name
stringlengths 14
14
| lean4_statement
stringlengths 75
1.48k
β | informal_statement
stringlengths 47
898
β | informal_solution
stringlengths 16
303
β | tags
stringclasses 36
values | coq_statement
stringlengths 91
1.51k
β | isabelle_statement
stringlengths 160
2.65k
β |
|---|---|---|---|---|---|---|
putnam_1981_a1 | abbrev putnam_1981_a1_solution : β := sorry
-- 1/8
theorem putnam_1981_a1
(P : β β β β Prop := fun n k : β => 5^k β£ β m in Finset.Icc 1 n, (m^m : β€))
(E : β β β)
(hE : β n β Ici 1, P n (E n) β§ β k : β, P n k β k β€ E n)
: Tendsto (fun n : β => ((E n) : β)/n^2) atTop (π putnam_1981_a1_solution) :=
sorry | Let $E(n)$ be the greatest integer $k$ such that $5^k$ divides $1^1 2^2 3^3 \cdots n^n$. Find $\lim_{n \rightarrow \infty} \frac{E(n)}{n^2}$. | The limit equals $\frac{1}{8}$. | ['analysis', 'number_theory'] | Section putnam_1981_a1.
Require Import Nat Reals Coquelicot.Coquelicot. From mathcomp Require Import div.
Definition putnam_1981_a1_solution := 1 / 8.
Theorem putnam_1981_a1
(prod_n : (nat -> nat) -> nat -> nat := fix prod_n (m: nat -> nat) (n : nat) :=
match n with
| O => m 0%nat
| S n' => mul (m n') (prod_n m n')
end)
(P : nat -> nat -> Prop := fun n k => 5 ^ k %| prod_n (fun m => Nat.pow m m) (S n) = true)
(f : nat -> nat)
(hf : forall (n: nat), gt n 1 -> P n (f n) /\ forall (k: nat), P n k -> le k (f n))
: Lim_seq (fun n => INR (f n) / INR n ^ 2) = putnam_1981_a1_solution.
Proof. Admitted.
End putnam_1981_a1. | theory putnam_1981_a1 imports Complex_Main
begin
definition putnam_1981_a1_solution::real where "putnam_1981_a1_solution \<equiv> undefined"
(* 1/8 *)
theorem putnam_1981_a1:
fixes P::"nat\<Rightarrow>nat\<Rightarrow>bool" and E::"nat\<Rightarrow>nat"
defines "P \<equiv> \<lambda>n. \<lambda>k. 5^k dvd (\<Prod>m=1..n. m^m)"
and "E \<equiv> \<lambda>n. (GREATEST k. P n k)"
shows "(\<lambda>n. (E n) / n^2) \<longlonglongrightarrow> putnam_1981_a1_solution"
sorry
end |
putnam_1981_a3 | abbrev putnam_1981_a3_solution : Prop := sorry
-- False
theorem putnam_1981_a3
(f : β β β := fun t : β => Real.exp (-t) * β« y in (Ico 0 t), β« x in (Ico 0 t), (Real.exp x - Real.exp y) / (x - y))
: (β L : β, Tendsto f atTop (π L)) β putnam_1981_a3_solution :=
sorry | Does the limit $$lim_{t \rightarrow \infty}e^{-t}\int_{0}^{t}\int_{0}^{t}\frac{e^x - e^y}{x - y} dx dy$$exist? | The limit does not exist. | ['analysis'] | Section putnam_1981_a3.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1981_a3_solution := 14.
Theorem putnam_1981_a3:
Lim_seq (fun k => exp (-1*INR k) * (RInt (fun x => (RInt (fun y => (exp x - exp y) / (x - y)) 0 (INR k))) 0 (INR k))) = putnam_1981_a3_solution.
Proof. Admitted.
End putnam_1981_a3. | theory putnam_1981_a3 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1981_a3_solution::bool where "putnam_1981_a3_solution \<equiv> undefined"
(* False *)
theorem putnam_1981_a3:
fixes f::"real\<Rightarrow>real"
defines "f \<equiv> \<lambda>t::real. exp (-t) * interval_lebesgue_integral lebesgue 0 t (\<lambda>y::real.
interval_lebesgue_integral lebesgue 0 t (\<lambda>x::real. (exp x - exp y) / (x - y)))"
shows "(\<exists>L::real. (f \<longlonglongrightarrow> L)) \<longleftrightarrow> putnam_1981_a3_solution"
sorry
end |
putnam_1981_a5 | abbrev putnam_1981_a5_solution : Prop := sorry
-- True
theorem putnam_1981_a5
(Q : Polynomial β β Polynomial β := fun P : Polynomial β => (X^2 + 1)*P*(derivative P) + X*(P^2 + (derivative P)^2))
(n : Polynomial β β β := fun P : Polynomial β => {x β Ioi 1 | P.eval x = 0}.ncard)
: (β P : Polynomial β, {x : β | (Q P).eval x = 0}.ncard β₯ 2*(n P) - 1) β putnam_1981_a5_solution :=
sorry | Let $P(x)$ be a polynomial with real coefficients; let $$Q(x) = (x^2 + 1)P(x)P'(x) + x((P(x))^2 + (P'(x))^2).$$
Given that $P$ has $n$ distinct real roots all greater than $1$, prove or disprove that $Q$ must have at least $2n - 1$ distinct real roots. | $Q(x)$ must have at least $2n - 1$ distinct real roots. | ['algebra'] | null | theory putnam_1981_a5 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1981_a5_solution::bool where "putnam_1981_a5_solution \<equiv> undefined"
(* True *)
theorem putnam_1981_a5:
fixes Q::"(real poly) \<Rightarrow> (real poly)" and n::"(real poly) \<Rightarrow> real"
defines "Q \<equiv> \<lambda>P. [: 1, 0, 1 :] * P * (pderiv P) + [: 0, 1 :] * (P^2 + (pderiv P)^2)"
and "n \<equiv> \<lambda>P. card {x::real. x > 1 \<and> (poly P x) = 0}"
shows "(\<forall>P. card {x::real. poly (Q P) x = 0} \<ge> 2 * (n P) - 1) \<longleftrightarrow> putnam_1981_a5_solution"
sorry
end |
putnam_1981_b1 | abbrev putnam_1981_b1_solution : β := sorry
-- -1
theorem putnam_1981_b1
(f : β β β := fun n : β => (1/n^5) * β h in Finset.Icc 1 n, β k in Finset.Icc 1 n, (5*(h : β)^4 - 18*h^2*k^2 + 5*k^4))
: Tendsto f atTop (π putnam_1981_b1_solution) :=
sorry | Find the value of $$\lim_{n \rightarrow \infty} \frac{1}{n^5}\sum_{h=1}^{n}\sum_{k=1}^{n}(5h^4 - 18h^2k^2 + 5k^4).$$ | The limit equals $-1$. | ['analysis'] | Section putnam_1981_b1.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1981_b1_solution := -1.
