ (Fin 2))) (r : Fin k → ℝ) (hr : ∀ i, r i > 0) (E : Set (EuclideanSpace ℝ (Fin 2))) (hE : E ⊆ ⋃ i, ball (c i) (r i)) : ∃ (n : ℕ) (t : Fin n → Fin k), (∀ i j, i ≠ j → (ball (c (t i)) (r (t i)) ∩ ball (c (t j)) (r (t j)) = ∅)) ∧ E ⊆ ⋃ i : Fin n, ball (c (t i)) (3 * (r (t i))) := sorry

Let $\mathcal F$ be a finite collection of open discs in $\mathbb R^2$ whose union contains a set $E\subseteq \mathbb R^2$. Show that there is a pairwise disjoint subcollection $D_1,\ldots, D_n$ in $\mathcal F$ such that \[E\subseteq \cup_{j=1}^n 3D_j.\] Here, if $D$ is the disc of radius $r$ and center $P$, then $3D$ is the disc of radius $3r$ and center $P$.

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putnam_1995_a6

2fcd9fcb-46ea-5256-b7a8-59fa62c551bf

train

theorem putnam_1995_a6 (SM : (n : ℕ) → Set (Matrix (Fin 3) (Fin n) ℤ)) (Srowsums : (n : ℕ) → Matrix (Fin 3) (Fin n) ℤ → Set ℤ) (sumsprog : (Set ℤ) → Prop) (hSM : ∀ n, SM n = {M : Matrix (Fin 3) (Fin n) ℤ | ∀ j : Fin n, {M i j | i : Fin 3} = {1, 2, 3}}) (hSrowsums : ∀ n M, Srowsums n M = {∑ j : Fin n, M i j | i : Fin 3}) (hsumsprog : ∀ sums, sumsprog sums ↔ sums.encard = 3 ∧ (∃ a b c : ℤ, {a, b, c} = sums ∧ b = a + 1 ∧ c = a + 2)) : ∃ n ≥ 1995, {M : (SM n) | sumsprog (Srowsums n M)}.encard ≥ 4 * {M : (SM n) | (Srowsums n M).encard = 1}.encard := sorry

import Mathlib open Filter Topology Real /-- Suppose that each of $n$ people writes down the numbers $1,2,3$ in random order in one column of a $3 \times n$ matrix, with all orders equally likely and with the orders for different columns independent of each other. Let the row sums $a,b,c$ of the resulting matrix be rearranged (if necessary) so that $a \leq b \leq c$. Show that for some $n \geq 1995$, it is at least four times as likely that both $b=a+1$ and $c=a+2$ as that $a=b=c$. -/ theorem putnam_1995_a6 (SM : (n : ℕ) → Set (Matrix (Fin 3) (Fin n) ℤ)) (Srowsums : (n : ℕ) → Matrix (Fin 3) (Fin n) ℤ → Set ℤ) (sumsprog : (Set ℤ) → Prop) (hSM : ∀ n, SM n = {M : Matrix (Fin 3) (Fin n) ℤ | ∀ j : Fin n, {M i j | i : Fin 3} = {1, 2, 3}}) (hSrowsums : ∀ n M, Srowsums n M = {∑ j : Fin n, M i j | i : Fin 3}) (hsumsprog : ∀ sums, sumsprog sums ↔ sums.encard = 3 ∧ (∃ a b c : ℤ, {a, b, c} = sums ∧ b = a + 1 ∧ c = a + 2)) : ∃ n ≥ 1995, {M : (SM n) | sumsprog (Srowsums n M)}.encard ≥ 4 * {M : (SM n) | (Srowsums n M).encard = 1}.encard := by

import Mathlib open Filter Topology Real /-- Suppose that each of $n$ people writes down the numbers $1,2,3$ in random order in one column of a $3 \times n$ matrix, with all orders equally likely and with the orders for different columns independent of each other. Let the row sums $a,b,c$ of the resulting matrix be rearranged (if necessary) so that $a \leq b \leq c$. Show that for some $n \geq 1995$, it is at least four times as likely that both $b=a+1$ and $c=a+2$ as that $a=b=c$. -/ theorem putnam_1995_a6 (SM : (n : ℕ) → Set (Matrix (Fin 3) (Fin n) ℤ)) (Srowsums : (n : ℕ) → Matrix (Fin 3) (Fin n) ℤ → Set ℤ) (sumsprog : (Set ℤ) → Prop) (hSM : ∀ n, SM n = {M : Matrix (Fin 3) (Fin n) ℤ | ∀ j : Fin n, {M i j | i : Fin 3} = {1, 2, 3}}) (hSrowsums : ∀ n M, Srowsums n M = {∑ j : Fin n, M i j | i : Fin 3}) (hsumsprog : ∀ sums, sumsprog sums ↔ sums.encard = 3 ∧ (∃ a b c : ℤ, {a, b, c} = sums ∧ b = a + 1 ∧ c = a + 2)) : ∃ n ≥ 1995, {M : (SM n) | sumsprog (Srowsums n M)}.encard ≥ 4 * {M : (SM n) | (Srowsums n M).encard = 1}.encard := sorry

Suppose that each of $n$ people writes down the numbers $1,2,3$ in random order in one column of a $3 \times n$ matrix, with all orders equally likely and with the orders for different columns independent of each other. Let the row sums $a,b,c$ of the resulting matrix be rearranged (if necessary) so that $a \leq b \leq c$. Show that for some $n \geq 1995$, it is at least four times as likely that both $b=a+1$ and $c=a+2$ as that $a=b=c$.

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putnam_2021_a6

39ac01e0-6ab1-57c5-82b2-bab484e8ef04

train

abbrev putnam_2021_a6_solution : Prop := sorry -- True /-- Let $P(x)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(x)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer? -/ theorem putnam_2021_a6 : (∀ P, (∀ n : ℕ, P.coeff n = 0 ∨ P.coeff n = 1) → (∃ Q R : Polynomial ℤ, Q.degree > 0 ∧ R.degree > 0 ∧ P = Q * R) → (P.eval 2 ≠ 0 ∧ P.eval 2 ≠ 1 ∧ ¬Prime (P.eval 2))) ↔ putnam_2021_a6_solution := sorry

import Mathlib open Filter Topology -- True /-- Let $P(x)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(x)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer? -/ theorem putnam_2021_a6 : (∀ P, (∀ n : ℕ, P.coeff n = 0 ∨ P.coeff n = 1) → (∃ Q R : Polynomial ℤ, Q.degree > 0 ∧ R.degree > 0 ∧ P = Q * R) → (P.eval 2 ≠ 0 ∧ P.eval 2 ≠ 1 ∧ ¬Prime (P.eval 2))) ↔ putnam_2021_a6_solution := by

import Mathlib open Filter Topology abbrev putnam_2021_a6_solution : Prop := sorry -- True /-- Let $P(x)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(x)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer? -/ theorem putnam_2021_a6 : (∀ P, (∀ n : ℕ, P.coeff n = 0 ∨ P.coeff n = 1) → (∃ Q R : Polynomial ℤ, Q.degree > 0 ∧ R.degree > 0 ∧ P = Q * R) → (P.eval 2 ≠ 0 ∧ P.eval 2 ≠ 1 ∧ ¬Prime (P.eval 2))) ↔ putnam_2021_a6_solution := sorry

Let $P(x)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(x)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer?

Show that it does follow that $P(2)$ is a composite integer.

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putnam_1972_b6

ce7cf111-3ea0-5ea4-bc90-c17ed42f145a

train

theorem putnam_1972_b6 (k : ℕ) (hk : k ≥ 1) (n : Fin k → ℤ) (hn : ∀ i : Fin k, n i > 0) (hn' : ∀ i j : Fin k, i < j → n i < n j) (zpoly : ℂ → ℂ) (hzpoly : zpoly = fun z => 1 + ∑ i : Fin k, z^(n i)) : ∀ z : ℂ, z ∈ ball 0 ((Real.sqrt 5 - 1)/2) → zpoly z ≠ 0 := sorry

import Mathlib open EuclideanGeometry Filter Topology Set MeasureTheory Metric /-- Let $n_1 < n_2 < \dots < n_k$ be a set of positive integers. Prove that the polynomail $1 + z^{n_1} + z^{n_2} + \dots + z^{n_k}$ has not roots inside the circle $|z| < (\frac{\sqrt{5}-1}{2}$. -/ theorem putnam_1972_b6 (k : ℕ) (hk : k ≥ 1) (n : Fin k → ℤ) (hn : ∀ i : Fin k, n i > 0) (hn' : ∀ i j : Fin k, i < j → n i < n j) (zpoly : ℂ → ℂ) (hzpoly : zpoly = fun z => 1 + ∑ i : Fin k, z^(n i)) : ∀ z : ℂ, z ∈ ball 0 ((Real.sqrt 5 - 1)/2) → zpoly z ≠ 0 := by

import Mathlib open EuclideanGeometry Filter Topology Set MeasureTheory Metric /-- Let $n_1 < n_2 < \dots < n_k$ be a set of positive integers. Prove that the polynomail $1 + z^{n_1} + z^{n_2} + \dots + z^{n_k}$ has not roots inside the circle $|z| < (\frac{\sqrt{5}-1}{2}$. -/ theorem putnam_1972_b6 (k : ℕ) (hk : k ≥ 1) (n : Fin k → ℤ) (hn : ∀ i : Fin k, n i > 0) (hn' : ∀ i j : Fin k, i < j → n i < n j) (zpoly : ℂ → ℂ) (hzpoly : zpoly = fun z => 1 + ∑ i : Fin k, z^(n i)) : ∀ z : ℂ, z ∈ ball 0 ((Real.sqrt 5 - 1)/2) → zpoly z ≠ 0 := sorry

Let $n_1 < n_2 < \dots < n_k$ be a set of positive integers. Prove that the polynomail $1 + z^{n_1} + z^{n_2} + \dots + z^{n_k}$ has not roots inside the circle $|z| < (\frac{\sqrt{5}-1}{2}$.

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putnam_1988_b4

f95e2134-7b38-5ee5-8868-503fc0946b22

train

theorem putnam_1988_b4 (a : ℕ → ℝ) (IsPosConv : (ℕ → ℝ) → Prop) (IsPosConv_def : ∀ a' : ℕ → ℝ, IsPosConv a' ↔ (∀ n ≥ 1, a' n > 0) ∧ (∃ s : ℝ, Tendsto (fun N : ℕ => ∑ n : Set.Icc 1 N, a' n) atTop (𝓝 s))) : (IsPosConv a) → IsPosConv (fun n : ℕ => (a n) ^ ((n : ℝ) / (n + 1))) := sorry

import Mathlib open Set Filter Topology /-- Prove that if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers, then so is $\sum_{n=1}^\infty (a_n)^{n/(n+1)}$. -/ theorem putnam_1988_b4 (a : ℕ → ℝ) (IsPosConv : (ℕ → ℝ) → Prop) (IsPosConv_def : ∀ a' : ℕ → ℝ, IsPosConv a' ↔ (∀ n ≥ 1, a' n > 0) ∧ (∃ s : ℝ, Tendsto (fun N : ℕ => ∑ n : Set.Icc 1 N, a' n) atTop (𝓝 s))) : (IsPosConv a) → IsPosConv (fun n : ℕ => (a n) ^ ((n : ℝ) / (n + 1))) := by

import Mathlib open Set Filter Topology /-- Prove that if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers, then so is $\sum_{n=1}^\infty (a_n)^{n/(n+1)}$. -/ theorem putnam_1988_b4 (a : ℕ → ℝ) (IsPosConv : (ℕ → ℝ) → Prop) (IsPosConv_def : ∀ a' : ℕ → ℝ, IsPosConv a' ↔ (∀ n ≥ 1, a' n > 0) ∧ (∃ s : ℝ, Tendsto (fun N : ℕ => ∑ n : Set.Icc 1 N, a' n) atTop (𝓝 s))) : (IsPosConv a) → IsPosConv (fun n : ℕ => (a n) ^ ((n : ℝ) / (n + 1))) := sorry

Prove that if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers, then so is $\sum_{n=1}^\infty (a_n)^{n/(n+1)}$.

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putnam_1996_a6

0b28e551-871f-53f9-bcf9-ed2c53fac06e

train

abbrev putnam_1996_a6_solution : ℝ → Set (ℝ → ℝ) := sorry -- (fun c : ℝ => if c ≤ 1 / 4 then {f : ℝ → ℝ | ∃ d : ℝ, ∀ x : ℝ, f x = d} else {f : ℝ → ℝ | ContinuousOn f (Set.Icc 0 c) ∧ f 0 = f c ∧ (∀ x > 0, f x = f (x ^ 2 + c)) ∧ (∀ x < 0, f x = f (-x))}) /-- Let $c>0$ be a constant. Give a complete description, with proof, of the set of all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)=f(x^2+c)$ for all $x \in \mathbb{R}$. -/ theorem putnam_1996_a6 (c : ℝ) (f : ℝ → ℝ) (cgt0 : c > 0) : (Continuous f ∧ ∀ x : ℝ, f x = f (x ^ 2 + c)) ↔ f ∈ putnam_1996_a6_solution c := sorry

import Mathlib open Function -- (fun c : ℝ => if c ≤ 1 / 4 then {f : ℝ → ℝ | ∃ d : ℝ, ∀ x : ℝ, f x = d} else {f : ℝ → ℝ | ContinuousOn f (Set.Icc 0 c) ∧ f 0 = f c ∧ (∀ x > 0, f x = f (x ^ 2 + c)) ∧ (∀ x < 0, f x = f (-x))}) /-- Let $c>0$ be a constant. Give a complete description, with proof, of the set of all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)=f(x^2+c)$ for all $x \in \mathbb{R}$. -/ theorem putnam_1996_a6 (c : ℝ) (f : ℝ → ℝ) (cgt0 : c > 0) : (Continuous f ∧ ∀ x : ℝ, f x = f (x ^ 2 + c)) ↔ f ∈ putnam_1996_a6_solution c := by

import Mathlib open Function abbrev putnam_1996_a6_solution : ℝ → Set (ℝ → ℝ) := sorry -- (fun c : ℝ => if c ≤ 1 / 4 then {f : ℝ → ℝ | ∃ d : ℝ, ∀ x : ℝ, f x = d} else {f : ℝ → ℝ | ContinuousOn f (Set.Icc 0 c) ∧ f 0 = f c ∧ (∀ x > 0, f x = f (x ^ 2 + c)) ∧ (∀ x < 0, f x = f (-x))}) /-- Let $c>0$ be a constant. Give a complete description, with proof, of the set of all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)=f(x^2+c)$ for all $x \in \mathbb{R}$. -/ theorem putnam_1996_a6 (c : ℝ) (f : ℝ → ℝ) (cgt0 : c > 0) : (Continuous f ∧ ∀ x : ℝ, f x = f (x ^ 2 + c)) ↔ f ∈ putnam_1996_a6_solution c := sorry

Let $c>0$ be a constant. Give a complete description, with proof, of the set of all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)=f(x^2+c)$ for all $x \in \mathbb{R}$.

Show that if $c \leq 1/4$ then $f$ must be constant, and if $c>1/4$ then $f$ can be defined on $[0,c]$ as any continuous function with equal values on the endpoints, then extended to $x>c$ by the relation $f(x)=f(x^2+c)$, then extended further to $x<0$ by the relation $f(x)=f(-x)$.

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putnam_2023_b5

1a5195a1-b018-5133-9c1e-3fcc1f16aa3c

train

abbrev putnam_2023_b5_solution : Set ℕ := sorry -- {n : ℕ | n = 1 ∨ n ≡ 2 [MOD 4]} /-- Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$. -/ theorem putnam_2023_b5 (n : ℕ) : n ∈ putnam_2023_b5_solution ↔ 0 < n ∧ (∀ m : ℤ, IsRelPrime m n → ∃ p : Equiv.Perm (Fin n), ∀ k : Fin n, (p (p k)).1 + 1 ≡ m * (k.1 + 1) [ZMOD n]) := sorry

import Mathlib open Nat Topology Filter -- {n : ℕ | n = 1 ∨ n ≡ 2 [MOD 4]} /-- Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$. -/ theorem putnam_2023_b5 (n : ℕ) : n ∈ putnam_2023_b5_solution ↔ 0 < n ∧ (∀ m : ℤ, IsRelPrime m n → ∃ p : Equiv.Perm (Fin n), ∀ k : Fin n, (p (p k)).1 + 1 ≡ m * (k.1 + 1) [ZMOD n]) := by

import Mathlib open Nat Topology Filter abbrev putnam_2023_b5_solution : Set ℕ := sorry -- {n : ℕ | n = 1 ∨ n ≡ 2 [MOD 4]} /-- Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$. -/ theorem putnam_2023_b5 (n : ℕ) : n ∈ putnam_2023_b5_solution ↔ 0 < n ∧ (∀ m : ℤ, IsRelPrime m n → ∃ p : Equiv.Perm (Fin n), ∀ k : Fin n, (p (p k)).1 + 1 ≡ m * (k.1 + 1) [ZMOD n]) := sorry

Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$.

Show that the desired property holds if and only if $n=1$ or $n \equiv 2 \pmod{4}$.

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putnam_2006_b2

9325a8ef-9b4b-5248-8ac1-eff8d3178da5

train

theorem putnam_2006_b2 (n : ℕ) (npos : n > 0) (X : Finset ℝ) (hXcard : X.card = n) : (∃ S ⊆ X, S ≠ ∅ ∧ ∃ m : ℤ, |m + ∑ s in S, s| ≤ 1 / (n + 1)) := sorry

import Mathlib /-- Prove that, for every set $X = \{x_1, x_2, \dots, x_n\}$ of $n$ real numbers, there exists a non-empty subset $S$ of $X$ and an integer $m$ such that \[ \left| m + \sum_{s \in S} s \right| \leq \frac{1}{n+1}. \] -/ theorem putnam_2006_b2 (n : ℕ) (npos : n > 0) (X : Finset ℝ) (hXcard : X.card = n) : (∃ S ⊆ X, S ≠ ∅ ∧ ∃ m : ℤ, |m + ∑ s in S, s| ≤ 1 / (n + 1)) := by

import Mathlib /-- Prove that, for every set $X = \{x_1, x_2, \dots, x_n\}$ of $n$ real numbers, there exists a non-empty subset $S$ of $X$ and an integer $m$ such that \[ \left| m + \sum_{s \in S} s \right| \leq \frac{1}{n+1}. \] -/ theorem putnam_2006_b2 (n : ℕ) (npos : n > 0) (X : Finset ℝ) (hXcard : X.card = n) : (∃ S ⊆ X, S ≠ ∅ ∧ ∃ m : ℤ, |m + ∑ s in S, s| ≤ 1 / (n + 1)) := sorry

Prove that, for every set $X = \{x_1, x_2, \dots, x_n\}$ of $n$ real numbers, there exists a non-empty subset $S$ of $X$ and an integer $m$ such that \[ \left| m + \sum_{s \in S} s \right| \leq \frac{1}{n+1}. \]

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putnam_2018_b3

a672c119-54e2-523e-b731-cb8e538615f0

train

abbrev putnam_2018_b3_solution : Set ℕ := sorry -- {2^2, 2^4, 2^16, 2^256} /-- Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$, and $n-2$ divides $2^n - 2$. -/ theorem putnam_2018_b3 (n : ℕ) (hn : n < 0) : (n < 10^100 ∧ ((n : ℤ) ∣ (2^n : ℤ) ∧ (n - 1 : ℤ) ∣ (2^n - 1 : ℤ) ∧ (n - 2 : ℤ) ∣ (2^n - 2 : ℤ))) ↔ n ∈ putnam_2018_b3_solution := sorry

import Mathlib -- {2^2, 2^4, 2^16, 2^256} /-- Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$, and $n-2$ divides $2^n - 2$. -/ theorem putnam_2018_b3 (n : ℕ) (hn : n < 0) : (n < 10^100 ∧ ((n : ℤ) ∣ (2^n : ℤ) ∧ (n - 1 : ℤ) ∣ (2^n - 1 : ℤ) ∧ (n - 2 : ℤ) ∣ (2^n - 2 : ℤ))) ↔ n ∈ putnam_2018_b3_solution := by

import Mathlib abbrev putnam_2018_b3_solution : Set ℕ := sorry -- {2^2, 2^4, 2^16, 2^256} /-- Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$, and $n-2$ divides $2^n - 2$. -/ theorem putnam_2018_b3 (n : ℕ) (hn : n < 0) : (n < 10^100 ∧ ((n : ℤ) ∣ (2^n : ℤ) ∧ (n - 1 : ℤ) ∣ (2^n - 1 : ℤ) ∧ (n - 2 : ℤ) ∣ (2^n - 2 : ℤ))) ↔ n ∈ putnam_2018_b3_solution := sorry

Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$, and $n-2$ divides $2^n - 2$.

Show that the solution is the set $\{2^2, 2^4, 2^8, 2^16\}$.