Theorem putnam_1981_b1:
Lim_seq (fun n => 1/(pow (INR n) 5) * (sum_n (fun r => (sum_n (fun s => 5 * pow (INR r) 4 - 18 * pow (INR r) 2 * pow (INR s) 2 + 5 * pow (INR s) 4) n)) n)) = putnam_1981_b1_solution.
Proof. Admitted.
End putnam_1981_b1. | theory putnam_1981_b1 imports Complex_Main
begin
definition putnam_1981_b1_solution::real where "putnam_1981_b1_solution \<equiv> undefined"
(* -1 *)
theorem putnam_1981_b1:
fixes f::"nat\<Rightarrow>real"
defines "f \<equiv> \<lambda>n. 1/n^5 * (\<Sum>h=1..n. (\<Sum>k=1..n. (5*h^4 -18*h^2*k^2 + 5*k^4)))"
shows "f \<longlonglongrightarrow> putnam_1981_b1_solution"
sorry
end |
putnam_1981_b2 | abbrev putnam_1981_b2_solution : β := sorry
-- 12 - 8 * Real.sqrt 2
theorem putnam_1981_b2
(P : β Γ β Γ β β Prop := fun (r, s, t) => 1 β€ r β§ r β€ s β§ s β€ t β§ t β€ 4)
(f : β Γ β Γ β β β := fun (r, s, t) => (r - 1)^2 + (s/r - 1)^2 + (t/s - 1)^2 + (4/t - 1)^2)
: (β r : β, β s : β, β t : β, P (r, s, t) β§ f (r, s, t) = putnam_1981_b2_solution) β§
β r : β, β s : β, β t : β, P (r, s, t) β f (r, s, t) β₯ putnam_1981_b2_solution :=
sorry | Determine the minimum value attained by $$(r - 1)^2 + (\frac{s}{r} - 1)^2 + (\frac{t}{s} - 1)^2 + (\frac{4}{t} - 1)^2$$ across all choices of real $r$, $s$, and $t$ that satisfy $1 \le r \le s \le t \le 4$. | The minimum is $12 - 8\sqrt{2}$. | ['algebra'] | Section putnam_1981_b2.
Require Import Reals.
Open Scope R.
Definition putnam_1981_b2_solution := 12 - 8 * sqrt 2.
Theorem putnam_1981_b2:
let f (a b c: R) := pow (a-1) 2 + pow (b / a - 1) 2 + pow (c / b - 1) 2 + pow (4 / c - 1) 2 in
(forall (a b c: R), 1 <= a /\ a <= b /\ b <= c /\ c <= 4 -> putnam_1981_b2_solution <= f a b c) /\
(exists (a b c: R), 1 <= a /\ a <= b /\ b <= c /\ c <= 4 -> putnam_1981_b2_solution = f a b c).
Proof. Admitted.
End putnam_1981_b2. | theory putnam_1981_b2 imports Complex_Main
begin
definition putnam_1981_b2_solution::real where "putnam_1981_b2_solution \<equiv> undefined"
(* 12 - 8 * sqrt 2 *)
theorem putnam_1981_b2:
fixes P::"real\<Rightarrow>real\<Rightarrow>real\<Rightarrow>bool" and f::"real\<Rightarrow>real\<Rightarrow>real\<Rightarrow>real"
defines "P \<equiv> \<lambda>r. \<lambda>s. \<lambda>t. 1 \<le> r \<and> r \<le> s \<and> s \<le> t \<and> t \<le> 4"
and "f \<equiv> \<lambda>r. \<lambda>s. \<lambda>t. (r-1)^2 + (s/r - 1)^2 + (t/s - 1)^2 + (4/t - 1)^2"
shows "putnam_1981_b2_solution = (LEAST z::real. \<exists>r. \<exists>s. \<exists>t. P r s t \<and> z = f r s t)"
sorry
end |
putnam_1981_b3 | theorem putnam_1981_b3
(P : β β Prop := fun n : β => β p : β, (Nat.Prime p β§ p β£ n^2 + 3) β β k : β, (p : β€) β£ (k : β€)^2 + 3 β§ k^2 < n)
: β n : β, β m : β, (m : β€) > n β§ P m :=
sorry | Prove that, for infinitely many positive integers $n$, all primes $p$ that divide $n^2 + 3$ also divide $k^2 + 3$ for some integer $k$ such that $k^2 < n$. | null | ['number_theory'] | Section putnam_1981_b3.
Require Import Nat ZArith Znumtheory. From mathcomp Require Import div.
Open Scope nat_scope.
Theorem putnam_1981_b3:
~ exists (N: nat),
forall (n: nat),
(forall (p: nat), prime (Z.of_nat p) /\ p %| pow n 2 + 3 = true -> exists (m: nat), p %| pow m 2 + 3 = true /\ pow m 2 < n) ->
n < N.
Proof. Admitted.
End putnam_1981_b3. | theory putnam_1981_b3 imports Complex_Main "HOL-Computational_Algebra.Primes"
begin
theorem putnam_1981_b3:
fixes P::"nat\<Rightarrow>bool"
defines "P \<equiv> \<lambda>n. \<forall>p::nat. (prime p \<and> p dvd (n^2 + 3)) \<longrightarrow> (\<exists>k::nat. k^2 < n \<and> p dvd (k^2 + 3))"
shows "\<forall>n::nat. \<exists>m::nat. m > n \<and> P m"
sorry
end |
putnam_1981_b4 | abbrev putnam_1981_b4_solution : Prop := sorry
-- False
theorem putnam_1981_b4
(VAB : Set (Matrix (Fin 5) (Fin 7) β) β Prop)
(Vrank : Set (Matrix (Fin 5) (Fin 7) β) β β β Prop)
(hVAB : β V : Set (Matrix (Fin 5) (Fin 7) β), VAB V = (β A β V, β B β V, β r s : β, r β’ A + s β’ B β V))
(hVrank : β (V : Set (Matrix (Fin 5) (Fin 7) β)) (k : β), Vrank V k = β A β V, A.rank = k)
: (β V : Set (Matrix (Fin 5) (Fin 7) β), (VAB V β§ Vrank V 0 β§ Vrank V 1 β§ Vrank V 2 β§ Vrank V 4 β§ Vrank V 5) β Vrank V 3) β putnam_1981_b4_solution :=
sorry | Let $V$ be a set of $5$ by $7$ matrices, with real entries and with the property that $rA+sB \in V$ whenever $A,B \in V$ and $r$ and $s$ are scalars (i.e., real numbers). \emph{Prove or disprove} the following assertion: If $V$ contains matrices of ranks $0$, $1$, $2$, $4$, and $5$, then it also contains a matrix of rank $3$. [The rank of a nonzero matrix $M$ is the largest $k$ such that the entries of some $k$ rows and some $k$ columns form a $k$ by $k$ matrix with a nonzero determinant.] | Show that the assertion is false. | ['linear_algebra'] | null | theory putnam_1981_b4 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Analysis.Cartesian_Space"
begin
definition putnam_1981_b4_solution :: bool where "putnam_1981_b4_solution \<equiv> undefined"
(* False *)
theorem putnam_1981_b4:
fixes VAB :: "(real^7^5) set \<Rightarrow> bool"
and Vrank :: "(real^7^5) set \<Rightarrow> nat \<Rightarrow> bool"
defines "VAB \<equiv> \<lambda> V :: (real^7^5) set. \<forall> A \<in> V. \<forall> B \<in> V. \<forall> r s :: real. r *\<^sub>R A + s *\<^sub>R B \<in> V"
and "Vrank \<equiv> \<lambda> (V :: (real^7^5) set) (k :: nat). \<exists> A \<in> V. rank A = k"
shows "(\<forall> V :: (real^7^5) set. (VAB V \<and> Vrank V 0 \<and> Vrank V 1 \<and> Vrank V 2 \<and> Vrank V 4 \<and> Vrank V 5) \<longrightarrow> Vrank V 3) \<longleftrightarrow> putnam_1981_b4_solution"
sorry
end |
putnam_1981_b5 | abbrev putnam_1981_b5_solution : Prop := sorry
-- True
theorem putnam_1981_b5
(sumbits : List β β β€)
(B : β β β€)
(hsumbits : β bits : List β, sumbits bits = β i : Fin bits.length, (bits[i] : β€))
(hB : β n > 0, B n = sumbits (Nat.digits 2 n))
: (β q : β, Real.exp (β' n : Set.Ici 1, B n / ((n : β) * ((n : β) + 1))) = q) β putnam_1981_b5_solution :=
sorry | Let $B(n)$ be the number of ones in the base two expression for the positive integer $n$. For example, $B(6)=B(110_2)=2$ and $B(15)=B(1111_2)=4$. Determine whether or not $\exp \left(\sum_{n=1}^\infty \frac{B(n)}{n(n+1)}\right)$ is a rational number. Here $\exp(x)$ denotes $e^x$. | Show that the expression is a rational number. | ['analysis', 'algebra'] | Section putnam_1981_b5.
Require Import BinNums Nat NArith Coquelicot.Coquelicot.
Definition putnam_1981_b5_solution := True.
Theorem putnam_1981_b5:
let f :=
fix count_ones (n : positive) : nat :=
match n with
| xH => 1
| xO n' => count_ones n'
| xI n' => 1 + count_ones n'
end in
let k := Series (fun n => Rdefinitions.Rdiv (Raxioms.INR (f (Pos.of_nat n))) (Raxioms.INR (n + pow n 2))) in
exists (a b: nat), Rtrigo_def.exp k = Rdefinitions.Rdiv (Raxioms.INR a) (Raxioms.INR b) <-> putnam_1981_b5_solution.
Proof. Admitted.
End putnam_1981_b5. | theory putnam_1981_b5 imports
Complex_Main
begin
definition putnam_1981_b5_solution :: bool where "putnam_1981_b5_solution \<equiv> undefined"
(* True *)
theorem putnam_1981_b5:
fixes B :: "nat \<Rightarrow> nat"
defines "B \<equiv> \<lambda> n. card {k :: nat. odd (n div (2 ^ k))}"
shows "(\<exists> q :: rat. exp (\<Sum> n :: nat. B (n + 1) / ((n + 1) * (n + 2))) = q) \<longleftrightarrow> putnam_1981_b5_solution"
sorry
end |
putnam_2007_a1 | abbrev putnam_2007_a1_solution : Set β := sorry
-- {2/3, 3/2, (13 + Real.sqrt 601)/12, (13 - Real.sqrt 601)/12}
def reflect_tangent (f g : β β β) := ContDiff β 1 f β§ ContDiff β 1 g β§ (β x y : β, f x = y β§ g y = x β§ (deriv f) x = 1 / (deriv g) y)
theorem putnam_2007_a1
: β a : β, reflect_tangent (fun x => a * x^2 + a * x + 1/24) (fun y => a * y^2 + a * y + 1/24) β a β putnam_2007_a1_solution :=
sorry | Find all values of $\alpha$ for which the curves $y = \alpha*x^2 + \alpha*x + 1/24$ and $x = \alpha*y^2 + \alpha*y + 1/24$ are tangent to each other. | Show that the solution is the set \{2/3, 3/2, (13 + \sqrt{601})/12, (13 - \sqrt{601})/12}. | ['algebra', 'geometry'] | null | theory putnam_2007_a1 imports Complex_Main
"HOL-Analysis.Derivative"
begin
(* Note: Modified definition of tangent to handle this, but this is a bit of cheating - You would have to know that this works *)
definition putnam_2007_a1_solution :: "real set" where "putnam_2007_a1_solution \<equiv> undefined"
(* {2/3, 3/2, (13 + sqrt 601)/12, (13 - sqrt 601)/12} *)
definition reflect_tangent :: "(real \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool" where "reflect_tangent \<equiv> (\<lambda>f g::real\<Rightarrow>real. f C1_differentiable_on UNIV \<and> g C1_differentiable_on UNIV \<and> (\<exists>x y::real. f x = y \<and> g y = x \<and> deriv f x = 1 / deriv g y))"
theorem putnam_2007_a1:
shows "\<forall>a::real. (reflect_tangent (\<lambda>x::real. a*x^2 + a*x + 1/24) (\<lambda>y::real. a*y^2 + a*y + 1/24) \<longleftrightarrow> a \<in> putnam_2007_a1_solution)"
sorry
end
|
putnam_2007_a2 | abbrev putnam_2007_a2_solution : ENNReal := sorry
-- 4
theorem putnam_2007_a2
(Sinterpos : Set (Fin 2 β β) β Prop)
(Sinterneg : Set (Fin 2 β β) β Prop)
(Sconv : Set (Fin 2 β β) β Prop)
(hSinterpos : β S : Set (Fin 2 β β), Sinterpos S = ((β p β S, p 0 > 0 β§ p 1 > 0 β§ p 0 * p 1 = 1) β§ (β p β S, p 0 < 0 β§ p 1 < 0 β§ p 0 * p 1 = 1)))
(hSinterneg : β S : Set (Fin 2 β β), Sinterneg S = ((β p β S, p 0 < 0 β§ p 1 > 0 β§ p 0 * p 1 = -1) β§ (β p β S, p 0 > 0 β§ p 1 < 0 β§ p 0 * p 1 = -1)))
(hSconv : β S : Set (Fin 2 β β), Sconv S = (Convex β S β§ Sinterpos S β§ Sinterneg S))
: (β S : Set (Fin 2 β β), Sconv S β§ MeasureTheory.volume S = putnam_2007_a2_solution) β§ (β S : Set (Fin 2 β β), Sconv S β MeasureTheory.volume S β₯ putnam_2007_a2_solution) :=
sorry | Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $xy=1$ and both branches of the hyperbola $xy=-1$. (A set $S$ in the plane is called \emph{convex} if for any two points in $S$ the line segment connecting them is contained in $S$.) | Show that the minimum is $4$. | ['geometry'] | null | theory putnam_2007_a2 imports Complex_Main
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_2007_a2_solution :: ennreal where "putnam_2007_a2_solution \<equiv> undefined"
(* 4 *)
theorem putnam_2007_a2:
fixes Sinterpos :: "((real^2) set) \<Rightarrow> bool"
and Sinterneg :: "((real^2) set) \<Rightarrow> bool"
and Sconv :: "((real^2) set) \<Rightarrow> bool"