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putnam_2008_b6

84945d81-b313-525f-a8c8-8960c12ac7bc

train

theorem putnam_2008_b6 (n k : ℕ) (hnk : n > 0 ∧ k > 0) : Odd (Set.ncard {s : Equiv.Perm (Fin n) | klimited k n s}) ↔ (n ≡ 0 [MOD 2*k+1] ∨ n ≡ 1 [MOD 2*k+1]) := sorry

import Mathlib open Filter Topology Set Nat def klimited (k n : ℕ) (s : Equiv.Perm (Fin n)) := ∀ i, |((s i) : ℤ) - i| ≤ k /-- Let $n$ and $k$ be positive integers. Say that a permutation $\sigma$ of $\{1,2,\dots,n\} is $k-limited$ if \|\sigma(i) - i\| \leq k$ for all $i$. Prove that the number of $k-limited$ permutations $\{1,2,\dots,n\}$ is odd if and only if $n \equiv 0$ or $1 (mod 2k+1)$. -/ theorem putnam_2008_b6 (n k : ℕ) (hnk : n > 0 ∧ k > 0) : Odd (Set.ncard {s : Equiv.Perm (Fin n) | klimited k n s}) ↔ (n ≡ 0 [MOD 2*k+1] ∨ n ≡ 1 [MOD 2*k+1]) := by

import Mathlib open Filter Topology Set Nat def klimited (k n : ℕ) (s : Equiv.Perm (Fin n)) := ∀ i, |((s i) : ℤ) - i| ≤ k /-- Let $n$ and $k$ be positive integers. Say that a permutation $\sigma$ of $\{1,2,\dots,n\} is $k-limited$ if \|\sigma(i) - i\| \leq k$ for all $i$. Prove that the number of $k-limited$ permutations $\{1,2,\dots,n\}$ is odd if and only if $n \equiv 0$ or $1 (mod 2k+1)$. -/ theorem putnam_2008_b6 (n k : ℕ) (hnk : n > 0 ∧ k > 0) : Odd (Set.ncard {s : Equiv.Perm (Fin n) | klimited k n s}) ↔ (n ≡ 0 [MOD 2*k+1] ∨ n ≡ 1 [MOD 2*k+1]) := sorry

Let $n$ and $k$ be positive integers. Say that a permutation $\sigma$ of $\{1,2,\dots,n\} is $k-limited$ if \|\sigma(i) - i\| \leq k$ for all $i$. Prove that the number of $k-limited$ permutations $\{1,2,\dots,n\}$ is odd if and only if $n \equiv 0$ or $1 (mod 2k+1)$.

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putnam_1989_a6

3c93059f-1f14-5c56-89fa-8dde3b383ef9

train

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

import Mathlib open Nat /-- Let $\alpha=1+a_1x+a_2x^2+\cdots$ be a formal power series with coefficients in the field of two elements. Let \[ a_n = \begin{cases} 1 & \parbox{2in}{if every block of zeros in the binary expansion of $n$ has an even number of zeros in the block} \\[.3in] 0 & \text{otherwise.} \end{cases} \] (For example, $a_{36}=1$ because $36=100100_2$ and $a_{20}=0$ because $20=10100_2.$) Prove that $\alpha^3+x\alpha+1=0.$ -/ theorem putnam_1989_a6 (F : Type*) [Field F] [Fintype F] (hF : Fintype.card F = 2) (α : PowerSeries F) (hα : ∀ n : ℕ, let bin := [1] ++ (digits 2 n) ++ [1]; PowerSeries.coeff F n α = ite (∀ i j : Fin bin.length, i < j → bin.get i = 1 → bin.get j = 1 → (∀ k, i < k → k < j → bin.get k = 0) → Even ((j : ℕ) - (i : ℕ) - 1)) 1 0) : (α ^ 3 + PowerSeries.X * α + 1 = 0) := by

import Mathlib open Nat /-- Let $\alpha=1+a_1x+a_2x^2+\cdots$ be a formal power series with coefficients in the field of two elements. Let \[ a_n = \begin{cases} 1 & \parbox{2in}{if every block of zeros in the binary expansion of $n$ has an even number of zeros in the block} \\[.3in] 0 & \text{otherwise.} \end{cases} \] (For example, $a_{36}=1$ because $36=100100_2$ and $a_{20}=0$ because $20=10100_2.$) Prove that $\alpha^3+x\alpha+1=0.$ -/ theorem putnam_1989_a6 (F : Type*) [Field F] [Fintype F] (hF : Fintype.card F = 2) (α : PowerSeries F) (hα : ∀ n : ℕ, let bin := [1] ++ (digits 2 n) ++ [1]; PowerSeries.coeff F n α = ite (∀ i j : Fin bin.length, i < j → bin.get i = 1 → bin.get j = 1 → (∀ k, i < k → k < j → bin.get k = 0) → Even ((j : ℕ) - (i : ℕ) - 1)) 1 0) : (α ^ 3 + PowerSeries.X * α + 1 = 0) := sorry

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

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putnam_1973_a4

0bba6054-4a50-5ba7-9bcd-cf26b7a96936

train

abbrev putnam_1973_a4_solution : ℕ := sorry -- 3 /-- How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line? -/ theorem putnam_1973_a4 (f : ℝ → ℝ) (hf : f = fun x => 2^x - 1 - x^2) : putnam_1973_a4_solution = {x : ℝ | f x = 0}.ncard := sorry

import Mathlib open Nat Set MeasureTheory Topology Filter -- 3 /-- How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line? -/ theorem putnam_1973_a4 (f : ℝ → ℝ) (hf : f = fun x => 2^x - 1 - x^2) : putnam_1973_a4_solution = {x : ℝ | f x = 0}.ncard := by

import Mathlib open Nat Set MeasureTheory Topology Filter abbrev putnam_1973_a4_solution : ℕ := sorry -- 3 /-- How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line? -/ theorem putnam_1973_a4 (f : ℝ → ℝ) (hf : f = fun x => 2^x - 1 - x^2) : putnam_1973_a4_solution = {x : ℝ | f x = 0}.ncard := sorry

How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line?

Show that the solution is 3.

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putnam_2014_a4

98772546-3ebb-5b60-959a-568073c9d2b6

train

abbrev putnam_2014_a4_solution : ℝ := sorry -- 1 / 3 /-- Suppose $X$ is a random variable that takes on only nonnegative integer values, with $E\left[X\right]=1$, $E\left[X^2\right]=2$, and $E\left[X^3\right]=5$. (Here $E\left[Y\right]$ denotes the expectation of the random variable $Y$.) Determine the smallest possible value of the probability of the event $X=0$. -/ theorem putnam_2014_a4 (Xrandvar : (ℕ → ℝ) → Prop) (hXrandvar : Xrandvar = (fun X : ℕ → ℝ => (∀ n : ℕ, X n ∈ Set.Icc 0 1) ∧ ∑' n : ℕ, X n = 1)) (E : (ℕ → ℝ) → (ℕ → ℝ) → ℝ) (hE : E = (fun (X : ℕ → ℝ) (f : ℕ → ℝ) => ∑' n : ℕ, f n * X n)) : sInf {X0 : ℝ | ∃ X : ℕ → ℝ, Xrandvar X ∧ E X (fun x : ℕ => x) = 1 ∧ E X (fun x : ℕ => x ^ 2) = 2 ∧ E X (fun x : ℕ => x ^ 3) = 5 ∧ X0 = X 0} = putnam_2014_a4_solution := sorry

import Mathlib open Topology Filter Nat -- 1 / 3 /-- Suppose $X$ is a random variable that takes on only nonnegative integer values, with $E\left[X\right]=1$, $E\left[X^2\right]=2$, and $E\left[X^3\right]=5$. (Here $E\left[Y\right]$ denotes the expectation of the random variable $Y$.) Determine the smallest possible value of the probability of the event $X=0$. -/ theorem putnam_2014_a4 (Xrandvar : (ℕ → ℝ) → Prop) (hXrandvar : Xrandvar = (fun X : ℕ → ℝ => (∀ n : ℕ, X n ∈ Set.Icc 0 1) ∧ ∑' n : ℕ, X n = 1)) (E : (ℕ → ℝ) → (ℕ → ℝ) → ℝ) (hE : E = (fun (X : ℕ → ℝ) (f : ℕ → ℝ) => ∑' n : ℕ, f n * X n)) : sInf {X0 : ℝ | ∃ X : ℕ → ℝ, Xrandvar X ∧ E X (fun x : ℕ => x) = 1 ∧ E X (fun x : ℕ => x ^ 2) = 2 ∧ E X (fun x : ℕ => x ^ 3) = 5 ∧ X0 = X 0} = putnam_2014_a4_solution := by

import Mathlib open Topology Filter Nat noncomputable abbrev putnam_2014_a4_solution : ℝ := sorry -- 1 / 3 /-- Suppose $X$ is a random variable that takes on only nonnegative integer values, with $E\left[X\right]=1$, $E\left[X^2\right]=2$, and $E\left[X^3\right]=5$. (Here $E\left[Y\right]$ denotes the expectation of the random variable $Y$.) Determine the smallest possible value of the probability of the event $X=0$. -/ theorem putnam_2014_a4 (Xrandvar : (ℕ → ℝ) → Prop) (hXrandvar : Xrandvar = (fun X : ℕ → ℝ => (∀ n : ℕ, X n ∈ Set.Icc 0 1) ∧ ∑' n : ℕ, X n = 1)) (E : (ℕ → ℝ) → (ℕ → ℝ) → ℝ) (hE : E = (fun (X : ℕ → ℝ) (f : ℕ → ℝ) => ∑' n : ℕ, f n * X n)) : sInf {X0 : ℝ | ∃ X : ℕ → ℝ, Xrandvar X ∧ E X (fun x : ℕ => x) = 1 ∧ E X (fun x : ℕ => x ^ 2) = 2 ∧ E X (fun x : ℕ => x ^ 3) = 5 ∧ X0 = X 0} = putnam_2014_a4_solution := sorry

Suppose $X$ is a random variable that takes on only nonnegative integer values, with $E\left[X\right]=1$, $E\left[X^2\right]=2$, and $E\left[X^3\right]=5$. (Here $E\left[Y\right]$ denotes the expectation of the random variable $Y$.) Determine the smallest possible value of the probability of the event $X=0$.

Show that the answer is $\frac{1}{3}$.

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putnam_2002_b6

decbcbc1-0884-52f7-b570-0108fe9b2d8d

train

theorem putnam_2002_b6 (p : ℕ) (hp : Nat.Prime p) (M : Matrix (Fin 3) (Fin 3) (MvPolynomial (Fin 3) ℤ)) (hM : M = fun (r c : Fin 3) => ((X c)^(p^(r : ℕ)) : MvPolynomial (Fin 3) ℤ)) (cong : ℕ → MvPolynomial (Fin 3) ℤ × MvPolynomial (Fin 3) ℤ → Prop) (hcong : cong = fun p : ℕ => fun (f, g) => ∀ n : Fin 3 →₀ ℕ, Int.ModEq p (f.coeff n) (g.coeff n)) : ∃ S : Finset (MvPolynomial (Fin 3) ℤ), cong p ((det M), (∏ s in S, s)) ∧ ∀ s ∈ S, (∃ a b c : ℤ, s = (C a)*(X 0) + (C b)*(X 1) + (C c)*(X 2)) := sorry

import Mathlib open Nat Set Topology Filter Matrix MvPolynomial /-- Let $p$ be a prime number. Prove that the determinant of the matrix \[ \begin{pmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{pmatrix} \] is congruent modulo $p$ to a product of polynomials of the form $ax+by+cz$, where $a,b,c$ are integers. (We say two integer polynomials are congruent modulo $p$ if corresponding coefficients are congruent modulo $p$.) -/ theorem putnam_2002_b6 (p : ℕ) (hp : Nat.Prime p) (M : Matrix (Fin 3) (Fin 3) (MvPolynomial (Fin 3) ℤ)) (hM : M = fun (r c : Fin 3) => ((X c)^(p^(r : ℕ)) : MvPolynomial (Fin 3) ℤ)) (cong : ℕ → MvPolynomial (Fin 3) ℤ × MvPolynomial (Fin 3) ℤ → Prop) (hcong : cong = fun p : ℕ => fun (f, g) => ∀ n : Fin 3 →₀ ℕ, Int.ModEq p (f.coeff n) (g.coeff n)) : ∃ S : Finset (MvPolynomial (Fin 3) ℤ), cong p ((det M), (∏ s in S, s)) ∧ ∀ s ∈ S, (∃ a b c : ℤ, s = (C a)*(X 0) + (C b)*(X 1) + (C c)*(X 2)) := by

import Mathlib open Nat Set Topology Filter Matrix MvPolynomial /-- Let $p$ be a prime number. Prove that the determinant of the matrix \[ \begin{pmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{pmatrix} \] is congruent modulo $p$ to a product of polynomials of the form $ax+by+cz$, where $a,b,c$ are integers. (We say two integer polynomials are congruent modulo $p$ if corresponding coefficients are congruent modulo $p$.) -/ theorem putnam_2002_b6 (p : ℕ) (hp : Nat.Prime p) (M : Matrix (Fin 3) (Fin 3) (MvPolynomial (Fin 3) ℤ)) (hM : M = fun (r c : Fin 3) => ((X c)^(p^(r : ℕ)) : MvPolynomial (Fin 3) ℤ)) (cong : ℕ → MvPolynomial (Fin 3) ℤ × MvPolynomial (Fin 3) ℤ → Prop) (hcong : cong = fun p : ℕ => fun (f, g) => ∀ n : Fin 3 →₀ ℕ, Int.ModEq p (f.coeff n) (g.coeff n)) : ∃ S : Finset (MvPolynomial (Fin 3) ℤ), cong p ((det M), (∏ s in S, s)) ∧ ∀ s ∈ S, (∃ a b c : ℤ, s = (C a)*(X 0) + (C b)*(X 1) + (C c)*(X 2)) := sorry

Let $p$ be a prime number. Prove that the determinant of the matrix \[ \begin{pmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{pmatrix} \] is congruent modulo $p$ to a product of polynomials of the form $ax+by+cz$, where $a,b,c$ are integers. (We say two integer polynomials are congruent modulo $p$ if corresponding coefficients are congruent modulo $p$.)

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putnam_1976_b5

9d59352d-d938-56b3-9aa6-946fad040835

train

abbrev putnam_1976_b5_solution : ℕ → Polynomial ℤ := sorry -- fun n => C (Nat.factorial n) /-- Find $$\sum_{k=0}^{n} (-1)^k {n \choose k} (x - k)^n.$$ -/ theorem putnam_1976_b5 : ∀ n : ℕ, ∑ k in Finset.range (n + 1), C ((-(1 : ℤ))^k * Nat.choose n k) * (X - (C (k : ℤ)))^n = putnam_1976_b5_solution n := sorry

import Mathlib open Polynomial Filter Topology ProbabilityTheory MeasureTheory -- fun n => C (Nat.factorial n) /-- Find $$\sum_{k=0}^{n} (-1)^k {n \choose k} (x - k)^n.$$ -/ theorem putnam_1976_b5 : ∀ n : ℕ, ∑ k in Finset.range (n + 1), C ((-(1 : ℤ))^k * Nat.choose n k) * (X - (C (k : ℤ)))^n = putnam_1976_b5_solution n := by

import Mathlib open Polynomial Filter Topology ProbabilityTheory MeasureTheory noncomputable abbrev putnam_1976_b5_solution : ℕ → Polynomial ℤ := sorry -- fun n => C (Nat.factorial n) /-- Find $$\sum_{k=0}^{n} (-1)^k {n \choose k} (x - k)^n.$$ -/ theorem putnam_1976_b5 : ∀ n : ℕ, ∑ k in Finset.range (n + 1), C ((-(1 : ℤ))^k * Nat.choose n k) * (X - (C (k : ℤ)))^n = putnam_1976_b5_solution n := sorry

Find $$\sum_{k=0}^{n} (-1)^k {n \choose k} (x - k)^n.$$

The sum equals $n!$.

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putnam_1974_b2

2bc1da42-a678-5228-b03b-df097d2a43ba

train

theorem putnam_1974_b2 (y : ℝ → ℝ) (ycontdiff : ContDiff ℝ 1 y) (limy : Tendsto (fun x : ℝ => (deriv y x)^2 + (y x)^3) atTop (𝓝 0)) : Tendsto y atTop (𝓝 0) ∧ Tendsto (deriv y) atTop (𝓝 0) := sorry

import Mathlib open Set Nat Polynomial Filter Topology /-- Let $y(x)$ be a continuously differentiable real-valued function of a real vairable $x$. Show that if $(y')^2 + y^3 \to 0$ as $x \to +\infty$, then $y(x)$ and $y'(x) \to 0$ as $x \to +\infty$. -/ theorem putnam_1974_b2 (y : ℝ → ℝ) (ycontdiff : ContDiff ℝ 1 y) (limy : Tendsto (fun x : ℝ => (deriv y x)^2 + (y x)^3) atTop (𝓝 0)) : Tendsto y atTop (𝓝 0) ∧ Tendsto (deriv y) atTop (𝓝 0) := by

import Mathlib open Set Nat Polynomial Filter Topology /-- Let $y(x)$ be a continuously differentiable real-valued function of a real vairable $x$. Show that if $(y')^2 + y^3 \to 0$ as $x \to +\infty$, then $y(x)$ and $y'(x) \to 0$ as $x \to +\infty$. -/ theorem putnam_1974_b2 (y : ℝ → ℝ) (ycontdiff : ContDiff ℝ 1 y) (limy : Tendsto (fun x : ℝ => (deriv y x)^2 + (y x)^3) atTop (𝓝 0)) : Tendsto y atTop (𝓝 0) ∧ Tendsto (deriv y) atTop (𝓝 0) := sorry

Let $y(x)$ be a continuously differentiable real-valued function of a real vairable $x$. Show that if $(y')^2 + y^3 \to 0$ as $x \to +\infty$, then $y(x)$ and $y'(x) \to 0$ as $x \to +\infty$.

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putnam_2001_a5

e335529b-36bd-54b2-b2ee-4a534369da4a

train

theorem putnam_2001_a5 : ∃! an : ℤ × ℕ, let (a, n) := an; a > 0 ∧ n > 0 ∧ a^(n+1) - (a+1)^n = 2001 := sorry

import Mathlib open Topology Filter Polynomial Set /-- Prove that there are unique positive integers $a$, $n$ such that $a^{n+1}-(a+1)^n=2001$. -/ theorem putnam_2001_a5 : ∃! an : ℤ × ℕ, let (a, n) := an; a > 0 ∧ n > 0 ∧ a^(n+1) - (a+1)^n = 2001 := by

import Mathlib open Topology Filter Polynomial Set /-- Prove that there are unique positive integers $a$, $n$ such that $a^{n+1}-(a+1)^n=2001$. -/ theorem putnam_2001_a5 : ∃! an : ℤ × ℕ, let (a, n) := an; a > 0 ∧ n > 0 ∧ a^(n+1) - (a+1)^n = 2001 := sorry

Prove that there are unique positive integers $a$, $n$ such that $a^{n+1}-(a+1)^n=2001$.

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putnam_2000_a6

02414c8b-71b9-50b9-b668-7c315c2b1b52

train

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

import Mathlib open Topology Filter /-- 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 $a_2=0$. -/ theorem putnam_2000_a6 (f : Polynomial ℤ) (a : ℕ → ℤ) (ha0 : a 0 = 0) (ha : ∀ n : ℕ, a (n + 1) = f.eval (a n)) : ((∃ m > 0, a m = 0) → (a 1 = 0 ∨ a 2 = 0)) := by

import Mathlib open Topology Filter /-- 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 $a_2=0$. -/ theorem putnam_2000_a6 (f : Polynomial ℤ) (a : ℕ → ℤ) (ha0 : a 0 = 0) (ha : ∀ n : ℕ, a (n + 1) = f.eval (a n)) : ((∃ m > 0, a m = 0) → (a 1 = 0 ∨ a 2 = 0)) := sorry

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

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putnam_1997_b2

5853bc8c-4356-5e10-b751-0910fa17c787

train

theorem putnam_1997_b2 (f g : ℝ → ℝ) (hg : ∀ x : ℝ, g x ≥ 0) (hfderiv1 : ContDiff ℝ 1 f) (hfderiv2 : Differentiable ℝ (deriv f)) (hfg : ∀ x : ℝ, f x + iteratedDeriv 2 f x = -x * g x * deriv f x) : IsBounded (range (fun x => |f x|)) := sorry

import Mathlib open Filter Topology Bornology Set /-- Let $f$ be a twice-differentiable real-valued function satisfying \[f(x)+f''(x)=-xg(x)f'(x),\] where $g(x)\geq 0$ for all real $x$. Prove that $|f(x)|$ is bounded. -/ theorem putnam_1997_b2 (f g : ℝ → ℝ) (hg : ∀ x : ℝ, g x ≥ 0) (hfderiv1 : ContDiff ℝ 1 f) (hfderiv2 : Differentiable ℝ (deriv f)) (hfg : ∀ x : ℝ, f x + iteratedDeriv 2 f x = -x * g x * deriv f x) : IsBounded (range (fun x => |f x|)) := by

import Mathlib open Filter Topology Bornology Set /-- Let $f$ be a twice-differentiable real-valued function satisfying \[f(x)+f''(x)=-xg(x)f'(x),\] where $g(x)\geq 0$ for all real $x$. Prove that $|f(x)|$ is bounded. -/ theorem putnam_1997_b2 (f g : ℝ → ℝ) (hg : ∀ x : ℝ, g x ≥ 0) (hfderiv1 : ContDiff ℝ 1 f) (hfderiv2 : Differentiable ℝ (deriv f)) (hfg : ∀ x : ℝ, f x + iteratedDeriv 2 f x = -x * g x * deriv f x) : IsBounded (range (fun x => |f x|)) := sorry

Let $f$ be a twice-differentiable real-valued function satisfying \[f(x)+f''(x)=-xg(x)f'(x),\] where $g(x)\geq 0$ for all real $x$. Prove that $|f(x)|$ is bounded.

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putnam_1992_b1

0a2d5787-4803-58b5-8d02-02248df7c00d

train

abbrev putnam_1992_b1_solution : ℕ → ℤ := sorry -- fun n ↦ 2 * n - 3 /-- Let $S$ be a set of $n$ distinct real numbers. Let $A_S$ be the set of numbers that occur as averages of two distinct elements of $S$. For a given $n \geq 2$, what is the smallest possible number of elements in $A_S$? -/ theorem putnam_1992_b1 (n : ℕ) (hn : n ≥ 2) (A : Finset ℝ → Set ℝ) (hA : A = fun S ↦ {x | ∃ a ∈ S, ∃ b ∈ S, a ≠ b ∧ (a + b) / 2 = x}) : IsLeast {k : ℤ | ∃ S : Finset ℝ, S.card = n ∧ k = (A S).ncard} (putnam_1992_b1_solution n) := sorry

import Mathlib open Topology Filter Nat Function -- fun n ↦ 2 * n - 3 /-- Let $S$ be a set of $n$ distinct real numbers. Let $A_S$ be the set of numbers that occur as averages of two distinct elements of $S$. For a given $n \geq 2$, what is the smallest possible number of elements in $A_S$? -/ theorem putnam_1992_b1 (n : ℕ) (hn : n ≥ 2) (A : Finset ℝ → Set ℝ) (hA : A = fun S ↦ {x | ∃ a ∈ S, ∃ b ∈ S, a ≠ b ∧ (a + b) / 2 = x}) : IsLeast {k : ℤ | ∃ S : Finset ℝ, S.card = n ∧ k = (A S).ncard} (putnam_1992_b1_solution n) := by

import Mathlib open Topology Filter Nat Function abbrev putnam_1992_b1_solution : ℕ → ℤ := sorry -- fun n ↦ 2 * n - 3 /-- Let $S$ be a set of $n$ distinct real numbers. Let $A_S$ be the set of numbers that occur as averages of two distinct elements of $S$. For a given $n \geq 2$, what is the smallest possible number of elements in $A_S$? -/ theorem putnam_1992_b1 (n : ℕ) (hn : n ≥ 2) (A : Finset ℝ → Set ℝ) (hA : A = fun S ↦ {x | ∃ a ∈ S, ∃ b ∈ S, a ≠ b ∧ (a + b) / 2 = x}) : IsLeast {k : ℤ | ∃ S : Finset ℝ, S.card = n ∧ k = (A S).ncard} (putnam_1992_b1_solution n) := sorry

Let $S$ be a set of $n$ distinct real numbers. Let $A_S$ be the set of numbers that occur as averages of two distinct elements of $S$. For a given $n \geq 2$, what is the smallest possible number of elements in $A_S$?

Show that the answer is $2n - 3$.

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putnam_1978_a3

b0e4432a-e03d-5188-a4ed-2a9ba72fb074

train

abbrev putnam_1978_a3_solution : ℕ := sorry -- 2 /-- Let $p(x) = 2(x^6 + 1) + 4(x^5 + x) + 3(x^4 + x^2) + 5x^3$. For $k$ with $0 < k < 5$, let \[ I_k = \int_0^{\infty} \frac{x^k}{p(x)} \, dx. \] For which $k$ is $I_k$ smallest? -/ theorem putnam_1978_a3 (p : Polynomial ℝ) (hp : p = 2 * (X ^ 6 + 1) + 4 * (X ^ 5 + X) + 3 * (X ^ 4 + X ^ 2) + 5 * X ^ 3) (I : ℕ → ℝ) (hI : I = fun k ↦ ∫ x in Ioi 0, x ^ k / p.eval x) : IsLeast {y | ∃ k ∈ Ioo 0 5, I k = y} putnam_1978_a3_solution := sorry

import Mathlib open Set Polynomial -- 2 /-- Let $p(x) = 2(x^6 + 1) + 4(x^5 + x) + 3(x^4 + x^2) + 5x^3$. For $k$ with $0 < k < 5$, let \[ I_k = \int_0^{\infty} \frac{x^k}{p(x)} \, dx. \] For which $k$ is $I_k$ smallest? -/ theorem putnam_1978_a3 (p : Polynomial ℝ) (hp : p = 2 * (X ^ 6 + 1) + 4 * (X ^ 5 + X) + 3 * (X ^ 4 + X ^ 2) + 5 * X ^ 3) (I : ℕ → ℝ) (hI : I = fun k ↦ ∫ x in Ioi 0, x ^ k / p.eval x) : IsLeast {y | ∃ k ∈ Ioo 0 5, I k = y} putnam_1978_a3_solution := by

import Mathlib open Set Polynomial abbrev putnam_1978_a3_solution : ℕ := sorry -- 2 /-- Let $p(x) = 2(x^6 + 1) + 4(x^5 + x) + 3(x^4 + x^2) + 5x^3$. For $k$ with $0 < k < 5$, let \[ I_k = \int_0^{\infty} \frac{x^k}{p(x)} \, dx. \] For which $k$ is $I_k$ smallest? -/ theorem putnam_1978_a3 (p : Polynomial ℝ) (hp : p = 2 * (X ^ 6 + 1) + 4 * (X ^ 5 + X) + 3 * (X ^ 4 + X ^ 2) + 5 * X ^ 3) (I : ℕ → ℝ) (hI : I = fun k ↦ ∫ x in Ioi 0, x ^ k / p.eval x) : IsLeast {y | ∃ k ∈ Ioo 0 5, I k = y} putnam_1978_a3_solution := sorry

Let $p(x) = 2(x^6 + 1) + 4(x^5 + x) + 3(x^4 + x^2) + 5x^3$. For $k$ with $0 < k < 5$, let \[ I_k = \int_0^{\infty} \frac{x^k}{p(x)} \, dx. \] For which $k$ is $I_k$ smallest?

Show that $I_k$ is smallest for $k = 2$.

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putnam_1967_b2

17cd65b4-6ac9-5243-9aa2-95c2fb3a17d6

train

theorem putnam_1967_b2 (p r A B C α β γ : ℝ) (prbound : 0 ≤ p ∧ p ≤ 1 ∧ 0 ≤ r ∧ r ≤ 1) (id1 : ∀ x y : ℝ, (p * x + (1 - p) * y) ^ 2 = A * x ^ 2 + B * x * y + C * y ^ 2) (id2 : ∀ x y : ℝ, (p * x + (1 - p) * y) * (r * x + (1 - r) * y) = α * x ^ 2 + β * x * y + γ * y ^ 2) : max (max A B) C ≥ 4 / 9 ∧ max (max α β) γ ≥ 4 / 9 := sorry

import Mathlib open Nat Topology Filter /-- Let $0 \leq p \leq 1$ and $0 \leq r \leq 1$ and consider the identities \begin{enumerate} \item[(a)] $(px+(1-p)y)^2=Ax^2+Bxy+Cy^2$, \item[(b)] $(px+(1-p)y)(rx+(1-r)y)=\alpha x^2+\beta xy+\gamma y^2$. \end{enumerate} Show that (with respect to $p$ and $r$) \begin{enumerate} \item[(a)] $\max\{A,B,C\} \geq 4/9$, \item[(b)] $\max\{\alpha,\beta,\gamma\} \geq 4/9$. \end{enumerate} -/ theorem putnam_1967_b2 (p r A B C α β γ : ℝ) (prbound : 0 ≤ p ∧ p ≤ 1 ∧ 0 ≤ r ∧ r ≤ 1) (id1 : ∀ x y : ℝ, (p * x + (1 - p) * y) ^ 2 = A * x ^ 2 + B * x * y + C * y ^ 2) (id2 : ∀ x y : ℝ, (p * x + (1 - p) * y) * (r * x + (1 - r) * y) = α * x ^ 2 + β * x * y + γ * y ^ 2) : max (max A B) C ≥ 4 / 9 ∧ max (max α β) γ ≥ 4 / 9 := by

import Mathlib open Nat Topology Filter /-- Let $0 \leq p \leq 1$ and $0 \leq r \leq 1$ and consider the identities \begin{enumerate} \item[(a)] $(px+(1-p)y)^2=Ax^2+Bxy+Cy^2$, \item[(b)] $(px+(1-p)y)(rx+(1-r)y)=\alpha x^2+\beta xy+\gamma y^2$. \end{enumerate} Show that (with respect to $p$ and $r$) \begin{enumerate} \item[(a)] $\max\{A,B,C\} \geq 4/9$, \item[(b)] $\max\{\alpha,\beta,\gamma\} \geq 4/9$. \end{enumerate} -/ theorem putnam_1967_b2 (p r A B C α β γ : ℝ) (prbound : 0 ≤ p ∧ p ≤ 1 ∧ 0 ≤ r ∧ r ≤ 1) (id1 : ∀ x y : ℝ, (p * x + (1 - p) * y) ^ 2 = A * x ^ 2 + B * x * y + C * y ^ 2) (id2 : ∀ x y : ℝ, (p * x + (1 - p) * y) * (r * x + (1 - r) * y) = α * x ^ 2 + β * x * y + γ * y ^ 2) : max (max A B) C ≥ 4 / 9 ∧ max (max α β) γ ≥ 4 / 9 := sorry

Let $0 \leq p \leq 1$ and $0 \leq r \leq 1$ and consider the identities \begin{enumerate} \item[(a)] $(px+(1-p)y)^2=Ax^2+Bxy+Cy^2$, \item[(b)] $(px+(1-p)y)(rx+(1-r)y)=\alpha x^2+\beta xy+\gamma y^2$. \end{enumerate} Show that (with respect to $p$ and $r$) \begin{enumerate} \item[(a)] $\max\{A,B,C\} \geq 4/9$, \item[(b)] $\max\{\alpha,\beta,\gamma\} \geq 4/9$. \end{enumerate}

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putnam_1995_b4

ff6da42a-7c4e-5d08-a542-6a1d101621e7

train

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

import Mathlib open Filter Topology Real Nat -- ⟨3,1,5,2⟩ /-- 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. -/ theorem putnam_1995_b4 (contfrac : ℝ) (hcontfrac : contfrac = 2207 - 1 / contfrac) (hcontfrac' : 1 < contfrac) : let ⟨a, b, c, d⟩ := putnam_1995_b4_solution contfrac ^ ((1 : ℝ) / 8) = (a + b * sqrt c) / d := by

import Mathlib open Filter Topology Real Nat abbrev putnam_1995_b4_solution : ℤ × ℤ × ℤ × ℤ := sorry -- ⟨3,1,5,2⟩ /-- 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. -/ theorem putnam_1995_b4 (contfrac : ℝ) (hcontfrac : contfrac = 2207 - 1 / contfrac) (hcontfrac' : 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.

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putnam_2008_a6

fef4518c-5305-524a-adf9-1cfed1923d57

train

theorem putnam_2008_a6 : ∃ c : ℝ, c > 0 ∧ ∀ G [Group G] (fing : Fintype G), fing.card > 1 → ∃ s : List G, s.length ≤ c * Real.log (fing.card : ℝ) ∧ ∀ g : G, ∃ t : List G, t.Sublist s ∧ t.prod = g := sorry

import Mathlib open Filter Topology Set /-- Prove that there exists a constant $c>0$ such that in every nontrivial finite group $G$ there exists a sequence of length at most $c \log |G|$ with the property that each element of $G$ equals the product of some subsequence. (The elements of $G$ in the sequence are not required to be distinct. A \emph{subsequence} of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but $2, 2, 4$ is not.) -/ theorem putnam_2008_a6 : ∃ c : ℝ, c > 0 ∧ ∀ G [Group G] (fing : Fintype G), fing.card > 1 → ∃ s : List G, s.length ≤ c * Real.log (fing.card : ℝ) ∧ ∀ g : G, ∃ t : List G, t.Sublist s ∧ t.prod = g := by

import Mathlib open Filter Topology Set /-- Prove that there exists a constant $c>0$ such that in every nontrivial finite group $G$ there exists a sequence of length at most $c \log |G|$ with the property that each element of $G$ equals the product of some subsequence. (The elements of $G$ in the sequence are not required to be distinct. A \emph{subsequence} of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but $2, 2, 4$ is not.) -/ theorem putnam_2008_a6 : ∃ c : ℝ, c > 0 ∧ ∀ G [Group G] (fing : Fintype G), fing.card > 1 → ∃ s : List G, s.length ≤ c * Real.log (fing.card : ℝ) ∧ ∀ g : G, ∃ t : List G, t.Sublist s ∧ t.prod = g := sorry

Prove that there exists a constant $c>0$ such that in every nontrivial finite group $G$ there exists a sequence of length at most $c \log |G|$ with the property that each element of $G$ equals the product of some subsequence. (The elements of $G$ in the sequence are not required to be distinct. A \emph{subsequence} of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but $2, 2, 4$ is not.)

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putnam_2009_b5

4e0ec67b-bff6-5a5d-9f02-fe9cd9bd4f63

train

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

import Mathlib open Topology MvPolynomial Filter Set Metric /-- 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 $\lim_{x \to \infty} f(x) = \infty$. -/ theorem putnam_2009_b5 (f : ℝ → ℝ) (hfdiff : DifferentiableOn ℝ f (Ioi 1)) (hf : ∀ x > 1, deriv f x = (x ^ 2 - (f x) ^ 2) / ((x ^ 2) * ((f x) ^ 2 + 1))) : (Tendsto f atTop atTop) := by

import Mathlib open Topology MvPolynomial Filter Set Metric /-- 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 $\lim_{x \to \infty} f(x) = \infty$. -/ theorem putnam_2009_b5 (f : ℝ → ℝ) (hfdiff : DifferentiableOn ℝ f (Ioi 1)) (hf : ∀ x > 1, deriv f x = (x ^ 2 - (f x) ^ 2) / ((x ^ 2) * ((f x) ^ 2 + 1))) : (Tendsto f atTop atTop) := 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 $\lim_{x \to \infty} f(x) = \infty$.

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putnam_2020_b6

d45de3d5-9f86-5a0e-bef4-90f27cf0bb73

train

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

import Mathlib open Filter Topology Set /-- Let $n$ be a positive integer. Prove that $\sum_{k=1}^n(-1)^{\lfloor k(\sqrt{2}-1) \rfloor} \geq 0$. -/ theorem putnam_2020_b6 (n : ℕ) (npos : n > 0) : ∑ k in Finset.Icc 1 n, ((-1) ^ Int.floor (k * (Real.sqrt 2 - 1)) : ℝ) ≥ 0 := by

import Mathlib open Filter Topology Set /-- Let $n$ be a positive integer. Prove that $\sum_{k=1}^n(-1)^{\lfloor k(\sqrt{2}-1) \rfloor} \geq 0$. -/ theorem putnam_2020_b6 (n : ℕ) (npos : n > 0) : ∑ k in Finset.Icc 1 n, ((-1) ^ Int.floor (k * (Real.sqrt 2 - 1)) : ℝ) ≥ 0 := sorry

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

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putnam_2018_a2

f635cde6-8968-5cfc-8f5d-c457bf1308a7

train

abbrev putnam_2018_a2_solution : ℕ → ℝ := sorry -- (fun n : ℕ => if n = 1 then 1 else -1) /-- Let \( S_1, S_2, \ldots, S_{2^n-1} \) be the nonempty subsets of \( \{1, 2, \ldots, n\} \) in some order, and let \( M \) be the \( (2^n - 1) \times (2^n - 1) \) matrix whose \((i, j)\) entry is $m_{ij} = \begin{cases} 0 & \text{if } S_i \cap S_j = \emptyset; \\ 1 & \text{otherwise}. \end{cases} $ Calculate the determinant of \( M \). -/ theorem putnam_2018_a2 (n : ℕ) (S : Fin (2 ^ n - 1) → Set ℕ) (M : Matrix (Fin (2 ^ n - 1)) (Fin (2 ^ n - 1)) ℝ) (npos : n ≥ 1) (hS : Set.range S = (Set.Icc 1 n).powerset \ {∅}) (hM : ∀ i j, M i j = if (S i ∩ S j = ∅) = True then 0 else 1) : M.det = putnam_2018_a2_solution n := sorry

import Mathlib -- (fun n : ℕ => if n = 1 then 1 else -1) /-- Let \( S_1, S_2, \ldots, S_{2^n-1} \) be the nonempty subsets of \( \{1, 2, \ldots, n\} \) in some order, and let \( M \) be the \( (2^n - 1) \times (2^n - 1) \) matrix whose \((i, j)\) entry is $m_{ij} = \begin{cases} 0 & \text{if } S_i \cap S_j = \emptyset; \\ 1 & \text{otherwise}. \end{cases} $ Calculate the determinant of \( M \). -/ theorem putnam_2018_a2 (n : ℕ) (S : Fin (2 ^ n - 1) → Set ℕ) (M : Matrix (Fin (2 ^ n - 1)) (Fin (2 ^ n - 1)) ℝ) (npos : n ≥ 1) (hS : Set.range S = (Set.Icc 1 n).powerset \ {∅}) (hM : ∀ i j, M i j = if (S i ∩ S j = ∅) = True then 0 else 1) : M.det = putnam_2018_a2_solution n := by

import Mathlib abbrev putnam_2018_a2_solution : ℕ → ℝ := sorry -- (fun n : ℕ => if n = 1 then 1 else -1) /-- Let \( S_1, S_2, \ldots, S_{2^n-1} \) be the nonempty subsets of \( \{1, 2, \ldots, n\} \) in some order, and let \( M \) be the \( (2^n - 1) \times (2^n - 1) \) matrix whose \((i, j)\) entry is $m_{ij} = \begin{cases} 0 & \text{if } S_i \cap S_j = \emptyset; \\ 1 & \text{otherwise}. \end{cases} $ Calculate the determinant of \( M \). -/ theorem putnam_2018_a2 (n : ℕ) (S : Fin (2 ^ n - 1) → Set ℕ) (M : Matrix (Fin (2 ^ n - 1)) (Fin (2 ^ n - 1)) ℝ) (npos : n ≥ 1) (hS : Set.range S = (Set.Icc 1 n).powerset \ {∅}) (hM : ∀ i j, M i j = if (S i ∩ S j = ∅) = True then 0 else 1) : M.det = putnam_2018_a2_solution n := sorry

Let \( S_1, S_2, \ldots, S_{2^n-1} \) be the nonempty subsets of \( \{1, 2, \ldots, n\} \) in some order, and let \( M \) be the \( (2^n - 1) \times (2^n - 1) \) matrix whose \((i, j)\) entry is $m_{ij} = \begin{cases} 0 & \text{if } S_i \cap S_j = \emptyset; \\ 1 & \text{otherwise}. \end{cases} $ Calculate the determinant of \( M \).

Show that the solution is 1 if n = 1, and otherwise -1.

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putnam_2014_b2

eb1eff11-97dc-504a-98ff-bc32dd98e280

train

abbrev putnam_2014_b2_solution : ℝ := sorry -- Real.log (4 / 3) /-- Suppose that \( f \) is a function on the interval \([1,3]\) such that \(-1 \leq f(x) \leq 1\) for all \( x \) and \( \int_{1}^{3} f(x) \, dx = 0 \). How large can \(\int_{1}^{3} \frac{f(x)}{x} \, dx \) be? -/ theorem putnam_2014_b2 : IsGreatest {t | ∃ f : ℝ → ℝ, (∀ x : Set.Icc (1 : ℝ) 3, -1 ≤ f x ∧ f x ≤ 1) ∧ (∫ x in Set.Ioo 1 3, f x = 0) ∧ (∫ x in Set.Ioo 1 3, (f x) / x) = t} putnam_2014_b2_solution := sorry

import Mathlib open Topology Filter Nat -- Note: uses (ℝ → ℝ) instead of (Set.Icc (1 : ℝ) 3 → ℝ) -- Real.log (4 / 3) /-- Suppose that \( f \) is a function on the interval \([1,3]\) such that \(-1 \leq f(x) \leq 1\) for all \( x \) and \( \int_{1}^{3} f(x) \, dx = 0 \). How large can \(\int_{1}^{3} \frac{f(x)}{x} \, dx \) be? -/ theorem putnam_2014_b2 : IsGreatest {t | ∃ f : ℝ → ℝ, (∀ x : Set.Icc (1 : ℝ) 3, -1 ≤ f x ∧ f x ≤ 1) ∧ (∫ x in Set.Ioo 1 3, f x = 0) ∧ (∫ x in Set.Ioo 1 3, (f x) / x) = t} putnam_2014_b2_solution := by

import Mathlib open Topology Filter Nat -- Note: uses (ℝ → ℝ) instead of (Set.Icc (1 : ℝ) 3 → ℝ) noncomputable abbrev putnam_2014_b2_solution : ℝ := sorry -- Real.log (4 / 3) /-- Suppose that \( f \) is a function on the interval \([1,3]\) such that \(-1 \leq f(x) \leq 1\) for all \( x \) and \( \int_{1}^{3} f(x) \, dx = 0 \). How large can \(\int_{1}^{3} \frac{f(x)}{x} \, dx \) be? -/ theorem putnam_2014_b2 : IsGreatest {t | ∃ f : ℝ → ℝ, (∀ x : Set.Icc (1 : ℝ) 3, -1 ≤ f x ∧ f x ≤ 1) ∧ (∫ x in Set.Ioo 1 3, f x = 0) ∧ (∫ x in Set.Ioo 1 3, (f x) / x) = t} putnam_2014_b2_solution := sorry

Suppose that \( f \) is a function on the interval \([1,3]\) such that \(-1 \leq f(x) \leq 1\) for all \( x \) and \( \int_{1}^{3} f(x) \, dx = 0 \). How large can \(\int_{1}^{3} \frac{f(x)}{x} \, dx \) be?

Show that the solution is log (4 / 3).