assumes hSinterpos: "\<forall>S::(real^2) set. Sinterpos S = ((\<exists>p\<in>S. p$1 > 0 \<and> p$2 > 0 \<and> p$1*p$2 = 1) \<and> (\<exists>p\<in>S. p$1 < 0 \<and> p$2 < 0 \<and> p$1*p$2 = 1))"
assumes hSinterneg: "\<forall>S::(real^2) set. Sinterneg S = ((\<exists>p\<in>S. p$1 < 0 \<and> p$2 > 0 \<and> p$1*p$2 = -1) \<and> (\<exists>p\<in>S. p$1 > 0 \<and> p$2 < 0 \<and> p$1*p$2 = -1))"
assumes hSconv: "\<forall>S::(real^2) set. Sconv S = (convex S \<and> Sinterpos S \<and> Sinterneg S)"
shows "(LEAST area::ennreal. (\<exists>S::(real^2) set. Sconv S \<and> emeasure lebesgue S = area)) = putnam_2007_a2_solution"
sorry
end
|
putnam_2007_a4 | abbrev putnam_2007_a4_solution : Set (Polynomial β) := sorry
-- {f : Polynomial β | β d : β, β c β₯ (1 : β€) - d, β n : β, f.eval n = (1 / 9) * ((10 ^ c) * (9 * n + 1) ^ d - 1)}
theorem putnam_2007_a4
(S : Set (Polynomial β))
(repunit : β β Prop := fun x β¦ x > 0 β§ x = floor x β§ β d β (digits 10 (floor x)), d = 1)
(hS : β f : Polynomial β, f β S β (β n : β, repunit n β repunit (f.eval n)))
: (S = putnam_2007_a4_solution) :=
sorry | A \emph{repunit} is a positive integer whose digits in base 10 are all ones. Find all polynomials $f$ with real coefficients such that if $n$ is a repunit, then so is $f(n)$. | Show that the desired polynomials $f$ are those of the form
\[
f(n) = \frac{1}{9}(10^c (9n+1)^d - 1)
\]
for integers $d \geq 0$ and $c \geq 1-d$. | ['analysis', 'algebra', 'number_theory'] | Section putnam_2007_a4.
Require Import Reals Zpower Coquelicot.Coquelicot.
Definition putnam_2007_a4_solution (f: R -> R) := exists (c d: Z), Z.ge d 0 /\ Z.ge c (1 - d) /\ f = (fun n => (IZR (Zpower 10 c) * IZR (Zpower (9 * floor n + 1) d) - 1) / 9).
Theorem putnam_2007_a4:
let repunit (n: R) := n = IZR (floor n) /\ n > 0 /\ exists (m: nat), n = sum_n (fun i => 10 ^ i) m in
forall (c: nat -> R) (n: nat),
let f (x: R) := sum_n (fun i => c i * x ^ i) (n + 1) in
repunit (INR n) -> repunit (f (INR n)) <->
putnam_2007_a4_solution f.
Proof. Admitted.
End putnam_2007_a4. | theory putnam_2007_a4 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_2007_a4_solution :: "(real poly) set" where "putnam_2007_a4_solution \<equiv> undefined"
(* {f :: real poly. \<exists> d :: nat. \<exists> c :: int. c \<ge> 1 - d \<and> (\<forall> n :: real. poly f n = (1 / 9) * ((10 powr c) * (9 * n + 1) ^ d - 1))} *)
theorem putnam_2007_a4:
fixes repunit :: "real \<Rightarrow> bool"
defines "repunit \<equiv> \<lambda> x :: real. x > 0 \<and> x = real_of_int (floor x) \<and> (\<forall> i :: nat. (10 ^ i \<le> floor x) \<longrightarrow> ((floor x) div (10 ^ i)) mod 10 = 1)"
shows "{f :: real poly. \<forall> n :: real. repunit n \<longrightarrow> repunit (poly f n)} = putnam_2007_a4_solution"
sorry
end |
putnam_2007_a5 | theorem putnam_2007_a5
(G : Type*) [Group G] [Fintype G]
(p : β)
(n : β)
(hp : Nat.Prime p)
(hn : n = {g : G | orderOf g = p}.encard)
: n = 0 β¨ p β£ (n + 1) :=
sorry | null | null | [] | null | theory putnam_2007_a5 imports Complex_Main
"HOL-Algebra.Multiplicative_Group"
"HOL-Computational_Algebra.Primes"
begin
theorem putnam_2007_a5:
fixes G (structure)
and p :: nat
and n :: nat
assumes hG: "group G \<and> finite (carrier G)"
and hp: "prime p"
and hn: "n \<equiv> card {g::'a. g \<in> carrier G \<and> (group.ord G) g = p}"
shows "n = 0 \<or> p dvd (n+1)"
sorry
end
|
putnam_2007_b1 | theorem putnam_2007_b1
(f : Polynomial β€)
(hf : β n : β, f.coeff n β₯ 0)
(hfnconst : β n : β, n > 0 β§ f.coeff n > 0)
(n : β€)
(hn : n > 0)
: f.eval n β£ f.eval (f.eval n + 1) β n = 1 :=
sorry | Let $f$ be a nonconstant polynomial with positive integer coefficients. Prove that if $n$ is a positive integer, then $f(n)$ divides $f(f(n) + 1)$ if and only if $n = 1$ | null | ['algebra'] | Section putnam_2007_b1.
Require Import Nat Reals Coquelicot.Coquelicot. From mathcomp Require Import div.
Theorem putnam_2007_b1:
forall (c: nat -> nat) (n: nat),
gt n 0 /\ forall (x: nat), gt (c x) 0 ->
let f (x: nat) := sum_n (fun i => INR (mul (c i) (x ^ i))) (n + 1) in
Z.to_nat (floor (f n)) %| Z.to_nat (floor (f (Z.to_nat (floor (f n + 1))))) = true <-> n = 1%nat.
Proof. Admitted.