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putnam_1987_a1

ec9f9a30-687f-5913-9bb9-9ace43f412c8

train

theorem putnam_1987_a1 (A B C D : Set (ℝ × ℝ)) (hA : A = {(x, y) : ℝ × ℝ | x ^ 2 + y ^ 2 ≠ 0 ∧ x ^ 2 - y ^ 2 = x / (x ^ 2 + y ^ 2)}) (hB : B = {(x, y) : ℝ × ℝ | x ^ 2 + y ^ 2 ≠ 0 ∧ 2 * x * y + y / (x ^ 2 + y ^ 2) = 3}) (hC : C = {(x, y) : ℝ × ℝ | x ^ 3 - 3 * x * y ^ 2 + 3 * y = 1}) (hD : D = {(x, y) : ℝ × ℝ | 3 * x ^ 2 * y - 3 * x - y ^ 3 = 0}) : A ∩ B = C ∩ D := sorry

import Mathlib /-- Curves $A$, $B$, $C$, and $D$ are defined in the plane as follows: \begin{align*} A&=\left\{ (x,y):x^2-y^2=\frac{x}{x^2+y^2} \right\}, \\ B&=\left\{ (x,y):2xy+\frac{y}{x^2+y^2}=3 \right\}, \\ C&=\left\{ (x,y):x^3-3xy^2+3y=1 \right\}, \\ D&=\left\{ (x,y):3x^2y-3x-y^3=0 \right\}. \end{align*} Prove that $A \cap B=C \cap D$. -/ theorem putnam_1987_a1 (A B C D : Set (ℝ × ℝ)) (hA : A = {(x, y) : ℝ × ℝ | x ^ 2 + y ^ 2 ≠ 0 ∧ x ^ 2 - y ^ 2 = x / (x ^ 2 + y ^ 2)}) (hB : B = {(x, y) : ℝ × ℝ | x ^ 2 + y ^ 2 ≠ 0 ∧ 2 * x * y + y / (x ^ 2 + y ^ 2) = 3}) (hC : C = {(x, y) : ℝ × ℝ | x ^ 3 - 3 * x * y ^ 2 + 3 * y = 1}) (hD : D = {(x, y) : ℝ × ℝ | 3 * x ^ 2 * y - 3 * x - y ^ 3 = 0}) : A ∩ B = C ∩ D := by

import Mathlib /-- Curves $A$, $B$, $C$, and $D$ are defined in the plane as follows: \begin{align*} A&=\left\{ (x,y):x^2-y^2=\frac{x}{x^2+y^2} \right\}, \\ B&=\left\{ (x,y):2xy+\frac{y}{x^2+y^2}=3 \right\}, \\ C&=\left\{ (x,y):x^3-3xy^2+3y=1 \right\}, \\ D&=\left\{ (x,y):3x^2y-3x-y^3=0 \right\}. \end{align*} Prove that $A \cap B=C \cap D$. -/ theorem putnam_1987_a1 (A B C D : Set (ℝ × ℝ)) (hA : A = {(x, y) : ℝ × ℝ | x ^ 2 + y ^ 2 ≠ 0 ∧ x ^ 2 - y ^ 2 = x / (x ^ 2 + y ^ 2)}) (hB : B = {(x, y) : ℝ × ℝ | x ^ 2 + y ^ 2 ≠ 0 ∧ 2 * x * y + y / (x ^ 2 + y ^ 2) = 3}) (hC : C = {(x, y) : ℝ × ℝ | x ^ 3 - 3 * x * y ^ 2 + 3 * y = 1}) (hD : D = {(x, y) : ℝ × ℝ | 3 * x ^ 2 * y - 3 * x - y ^ 3 = 0}) : A ∩ B = C ∩ D := sorry

Curves $A$, $B$, $C$, and $D$ are defined in the plane as follows: \begin{align*} A&=\left\{ (x,y):x^2-y^2=\frac{x}{x^2+y^2} \right\}, \\ B&=\left\{ (x,y):2xy+\frac{y}{x^2+y^2}=3 \right\}, \\ C&=\left\{ (x,y):x^3-3xy^2+3y=1 \right\}, \\ D&=\left\{ (x,y):3x^2y-3x-y^3=0 \right\}. \end{align*} Prove that $A \cap B=C \cap D$.

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putnam_1996_b3

3e3bc971-6258-57f9-9978-2613ff827f51

train

abbrev putnam_1996_b3_solution : ℕ → ℕ := sorry -- (fun n : ℕ => (2 * n ^ 3 + 3 * n ^ 2 - 11 * n + 18) / 6) /-- Given that $\{x_1,x_2,\ldots,x_n\}=\{1,2,\ldots,n\}$, find, with proof, the largest possible value, as a function of $n$ (with $n \geq 2$), of $x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1$. -/ theorem putnam_1996_b3 (n : ℕ) (hn : n ≥ 2) : IsGreatest {k | ∃ x : ℕ → ℤ, (x '' (Finset.range n) = Set.Icc (1 : ℤ) n) ∧ ∑ i : Fin n, x i * x ((i + 1) % n) = k} (putnam_1996_b3_solution n) := sorry

import Mathlib open Function -- Note: uses (ℕ → ℕ) instead of (Fin n → ℕ) -- (fun n : ℕ => (2 * n ^ 3 + 3 * n ^ 2 - 11 * n + 18) / 6) /-- Given that $\{x_1,x_2,\ldots,x_n\}=\{1,2,\ldots,n\}$, find, with proof, the largest possible value, as a function of $n$ (with $n \geq 2$), of $x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1$. -/ theorem putnam_1996_b3 (n : ℕ) (hn : n ≥ 2) : IsGreatest {k | ∃ x : ℕ → ℤ, (x '' (Finset.range n) = Set.Icc (1 : ℤ) n) ∧ ∑ i : Fin n, x i * x ((i + 1) % n) = k} (putnam_1996_b3_solution n) := by

import Mathlib open Function -- Note: uses (ℕ → ℕ) instead of (Fin n → ℕ) abbrev putnam_1996_b3_solution : ℕ → ℕ := sorry -- (fun n : ℕ => (2 * n ^ 3 + 3 * n ^ 2 - 11 * n + 18) / 6) /-- Given that $\{x_1,x_2,\ldots,x_n\}=\{1,2,\ldots,n\}$, find, with proof, the largest possible value, as a function of $n$ (with $n \geq 2$), of $x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1$. -/ theorem putnam_1996_b3 (n : ℕ) (hn : n ≥ 2) : IsGreatest {k | ∃ x : ℕ → ℤ, (x '' (Finset.range n) = Set.Icc (1 : ℤ) n) ∧ ∑ i : Fin n, x i * x ((i + 1) % n) = k} (putnam_1996_b3_solution n) := sorry

Given that $\{x_1,x_2,\ldots,x_n\}=\{1,2,\ldots,n\}$, find, with proof, the largest possible value, as a function of $n$ (with $n \geq 2$), of $x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1$.

Show that the maximum is $(2n^3+3n^2-11n+18)/6$.

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putnam_1994_b4

5f231a3b-9df6-566f-a4f9-33ffdb53c327

train

theorem putnam_1994_b4 (matgcd : Matrix (Fin 2) (Fin 2) ℤ → ℤ) (A : Matrix (Fin 2) (Fin 2) ℤ) (d : ℕ → ℤ) (hmatgcd : ∀ M, matgcd M = Int.gcd (Int.gcd (Int.gcd (M 0 0) (M 0 1)) (M 1 0)) (M 1 1)) (hA : A 0 0 = 3 ∧ A 0 1 = 2 ∧ A 1 0 = 4 ∧ A 1 1 = 3) (hd : ∀ n ≥ 1, d n = matgcd (A ^ n - 1)) : Tendsto d atTop atTop := sorry

import Mathlib open Filter Topology /-- For $n \geq 1$, let $d_n$ be the greatest common divisor of the entries of $A^n-I$, where $A=\begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ and $I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Show that $\lim_{n \to \infty} d_n=\infty$. -/ theorem putnam_1994_b4 (matgcd : Matrix (Fin 2) (Fin 2) ℤ → ℤ) (A : Matrix (Fin 2) (Fin 2) ℤ) (d : ℕ → ℤ) (hmatgcd : ∀ M, matgcd M = Int.gcd (Int.gcd (Int.gcd (M 0 0) (M 0 1)) (M 1 0)) (M 1 1)) (hA : A 0 0 = 3 ∧ A 0 1 = 2 ∧ A 1 0 = 4 ∧ A 1 1 = 3) (hd : ∀ n ≥ 1, d n = matgcd (A ^ n - 1)) : Tendsto d atTop atTop := by

import Mathlib open Filter Topology /-- For $n \geq 1$, let $d_n$ be the greatest common divisor of the entries of $A^n-I$, where $A=\begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ and $I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Show that $\lim_{n \to \infty} d_n=\infty$. -/ theorem putnam_1994_b4 (matgcd : Matrix (Fin 2) (Fin 2) ℤ → ℤ) (A : Matrix (Fin 2) (Fin 2) ℤ) (d : ℕ → ℤ) (hmatgcd : ∀ M, matgcd M = Int.gcd (Int.gcd (Int.gcd (M 0 0) (M 0 1)) (M 1 0)) (M 1 1)) (hA : A 0 0 = 3 ∧ A 0 1 = 2 ∧ A 1 0 = 4 ∧ A 1 1 = 3) (hd : ∀ n ≥ 1, d n = matgcd (A ^ n - 1)) : Tendsto d atTop atTop := sorry

For $n \geq 1$, let $d_n$ be the greatest common divisor of the entries of $A^n-I$, where $A=\begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ and $I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Show that $\lim_{n \to \infty} d_n=\infty$.

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putnam_2004_a4

f293844d-e392-5bb6-a90b-7da7ee04be79

train

theorem putnam_2004_a4 (n : ℕ) (npos : n > 0) (x : Fin n → ℝ) (avals : ℕ → (ℕ → Fin n → ℝ) → Prop) (havals : ∀ N a, avals N a ↔ ∀ (i : Fin N) (j : Fin n), (a i j = -1 ∨ a i j = 0 ∨ a i j = 1)) : ∃ (N : ℕ) (c : Fin N → ℚ) (a : ℕ → Fin n → ℝ), avals N a ∧ (∏ i : Fin n, x i) = ∑ i : Fin N, c i * (∑ j : Fin n, a i j * x j) ^ n := sorry

import Mathlib open Nat Topology Filter -- Note: uses (ℕ → Fin n → ℝ) instead of (Fin N → Fin n → ℝ) /-- Show that for any positive integer $n$ there is an integer $N$ such that the product $x_1x_2 \cdots x_n$ can be expressed identically in the form $x_1x_2 \cdots x_n=\sum_{i=1}^Nc_i(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^n$ where the $c_i$ are rational numbers and each $a_{ij}$ is one of the numbers $-1,0,1$. -/ theorem putnam_2004_a4 (n : ℕ) (npos : n > 0) (x : Fin n → ℝ) (avals : ℕ → (ℕ → Fin n → ℝ) → Prop) (havals : ∀ N a, avals N a ↔ ∀ (i : Fin N) (j : Fin n), (a i j = -1 ∨ a i j = 0 ∨ a i j = 1)) : ∃ (N : ℕ) (c : Fin N → ℚ) (a : ℕ → Fin n → ℝ), avals N a ∧ (∏ i : Fin n, x i) = ∑ i : Fin N, c i * (∑ j : Fin n, a i j * x j) ^ n := by

import Mathlib open Nat Topology Filter -- Note: uses (ℕ → Fin n → ℝ) instead of (Fin N → Fin n → ℝ) /-- Show that for any positive integer $n$ there is an integer $N$ such that the product $x_1x_2 \cdots x_n$ can be expressed identically in the form $x_1x_2 \cdots x_n=\sum_{i=1}^Nc_i(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^n$ where the $c_i$ are rational numbers and each $a_{ij}$ is one of the numbers $-1,0,1$. -/ theorem putnam_2004_a4 (n : ℕ) (npos : n > 0) (x : Fin n → ℝ) (avals : ℕ → (ℕ → Fin n → ℝ) → Prop) (havals : ∀ N a, avals N a ↔ ∀ (i : Fin N) (j : Fin n), (a i j = -1 ∨ a i j = 0 ∨ a i j = 1)) : ∃ (N : ℕ) (c : Fin N → ℚ) (a : ℕ → Fin n → ℝ), avals N a ∧ (∏ i : Fin n, x i) = ∑ i : Fin N, c i * (∑ j : Fin n, a i j * x j) ^ n := sorry

Show that for any positive integer $n$ there is an integer $N$ such that the product $x_1x_2 \cdots x_n$ can be expressed identically in the form $x_1x_2 \cdots x_n=\sum_{i=1}^Nc_i(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^n$ where the $c_i$ are rational numbers and each $a_{ij}$ is one of the numbers $-1,0,1$.

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putnam_2012_a2

4b49224b-73e8-584a-adb4-78e7d8776c9b

train

theorem putnam_2012_a2 (S : Type*) [CommSemigroup S] (a b c : S) (hS : ∀ x y : S, ∃ z : S, x * z = y) (habc : a * c = b * c) : a = b := sorry

import Mathlib open Matrix /-- Let $*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$. (This $z$ may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and $a*c=b*c$, then $a=b$. -/ theorem putnam_2012_a2 (S : Type*) [CommSemigroup S] (a b c : S) (hS : ∀ x y : S, ∃ z : S, x * z = y) (habc : a * c = b * c) : a = b := by

import Mathlib open Matrix /-- Let $*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$. (This $z$ may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and $a*c=b*c$, then $a=b$. -/ theorem putnam_2012_a2 (S : Type*) [CommSemigroup S] (a b c : S) (hS : ∀ x y : S, ∃ z : S, x * z = y) (habc : a * c = b * c) : a = b := sorry

Let $*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x*z=y$. (This $z$ may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and $a*c=b*c$, then $a=b$.

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putnam_1990_b3

121f7794-1ff4-5198-a15e-1c4ab9fe92fb

train

theorem putnam_1990_b3 (S : Set (Matrix (Fin 2) (Fin 2) ℕ)) (hS : ∀ A ∈ S, ∀ i j : Fin 2, (∃ x : ℤ, A i j = x ^ 2) ∧ A i j ≤ 200) : (S.encard > 50387) → (∃ A ∈ S, ∃ B ∈ S, A ≠ B ∧ A * B = B * A) := sorry

import Mathlib open Filter Topology Nat /-- Let $S$ be a set of $2 \times 2$ integer matrices whose entries $a_{ij}$ (1) are all squares of integers, and, (2) satisfy $a_{ij} \leq 200$. Show that if $S$ has more than $50387$ ($=15^4-15^2-15+2$) elements, then it has two elements that commute. -/ theorem putnam_1990_b3 (S : Set (Matrix (Fin 2) (Fin 2) ℕ)) (hS : ∀ A ∈ S, ∀ i j : Fin 2, (∃ x : ℤ, A i j = x ^ 2) ∧ A i j ≤ 200) : (S.encard > 50387) → (∃ A ∈ S, ∃ B ∈ S, A ≠ B ∧ A * B = B * A) := by

import Mathlib open Filter Topology Nat /-- Let $S$ be a set of $2 \times 2$ integer matrices whose entries $a_{ij}$ (1) are all squares of integers, and, (2) satisfy $a_{ij} \leq 200$. Show that if $S$ has more than $50387$ ($=15^4-15^2-15+2$) elements, then it has two elements that commute. -/ theorem putnam_1990_b3 (S : Set (Matrix (Fin 2) (Fin 2) ℕ)) (hS : ∀ A ∈ S, ∀ i j : Fin 2, (∃ x : ℤ, A i j = x ^ 2) ∧ A i j ≤ 200) : (S.encard > 50387) → (∃ A ∈ S, ∃ B ∈ S, A ≠ B ∧ A * B = B * A) := sorry

Let $S$ be a set of $2 \times 2$ integer matrices whose entries $a_{ij}$ (1) are all squares of integers, and, (2) satisfy $a_{ij} \leq 200$. Show that if $S$ has more than $50387$ ($=15^4-15^2-15+2$) elements, then it has two elements that commute.

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putnam_2019_a3

ea19e0cf-e487-52fe-8964-7badda686f1f

train

abbrev putnam_2019_a3_solution : ℝ := sorry -- 2019^(-(1:ℝ)/2019) /-- Given real numbers $b_0, b_1, \dots, b_{2019}$ with $b_{2019} \neq 0$, let $z_1,z_2,\dots,z_{2019}$ be the roots in the complex plane of the polynomial \[ P(z) = \sum_{k=0}^{2019} b_k z^k. \] Let $\mu = (|z_1| + \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\dots,z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu \geq M$ for all choices of $b_0,b_1,\dots, b_{2019}$ that satisfy \[ 1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019. \] -/ theorem putnam_2019_a3 (v : Polynomial ℂ → Prop) (hv : v = fun b => b.degree = 2019 ∧ 1 ≤ (b.coeff 0).re ∧ (b.coeff 2019).re ≤ 2019 ∧ (∀ i : Fin 2020, (b.coeff i).im = 0) ∧ (∀ i : Fin 2019, (b.coeff i).re < (b.coeff (i + 1)).re)) (μ : Polynomial ℂ → ℝ) (hμ : μ = fun b => (Multiset.map (fun ω : ℂ => ‖ω‖) (Polynomial.roots b)).sum/2019) : IsGreatest {M : ℝ | ∀ b, v b → μ b ≥ M} putnam_2019_a3_solution := sorry

import Mathlib open Topology Filter -- 2019^(-(1:ℝ)/2019) /-- Given real numbers $b_0, b_1, \dots, b_{2019}$ with $b_{2019} \neq 0$, let $z_1,z_2,\dots,z_{2019}$ be the roots in the complex plane of the polynomial \[ P(z) = \sum_{k=0}^{2019} b_k z^k. \] Let $\mu = (|z_1| + \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\dots,z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu \geq M$ for all choices of $b_0,b_1,\dots, b_{2019}$ that satisfy \[ 1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019. \] -/ theorem putnam_2019_a3 (v : Polynomial ℂ → Prop) (hv : v = fun b => b.degree = 2019 ∧ 1 ≤ (b.coeff 0).re ∧ (b.coeff 2019).re ≤ 2019 ∧ (∀ i : Fin 2020, (b.coeff i).im = 0) ∧ (∀ i : Fin 2019, (b.coeff i).re < (b.coeff (i + 1)).re)) (μ : Polynomial ℂ → ℝ) (hμ : μ = fun b => (Multiset.map (fun ω : ℂ => ‖ω‖) (Polynomial.roots b)).sum/2019) : IsGreatest {M : ℝ | ∀ b, v b → μ b ≥ M} putnam_2019_a3_solution := by

import Mathlib open Topology Filter noncomputable abbrev putnam_2019_a3_solution : ℝ := sorry -- 2019^(-(1:ℝ)/2019) /-- Given real numbers $b_0, b_1, \dots, b_{2019}$ with $b_{2019} \neq 0$, let $z_1,z_2,\dots,z_{2019}$ be the roots in the complex plane of the polynomial \[ P(z) = \sum_{k=0}^{2019} b_k z^k. \] Let $\mu = (|z_1| + \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\dots,z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu \geq M$ for all choices of $b_0,b_1,\dots, b_{2019}$ that satisfy \[ 1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019. \] -/ theorem putnam_2019_a3 (v : Polynomial ℂ → Prop) (hv : v = fun b => b.degree = 2019 ∧ 1 ≤ (b.coeff 0).re ∧ (b.coeff 2019).re ≤ 2019 ∧ (∀ i : Fin 2020, (b.coeff i).im = 0) ∧ (∀ i : Fin 2019, (b.coeff i).re < (b.coeff (i + 1)).re)) (μ : Polynomial ℂ → ℝ) (hμ : μ = fun b => (Multiset.map (fun ω : ℂ => ‖ω‖) (Polynomial.roots b)).sum/2019) : IsGreatest {M : ℝ | ∀ b, v b → μ b ≥ M} putnam_2019_a3_solution := sorry

Given real numbers $b_0, b_1, \dots, b_{2019}$ with $b_{2019} \neq 0$, let $z_1,z_2,\dots,z_{2019}$ be the roots in the complex plane of the polynomial \[ P(z) = \sum_{k=0}^{2019} b_k z^k. \] Let $\mu = (|z_1| + \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\dots,z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu \geq M$ for all choices of $b_0,b_1,\dots, b_{2019}$ that satisfy \[ 1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019. \]

The answer is $M = 2019^{-1/2019}$.

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putnam_1988_b3

f7f87ddf-92fc-51c1-bebe-d611c193f397

train

abbrev putnam_1988_b3_solution : ℝ := sorry -- (1 + Real.sqrt 3) / 2 /-- For every $n$ in the set $N=\{1,2,\dots\}$ of positive integers, let $r_n$ be the minimum value of $|c-d \sqrt{3}|$ for all nonnegative integers $c$ and $d$ with $c+d=n$. Find, with proof, the smallest positive real number $g$ with $r_n \leq g$ for all $n \in N$. -/ theorem putnam_1988_b3 (r : ℤ → ℝ) (hr : ∀ n ≥ 1, (∃ c d : ℤ, (c ≥ 0 ∧ d ≥ 0) ∧ c + d = n ∧ r n = |c - d * Real.sqrt 3|) ∧ (∀ c d : ℤ, (c ≥ 0 ∧ d ≥ 0 ∧ c + d = n) → |c - d * Real.sqrt 3| ≥ r n)) : IsLeast {g : ℝ | g > 0 ∧ (∀ n : ℤ, n ≥ 1 → r n ≤ g)} putnam_1988_b3_solution := sorry

import Mathlib open Set Filter Topology -- (1 + Real.sqrt 3) / 2 /-- For every $n$ in the set $N=\{1,2,\dots\}$ of positive integers, let $r_n$ be the minimum value of $|c-d \sqrt{3}|$ for all nonnegative integers $c$ and $d$ with $c+d=n$. Find, with proof, the smallest positive real number $g$ with $r_n \leq g$ for all $n \in N$. -/ theorem putnam_1988_b3 (r : ℤ → ℝ) (hr : ∀ n ≥ 1, (∃ c d : ℤ, (c ≥ 0 ∧ d ≥ 0) ∧ c + d = n ∧ r n = |c - d * Real.sqrt 3|) ∧ (∀ c d : ℤ, (c ≥ 0 ∧ d ≥ 0 ∧ c + d = n) → |c - d * Real.sqrt 3| ≥ r n)) : IsLeast {g : ℝ | g > 0 ∧ (∀ n : ℤ, n ≥ 1 → r n ≤ g)} putnam_1988_b3_solution := by

import Mathlib open Set Filter Topology noncomputable abbrev putnam_1988_b3_solution : ℝ := sorry -- (1 + Real.sqrt 3) / 2 /-- For every $n$ in the set $N=\{1,2,\dots\}$ of positive integers, let $r_n$ be the minimum value of $|c-d \sqrt{3}|$ for all nonnegative integers $c$ and $d$ with $c+d=n$. Find, with proof, the smallest positive real number $g$ with $r_n \leq g$ for all $n \in N$. -/ theorem putnam_1988_b3 (r : ℤ → ℝ) (hr : ∀ n ≥ 1, (∃ c d : ℤ, (c ≥ 0 ∧ d ≥ 0) ∧ c + d = n ∧ r n = |c - d * Real.sqrt 3|) ∧ (∀ c d : ℤ, (c ≥ 0 ∧ d ≥ 0 ∧ c + d = n) → |c - d * Real.sqrt 3| ≥ r n)) : IsLeast {g : ℝ | g > 0 ∧ (∀ n : ℤ, n ≥ 1 → r n ≤ g)} putnam_1988_b3_solution := sorry

For every $n$ in the set $N=\{1,2,\dots\}$ of positive integers, let $r_n$ be the minimum value of $|c-d \sqrt{3}|$ for all nonnegative integers $c$ and $d$ with $c+d=n$. Find, with proof, the smallest positive real number $g$ with $r_n \leq g$ for all $n \in N$.