End putnam_2007_b1. | theory putnam_2007_b1 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
(* Note: Since the coefficient is defined for all natural numbers, the condition here is modified to nonnegative which is what makes sense *)
theorem putnam_2007_b1:
fixes f :: "int poly"
and n :: int
assumes hf: "\<forall>n'::nat. poly.coeff f n' \<ge> 0"
and hfnconst: "\<exists>n'::nat. n' > 0 \<and> poly.coeff f n' > 0"
and hn: "n > 0"
shows "((poly f n) dvd (poly f (poly f n + 1))) \<longleftrightarrow> n = 1"
sorry
end
|
putnam_2007_b2 | theorem putnam_2007_b2
(f : β β β)
(hf : ContDiffOn β 1 f (Icc 0 1))
(hfint : β« x in (0)..1, f x = 0)
(max : β)
(heqmax : β x β Icc (0 : β) 1, |deriv f x| = max)
(hmaxub : β x β Icc (0 : β) 1, |deriv f x| β€ max)
: (β Ξ± β (Ioo (0 : β) 1), |β« x in (0)..Ξ±, f x| β€ (1 / 8) * max) :=
sorry | Suppose that $f: [0,1] \to \mathbb{R}$ has a continuous derivative and that $\int_0^1 f(x)\,dx = 0$. Prove that for every $\alpha \in (0,1)$,
\[
\left| \int_0^\alpha f(x)\,dx \right| \leq \frac{1}{8} \max_{0 \leq x
\leq 1} |f'(x)|.
\] | null | ['analysis'] | Section putnam_2007_b2.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2007_b2:
forall (f: R -> R), (forall (x: R), 0 <= x <= 1 /\ continuity_pt f x /\ ex_derive f x /\ RInt f 0 1 = 0) ->
exists (max_f_prime_abs: R), (forall (x: R), 0 <= x <= 1 -> max_f_prime_abs >= Rabs (Derive f x)) /\ (exists (x: R), 0 <= x <= 1 -> max_f_prime_abs = Rabs (Derive f x)) /\
forall (a: R), 0 < a < 1 -> Rabs (RInt f 0 a) = max_f_prime_abs / 8.
Proof. Admitted.
End putnam_2007_b2. | theory putnam_2007_b2 imports
Complex_Main
"HOL-Analysis.Derivative"
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
theorem putnam_2007_b2:
fixes f :: "real \<Rightarrow> real"
assumes hf: "f C1_differentiable_on {0..1}"
and hfint: "set_lebesgue_integral lebesgue {0..1} f = 0"
shows "\<forall> \<alpha> \<in> {0<..<1}. \<bar>set_lebesgue_integral lebesgue {0..\<alpha>} f\<bar> \<le> (1 / 8) * (GREATEST y. \<exists> x \<in> {0..1}. \<bar>deriv f x\<bar> = y)"
sorry
end |
putnam_2007_b3 | abbrev putnam_2007_b3_solution : β := sorry
-- (2 ^ 2006 / Real.sqrt 5) * (((1 + Real.sqrt 5) / 2) ^ 3997 - ((1 + Real.sqrt 5) / 2) ^ (-3997 : β€))
theorem putnam_2007_b3
(x : β β β)
(hx0 : x 0 = 1)
(hx : β n : β, x (n + 1) = 3 * (x n) + β(x n) * Real.sqrt 5β)
: (x 2007 = putnam_2007_b3_solution) :=
sorry | Let $x_0 = 1$ and for $n \geq 0$, let $x_{n+1} = 3x_n + \lfloor x_n \sqrt{5} \rfloor$. In particular, $x_1 = 5$, $x_2 = 26$, $x_3 = 136$, $x_4 = 712$. Find a closed-form expression for $x_{2007}$. ($\lfloor a \rfloor$ means the largest integer $\leq a$.) | Prove that $x_{2007} = \frac{2^{2006}}{\sqrt{5}}(\alpha^{3997}-\alpha^{-3997})$, where $\alpha = \frac{1+\sqrt{5}}{2}$. | ['analysis'] | Section putnam_2007_b3.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2007_b3_solution := let a := (1 + sqrt 5) / 2 in 2 ^ 2006 / sqrt 5 * (a ^ 3997 - Rpower a (-3997)).
Theorem putnam_2007_b3:
let fix x (n: nat) :=
match n with
| O => 1
| S n' => 3 * x n' + IZR (floor (x n' * sqrt (INR n)))
end in
x 2007%nat = putnam_2007_b3_solution.
Proof. Admitted.
End putnam_2007_b3. | theory putnam_2007_b3 imports
Complex_Main
begin
definition putnam_2007_b3_solution :: real where "putnam_2007_b3_solution \<equiv> undefined"
(* (2 ^ 2006 / sqrt 5) * ((((1 + sqrt 5) / 2) powr 3997) - (((1 + sqrt 5) / 2) powr -3997)) *)
theorem putnam_2007_b3:
fixes x :: "nat \<Rightarrow> real"
assumes hx0: "x 0 = 1"
and hx: "\<forall> n :: nat. x (n + 1) = 3 * (x n) + floor (x n + sqrt 5)"
shows "x 2007 = putnam_2007_b3_solution"
sorry
end |
putnam_2007_b4 | abbrev putnam_2007_b4_solution : β β β := sorry
-- fun n β¦ 2 ^ (n + 1)
theorem putnam_2007_b4
(n : β)
(npos : n > 0)
: ({(P, Q) : (Polynomial β) Γ (Polynomial β) | P ^ 2 + Q ^ 2 = Polynomial.X ^ (2 * n) + 1 β§ P.degree > Q.degree}.ncard = putnam_2007_b4_solution n) :=
sorry | Let $n$ be a positive integer. Find the number of pairs $P, Q$ of polynomials with real coefficients such that
\[
(P(X))^2 + (Q(X))^2 = X^{2n} + 1
\]
and $\deg P > \deg Q$. | Show that the number of pairs is $2^{n+1}$. | ['algebra'] | null | theory putnam_2007_b4 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_2007_b4_solution :: "nat \<Rightarrow> nat" where "putnam_2007_b4_solution \<equiv> undefined"
(* \<lambda> n. 2 ^ (n + 1) *)
theorem putnam_2007_b4:
fixes n :: nat
assumes npos: "n > 0"
shows "card {(P :: real poly, Q :: real poly). P ^ 2 + Q ^ 2 = (monom 1 (2 * n)) + 1 \<and> degree P > degree Q} = putnam_2007_b4_solution n"
sorry
end |
putnam_2007_b5 | theorem putnam_2007_b5
(k : β)
(kpos : k > 0)
: (β P : Finset.range k β Polynomial β, β n : β€, β(n : β) / kβ ^ k = β i : Finset.range k, (P i).eval (n : β) * β(n : β) / kβ ^ (i : β)) :=
sorry | Let $k$ be a positive integer. Prove that there exist polynomials $P_0(n), P_1(n), \dots, P_{k-1}(n)$ (which may depend on $k$) such that for any integer $n$,
\[
\left\lfloor \frac{n}{k} \right\rfloor^k = P_0(n) + P_1(n) \left\lfloor
\frac{n}{k} \right\rfloor + \cdots + P_{k-1}(n) \left\lfloor \frac{n}{k}
\right\rfloor^{k-1}.