Show that the smallest such $g$ is $(1+\sqrt{3})/2$.

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putnam_1998_b6

59a6c8ff-b1d2-518b-90f7-8cc2887d3acf

train

theorem putnam_1998_b6 : ∀ a b c : ℤ, ∃ n : ℤ, n > 0 ∧ ¬(∃ k : ℤ, k = Real.sqrt (n^3 + a * n^2 + b * n + c)) := sorry

import Mathlib open Set Function Metric /-- Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer. -/ theorem putnam_1998_b6 : ∀ a b c : ℤ, ∃ n : ℤ, n > 0 ∧ ¬(∃ k : ℤ, k = Real.sqrt (n^3 + a * n^2 + b * n + c)) := by

import Mathlib open Set Function Metric /-- Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer. -/ theorem putnam_1998_b6 : ∀ a b c : ℤ, ∃ n : ℤ, n > 0 ∧ ¬(∃ k : ℤ, k = Real.sqrt (n^3 + a * n^2 + b * n + c)) := sorry

Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer.

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putnam_2013_b3

79328774-9835-59c0-8354-c438fce75ad9

train

abbrev putnam_2013_b3_solution : Prop := sorry -- True /-- Let $\mathcal{P}$ be a nonempty collection of subsets of $\{1,\dots, n\}$ such that: \begin{enumerate} \item[(i)] if $S, S' \in \mathcal{P}$, then $S \cup S' \in \mathcal{P}$ and $S \cap S' \in \mathcal{P}$, and \item[(ii)] if $S \in \mathcal{P}$ and $S \neq \emptyset$, then there is a subset $T \subset S$ such that $T \in \mathcal{P}$ and $T$ contains exactly one fewer element than $S$. \end{enumerate} Suppose that $f: \mathcal{P} \to \mathbb{R}$ is a function such that $f(\emptyset) = 0$ and \[f(S \cup S') = f(S) + f(S') - f(S \cap S') \mbox{ for all $S,S' \in \mathcal{P}$.} \] Must there exist real numbers $f_1,\dots,f_n$ such that\[f(S) = \sum_{i \in S} f_i\] \n for every $S \in \mathcal{P}$? -/ theorem putnam_2013_b3 : (∀ (n : ℕ) (P : Finset (Finset (Fin n))) (f : Finset (Fin n) → ℝ), P ≠ ⊥ → (∀ S ∈ P, ∀ S' ∈ P, S ∪ S' ∈ P ∧ S ∩ S' ∈ P) → (∀ S ∈ P, S ≠ ⊥ → ∃ T ∈ P, T ⊂ S ∧ Finset.card T + 1 = Finset.card S) → f ⊥ = 0 → (∀ S ∈ P, ∀ S' ∈ P, f (S ∪ S') = f S + f S' - f (S ∩ S')) → ∃ r : Fin n → ℝ, ∀ S ∈ P, f S = ∑ i in S, r i) ↔ putnam_2013_b3_solution := sorry

import Mathlib open Function Set -- True /-- Let $\mathcal{P}$ be a nonempty collection of subsets of $\{1,\dots, n\}$ such that: \begin{enumerate} \item[(i)] if $S, S' \in \mathcal{P}$, then $S \cup S' \in \mathcal{P}$ and $S \cap S' \in \mathcal{P}$, and \item[(ii)] if $S \in \mathcal{P}$ and $S \neq \emptyset$, then there is a subset $T \subset S$ such that $T \in \mathcal{P}$ and $T$ contains exactly one fewer element than $S$. \end{enumerate} Suppose that $f: \mathcal{P} \to \mathbb{R}$ is a function such that $f(\emptyset) = 0$ and \[f(S \cup S') = f(S) + f(S') - f(S \cap S') \mbox{ for all $S,S' \in \mathcal{P}$.} \] Must there exist real numbers $f_1,\dots,f_n$ such that\[f(S) = \sum_{i \in S} f_i\] \n for every $S \in \mathcal{P}$? -/ theorem putnam_2013_b3 : (∀ (n : ℕ) (P : Finset (Finset (Fin n))) (f : Finset (Fin n) → ℝ), P ≠ ⊥ → (∀ S ∈ P, ∀ S' ∈ P, S ∪ S' ∈ P ∧ S ∩ S' ∈ P) → (∀ S ∈ P, S ≠ ⊥ → ∃ T ∈ P, T ⊂ S ∧ Finset.card T + 1 = Finset.card S) → f ⊥ = 0 → (∀ S ∈ P, ∀ S' ∈ P, f (S ∪ S') = f S + f S' - f (S ∩ S')) → ∃ r : Fin n → ℝ, ∀ S ∈ P, f S = ∑ i in S, r i) ↔ putnam_2013_b3_solution := by

import Mathlib open Function Set abbrev putnam_2013_b3_solution : Prop := sorry -- True /-- Let $\mathcal{P}$ be a nonempty collection of subsets of $\{1,\dots, n\}$ such that: \begin{enumerate} \item[(i)] if $S, S' \in \mathcal{P}$, then $S \cup S' \in \mathcal{P}$ and $S \cap S' \in \mathcal{P}$, and \item[(ii)] if $S \in \mathcal{P}$ and $S \neq \emptyset$, then there is a subset $T \subset S$ such that $T \in \mathcal{P}$ and $T$ contains exactly one fewer element than $S$. \end{enumerate} Suppose that $f: \mathcal{P} \to \mathbb{R}$ is a function such that $f(\emptyset) = 0$ and \[f(S \cup S') = f(S) + f(S') - f(S \cap S') \mbox{ for all $S,S' \in \mathcal{P}$.} \] Must there exist real numbers $f_1,\dots,f_n$ such that\[f(S) = \sum_{i \in S} f_i\] \n for every $S \in \mathcal{P}$? -/ theorem putnam_2013_b3 : (∀ (n : ℕ) (P : Finset (Finset (Fin n))) (f : Finset (Fin n) → ℝ), P ≠ ⊥ → (∀ S ∈ P, ∀ S' ∈ P, S ∪ S' ∈ P ∧ S ∩ S' ∈ P) → (∀ S ∈ P, S ≠ ⊥ → ∃ T ∈ P, T ⊂ S ∧ Finset.card T + 1 = Finset.card S) → f ⊥ = 0 → (∀ S ∈ P, ∀ S' ∈ P, f (S ∪ S') = f S + f S' - f (S ∩ S')) → ∃ r : Fin n → ℝ, ∀ S ∈ P, f S = ∑ i in S, r i) ↔ putnam_2013_b3_solution := sorry

Let $\mathcal{P}$ be a nonempty collection of subsets of $\{1,\dots, n\}$ such that: \begin{enumerate} \item[(i)] if $S, S' \in \mathcal{P}$, then $S \cup S' \in \mathcal{P}$ and $S \cap S' \in \mathcal{P}$, and \item[(ii)] if $S \in \mathcal{P}$ and $S \neq \emptyset$, then there is a subset $T \subset S$ such that $T \in \mathcal{P}$ and $T$ contains exactly one fewer element than $S$. \end{enumerate} Suppose that $f: \mathcal{P} \to \mathbb{R}$ is a function such that $f(\emptyset) = 0$ and \[f(S \cup S') = f(S) + f(S') - f(S \cap S') \mbox{ for all $S,S' \in \mathcal{P}$.} \] Must there exist real numbers $f_1,\dots,f_n$ such that\[f(S) = \sum_{i \in S} f_i\] \n for every $S \in \mathcal{P}$?

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putnam_1987_b5

dc6dbfc1-9825-596f-9cf4-2d29e9073921

train

theorem putnam_1987_b5 (n : ℕ) (npos : n > 0) (M : Matrix (Fin (2 * n)) (Fin n) ℂ) (hM : ∀ z : Matrix (Fin 1) (Fin (2 * n)) ℂ, z * M = 0 → (¬∀ i : Fin (2 * n), z 0 i = 0) → ∃ i : Fin (2 * n), (z 0 i).im ≠ 0) : (∀ r : Matrix (Fin (2 * n)) (Fin 1) ℝ, ∃ w : Matrix (Fin n) (Fin 1) ℂ, ∀ i : (Fin (2 * n)), ((M * w) i 0).re = r i 0) := sorry

import Mathlib open MvPolynomial Real Nat Filter Topology /-- Let $O_n$ be the $n$-dimensional vector $(0,0,\cdots, 0)$. Let $M$ be a $2n \times n$ matrix of complex numbers such that whenever $(z_1, z_2, \dots, z_{2n})M = O_n$, with complex $z_i$, not all zero, then at least one of the $z_i$ is not real. Prove that for arbitrary real numbers $r_1, r_2, \dots, r_{2n}$, there are complex numbers $w_1, w_2, \dots, w_n$ such that \[ \mathrm{re}\left[ M \left( \begin{array}{c} w_1 \\ \vdots \\ w_n \end{array} \right) \right] = \left( \begin{array}{c} r_1 \\ \vdots \\ r_{2n} \end{array} \right). \] (Note: if $C$ is a matrix of complex numbers, $\mathrm{re}(C)$ is the matrix whose entries are the real parts of the entries of $C$.) -/ theorem putnam_1987_b5 (n : ℕ) (npos : n > 0) (M : Matrix (Fin (2 * n)) (Fin n) ℂ) (hM : ∀ z : Matrix (Fin 1) (Fin (2 * n)) ℂ, z * M = 0 → (¬∀ i : Fin (2 * n), z 0 i = 0) → ∃ i : Fin (2 * n), (z 0 i).im ≠ 0) : (∀ r : Matrix (Fin (2 * n)) (Fin 1) ℝ, ∃ w : Matrix (Fin n) (Fin 1) ℂ, ∀ i : (Fin (2 * n)), ((M * w) i 0).re = r i 0) := by

import Mathlib open MvPolynomial Real Nat Filter Topology /-- Let $O_n$ be the $n$-dimensional vector $(0,0,\cdots, 0)$. Let $M$ be a $2n \times n$ matrix of complex numbers such that whenever $(z_1, z_2, \dots, z_{2n})M = O_n$, with complex $z_i$, not all zero, then at least one of the $z_i$ is not real. Prove that for arbitrary real numbers $r_1, r_2, \dots, r_{2n}$, there are complex numbers $w_1, w_2, \dots, w_n$ such that \[ \mathrm{re}\left[ M \left( \begin{array}{c} w_1 \\ \vdots \\ w_n \end{array} \right) \right] = \left( \begin{array}{c} r_1 \\ \vdots \\ r_{2n} \end{array} \right). \] (Note: if $C$ is a matrix of complex numbers, $\mathrm{re}(C)$ is the matrix whose entries are the real parts of the entries of $C$.) -/ theorem putnam_1987_b5 (n : ℕ) (npos : n > 0) (M : Matrix (Fin (2 * n)) (Fin n) ℂ) (hM : ∀ z : Matrix (Fin 1) (Fin (2 * n)) ℂ, z * M = 0 → (¬∀ i : Fin (2 * n), z 0 i = 0) → ∃ i : Fin (2 * n), (z 0 i).im ≠ 0) : (∀ r : Matrix (Fin (2 * n)) (Fin 1) ℝ, ∃ w : Matrix (Fin n) (Fin 1) ℂ, ∀ i : (Fin (2 * n)), ((M * w) i 0).re = r i 0) := sorry

Let $O_n$ be the $n$-dimensional vector $(0,0,\cdots, 0)$. Let $M$ be a $2n \times n$ matrix of complex numbers such that whenever $(z_1, z_2, \dots, z_{2n})M = O_n$, with complex $z_i$, not all zero, then at least one of the $z_i$ is not real. Prove that for arbitrary real numbers $r_1, r_2, \dots, r_{2n}$, there are complex numbers $w_1, w_2, \dots, w_n$ such that \[ \mathrm{re}\left[ M \left( \begin{array}{c} w_1 \\ \vdots \\ w_n \end{array} \right) \right] = \left( \begin{array}{c} r_1 \\ \vdots \\ r_{2n} \end{array} \right). \] (Note: if $C$ is a matrix of complex numbers, $\mathrm{re}(C)$ is the matrix whose entries are the real parts of the entries of $C$.)

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putnam_2010_a2

a2e969dc-ca3c-508c-a385-55d08bf3087e

train

abbrev putnam_2010_a2_solution : Set (ℝ → ℝ) := sorry -- {f : ℝ → ℝ | ∃ c d : ℝ, ∀ x : ℝ, f x = c*x + d} /-- Find all differentiable functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f'(x) = \frac{f(x+n)-f(x)}{n} \] for all real numbers $x$ and all positive integers $n$. -/ theorem putnam_2010_a2 : {f : ℝ → ℝ | Differentiable ℝ f ∧ ∀ x : ℝ, ∀ n : ℤ, n > 0 → deriv f x = (f (x + n) - f x)/n} = putnam_2010_a2_solution := sorry

import Mathlib -- {f : ℝ → ℝ | ∃ c d : ℝ, ∀ x : ℝ, f x = c*x + d} /-- Find all differentiable functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f'(x) = \frac{f(x+n)-f(x)}{n} \] for all real numbers $x$ and all positive integers $n$. -/ theorem putnam_2010_a2 : {f : ℝ → ℝ | Differentiable ℝ f ∧ ∀ x : ℝ, ∀ n : ℤ, n > 0 → deriv f x = (f (x + n) - f x)/n} = putnam_2010_a2_solution := by

import Mathlib abbrev putnam_2010_a2_solution : Set (ℝ → ℝ) := sorry -- {f : ℝ → ℝ | ∃ c d : ℝ, ∀ x : ℝ, f x = c*x + d} /-- Find all differentiable functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f'(x) = \frac{f(x+n)-f(x)}{n} \] for all real numbers $x$ and all positive integers $n$. -/ theorem putnam_2010_a2 : {f : ℝ → ℝ | Differentiable ℝ f ∧ ∀ x : ℝ, ∀ n : ℤ, n > 0 → deriv f x = (f (x + n) - f x)/n} = putnam_2010_a2_solution := sorry

Find all differentiable functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f'(x) = \frac{f(x+n)-f(x)}{n} \] for all real numbers $x$ and all positive integers $n$.

The solution consists of all functions of the form $f(x) = cx+d$ for some real numbers $c,d$.

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putnam_2023_b3

96031ff9-84d6-5729-8e44-f48b95374c5b

train

abbrev putnam_2023_b3_solution : ℕ → ℝ := sorry -- (fun n : ℕ ↦ (2 * (n : ℝ) + 2) / 3) /-- A sequence $y_1, y_2, \ldots, y_k$ of real numbers is called zigzag if $k = 1$, or if $y_2 - y_1, y_3 - y_2, \ldots, y_k - y_{k-1}$ are nonzero and alternate in sign. Let $X_1, X_2,\ldots, X_n$ be chosen independently from the uniform distribution on $[0,1]$. Let $a(X_1, X_2, \ldots, X_n)$ be the largest value of $k$ for which there exists an increasing sequence of integers $i_1, i_2, \ldots, i_k$ such that $X_{i_1}, X_{i_2}, \ldots, X_{i_k}$ is zigzag. Find the expected value of $a(X_1, X_2, \ldots, X_n)$ for $n \ge 2$. -/ theorem putnam_2023_b3 (IsZigZag : {k : ℕ} → (Fin k → ℝ) → Prop) (IsZigZag_def : ∀ (k : ℕ) [NeZero k] (y : Fin k → ℝ), IsZigZag y ↔ k = 1 ∨ ((∀ i, i + 1 < k → y (i + 1) ≠ y i)) ∧ (∀ i, i + 2 < k → (y (i + 2) < y (i + 1) ↔ y i < y (i + 1)))) (n : ℕ) (hn : 2 ≤ n) (a : (Fin n → Icc (0 : ℝ) 1) → ℕ) (ha : ∀ x, IsGreatest {k | ∃ i : Fin k ↪o Fin n, IsZigZag ((↑) ∘ x ∘ i)} (a x)) : 𝔼[(↑) ∘ a] = putnam_2023_b3_solution n := sorry

import Mathlib open ProbabilityTheory Set -- (fun n : ℕ ↦ (2 * (n : ℝ) + 2) / 3) /-- A sequence $y_1, y_2, \ldots, y_k$ of real numbers is called zigzag if $k = 1$, or if $y_2 - y_1, y_3 - y_2, \ldots, y_k - y_{k-1}$ are nonzero and alternate in sign. Let $X_1, X_2,\ldots, X_n$ be chosen independently from the uniform distribution on $[0,1]$. Let $a(X_1, X_2, \ldots, X_n)$ be the largest value of $k$ for which there exists an increasing sequence of integers $i_1, i_2, \ldots, i_k$ such that $X_{i_1}, X_{i_2}, \ldots, X_{i_k}$ is zigzag. Find the expected value of $a(X_1, X_2, \ldots, X_n)$ for $n \ge 2$. -/ theorem putnam_2023_b3 (IsZigZag : {k : ℕ} → (Fin k → ℝ) → Prop) (IsZigZag_def : ∀ (k : ℕ) [NeZero k] (y : Fin k → ℝ), IsZigZag y ↔ k = 1 ∨ ((∀ i, i + 1 < k → y (i + 1) ≠ y i)) ∧ (∀ i, i + 2 < k → (y (i + 2) < y (i + 1) ↔ y i < y (i + 1)))) (n : ℕ) (hn : 2 ≤ n) (a : (Fin n → Icc (0 : ℝ) 1) → ℕ) (ha : ∀ x, IsGreatest {k | ∃ i : Fin k ↪o Fin n, IsZigZag ((↑) ∘ x ∘ i)} (a x)) : 𝔼[(↑) ∘ a] = putnam_2023_b3_solution n := by

import Mathlib open ProbabilityTheory Set noncomputable abbrev putnam_2023_b3_solution : ℕ → ℝ := sorry -- (fun n : ℕ ↦ (2 * (n : ℝ) + 2) / 3) /-- A sequence $y_1, y_2, \ldots, y_k$ of real numbers is called zigzag if $k = 1$, or if $y_2 - y_1, y_3 - y_2, \ldots, y_k - y_{k-1}$ are nonzero and alternate in sign. Let $X_1, X_2,\ldots, X_n$ be chosen independently from the uniform distribution on $[0,1]$. Let $a(X_1, X_2, \ldots, X_n)$ be the largest value of $k$ for which there exists an increasing sequence of integers $i_1, i_2, \ldots, i_k$ such that $X_{i_1}, X_{i_2}, \ldots, X_{i_k}$ is zigzag. Find the expected value of $a(X_1, X_2, \ldots, X_n)$ for $n \ge 2$. -/ theorem putnam_2023_b3 (IsZigZag : {k : ℕ} → (Fin k → ℝ) → Prop) (IsZigZag_def : ∀ (k : ℕ) [NeZero k] (y : Fin k → ℝ), IsZigZag y ↔ k = 1 ∨ ((∀ i, i + 1 < k → y (i + 1) ≠ y i)) ∧ (∀ i, i + 2 < k → (y (i + 2) < y (i + 1) ↔ y i < y (i + 1)))) (n : ℕ) (hn : 2 ≤ n) (a : (Fin n → Icc (0 : ℝ) 1) → ℕ) (ha : ∀ x, IsGreatest {k | ∃ i : Fin k ↪o Fin n, IsZigZag ((↑) ∘ x ∘ i)} (a x)) : 𝔼[(↑) ∘ a] = putnam_2023_b3_solution n := sorry

A sequence $y_1, y_2, \ldots, y_k$ of real numbers is called zigzag if $k = 1$, or if $y_2 - y_1, y_3 - y_2, \ldots, y_k - y_{k-1}$ are nonzero and alternate in sign. Let $X_1, X_2,\ldots, X_n$ be chosen independently from the uniform distribution on $[0,1]$. Let $a(X_1, X_2, \ldots, X_n)$ be the largest value of $k$ for which there exists an increasing sequence of integers $i_1, i_2, \ldots, i_k$ such that $X_{i_1}, X_{i_2}, \ldots, X_{i_k}$ is zigzag. Find the expected value of $a(X_1, X_2, \ldots, X_n)$ for $n \ge 2$.

Show that the expected value is \frac{2n + 2}{3}.

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putnam_1993_b4

400ee065-c782-5576-aece-75893ca56ba0

train

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

import Mathlib /-- The function $K(x,y)$ is positive and continuous for $0 \leq x \leq 1,0 \leq y \leq 1$, and the functions $f(x)$ and $g(x)$ are positive and continuous for $0 \leq x \leq 1$. Suppose that for all $x$, $0 \leq x \leq 1$, $\int_0^1 f(y)K(x,y)\,dy=g(x)$ and $\int_0^1 g(y)K(x,y)\,dy=f(x)$. Show that $f(x)=g(x)$ for $0 \leq x \leq 1$. -/ theorem putnam_1993_b4 (K : ℝ × ℝ → ℝ) (f g : ℝ → ℝ) (Kpos : ∀ x y : Set.Icc (0 : ℝ) 1, K (x, y) > 0) (Kcont : ContinuousOn K {(x, y) : ℝ × ℝ | x ∈ Set.Icc 0 1 ∧ y ∈ Set.Icc 0 1}) (fgpos : ∀ x : Set.Icc (0 : ℝ) 1, f x > 0 ∧ g x > 0) (fgcont : ContinuousOn f (Set.Icc 0 1) ∧ ContinuousOn g (Set.Icc 0 1)) (fgint : ∀ x : Set.Icc (0 : ℝ) 1, (∫ y in Set.Ioo 0 1, f y * K (x, y)) = g x ∧ (∫ y in Set.Ioo 0 1, g y * K (x, y)) = f x) : ∀ x : Set.Icc (0 : ℝ) 1, f x = g x := by

import Mathlib /-- The function $K(x,y)$ is positive and continuous for $0 \leq x \leq 1,0 \leq y \leq 1$, and the functions $f(x)$ and $g(x)$ are positive and continuous for $0 \leq x \leq 1$. Suppose that for all $x$, $0 \leq x \leq 1$, $\int_0^1 f(y)K(x,y)\,dy=g(x)$ and $\int_0^1 g(y)K(x,y)\,dy=f(x)$. Show that $f(x)=g(x)$ for $0 \leq x \leq 1$. -/ theorem putnam_1993_b4 (K : ℝ × ℝ → ℝ) (f g : ℝ → ℝ) (Kpos : ∀ x y : Set.Icc (0 : ℝ) 1, K (x, y) > 0) (Kcont : ContinuousOn K {(x, y) : ℝ × ℝ | x ∈ Set.Icc 0 1 ∧ y ∈ Set.Icc 0 1}) (fgpos : ∀ x : Set.Icc (0 : ℝ) 1, f x > 0 ∧ g x > 0) (fgcont : ContinuousOn f (Set.Icc 0 1) ∧ ContinuousOn g (Set.Icc 0 1)) (fgint : ∀ x : Set.Icc (0 : ℝ) 1, (∫ y in Set.Ioo 0 1, f y * K (x, y)) = g x ∧ (∫ y in Set.Ioo 0 1, g y * K (x, y)) = f x) : ∀ x : Set.Icc (0 : ℝ) 1, f x = g x := sorry

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

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putnam_1963_b3

9c91f28c-bfef-5481-8758-b0f50e30eecf

train

abbrev putnam_1963_b3_solution : Set (ℝ → ℝ) := sorry -- {(fun u : ℝ => A * Real.sinh (k * u)) | (A : ℝ) (k : ℝ)} ∪ {(fun u : ℝ => A * u) | A : ℝ} ∪ {(fun u : ℝ => A * Real.sin (k * u)) | (A : ℝ) (k : ℝ)} /-- Find every twice-differentiable real-valued function $f$ with domain the set of all real numbers and satisfying the functional equation $(f(x))^2-(f(y))^2=f(x+y)f(x-y)$ for all real numbers $x$ and $y$. -/ theorem putnam_1963_b3 (f : ℝ → ℝ) : f ∈ putnam_1963_b3_solution ↔ (ContDiff ℝ 1 f ∧ Differentiable ℝ (deriv f) ∧ ∀ x y : ℝ, (f x) ^ 2 - (f y) ^ 2 = f (x + y) * f (x - y)) := sorry

import Mathlib open Topology Filter Polynomial -- {(fun u : ℝ => A * Real.sinh (k * u)) | (A : ℝ) (k : ℝ)} ∪ {(fun u : ℝ => A * u) | A : ℝ} ∪ {(fun u : ℝ => A * Real.sin (k * u)) | (A : ℝ) (k : ℝ)} /-- Find every twice-differentiable real-valued function $f$ with domain the set of all real numbers and satisfying the functional equation $(f(x))^2-(f(y))^2=f(x+y)f(x-y)$ for all real numbers $x$ and $y$. -/ theorem putnam_1963_b3 (f : ℝ → ℝ) : f ∈ putnam_1963_b3_solution ↔ (ContDiff ℝ 1 f ∧ Differentiable ℝ (deriv f) ∧ ∀ x y : ℝ, (f x) ^ 2 - (f y) ^ 2 = f (x + y) * f (x - y)) := by

import Mathlib open Topology Filter Polynomial abbrev putnam_1963_b3_solution : Set (ℝ → ℝ) := sorry -- {(fun u : ℝ => A * Real.sinh (k * u)) | (A : ℝ) (k : ℝ)} ∪ {(fun u : ℝ => A * u) | A : ℝ} ∪ {(fun u : ℝ => A * Real.sin (k * u)) | (A : ℝ) (k : ℝ)} /-- Find every twice-differentiable real-valued function $f$ with domain the set of all real numbers and satisfying the functional equation $(f(x))^2-(f(y))^2=f(x+y)f(x-y)$ for all real numbers $x$ and $y$. -/ theorem putnam_1963_b3 (f : ℝ → ℝ) : f ∈ putnam_1963_b3_solution ↔ (ContDiff ℝ 1 f ∧ Differentiable ℝ (deriv f) ∧ ∀ x y : ℝ, (f x) ^ 2 - (f y) ^ 2 = f (x + y) * f (x - y)) := sorry

Find every twice-differentiable real-valued function $f$ with domain the set of all real numbers and satisfying the functional equation $(f(x))^2-(f(y))^2=f(x+y)f(x-y)$ for all real numbers $x$ and $y$.

Show that the solution is the sets of functions $f(u)=A\sinh ku$, $f(u)=Au$, and $f(u)=A\sin ku$ with $A,k \in \mathbb{R}$.

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putnam_2009_a4

1f52368c-4ab9-5a48-8906-e619618d50b1

train

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

import Mathlib open Topology MvPolynomial Filter Set -- False /-- 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? -/ theorem putnam_2009_a4 : ((∀ S : Set ℚ, 0 ∈ S → (∀ x ∈ S, x + 1 ∈ S ∧ x - 1 ∈ S) → (∀ x ∈ S, x ∉ ({0, 1} : Set ℚ) → 1 / (x * (x - 1)) ∈ S) → ∀ r : ℚ, r ∈ S) ↔ putnam_2009_a4_solution) := by

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

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

Prove that $S$ need not contain all rationals.

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putnam_1978_a2

7327b79e-c562-5cc9-b082-e019e44c509c

train

theorem putnam_1978_a2 (n : ℕ) (npos : n > 0) (a b : ℝ) (hab : a ≠ b) (c : Fin n → ℝ) (A : Matrix (Fin n) (Fin n) ℝ) (hA : A = fun i j ↦ ite (i < j) a (ite (i > j) b (c i))) (p : ℝ → ℝ) (hp : p = fun x ↦ ∏ i : Fin n, (c i - x)) : (A.det = ((b * p a - a * p b) / (b - a))) := sorry

import Mathlib /-- Let $A$ be the real $n \times n$ matrix $(a_{ij})$ where $a_{ij} = a$ for $i < j$, $b \; (\neq a)$ for $i > j$, and $c_i$ for $i = j$. Show that $\det A = \frac{b p(a) - a p(b)}{b - a}$, where $p(x) = \prod_{i=1}^n (c_i - x)$. -/ theorem putnam_1978_a2 (n : ℕ) (npos : n > 0) (a b : ℝ) (hab : a ≠ b) (c : Fin n → ℝ) (A : Matrix (Fin n) (Fin n) ℝ) (hA : A = fun i j ↦ ite (i < j) a (ite (i > j) b (c i))) (p : ℝ → ℝ) (hp : p = fun x ↦ ∏ i : Fin n, (c i - x)) : (A.det = ((b * p a - a * p b) / (b - a))) := by

import Mathlib /-- Let $A$ be the real $n \times n$ matrix $(a_{ij})$ where $a_{ij} = a$ for $i < j$, $b \; (\neq a)$ for $i > j$, and $c_i$ for $i = j$. Show that $\det A = \frac{b p(a) - a p(b)}{b - a}$, where $p(x) = \prod_{i=1}^n (c_i - x)$. -/ theorem putnam_1978_a2 (n : ℕ) (npos : n > 0) (a b : ℝ) (hab : a ≠ b) (c : Fin n → ℝ) (A : Matrix (Fin n) (Fin n) ℝ) (hA : A = fun i j ↦ ite (i < j) a (ite (i > j) b (c i))) (p : ℝ → ℝ) (hp : p = fun x ↦ ∏ i : Fin n, (c i - x)) : (A.det = ((b * p a - a * p b) / (b - a))) := sorry

Let $A$ be the real $n \times n$ matrix $(a_{ij})$ where $a_{ij} = a$ for $i < j$, $b \; (\neq a)$ for $i > j$, and $c_i$ for $i = j$. Show that $\det A = \frac{b p(a) - a p(b)}{b - a}$, where $p(x) = \prod_{i=1}^n (c_i - x)$.

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putnam_1980_a2

dbcac05e-f56d-5ec0-b6d5-ff3207141412

train

abbrev putnam_1980_a2_solution : ℕ → ℕ → ℕ := sorry -- (fun r s : ℕ => (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2)) /-- Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a,b,c,d)$ of positive integers such that $3^r \cdot 7^s=\text{lcm}[a,b,c]=\text{lcm}[a,b,d]=\text{lcm}[a,c,d]=\text{lcm}[b,c,d]$. The answer should be a function of $r$ and $s$. (Note that $\text{lcm}[x,y,z]$ denotes the least common multiple of $x,y,z$.) -/ theorem putnam_1980_a2 (r s : ℕ) (abcdlcm : ℕ → ℕ → ℕ → ℕ → Prop) (rspos : r > 0 ∧ s > 0) (habcdlcm : ∀ a b c d : ℕ, abcdlcm a b c d ↔ (a > 0 ∧ b > 0 ∧ c > 0 ∧ d > 0 ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) c) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a c) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm b c) d))) : {(a, b, c, d) : ℕ × ℕ × ℕ × ℕ | abcdlcm a b c d}.encard = putnam_1980_a2_solution r s := sorry

import Mathlib -- (fun r s : ℕ => (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2)) /-- Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a,b,c,d)$ of positive integers such that $3^r \cdot 7^s=\text{lcm}[a,b,c]=\text{lcm}[a,b,d]=\text{lcm}[a,c,d]=\text{lcm}[b,c,d]$. The answer should be a function of $r$ and $s$. (Note that $\text{lcm}[x,y,z]$ denotes the least common multiple of $x,y,z$.) -/ theorem putnam_1980_a2 (r s : ℕ) (abcdlcm : ℕ → ℕ → ℕ → ℕ → Prop) (rspos : r > 0 ∧ s > 0) (habcdlcm : ∀ a b c d : ℕ, abcdlcm a b c d ↔ (a > 0 ∧ b > 0 ∧ c > 0 ∧ d > 0 ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) c) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a c) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm b c) d))) : {(a, b, c, d) : ℕ × ℕ × ℕ × ℕ | abcdlcm a b c d}.encard = putnam_1980_a2_solution r s := by

import Mathlib abbrev putnam_1980_a2_solution : ℕ → ℕ → ℕ := sorry -- (fun r s : ℕ => (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2)) /-- Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a,b,c,d)$ of positive integers such that $3^r \cdot 7^s=\text{lcm}[a,b,c]=\text{lcm}[a,b,d]=\text{lcm}[a,c,d]=\text{lcm}[b,c,d]$. The answer should be a function of $r$ and $s$. (Note that $\text{lcm}[x,y,z]$ denotes the least common multiple of $x,y,z$.) -/ theorem putnam_1980_a2 (r s : ℕ) (abcdlcm : ℕ → ℕ → ℕ → ℕ → Prop) (rspos : r > 0 ∧ s > 0) (habcdlcm : ∀ a b c d : ℕ, abcdlcm a b c d ↔ (a > 0 ∧ b > 0 ∧ c > 0 ∧ d > 0 ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) c) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a c) d) ∧ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm b c) d))) : {(a, b, c, d) : ℕ × ℕ × ℕ × ℕ | abcdlcm a b c d}.encard = putnam_1980_a2_solution r s := sorry

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

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

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putnam_2015_a5

ca6fa0b3-64e9-57cf-8c9c-cf2996f4e5bb

train

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

import Mathlib /-- Let $q$ be an odd positive integer, and let $N_q$ denote the number of integers $a$ such that $0<a<q/4$ and $\gcd(a,q)=1$. Show that $N_q$ is odd if and only if $q$ is of the form $p^k$ with $k$ a positive integer and $p$ a prime congruent to $5$ or $7$ modulo $8$. -/ theorem putnam_2015_a5 (q : ℕ) (Nq : ℕ) (qodd : Odd q) (qpos : q > 0) (hNq : Nq = {a : ℕ | 0 < a ∧ a < (q : ℝ) / 4 ∧ Nat.gcd a q = 1}.encard) : Odd Nq ↔ ∃ p k : ℕ, q = p ^ k ∧ k > 0 ∧ p.Prime ∧ (p % 8 = 5 ∨ p % 8 = 7) := by

import Mathlib /-- Let $q$ be an odd positive integer, and let $N_q$ denote the number of integers $a$ such that $0<a<q/4$ and $\gcd(a,q)=1$. Show that $N_q$ is odd if and only if $q$ is of the form $p^k$ with $k$ a positive integer and $p$ a prime congruent to $5$ or $7$ modulo $8$. -/ theorem putnam_2015_a5 (q : ℕ) (Nq : ℕ) (qodd : Odd q) (qpos : q > 0) (hNq : Nq = {a : ℕ | 0 < a ∧ a < (q : ℝ) / 4 ∧ Nat.gcd a q = 1}.encard) : Odd Nq ↔ ∃ p k : ℕ, q = p ^ k ∧ k > 0 ∧ p.Prime ∧ (p % 8 = 5 ∨ p % 8 = 7) := sorry

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

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putnam_2011_a5

c03f3f10-59d2-57b7-b28f-066387bf32e7

train

theorem putnam_2011_a5 (F : (Fin 2 → ℝ) → ℝ) (g : ℝ → ℝ) (vec : ℝ → ℝ → (Fin 2 → ℝ)) (Fgrad : (Fin 2 → ℝ) → (Fin 2 → ℝ)) (parallel : (Fin 2 → ℝ) → (Fin 2 → ℝ) → Prop) (hparallel : parallel = (fun u v : Fin 2 → ℝ => ∃ c : ℝ, u = c • v)) (Fgdiff : ContDiff ℝ 2 F ∧ ContDiff ℝ 2 g) (prop1 : ∀ uu : Fin 2 → ℝ, uu 0 = uu 1 → F uu = 0) (prop2 : ∀ x : ℝ, g x > 0 ∧ x ^ 2 * g x ≤ 1) (hvec : ∀ x y : ℝ, (vec x y) 0 = x ∧ (vec x y) 1 = y) (hFgrad : ∀ uv : Fin 2 → ℝ, Fgrad uv 0 = deriv (fun x : ℝ => F (vec x (uv 1))) (uv 0) ∧ Fgrad uv 1 = deriv (fun y : ℝ => F (vec (uv 0) y)) (uv 1)) (prop3 : ∀ uv : Fin 2 → ℝ, Fgrad uv = 0 ∨ parallel (Fgrad uv) (vec (g (uv 0)) (-g (uv 1)))) : ∃ C : ℝ, ∀ n ≥ 2, ∀ x : Fin (n + 1) → ℝ, sInf {Fxx : ℝ | ∃ i j : Fin (n + 1), i ≠ j ∧ Fxx = |F (vec (x i) (x j))|} ≤ C / n := sorry

import Mathlib open Topology Filter Matrix /-- Let $F:\mathbb{R}^2 \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ be twice continuously differentiable functions with the following properties: \begin{itemize} \item $F(u,u)=0$ for every $u \in \mathbb{R}$; \item for every $x \in \mathbb{R}$, $g(x)>0$ and $x^2g(x) \leq 1$; \item for every $(u,v) \in \mathbb{R}^2$, the vector $\nabla F(u,v)$ is either $\mathbf{0}$ or parallel to the vector $\langle g(u),-g(v) \rangle$. \end{itemize} Prove that there exists a constant $C$ such that for every $n \geq 2$ and any $x_1,\dots,x_{n+1} \in \mathbb{R}$, we have $\min_{i \neq j} |F(x_i,x_j)| \leq \frac{C}{n}$. -/ theorem putnam_2011_a5 (F : (Fin 2 → ℝ) → ℝ) (g : ℝ → ℝ) (vec : ℝ → ℝ → (Fin 2 → ℝ)) (Fgrad : (Fin 2 → ℝ) → (Fin 2 → ℝ)) (parallel : (Fin 2 → ℝ) → (Fin 2 → ℝ) → Prop) (hparallel : parallel = (fun u v : Fin 2 → ℝ => ∃ c : ℝ, u = c • v)) (Fgdiff : ContDiff ℝ 2 F ∧ ContDiff ℝ 2 g) (prop1 : ∀ uu : Fin 2 → ℝ, uu 0 = uu 1 → F uu = 0) (prop2 : ∀ x : ℝ, g x > 0 ∧ x ^ 2 * g x ≤ 1) (hvec : ∀ x y : ℝ, (vec x y) 0 = x ∧ (vec x y) 1 = y) (hFgrad : ∀ uv : Fin 2 → ℝ, Fgrad uv 0 = deriv (fun x : ℝ => F (vec x (uv 1))) (uv 0) ∧ Fgrad uv 1 = deriv (fun y : ℝ => F (vec (uv 0) y)) (uv 1)) (prop3 : ∀ uv : Fin 2 → ℝ, Fgrad uv = 0 ∨ parallel (Fgrad uv) (vec (g (uv 0)) (-g (uv 1)))) : ∃ C : ℝ, ∀ n ≥ 2, ∀ x : Fin (n + 1) → ℝ, sInf {Fxx : ℝ | ∃ i j : Fin (n + 1), i ≠ j ∧ Fxx = |F (vec (x i) (x j))|} ≤ C / n := by

import Mathlib open Topology Filter Matrix /-- Let $F:\mathbb{R}^2 \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ be twice continuously differentiable functions with the following properties: \begin{itemize} \item $F(u,u)=0$ for every $u \in \mathbb{R}$; \item for every $x \in \mathbb{R}$, $g(x)>0$ and $x^2g(x) \leq 1$; \item for every $(u,v) \in \mathbb{R}^2$, the vector $\nabla F(u,v)$ is either $\mathbf{0}$ or parallel to the vector $\langle g(u),-g(v) \rangle$. \end{itemize} Prove that there exists a constant $C$ such that for every $n \geq 2$ and any $x_1,\dots,x_{n+1} \in \mathbb{R}$, we have $\min_{i \neq j} |F(x_i,x_j)| \leq \frac{C}{n}$. -/ theorem putnam_2011_a5 (F : (Fin 2 → ℝ) → ℝ) (g : ℝ → ℝ) (vec : ℝ → ℝ → (Fin 2 → ℝ)) (Fgrad : (Fin 2 → ℝ) → (Fin 2 → ℝ)) (parallel : (Fin 2 → ℝ) → (Fin 2 → ℝ) → Prop) (hparallel : parallel = (fun u v : Fin 2 → ℝ => ∃ c : ℝ, u = c • v)) (Fgdiff : ContDiff ℝ 2 F ∧ ContDiff ℝ 2 g) (prop1 : ∀ uu : Fin 2 → ℝ, uu 0 = uu 1 → F uu = 0) (prop2 : ∀ x : ℝ, g x > 0 ∧ x ^ 2 * g x ≤ 1) (hvec : ∀ x y : ℝ, (vec x y) 0 = x ∧ (vec x y) 1 = y) (hFgrad : ∀ uv : Fin 2 → ℝ, Fgrad uv 0 = deriv (fun x : ℝ => F (vec x (uv 1))) (uv 0) ∧ Fgrad uv 1 = deriv (fun y : ℝ => F (vec (uv 0) y)) (uv 1)) (prop3 : ∀ uv : Fin 2 → ℝ, Fgrad uv = 0 ∨ parallel (Fgrad uv) (vec (g (uv 0)) (-g (uv 1)))) : ∃ C : ℝ, ∀ n ≥ 2, ∀ x : Fin (n + 1) → ℝ, sInf {Fxx : ℝ | ∃ i j : Fin (n + 1), i ≠ j ∧ Fxx = |F (vec (x i) (x j))|} ≤ C / n := sorry

Let $F:\mathbb{R}^2 \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ be twice continuously differentiable functions with the following properties: \begin{itemize} \item $F(u,u)=0$ for every $u \in \mathbb{R}$; \item for every $x \in \mathbb{R}$, $g(x)>0$ and $x^2g(x) \leq 1$; \item for every $(u,v) \in \mathbb{R}^2$, the vector $\nabla F(u,v)$ is either $\mathbf{0}$ or parallel to the vector $\langle g(u),-g(v) \rangle$. \end{itemize} Prove that there exists a constant $C$ such that for every $n \geq 2$ and any $x_1,\dots,x_{n+1} \in \mathbb{R}$, we have $\min_{i \neq j} |F(x_i,x_j)| \leq \frac{C}{n}$.