\]
($\lfloor a \rfloor$ means the largest integer $\leq a$.) | null | ['algebra'] | null | theory putnam_2007_b5 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_2007_b5:
fixes k :: nat
assumes kpos: "k > 0"
shows "\<exists> P :: real poly list. length P = k \<and> (\<forall> n :: int. \<lfloor>n / k\<rfloor> ^ k = (\<Sum> i = 0..(k - 1). poly (P!i) n * \<lfloor>n / k\<rfloor> ^ i))"
sorry
end |
putnam_1980_a2 | abbrev putnam_1980_a2_solution : β β β β β := sorry
-- (fun r s : β => (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2))
theorem putnam_1980_a2
(r s : β)
(abcdlcm : β β β β β β β β Prop)
(rspos : r > 0 β§ s > 0)
(habcdlcm : β a b c d : β, abcdlcm a b c d = (a > 0 β§ b > 0 β§ c > 0 β§ d > 0 β§ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) c) β§ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a b) d) β§ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm a c) d) β§ (3 ^ r * 7 ^ s = Nat.lcm (Nat.lcm b c) d)))
: {(a, b, c, d) : β Γ β Γ β Γ β | abcdlcm a b c d}.encard = putnam_1980_a2_solution r s :=
sorry | Let $r$ and $s$ be positive integers. Derive a formula for the number of ordered quadruples $(a,b,c,d)$ of positive integers such that $3^r \cdot 7^s=\text{lcm}[a,b,c]=\text{lcm}[a,b,d]=\text{lcm}[a,c,d]=\text{lcm}[b,c,d]$. The answer should be a function of $r$ and $s$. (Note that $\text{lcm}[x,y,z]$ denotes the least common multiple of $x,y,z$.) | Show that the number is $(1+4r+6r^2)(1+4s+6s^2)$. | ['number_theory'] | Section putnam_1980_a2.
Require Import Nat Ensembles Finite_sets.
Definition putnam_1980_a2_solution (m n: nat) := (6*m*m + 3*m + 1) * (6*n*n + 3*n + 1).
Theorem putnam_1980_a2:
let gcd3 (a b c: nat) := gcd (gcd a b) c in
exists (f: Ensemble (nat*nat)),
forall (m n: nat) (a b c d: nat),
(f (m, n) <-> gcd3 a b c = 3 ^ m * 7 ^ n \/ gcd3 a b d = 3 ^ m * 7 ^ n \/ gcd3 a c d = 3 ^ m * 7 ^ n \/ gcd3 b c d = 3 ^ m * 7 ^ n) ->
cardinal (nat*nat) f (putnam_1980_a2_solution m n).
Proof. Admitted.
End putnam_1980_a2. | theory putnam_1980_a2 imports
Complex_Main
begin
definition putnam_1980_a2_solution :: "nat \<Rightarrow> nat \<Rightarrow> nat" where "putnam_1980_a2_solution \<equiv> undefined"
(* \<lambda> r s :: nat. (1 + 4 * r + 6 * r ^ 2) * (1 + 4 * s + 6 * s ^ 2) *)
theorem putnam_1980_a2:
fixes r s :: nat
and abcdlcm :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool"
defines "abcdlcm \<equiv> \<lambda> a b c d :: nat. a > 0 \<and> b > 0 \<and> c > 0 \<and> d > 0 \<and> 3 ^ r * 7 ^ s = lcm (lcm a b) c \<and> 3 ^ r * 7 ^ s = lcm (lcm a b) d \<and> 3 ^ r * 7 ^ s = lcm (lcm a c) d \<and> 3 ^ r * 7 ^ s = lcm (lcm b c) d"
assumes rspos: "r > 0 \<and> s > 0"
shows "card {(a, b, c, d). abcdlcm a b c d} = putnam_1980_a2_solution r s"
sorry
end |
putnam_1980_a3 | abbrev putnam_1980_a3_solution : β := sorry
-- Real.pi / 4
theorem putnam_1980_a3
: β« x in Set.Ioo 0 (Real.pi / 2), 1 / (1 + (Real.tan x) ^ (Real.sqrt 2)) = putnam_1980_a3_solution :=
sorry | Evaluate $\int_0^{\pi/2}\frac{dx}{1+(\tan x)^{\sqrt{2}}}$. | Show that the integral is $\pi/4$. | ['analysis'] | Section putnam_1980_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1980_a3_solution := PI/4.
Theorem putnam_1980_a3:
let f (x: R) := 1/(1 + Rpower (tan x) (sqrt 2)) in
RInt f 0 PI/2 = putnam_1980_a3_solution.
Proof. Admitted.
End putnam_1980_a3. | theory putnam_1980_a3 imports
Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1980_a3_solution :: real where "putnam_1980_a3_solution \<equiv> undefined"
(* pi / 4 *)
theorem putnam_1980_a3:
shows "interval_lebesgue_integral lebesgue 0 (pi / 2) (\<lambda> x :: real. 1 / (1 + (tan x) powr (sqrt 2))) = putnam_1980_a3_solution"
sorry
end |
putnam_1980_a4 | theorem putnam_1980_a4
(abcvals : β€ β β€ β β€ β Prop)
(habcvals : β a b c : β€, abcvals a b c = (Β¬(a = 0 β§ b = 0 β§ c = 0) β§ |a| < 1000000 β§ |b| < 1000000 β§ |c| < 1000000))
: (β a b c : β€, abcvals a b c β§ |a + b * Real.sqrt 2 + c * Real.sqrt 3| < 10 ^ (-(11 : β))) β§ (β a b c : β€, abcvals a b c β |a + b * Real.sqrt 2 + c * Real.sqrt 3| > 10 ^ (-(21 : β))) :=
sorry | \begin{enumerate}
\item[(a)] Prove that there exist integers $a,b,c$, not all zero and each of absolute value less than one million, such that $|a+b\sqrt{2}+c\sqrt{3}|<10^{-11}$.
\item[(b)] Let $a,b,c$ be integers, not all zero and each of absolute value less than one million. Prove that $|a+b\sqrt{2}+c\sqrt{3}|>10^{-21}$.
\end{enumerate} | null | ['algebra'] | Section putnam_1980_a4.
Require Import Reals BinInt.
Open Scope Z.
Theorem putnam_1980_a4:
(forall (a b c: Z), (~ (a = 0 \/ b = 0 /\ c = 0) /\ Z.abs a < 10^6 /\ Z.abs b < 10^6 /\ Z.abs c < 10^6) ->
Rgt (Rabs (Rplus (Rplus (IZR a) (Rmult (IZR b) (sqrt 2))) (Rmult (IZR c) (sqrt 3)))) (Rpower 10 (-21)) )
/\
(exists (a b c: Z),
Rlt (Rabs (Rplus (IZR a) (Rplus (Rmult (IZR b) (sqrt 2)) (Rmult (IZR c) (sqrt 3))))) (Rpower 10 (-11)) ).
Proof. Admitted.