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putnam_1995_b6

54fa8ca2-9eba-58bc-bd96-2d846016507d

train

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

import Mathlib open Filter Topology Real Nat /-- 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 $\leq x$.] -/ theorem putnam_1995_b6 (S : ℝ → Set ℕ) (hS : S = fun (α : ℝ) => {x : ℕ | ∃ n : ℕ, n ≥ 1 ∧ x = floor (n * α)}) : ¬ ∃ α β γ, α > 0 ∧ β > 0 ∧ γ > 0 ∧ (S α) ∩ (S β) = ∅ ∧ (S β) ∩ (S γ) = ∅ ∧ (S α) ∩ (S γ) = ∅ ∧ Set.Ici 1 = (S α) ∪ (S β) ∪ (S γ) := by

import Mathlib open Filter Topology Real Nat /-- 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 $\leq x$.] -/ theorem putnam_1995_b6 (S : ℝ → Set ℕ) (hS : S = fun (α : ℝ) => {x : ℕ | ∃ n : ℕ, n ≥ 1 ∧ x = floor (n * α)}) : ¬ ∃ α β γ, α > 0 ∧ β > 0 ∧ γ > 0 ∧ (S α) ∩ (S β) = ∅ ∧ (S β) ∩ (S γ) = ∅ ∧ (S α) ∩ (S γ) = ∅ ∧ Set.Ici 1 = (S α) ∪ (S β) ∪ (S γ) := sorry

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

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putnam_1979_a6

01fc52aa-7118-561e-8672-d6109fdf5adf

train

theorem putnam_1979_a6 (n : ℕ) (p : ℕ → ℝ) (hp : ∀ i ∈ Finset.range n, p i ∈ Icc 0 1) : ∃ x ∈ Icc 0 1, (∀ i ∈ Finset.range n, x ≠ p i) ∧ ∑ i in Finset.range n, 1/|x - p i| ≤ 8*n*∑ i in Finset.range n, (1 : ℝ)/(2*i + 1) := sorry

import Mathlib open Set /-- For all $i \in \{0, 1, \dots, n - 1\}$, let $p_i \in [0, 1]$. Prove that there exists some $x \in [0, 1]$ such that $$\sum_{i = 0}^{n - 1} \frac{1}{|x - p_i|} \le 8n\left(\sum_{i = 0}^{n-1} \frac{1}{2i + 1}\right).$$ -/ theorem putnam_1979_a6 (n : ℕ) (p : ℕ → ℝ) (hp : ∀ i ∈ Finset.range n, p i ∈ Icc 0 1) : ∃ x ∈ Icc 0 1, (∀ i ∈ Finset.range n, x ≠ p i) ∧ ∑ i in Finset.range n, 1/|x - p i| ≤ 8*n*∑ i in Finset.range n, (1 : ℝ)/(2*i + 1) := by

import Mathlib open Set /-- For all $i \in \{0, 1, \dots, n - 1\}$, let $p_i \in [0, 1]$. Prove that there exists some $x \in [0, 1]$ such that $$\sum_{i = 0}^{n - 1} \frac{1}{|x - p_i|} \le 8n\left(\sum_{i = 0}^{n-1} \frac{1}{2i + 1}\right).$$ -/ theorem putnam_1979_a6 (n : ℕ) (p : ℕ → ℝ) (hp : ∀ i ∈ Finset.range n, p i ∈ Icc 0 1) : ∃ x ∈ Icc 0 1, (∀ i ∈ Finset.range n, x ≠ p i) ∧ ∑ i in Finset.range n, 1/|x - p i| ≤ 8*n*∑ i in Finset.range n, (1 : ℝ)/(2*i + 1) := sorry

For all $i \in \{0, 1, \dots, n - 1\}$, let $p_i \in [0, 1]$. Prove that there exists some $x \in [0, 1]$ such that $$\sum_{i = 0}^{n - 1} \frac{1}{|x - p_i|} \le 8n\left(\sum_{i = 0}^{n-1} \frac{1}{2i + 1}\right).$$

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putnam_1997_b5

b5afb8da-96d6-586b-947b-dec418d74657

train

theorem putnam_1997_b5 (n : ℕ) (hn : n ≥ 2) : tetration 2 n ≡ tetration 2 (n-1) [MOD n] := sorry

import Mathlib def tetration : ℕ → ℕ → ℕ | _, 0 => 1 | b, (m + 1) => b^(tetration b m) /-- Prove that for $n\geq 2$, \[\overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n$ terms}} \equiv \overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n-1$ terms}} \quad \pmod{n}.\] -/ theorem putnam_1997_b5 (n : ℕ) (hn : n ≥ 2) : tetration 2 n ≡ tetration 2 (n-1) [MOD n] := by

import Mathlib def tetration : ℕ → ℕ → ℕ | _, 0 => 1 | b, (m + 1) => b^(tetration b m) /-- Prove that for $n\geq 2$, \[\overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n$ terms}} \equiv \overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n-1$ terms}} \quad \pmod{n}.\] -/ theorem putnam_1997_b5 (n : ℕ) (hn : n ≥ 2) : tetration 2 n ≡ tetration 2 (n-1) [MOD n] := sorry

Prove that for $n\geq 2$, \[\overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n$ terms}} \equiv \overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n-1$ terms}} \quad \pmod{n}.\]

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putnam_1983_b5

d15768c8-2b6a-5a31-bc35-216c261e04ee

train

abbrev putnam_1983_b5_solution : ℝ := sorry -- log (4 / Real.pi) /-- Define $\left\lVert x \right\rVert$ as the distance from $x$ to the nearest integer. Find $\lim_{n \to \infty} \frac{1}{n} \int_{1}^{n} \left\lVert \frac{n}{x} \right\rVert \, dx$. You may assume that $\prod_{n=1}^{\infty} \frac{2n}{(2n-1)} \cdot \frac{2n}{(2n+1)} = \frac{\pi}{2}$. -/ theorem putnam_1983_b5 (dist_fun : ℝ → ℝ) (hdist_fun : dist_fun = fun (x : ℝ) ↦ min (x - ⌊x⌋) (⌈x⌉ - x)) (fact : Tendsto (fun N ↦ ∏ n in Finset.Icc 1 N, (2 * n / (2 * n - 1)) * (2 * n / (2 * n + 1)) : ℕ → ℝ) atTop (𝓝 (Real.pi / 2))) : (Tendsto (fun n ↦ (1 / n) * ∫ x in (1)..n, dist_fun (n / x) : ℕ → ℝ) atTop (𝓝 putnam_1983_b5_solution)) := sorry

import Mathlib open Nat Filter Topology Real -- log (4 / Real.pi) /-- Define $\left\lVert x \right\rVert$ as the distance from $x$ to the nearest integer. Find $\lim_{n \to \infty} \frac{1}{n} \int_{1}^{n} \left\lVert \frac{n}{x} \right\rVert \, dx$. You may assume that $\prod_{n=1}^{\infty} \frac{2n}{(2n-1)} \cdot \frac{2n}{(2n+1)} = \frac{\pi}{2}$. -/ theorem putnam_1983_b5 (dist_fun : ℝ → ℝ) (hdist_fun : dist_fun = fun (x : ℝ) ↦ min (x - ⌊x⌋) (⌈x⌉ - x)) (fact : Tendsto (fun N ↦ ∏ n in Finset.Icc 1 N, (2 * n / (2 * n - 1)) * (2 * n / (2 * n + 1)) : ℕ → ℝ) atTop (𝓝 (Real.pi / 2))) : (Tendsto (fun n ↦ (1 / n) * ∫ x in (1)..n, dist_fun (n / x) : ℕ → ℝ) atTop (𝓝 putnam_1983_b5_solution)) := by

import Mathlib open Nat Filter Topology Real noncomputable abbrev putnam_1983_b5_solution : ℝ := sorry -- log (4 / Real.pi) /-- Define $\left\lVert x \right\rVert$ as the distance from $x$ to the nearest integer. Find $\lim_{n \to \infty} \frac{1}{n} \int_{1}^{n} \left\lVert \frac{n}{x} \right\rVert \, dx$. You may assume that $\prod_{n=1}^{\infty} \frac{2n}{(2n-1)} \cdot \frac{2n}{(2n+1)} = \frac{\pi}{2}$. -/ theorem putnam_1983_b5 (dist_fun : ℝ → ℝ) (hdist_fun : dist_fun = fun (x : ℝ) ↦ min (x - ⌊x⌋) (⌈x⌉ - x)) (fact : Tendsto (fun N ↦ ∏ n in Finset.Icc 1 N, (2 * n / (2 * n - 1)) * (2 * n / (2 * n + 1)) : ℕ → ℝ) atTop (𝓝 (Real.pi / 2))) : (Tendsto (fun n ↦ (1 / n) * ∫ x in (1)..n, dist_fun (n / x) : ℕ → ℝ) atTop (𝓝 putnam_1983_b5_solution)) := sorry

Define $\left\lVert x \right\rVert$ as the distance from $x$ to the nearest integer. Find $\lim_{n \to \infty} \frac{1}{n} \int_{1}^{n} \left\lVert \frac{n}{x} \right\rVert \, dx$. You may assume that $\prod_{n=1}^{\infty} \frac{2n}{(2n-1)} \cdot \frac{2n}{(2n+1)} = \frac{\pi}{2}$.

Show that the limit equals $\ln \left( \frac{4}{\pi} \right)$.

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putnam_1975_b5

eef33668-1bab-501d-8704-ec9e2c4f7897

train

theorem putnam_1975_b5 (e : ℝ) (he : e = Real.exp 1) (f : ℕ → ℝ → ℝ) (h0 : ∀ x : ℝ, f 0 x = e^x) (hf : ∀ n : ℕ, ∀ x : ℝ, f (n + 1) x = x * (deriv (f n) x)) : ∑' n : ℕ, (f n 1)/(Nat.factorial n) = e^e := sorry

import Mathlib open Polynomial Real Complex Matrix Filter Topology Multiset /-- Let $f_0(x) = e^x$ and $f_{n+1}(x) = xf'_n(x)$ for all $n \ge 0$. Prove that $$\sum_{n=0}^{\infty} \frac{f_n(1)}{n!} = e^e.$$ -/ theorem putnam_1975_b5 (e : ℝ) (he : e = Real.exp 1) (f : ℕ → ℝ → ℝ) (h0 : ∀ x : ℝ, f 0 x = e^x) (hf : ∀ n : ℕ, ∀ x : ℝ, f (n + 1) x = x * (deriv (f n) x)) : ∑' n : ℕ, (f n 1)/(Nat.factorial n) = e^e := by

import Mathlib open Polynomial Real Complex Matrix Filter Topology Multiset /-- Let $f_0(x) = e^x$ and $f_{n+1}(x) = xf'_n(x)$ for all $n \ge 0$. Prove that $$\sum_{n=0}^{\infty} \frac{f_n(1)}{n!} = e^e.$$ -/ theorem putnam_1975_b5 (e : ℝ) (he : e = Real.exp 1) (f : ℕ → ℝ → ℝ) (h0 : ∀ x : ℝ, f 0 x = e^x) (hf : ∀ n : ℕ, ∀ x : ℝ, f (n + 1) x = x * (deriv (f n) x)) : ∑' n : ℕ, (f n 1)/(Nat.factorial n) = e^e := sorry

Let $f_0(x) = e^x$ and $f_{n+1}(x) = xf'_n(x)$ for all $n \ge 0$. Prove that $$\sum_{n=0}^{\infty} \frac{f_n(1)}{n!} = e^e.$$

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putnam_2008_b1

92596519-5bdc-577d-952a-0d75264935ea

train

abbrev putnam_2008_b1_solution : ℕ := sorry -- 2 def is_rational_point (p : Fin 2 → ℝ) : Prop := ∃ (a b : ℚ), a = p 0 ∧ b = p 1 def real_circle (c : EuclideanSpace ℝ (Fin 2)) (r : ℝ) : Set (EuclideanSpace ℝ (Fin 2)) := {p | dist p c = r} /-- What is the maximum number of rational points that can lie on a circle in $\mathbb{R}^2$ whose center is not a rational point? (A \emph{rational point} is a point both of whose coordinates are rational numbers.) -/ theorem putnam_2008_b1 : IsGreatest {n : ℕ | ∃ (c : EuclideanSpace ℝ (Fin 2)) (r : ℝ), ¬ is_rational_point c ∧ (Set.ncard {p : EuclideanSpace ℝ (Fin 2) | p ∈ real_circle c r ∧ is_rational_point p} = n)} putnam_2008_b1_solution := sorry

import Mathlib open Filter Topology Set -- 2 def is_rational_point (p : Fin 2 → ℝ) : Prop := ∃ (a b : ℚ), a = p 0 ∧ b = p 1 def real_circle (c : EuclideanSpace ℝ (Fin 2)) (r : ℝ) : Set (EuclideanSpace ℝ (Fin 2)) := {p | dist p c = r} /-- What is the maximum number of rational points that can lie on a circle in $\mathbb{R}^2$ whose center is not a rational point? (A \emph{rational point} is a point both of whose coordinates are rational numbers.) -/ theorem putnam_2008_b1 : IsGreatest {n : ℕ | ∃ (c : EuclideanSpace ℝ (Fin 2)) (r : ℝ), ¬ is_rational_point c ∧ (Set.ncard {p : EuclideanSpace ℝ (Fin 2) | p ∈ real_circle c r ∧ is_rational_point p} = n)} putnam_2008_b1_solution := by

import Mathlib open Filter Topology Set abbrev putnam_2008_b1_solution : ℕ := sorry -- 2 def is_rational_point (p : Fin 2 → ℝ) : Prop := ∃ (a b : ℚ), a = p 0 ∧ b = p 1 def real_circle (c : EuclideanSpace ℝ (Fin 2)) (r : ℝ) : Set (EuclideanSpace ℝ (Fin 2)) := {p | dist p c = r} /-- What is the maximum number of rational points that can lie on a circle in $\mathbb{R}^2$ whose center is not a rational point? (A \emph{rational point} is a point both of whose coordinates are rational numbers.) -/ theorem putnam_2008_b1 : IsGreatest {n : ℕ | ∃ (c : EuclideanSpace ℝ (Fin 2)) (r : ℝ), ¬ is_rational_point c ∧ (Set.ncard {p : EuclideanSpace ℝ (Fin 2) | p ∈ real_circle c r ∧ is_rational_point p} = n)} putnam_2008_b1_solution := sorry

What is the maximum number of rational points that can lie on a circle in $\mathbb{R}^2$ whose center is not a rational point? (A \emph{rational point} is a point both of whose coordinates are rational numbers.)

Show that the maximum number is $2$.

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putnam_2008_a5

4c40e1cf-79d1-5b36-aa45-33db461e7c7a

train

theorem putnam_2008_a5 (n : ℕ) (nge3 : n ≥ 3) (f g : Polynomial ℝ) (hfg : ∃ O z : ℂ, z ≠ 0 ∧ ∀ k : ℕ, k ∈ Icc 1 n → (f.eval (k : ℝ)) + Complex.I * (g.eval (k : ℝ)) = O + z * Complex.exp (Complex.I * 2 * Real.pi * k / n)) : (f.natDegree ≥ n - 1 ∨ g.natDegree ≥ n - 1) := sorry

import Mathlib open Filter Topology Set /-- Let $n \geq 3$ be an integer. Let $f(x)$ and $g(x)$ be polynomials with real coefficients such that the points $(f(1), g(1)), (f(2), g(2)), \dots, (f(n), g(n))$ in $\mathbb{R}^2$ are the vertices of a regular $n$-gon in counterclockwise order. Prove that at least one of $f(x)$ and $g(x)$ has degree greater than or equal to $n-1$. -/ theorem putnam_2008_a5 (n : ℕ) (nge3 : n ≥ 3) (f g : Polynomial ℝ) (hfg : ∃ O z : ℂ, z ≠ 0 ∧ ∀ k : ℕ, k ∈ Icc 1 n → (f.eval (k : ℝ)) + Complex.I * (g.eval (k : ℝ)) = O + z * Complex.exp (Complex.I * 2 * Real.pi * k / n)) : (f.natDegree ≥ n - 1 ∨ g.natDegree ≥ n - 1) := by

import Mathlib open Filter Topology Set /-- Let $n \geq 3$ be an integer. Let $f(x)$ and $g(x)$ be polynomials with real coefficients such that the points $(f(1), g(1)), (f(2), g(2)), \dots, (f(n), g(n))$ in $\mathbb{R}^2$ are the vertices of a regular $n$-gon in counterclockwise order. Prove that at least one of $f(x)$ and $g(x)$ has degree greater than or equal to $n-1$. -/ theorem putnam_2008_a5 (n : ℕ) (nge3 : n ≥ 3) (f g : Polynomial ℝ) (hfg : ∃ O z : ℂ, z ≠ 0 ∧ ∀ k : ℕ, k ∈ Icc 1 n → (f.eval (k : ℝ)) + Complex.I * (g.eval (k : ℝ)) = O + z * Complex.exp (Complex.I * 2 * Real.pi * k / n)) : (f.natDegree ≥ n - 1 ∨ g.natDegree ≥ n - 1) := sorry

Let $n \geq 3$ be an integer. Let $f(x)$ and $g(x)$ be polynomials with real coefficients such that the points $(f(1), g(1)), (f(2), g(2)), \dots, (f(n), g(n))$ in $\mathbb{R}^2$ are the vertices of a regular $n$-gon in counterclockwise order. Prove that at least one of $f(x)$ and $g(x)$ has degree greater than or equal to $n-1$.

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putnam_1969_b2

1e680b8e-4938-51f3-af09-dc9d5dd0dbad

train

abbrev putnam_1969_b2_solution : Prop := sorry -- False /-- Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if 'two' is replaced by 'three'? -/ theorem putnam_1969_b2 (P : ℕ → Prop) (P_def : ∀ n, P n ↔ ∀ (G : Type) [Group G] [Finite G], ∀ H : Fin n → Subgroup G, (∀ i, H i < ⊤) → ⋃ i, (H i : Set G) < ⊤) : P 2 ∧ (P 3 ↔ putnam_1969_b2_solution) := sorry

import Mathlib open Matrix Filter Topology Set Nat -- False /-- Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if 'two' is replaced by 'three'? -/ theorem putnam_1969_b2 (P : ℕ → Prop) (P_def : ∀ n, P n ↔ ∀ (G : Type) [Group G] [Finite G], ∀ H : Fin n → Subgroup G, (∀ i, H i < ⊤) → ⋃ i, (H i : Set G) < ⊤) : P 2 ∧ (P 3 ↔ putnam_1969_b2_solution) := by

import Mathlib open Matrix Filter Topology Set Nat abbrev putnam_1969_b2_solution : Prop := sorry -- False /-- Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if 'two' is replaced by 'three'? -/ theorem putnam_1969_b2 (P : ℕ → Prop) (P_def : ∀ n, P n ↔ ∀ (G : Type) [Group G] [Finite G], ∀ H : Fin n → Subgroup G, (∀ i, H i < ⊤) → ⋃ i, (H i : Set G) < ⊤) : P 2 ∧ (P 3 ↔ putnam_1969_b2_solution) := sorry

Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if 'two' is replaced by 'three'?

Show that the statement is no longer true if 'two' is replaced by 'three'.

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putnam_2023_b6

12195118-b6f7-5dfa-b81a-3b4ce677e02c

train

abbrev putnam_2023_b6_solution : ℕ → ℤ := sorry -- (fun n : ℕ => (-1) ^ (⌈(n / 2 : ℚ)⌉₊ + 1) * 2 * ⌈(n / 2 : ℚ)⌉₊) /-- Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai+bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix} 6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2 \end{bmatrix}$. Compute the determinant of $S$. -/ theorem putnam_2023_b6 (n : ℕ) (S : Matrix (Fin n) (Fin n) ℤ) (npos : n > 0) (hS : ∀ i j : Fin n, S i j = ∑' a : ℕ, ∑' b : ℕ, if a * (i.1 + 1) + b * (j.1 + 1) = (n : ℤ) then 1 else 0) : S.det = putnam_2023_b6_solution n := sorry

import Mathlib open Nat Topology Filter -- (fun n : ℕ => (-1) ^ (⌈(n / 2 : ℚ)⌉₊ + 1) * 2 * ⌈(n / 2 : ℚ)⌉₊) /-- Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai+bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix} 6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2 \end{bmatrix}$. Compute the determinant of $S$. -/ theorem putnam_2023_b6 (n : ℕ) (S : Matrix (Fin n) (Fin n) ℤ) (npos : n > 0) (hS : ∀ i j : Fin n, S i j = ∑' a : ℕ, ∑' b : ℕ, if a * (i.1 + 1) + b * (j.1 + 1) = (n : ℤ) then 1 else 0) : S.det = putnam_2023_b6_solution n := by

import Mathlib open Nat Topology Filter abbrev putnam_2023_b6_solution : ℕ → ℤ := sorry -- (fun n : ℕ => (-1) ^ (⌈(n / 2 : ℚ)⌉₊ + 1) * 2 * ⌈(n / 2 : ℚ)⌉₊) /-- Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai+bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix} 6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2 \end{bmatrix}$. Compute the determinant of $S$. -/ theorem putnam_2023_b6 (n : ℕ) (S : Matrix (Fin n) (Fin n) ℤ) (npos : n > 0) (hS : ∀ i j : Fin n, S i j = ∑' a : ℕ, ∑' b : ℕ, if a * (i.1 + 1) + b * (j.1 + 1) = (n : ℤ) then 1 else 0) : S.det = putnam_2023_b6_solution n := sorry

Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai+bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix} 6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2 \end{bmatrix}$. Compute the determinant of $S$.

Show that the determinant equals $(-1)^{\lceil n/2 \rceil-1}2\lceil\frac{n}{2}\rceil$.

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putnam_1982_b5

83898fab-503f-5278-914f-37d217acbab1

train

theorem putnam_1982_b5 (T : Set ℝ) (hT : T = Ioi (Real.exp (Real.exp 1))) (S : ℝ → ℕ → ℝ) (hS : ∀ x ∈ T, S x 0 = (Real.exp 1) ∧ ∀ n : ℕ, S x (n + 1) = Real.logb (S x n) x) (g : ℝ → ℝ) : ∀ x ∈ T, (∃ L : ℝ, Tendsto (S x) atTop (𝓝 L)) ∧ (∀ x ∈ T, Tendsto (S x) atTop (𝓝 (g x))) → ContinuousOn g T := sorry

import Mathlib open Set Function Filter Topology Polynomial Real /-- For all $x > e^e$, let $S = u_0, u_1, \dots$ be a recursively defined sequence with $u_0 = e$ and $u_{n+1} = \log_{u_n} x$ for all $n \ge 0$. Prove that $S_x$ converges to some real number $g(x)$ and that this function $g$ is continuous for $x > e^e$. -/ theorem putnam_1982_b5 (T : Set ℝ) (hT : T = Ioi (Real.exp (Real.exp 1))) (S : ℝ → ℕ → ℝ) (hS : ∀ x ∈ T, S x 0 = (Real.exp 1) ∧ ∀ n : ℕ, S x (n + 1) = Real.logb (S x n) x) (g : ℝ → ℝ) : ∀ x ∈ T, (∃ L : ℝ, Tendsto (S x) atTop (𝓝 L)) ∧ (∀ x ∈ T, Tendsto (S x) atTop (𝓝 (g x))) → ContinuousOn g T := by

import Mathlib open Set Function Filter Topology Polynomial Real /-- For all $x > e^e$, let $S = u_0, u_1, \dots$ be a recursively defined sequence with $u_0 = e$ and $u_{n+1} = \log_{u_n} x$ for all $n \ge 0$. Prove that $S_x$ converges to some real number $g(x)$ and that this function $g$ is continuous for $x > e^e$. -/ theorem putnam_1982_b5 (T : Set ℝ) (hT : T = Ioi (Real.exp (Real.exp 1))) (S : ℝ → ℕ → ℝ) (hS : ∀ x ∈ T, S x 0 = (Real.exp 1) ∧ ∀ n : ℕ, S x (n + 1) = Real.logb (S x n) x) (g : ℝ → ℝ) : ∀ x ∈ T, (∃ L : ℝ, Tendsto (S x) atTop (𝓝 L)) ∧ (∀ x ∈ T, Tendsto (S x) atTop (𝓝 (g x))) → ContinuousOn g T := sorry

For all $x > e^e$, let $S = u_0, u_1, \dots$ be a recursively defined sequence with $u_0 = e$ and $u_{n+1} = \log_{u_n} x$ for all $n \ge 0$. Prove that $S_x$ converges to some real number $g(x)$ and that this function $g$ is continuous for $x > e^e$.

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putnam_2003_a3

10e1dec8-2b3c-5535-962f-c20c7ca275f0

train

abbrev putnam_2003_a3_solution : ℝ := sorry -- 2 * Real.sqrt 2 - 1 /-- Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$. -/ theorem putnam_2003_a3 (f : ℝ → ℝ) (hf : ∀ x : ℝ, f x = |Real.sin x + Real.cos x + Real.tan x + 1 / Real.tan x + 1 / Real.cos x + 1 / Real.sin x|) : IsLeast (Set.range f) putnam_2003_a3_solution := sorry

import Mathlib open Set -- 2 * Real.sqrt 2 - 1 /-- Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$. -/ theorem putnam_2003_a3 (f : ℝ → ℝ) (hf : ∀ x : ℝ, f x = |Real.sin x + Real.cos x + Real.tan x + 1 / Real.tan x + 1 / Real.cos x + 1 / Real.sin x|) : IsLeast (Set.range f) putnam_2003_a3_solution := by

import Mathlib open Set noncomputable abbrev putnam_2003_a3_solution : ℝ := sorry -- 2 * Real.sqrt 2 - 1 /-- Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$. -/ theorem putnam_2003_a3 (f : ℝ → ℝ) (hf : ∀ x : ℝ, f x = |Real.sin x + Real.cos x + Real.tan x + 1 / Real.tan x + 1 / Real.cos x + 1 / Real.sin x|) : IsLeast (Set.range f) putnam_2003_a3_solution := sorry

Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$.

Show that the minimum is $2\sqrt{2}-1$.

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putnam_1981_b3

373df209-5160-5dc3-b61b-828c7fd6c6ee

train

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

import Mathlib open Topology Filter Set Polynomial Function /-- 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 $k^2 < n$. -/ theorem putnam_1981_b3 (P : ℕ → Prop) (hP : ∀ n, P n ↔ ∀ p : ℕ, (Nat.Prime p ∧ p ∣ n^2 + 3) → ∃ k : ℕ, (p : ℤ) ∣ (k : ℤ)^2 + 3 ∧ k^2 < n) : ∀ n : ℕ, ∃ m : ℕ, (m : ℤ) > n ∧ P m := by

import Mathlib open Topology Filter Set Polynomial Function /-- 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 $k^2 < n$. -/ theorem putnam_1981_b3 (P : ℕ → Prop) (hP : ∀ n, P n ↔ ∀ p : ℕ, (Nat.Prime p ∧ p ∣ n^2 + 3) → ∃ k : ℕ, (p : ℤ) ∣ (k : ℤ)^2 + 3 ∧ k^2 < n) : ∀ n : ℕ, ∃ m : ℕ, (m : ℤ) > n ∧ P m := sorry

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

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putnam_1993_a2

ba08f679-e7b0-52e2-8036-e28357c76aba

train

theorem putnam_1993_a2 (x : ℕ → ℝ) (xnonzero : ∀ n : ℕ, x n ≠ 0) (hx : ∀ n ≥ 1, (x n) ^ 2 - x (n - 1) * x (n + 1) = 1) : ∃ a : ℝ, ∀ n ≥ 1, x (n + 1) = a * x n - x (n - 1) := sorry

import Mathlib /-- Let $(x_n)_{n \geq 0}$ be a sequence of nonzero real numbers such that $x_n^2-x_{n-1}x_{n+1}=1$ for $n=1,2,3,\dots$. Prove there exists a real number $a$ such that $x_{n+1}=ax_n-x_{n-1}$ for all $n \geq 1$. -/ theorem putnam_1993_a2 (x : ℕ → ℝ) (xnonzero : ∀ n : ℕ, x n ≠ 0) (hx : ∀ n ≥ 1, (x n) ^ 2 - x (n - 1) * x (n + 1) = 1) : ∃ a : ℝ, ∀ n ≥ 1, x (n + 1) = a * x n - x (n - 1) := by

import Mathlib /-- Let $(x_n)_{n \geq 0}$ be a sequence of nonzero real numbers such that $x_n^2-x_{n-1}x_{n+1}=1$ for $n=1,2,3,\dots$. Prove there exists a real number $a$ such that $x_{n+1}=ax_n-x_{n-1}$ for all $n \geq 1$. -/ theorem putnam_1993_a2 (x : ℕ → ℝ) (xnonzero : ∀ n : ℕ, x n ≠ 0) (hx : ∀ n ≥ 1, (x n) ^ 2 - x (n - 1) * x (n + 1) = 1) : ∃ a : ℝ, ∀ n ≥ 1, x (n + 1) = a * x n - x (n - 1) := sorry

Let $(x_n)_{n \geq 0}$ be a sequence of nonzero real numbers such that $x_n^2-x_{n-1}x_{n+1}=1$ for $n=1,2,3,\dots$. Prove there exists a real number $a$ such that $x_{n+1}=ax_n-x_{n-1}$ for all $n \geq 1$.

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putnam_1991_a4

081fb2f1-5e09-5bc7-a974-c2b49b787196

train

abbrev putnam_1991_a4_solution : Prop := sorry -- True /-- Does there exist an infinite sequence of closed discs $D_1,D_2,D_3,\dots$ in the plane, with centers $c_1,c_2,c_3,\dots$, respectively, such that \begin{enumerate} \item the $c_i$ have no limit point in the finite plane, \item the sum of the areas of the $D_i$ is finite, and \item every line in the plane intersects at least one of the $D_i$? \end{enumerate} -/ theorem putnam_1991_a4 : (∃ (c : ℕ → EuclideanSpace ℝ (Fin 2)) (r : ℕ → ℝ), (¬ ∃ p, MapClusterPt p atTop c) ∧ (Summable <| fun i ↦ r i ^ 2) ∧ (∀ L : AffineSubspace ℝ (EuclideanSpace ℝ (Fin 2)), finrank ℝ L.direction = 1 → ∃ i, (↑L ∩ closedBall (c i) (r i)).Nonempty)) ↔ putnam_1991_a4_solution := sorry

import Mathlib open Filter FiniteDimensional Metric Topology -- True /-- Does there exist an infinite sequence of closed discs $D_1,D_2,D_3,\dots$ in the plane, with centers $c_1,c_2,c_3,\dots$, respectively, such that \begin{enumerate} \item the $c_i$ have no limit point in the finite plane, \item the sum of the areas of the $D_i$ is finite, and \item every line in the plane intersects at least one of the $D_i$? \end{enumerate} -/ theorem putnam_1991_a4 : (∃ (c : ℕ → EuclideanSpace ℝ (Fin 2)) (r : ℕ → ℝ), (¬ ∃ p, MapClusterPt p atTop c) ∧ (Summable <| fun i ↦ r i ^ 2) ∧ (∀ L : AffineSubspace ℝ (EuclideanSpace ℝ (Fin 2)), finrank ℝ L.direction = 1 → ∃ i, (↑L ∩ closedBall (c i) (r i)).Nonempty)) ↔ putnam_1991_a4_solution := by

import Mathlib open Filter FiniteDimensional Metric Topology abbrev putnam_1991_a4_solution : Prop := sorry -- True /-- Does there exist an infinite sequence of closed discs $D_1,D_2,D_3,\dots$ in the plane, with centers $c_1,c_2,c_3,\dots$, respectively, such that \begin{enumerate} \item the $c_i$ have no limit point in the finite plane, \item the sum of the areas of the $D_i$ is finite, and \item every line in the plane intersects at least one of the $D_i$? \end{enumerate} -/ theorem putnam_1991_a4 : (∃ (c : ℕ → EuclideanSpace ℝ (Fin 2)) (r : ℕ → ℝ), (¬ ∃ p, MapClusterPt p atTop c) ∧ (Summable <| fun i ↦ r i ^ 2) ∧ (∀ L : AffineSubspace ℝ (EuclideanSpace ℝ (Fin 2)), finrank ℝ L.direction = 1 → ∃ i, (↑L ∩ closedBall (c i) (r i)).Nonempty)) ↔ putnam_1991_a4_solution := sorry

Does there exist an infinite sequence of closed discs $D_1,D_2,D_3,\dots$ in the plane, with centers $c_1,c_2,c_3,\dots$, respectively, such that \begin{enumerate} \item the $c_i$ have no limit point in the finite plane, \item the sum of the areas of the $D_i$ is finite, and \item every line in the plane intersects at least one of the $D_i$? \end{enumerate}

Show that the answer is yes, such a sequence of closed discs exists.

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putnam_1987_a6

330228f0-9ab5-52ea-9a21-5c5d6495ebab

train

abbrev putnam_1987_a6_solution : Set ℝ := sorry -- {x : ℝ | x > 0 ∧ x < 25} /-- For each positive integer $n$, let $a(n)$ be the number of zeroes in the base $3$ representation of $n$. For which positive real numbers $x$ does the series \[ \sum_{n=1}^\infty \frac{x^{a(n)}}{n^3} \] converge? -/ theorem putnam_1987_a6 (a : ℕ → ℕ) (ha : a = fun n ↦ {i | (digits 3 n).get i = 0}.ncard) : ({x : ℝ | x > 0 ∧ Summable (fun n ↦ x ^ (a n) / (n ^ 3))} = putnam_1987_a6_solution) := sorry

import Mathlib open MvPolynomial Real Nat -- {x : ℝ | x > 0 ∧ x < 25} /-- For each positive integer $n$, let $a(n)$ be the number of zeroes in the base $3$ representation of $n$. For which positive real numbers $x$ does the series \[ \sum_{n=1}^\infty \frac{x^{a(n)}}{n^3} \] converge? -/ theorem putnam_1987_a6 (a : ℕ → ℕ) (ha : a = fun n ↦ {i | (digits 3 n).get i = 0}.ncard) : ({x : ℝ | x > 0 ∧ Summable (fun n ↦ x ^ (a n) / (n ^ 3))} = putnam_1987_a6_solution) := by

import Mathlib open MvPolynomial Real Nat abbrev putnam_1987_a6_solution : Set ℝ := sorry -- {x : ℝ | x > 0 ∧ x < 25} /-- For each positive integer $n$, let $a(n)$ be the number of zeroes in the base $3$ representation of $n$. For which positive real numbers $x$ does the series \[ \sum_{n=1}^\infty \frac{x^{a(n)}}{n^3} \] converge? -/ theorem putnam_1987_a6 (a : ℕ → ℕ) (ha : a = fun n ↦ {i | (digits 3 n).get i = 0}.ncard) : ({x : ℝ | x > 0 ∧ Summable (fun n ↦ x ^ (a n) / (n ^ 3))} = putnam_1987_a6_solution) := sorry

For each positive integer $n$, let $a(n)$ be the number of zeroes in the base $3$ representation of $n$. For which positive real numbers $x$ does the series \[ \sum_{n=1}^\infty \frac{x^{a(n)}}{n^3} \] converge?

Show that for positive $x$, the series converges if and only if $x < 25$.

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putnam_1965_a6

23c83676-ac0d-5b49-abd8-1155814b3350

train

theorem putnam_1965_a6 (u v m : ℝ) (hu : 0 < u) (hv : 0 < v) (hm : 1 < m) : (∃ᵉ (x > 0) (y > 0), u * x + v * y = 1 ∧ x ^ m + y ^ m = 1 ∧ u = x ^ (m - 1) ∧ v = y ^ (m - 1)) ↔ ∃ n, u ^ n + v ^ n = 1 ∧ m⁻¹ + n⁻¹ = 1 := sorry

import Mathlib open EuclideanGeometry Topology Filter Complex /-- Prove that the line $ux + vy = 1$ (where $u \ge 0$ and $v \ge 0$) will lie tangent to the curve $x^m + y^m = 1$ (where $m > 1$) if and only if $u^n + v^n = 1$ for some $n$ such that $m^{-1} + n^{-1} = 1$. -/ theorem putnam_1965_a6 (u v m : ℝ) (hu : 0 < u) (hv : 0 < v) (hm : 1 < m) : (∃ᵉ (x > 0) (y > 0), u * x + v * y = 1 ∧ x ^ m + y ^ m = 1 ∧ u = x ^ (m - 1) ∧ v = y ^ (m - 1)) ↔ ∃ n, u ^ n + v ^ n = 1 ∧ m⁻¹ + n⁻¹ = 1 := by

import Mathlib open EuclideanGeometry Topology Filter Complex /-- Prove that the line $ux + vy = 1$ (where $u \ge 0$ and $v \ge 0$) will lie tangent to the curve $x^m + y^m = 1$ (where $m > 1$) if and only if $u^n + v^n = 1$ for some $n$ such that $m^{-1} + n^{-1} = 1$. -/ theorem putnam_1965_a6 (u v m : ℝ) (hu : 0 < u) (hv : 0 < v) (hm : 1 < m) : (∃ᵉ (x > 0) (y > 0), u * x + v * y = 1 ∧ x ^ m + y ^ m = 1 ∧ u = x ^ (m - 1) ∧ v = y ^ (m - 1)) ↔ ∃ n, u ^ n + v ^ n = 1 ∧ m⁻¹ + n⁻¹ = 1 := sorry

Prove that the line $ux + vy = 1$ (where $u \ge 0$ and $v \ge 0$) will lie tangent to the curve $x^m + y^m = 1$ (where $m > 1$) if and only if $u^n + v^n = 1$ for some $n$ such that $m^{-1} + n^{-1} = 1$.

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putnam_1972_a2

762440ae-b383-559d-85b4-b2b6596a776a

train

theorem putnam_1972_a2 : (∀ (S : Type*) (_ : Mul S), (∀ x y : S, x * (x * y) = y ∧ ((y * x) * x) = y) → (∀ x y : S, x * y = y * x)) ∧ ∃ (S : Type*) (_ : Mul S), (∀ x y : S, x * (x * y) = y ∧ ((y * x) * x) = y) ∧ ¬(∀ x y z : S, x * (y * z) = (x * y) * z) := sorry

import Mathlib open EuclideanGeometry Filter Topology Set /-- Let $S$ be a set and $\cdot$ be a binary operation on $S$ satisfying: (1) for all $x,y$ in $S$, $x \cdot (x \cdot y) = y$ (2) for all $x,y$ in $S$, $(y \cdot x) \cdot x = y$. Show that $\cdot$ is commutative but not necessarily associative. -/ theorem putnam_1972_a2 : (∀ (S : Type*) (_ : Mul S), (∀ x y : S, x * (x * y) = y ∧ ((y * x) * x) = y) → (∀ x y : S, x * y = y * x)) ∧ ∃ (S : Type*) (_ : Mul S), (∀ x y : S, x * (x * y) = y ∧ ((y * x) * x) = y) ∧ ¬(∀ x y z : S, x * (y * z) = (x * y) * z) := by

import Mathlib open EuclideanGeometry Filter Topology Set /-- Let $S$ be a set and $\cdot$ be a binary operation on $S$ satisfying: (1) for all $x,y$ in $S$, $x \cdot (x \cdot y) = y$ (2) for all $x,y$ in $S$, $(y \cdot x) \cdot x = y$. Show that $\cdot$ is commutative but not necessarily associative. -/ theorem putnam_1972_a2 : (∀ (S : Type*) (_ : Mul S), (∀ x y : S, x * (x * y) = y ∧ ((y * x) * x) = y) → (∀ x y : S, x * y = y * x)) ∧ ∃ (S : Type*) (_ : Mul S), (∀ x y : S, x * (x * y) = y ∧ ((y * x) * x) = y) ∧ ¬(∀ x y z : S, x * (y * z) = (x * y) * z) := sorry

Let $S$ be a set and $\cdot$ be a binary operation on $S$ satisfying: (1) for all $x,y$ in $S$, $x \cdot (x \cdot y) = y$ (2) for all $x,y$ in $S$, $(y \cdot x) \cdot x = y$. Show that $\cdot$ is commutative but not necessarily associative.

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putnam_1977_a6

1e6e09d0-9de2-5cdd-b73a-9d253d509438

train

abbrev putnam_1977_a6_solution : Prop := sorry -- True /-- Let $X$ be the square $[0, 1] \times [0, 1]$, and let $f : X \to \mathbb{R}$ be continuous. If $\int_Y f(x, y) \, dx \, dy = 0$ for all squares $Y$ such that \begin{itemize} \item[(1)] $Y \subseteq X$, \item[(2)] $Y$ has sides parallel to those of $X$, \item[(3)] at least one of $Y$'s sides is contained in the boundary of $X$, \end{itemize} is it true that $f(x, y) = 0$ for all $x, y$? -/ theorem putnam_1977_a6 (X : Set (ℝ × ℝ)) (hX : X = Set.prod (Icc 0 1) (Icc 0 1)) (room : (ℝ × ℝ) → ℝ) (hroom : room = fun (a,b) ↦ min (min a (1 - a)) (min b (1 - b))) : ((∀ f : (ℝ × ℝ) → ℝ, Continuous f → (∀ P ∈ X, ∫ x in (P.1 - room P)..(P.1 + room P), ∫ y in (P.2 - room P)..(P.2 + room P), f (x, y) = 0) → (∀ P ∈ X, f P = 0)) ↔ putnam_1977_a6_solution) := sorry

import Mathlib open RingHom Set Nat -- True /-- Let $X$ be the square $[0, 1] \times [0, 1]$, and let $f : X \to \mathbb{R}$ be continuous. If $\int_Y f(x, y) \, dx \, dy = 0$ for all squares $Y$ such that \begin{itemize} \item[(1)] $Y \subseteq X$, \item[(2)] $Y$ has sides parallel to those of $X$, \item[(3)] at least one of $Y$'s sides is contained in the boundary of $X$, \end{itemize} is it true that $f(x, y) = 0$ for all $x, y$? -/ theorem putnam_1977_a6 (X : Set (ℝ × ℝ)) (hX : X = Set.prod (Icc 0 1) (Icc 0 1)) (room : (ℝ × ℝ) → ℝ) (hroom : room = fun (a,b) ↦ min (min a (1 - a)) (min b (1 - b))) : ((∀ f : (ℝ × ℝ) → ℝ, Continuous f → (∀ P ∈ X, ∫ x in (P.1 - room P)..(P.1 + room P), ∫ y in (P.2 - room P)..(P.2 + room P), f (x, y) = 0) → (∀ P ∈ X, f P = 0)) ↔ putnam_1977_a6_solution) := by

import Mathlib open RingHom Set Nat abbrev putnam_1977_a6_solution : Prop := sorry -- True /-- Let $X$ be the square $[0, 1] \times [0, 1]$, and let $f : X \to \mathbb{R}$ be continuous. If $\int_Y f(x, y) \, dx \, dy = 0$ for all squares $Y$ such that \begin{itemize} \item[(1)] $Y \subseteq X$, \item[(2)] $Y$ has sides parallel to those of $X$, \item[(3)] at least one of $Y$'s sides is contained in the boundary of $X$, \end{itemize} is it true that $f(x, y) = 0$ for all $x, y$? -/ theorem putnam_1977_a6 (X : Set (ℝ × ℝ)) (hX : X = Set.prod (Icc 0 1) (Icc 0 1)) (room : (ℝ × ℝ) → ℝ) (hroom : room = fun (a,b) ↦ min (min a (1 - a)) (min b (1 - b))) : ((∀ f : (ℝ × ℝ) → ℝ, Continuous f → (∀ P ∈ X, ∫ x in (P.1 - room P)..(P.1 + room P), ∫ y in (P.2 - room P)..(P.2 + room P), f (x, y) = 0) → (∀ P ∈ X, f P = 0)) ↔ putnam_1977_a6_solution) := sorry

Let $X$ be the square $[0, 1] \times [0, 1]$, and let $f : X \to \mathbb{R}$ be continuous. If $\int_Y f(x, y) \, dx \, dy = 0$ for all squares $Y$ such that \begin{itemize} \item[(1)] $Y \subseteq X$, \item[(2)] $Y$ has sides parallel to those of $X$, \item[(3)] at least one of $Y$'s sides is contained in the boundary of $X$, \end{itemize} is it true that $f(x, y) = 0$ for all $x, y$?

Prove that $f(x,y)$ must be identically zero.

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putnam_1963_b1

f0a91301-efc9-56eb-9fb2-1c2902d9c7f5

train

abbrev putnam_1963_b1_solution : ℤ := sorry -- 2 /-- For what integer $a$ does $x^2-x+a$ divide $x^{13}+x+90$? -/ theorem putnam_1963_b1 : ∀ a : ℤ, (X^2 - X + (C a)) ∣ (X ^ 13 + X + (C 90)) ↔ a = putnam_1963_b1_solution := sorry

import Mathlib open Topology Filter Polynomial -- 2 /-- For what integer $a$ does $x^2-x+a$ divide $x^{13}+x+90$? -/ theorem putnam_1963_b1 : ∀ a : ℤ, (X^2 - X + (C a)) ∣ (X ^ 13 + X + (C 90)) ↔ a = putnam_1963_b1_solution := by

import Mathlib open Topology Filter Polynomial abbrev putnam_1963_b1_solution : ℤ := sorry -- 2 /-- For what integer $a$ does $x^2-x+a$ divide $x^{13}+x+90$? -/ theorem putnam_1963_b1 : ∀ a : ℤ, (X^2 - X + (C a)) ∣ (X ^ 13 + X + (C 90)) ↔ a = putnam_1963_b1_solution := sorry

For what integer $a$ does $x^2-x+a$ divide $x^{13}+x+90$?

Show that $a=2$.

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