End putnam_1980_a4. | theory putnam_1980_a4 imports
Complex_Main
begin
theorem putnam_1980_a4:
fixes abcvals :: "int \<Rightarrow> int \<Rightarrow> int \<Rightarrow> bool"
defines "abcvals \<equiv> \<lambda> a b c :: int. \<not>(a = 0 \<and> b = 0 \<and> c = 0) \<and> \<bar>a\<bar> < 1000000 \<and> \<bar>b\<bar> < 1000000 \<and> \<bar>c\<bar> < 1000000"
shows "(\<exists> a b c :: int. abcvals a b c \<and> \<bar>a + b * sqrt 2 + c * sqrt 3\<bar> < 10 powi -11) \<and> (\<forall> a b c :: int. abcvals a b c \<longrightarrow> \<bar>a + b * sqrt 2 + c * sqrt 3\<bar> > 10 powi -21)"
sorry
end
|
putnam_1980_a5 | theorem putnam_1980_a5
(P : Polynomial β)
(xeqs : β β Prop)
(Pnonconst : P.degree > 0)
(hxeqs : β x : β, xeqs x = (0 = (β« t in (0)..x, P.eval t * Real.sin t) β§ 0 = (β« t in (0)..x, P.eval t * Real.cos t)))
: {x : β | xeqs x}.Finite :=
sorry | Let $P(t)$ be a nonconstant polynomial with real coefficients. Prove that the system of simultaneous equations $0=\int_0^xP(t)\sin t\,dt=\int_0^xP(t)\cos t\,dt$ has only finitely many real solutions $x$. | null | ['analysis'] | Section putnam_1980_a5.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_1980_a5
(n : nat)
(npos : gt n 0)
(coeff : nat -> R)
(hcoeff : coeff n <> 0)
(p : R -> R := fun x => sum_n (fun i => coeff i * x ^ i) (S n))
(h1 : nat -> Prop := fun a => RInt (fun x => p x * sin x) 0 (INR a) = 0)
(h2 : nat -> Prop := fun a => RInt (fun x => p x * cos x) 0 (INR a) = 0)
: exists (m: nat), forall (b: nat), h1 b /\ h2 b -> lt b m.
Proof. Admitted.
End putnam_1980_a5. | theory putnam_1980_a5 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
"HOL-Analysis.Interval_Integral"
begin
theorem putnam_1980_a5:
fixes P :: "real poly"
and xeqs :: "real \<Rightarrow> bool"
defines "xeqs \<equiv> \<lambda> x :: real. 0 = interval_lebesgue_integral lebesgue 0 x (\<lambda> t. poly P t * sin t) \<and> 0 = interval_lebesgue_integral lebesgue 0 x (\<lambda> t. poly P t * cos t)"
assumes Pnonconst: "degree P > 0"
shows "finite {x :: real. xeqs x}"
sorry
end |
putnam_1980_a6 | abbrev putnam_1980_a6_solution : β := sorry
-- 1 / Real.exp 1
theorem putnam_1980_a6
(C : Set (β β β))
(uleint : β β Prop)
(hC : C = {f : β β β | ContDiffOn β 1 f (Set.Icc 0 1) β§ f 0 = 0 β§ f 1 = 1})
(huleint : β u : β, uleint u = β f β C, u β€ (β« x in Set.Ioo 0 1, |deriv f x - f x|))
: uleint putnam_1980_a6_solution β§ (β u : β, uleint u β u β€ putnam_1980_a6_solution) :=
sorry | Let $C$ be the class of all real valued continuously differentiable functions $f$ on the interval $0 \leq x \leq 1$ with $f(0)=0$ and $f(1)=1$. Determine the largest real number $u$ such that $u \leq \int_0^1|f'(x)-f(x)|\,dx$ for all $f$ in $C$. | Show that $u=1/e$. | ['analysis'] | null | theory putnam_1980_a6 imports
Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1980_a6_solution :: real where "putnam_1980_a6_solution \<equiv> undefined"
(* 1 / exp 1 *)
theorem putnam_1980_a6:
fixes C :: "(real \<Rightarrow> real) set"
and uleint :: "real \<Rightarrow> bool"
defines "C \<equiv> {f :: real \<Rightarrow> real. f C1_differentiable_on {0..1} \<and> f 0 = 0 \<and> f 1 = 1}"
and "uleint \<equiv> \<lambda> u :: real. \<forall> f \<in> C. u \<le> interval_lebesgue_integral lebesgue 0 1 (\<lambda> x. \<bar>deriv f x - f x\<bar>)"
shows "(GREATEST u. uleint u) = putnam_1980_a6_solution"
sorry
end |
putnam_1980_b1 | abbrev putnam_1980_b1_solution : Set β := sorry
-- {c : β | c β₯ 1 / 2}
theorem putnam_1980_b1
(c : β)
: (β x : β, (Real.exp x + Real.exp (-x)) / 2 β€ Real.exp (c * x ^ 2)) β c β putnam_1980_b1_solution :=
sorry | For which real numbers $c$ is $(e^x+e^{-x})/2 \leq e^{cx^2}$ for all real $x$? | Show that the inequality holds if and only if $c \geq 1/2$. | ['analysis'] | Section putnam_1980_b1.
Require Import Reals Rtrigo_def.
Open Scope R.
Definition putnam_1980_b1_solution (k: R) := k >= 1/2.
Theorem putnam_1980_b1:
forall (k: R), forall (x: R), cosh x <= exp (k*x*x) <-> putnam_1980_b1_solution k.
Proof. Admitted.
End putnam_1980_b1. | theory putnam_1980_b1 imports
Complex_Main
begin
definition putnam_1980_b1_solution :: "real set" where "putnam_1980_b1_solution \<equiv> undefined"
(* {c :: real. c \<ge> 1 / 2} *)
theorem putnam_1980_b1:
fixes c :: real
shows "(\<forall> x :: real. (exp x + exp (-x)) / 2 \<le> exp (c * x ^ 2)) \<longleftrightarrow> c \<in> putnam_1980_b1_solution"
sorry
end |
putnam_1980_b3 | abbrev putnam_1980_b3_solution : Set β := sorry
-- {a : β | a β₯ 3}
theorem putnam_1980_b3
(a : β)
(u : β β β)
(hu : u 0 = a β§ (β n : β, u (n + 1) = 2 * u n - n ^ 2))
: (β n : β, u n > 0) β a β putnam_1980_b3_solution :=
sorry | For which real numbers $a$ does the sequence defined by the initial condition $u_0=a$ and the recursion $u_{n+1}=2u_n-n^2$ have $u_n>0$ for all $n \geq 0$? (Express the answer in the simplest form.) | Show that $u_n>0$ for all $n \geq 0$ if and only if $a \geq 3$. | ['algebra'] | Section putnam_1980_b3.
Require Import Reals.
Open Scope R.
Definition putnam_1980_b3_solution (b: R) := b >= 3.
Theorem putnam_1980_b3:
forall (b: R),
let A :=
fix a (n: nat) : R :=
match n with
| O => b
| S n' => 2 * a n' - INR (n' * n')
end in
forall (n: nat), A n > 0 <-> putnam_1980_b3_solution b.
Proof. Admitted.
End putnam_1980_b3. | theory putnam_1980_b3 imports
Complex_Main
begin
definition putnam_1980_b3_solution :: "real set" where "putnam_1980_b3_solution \<equiv> undefined"
(* {a :: real. a \<ge> 3} *)
theorem putnam_1980_a3:
fixes a :: real
and u :: "nat \<Rightarrow> real"
assumes hu: "u 0 = a \<and> (\<forall> n :: nat. u (n + 1) = 2 * (u n) - n ^ 2)"
shows "(\<forall> n :: nat. u n > 0) \<longleftrightarrow> a \<in> putnam_1980_b3_solution"
sorry
end |
putnam_1980_b4 | theorem putnam_1980_b4
{T : Type}
(X : Finset T)
(A : Fin 1066 β Finset T)
(hX : X.card β₯ 10)
(hA : β i : Fin 1066, A i β X β§ (A i).card > ((1 : β)/2) * X.card)
: β Y : Finset T, Y β X β§ Y.card = 10 β§ β i : Fin 1066, β y β Y, y β A i :=
sorry | Let $X$ be a finite set with at least $10$ elements; for each $i \in \{0, 1, ..., 1065\}$, let $A_i \subseteq X$ satisfy $|A_i| > \frac{1}{2}|X|$. Prove that there exist $10$ elements $x_1, x_2, \dots, x_{10} \in X$ such that each $A_i$ contains at least one of $x_1, x_2, \dots, x_{10}$. | null | ['set_theory', 'combinatorics'] | null | theory putnam_1980_b4 imports
Complex_Main
begin
theorem putnam_1980_b4:
fixes X :: "'t set"
and A :: "nat \<Rightarrow> 't set"
assumes Xfin: "finite X"
and hX: "card X \<ge> 10"
and hA: "\<forall> i \<in> {1 .. 1066}. A i \<subseteq> X \<and> card (A i) > (1 / 2) * card X"
shows "\<exists> Y :: 't set. Y \<subseteq> X \<and> card Y = 10 \<and> (\<forall> i \<in> {1 .. 1066}. \<exists> y \<in> Y. y \<in> A i)"
sorry
end |
putnam_1980_b5 | abbrev putnam_1980_b5_solution : β β Prop := sorry
-- fun t : β => 1 β₯ t
theorem putnam_1980_b5
(T : Set β := Icc 0 1)
(P : β β (β β β) β Prop := fun t : β => fun f : β β β => f 1 - 2*f (2/3) + f (1/3) β₯ t*(f (2/3) - 2*f (1/3) + f 0))
(Convex : (β β β) β Prop := fun f : β β β => β u β T, β v β T, β s β T, f (s*u + (1 - s)*v) β€ s*(f u) + (1 - s)*(f v))
(S : β β Set (β β β) := fun t : β => {f : β β β | (β x β T, f x β₯ 0) β§ StrictMonoOn f T β§ Convex f β§ ContinuousOn f T β§ P t f})
: β t : β, t β₯ 0 β ((β f β S t, β g β S t, f * g β S t) β putnam_1980_b5_solution t) :=
sorry | A function $f$ is convex on $[0, 1]$ if and only if $$f(su + (1-s)v) \le sf(u) + (1 - s)f(v)$$ for all $s \in [0, 1]$.
Let $S_t$ denote the set of all nonnegative increasing convex continuous functions $f : [0, 1] \rightarrow \mathbb{R}$ such that $$f(1) - 2f\left(\frac{2}{3}\right) + f\left(\frac{1}{3}\right) \ge t\left(f\left(\frac{2}{3}\right) - 2f\left(\frac{1}{3}\right) + f(0)\right).$$
For which real numbers $t \ge 0$ is $S_t$ closed under multiplication? | $S_t$ is closed under multiplication if and only if $1 \ge t$. | ['analysis', 'algebra'] | null | theory putnam_1980_b5 imports
Complex_Main
begin
definition putnam_1980_b5_solution :: "real \<Rightarrow> bool" where "putnam_1980_b5_solution \<equiv> undefined"
(* \<lambda> t :: real. 1 \<ge> t *)
theorem putnam_1980_b5:
fixes T :: "real set"
and P :: "real \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool"
and Convex :: "(real \<Rightarrow> real) \<Rightarrow> bool"
and S :: "real \<Rightarrow> (real \<Rightarrow> real) set"
defines "T \<equiv> {0..1}"
and "P \<equiv> \<lambda> (t :: real) (f :: real \<Rightarrow> real). f 1 - 2 * f (2 / 3) + f (1 / 3) \<ge> t * (f (2 / 3) - 2 * f (1 / 3) + f 0)"
and "Convex \<equiv> \<lambda> f :: real \<Rightarrow> real. \<forall> u \<in> T. \<forall> v \<in> T. \<forall> s \<in> T. f (s * u + (1 - s) * v) \<le> s * (f u) + (1 - s) * (f v)"
and "S \<equiv> \<lambda> t :: real. {f :: real \<Rightarrow> real. (\<forall> x \<in> T. f x \<ge> 0) \<and> strict_mono_on T f \<and> Convex f \<and> continuous_on T f \<and> P t f}"
shows "\<forall> t :: real. t \<ge> 0 \<longrightarrow> ((\<forall> f \<in> S t. \<forall> g \<in> S t. (\<lambda> x. f x * g x) \<in> S t) \<longleftrightarrow> putnam_1980_b5_solution t)"
sorry
end |
putnam_1980_b6 | theorem putnam_1980_b6
(G : β€ Γ β€ β β)
(hG : β d n : β, d β€ n β (d = 1 β G (d, n) = 1/(n : β)) β§ (d > 1 β G (d, n) = (d/(n : β))*β i in Finset.Icc d n, G ((d : β€) - 1, (i : β€) - 1)))
: β d p : β, 1 < d β§ d β€ p β§ Prime p β Β¬p β£ (G (d, p)).den :=
sorry | For integers $d, n$ with $1 \le d \le n$, let $G(1, n) = \frac{1}{n}$ and $G(d, n) = \frac{d}{n}\sum_{i=d}^{n}G(d - 1, i - 1)$ for all $d > 1$. If $1 < d \le p$ for some prime $p$, prove that the reduced denominator of $G(d, p)$ is not divisible by $p$. | null | ['number_theory', 'algebra'] | Section putnam_1980_b6.
Require Import Reals Nat Znumtheory QArith Coquelicot.Coquelicot. From mathcomp Require Import div.
Theorem putnam_1980_b6:
let A :=
fix f (n i: nat) :=
match (n,i) with
| (O,i') => 1/INR i'
| (S n', i') => (INR n' + 1)/(INR i') * sum_n (fun x => f n' (Nat.add n' x)) (Nat.sub i' n')
end in
forall (n p: nat), and (gt p n) (gt n 1) /\ prime (Z.of_nat p) -> exists (a b: nat), A n p = INR a/INR b /\ p %| b/(gcd a b) = false.
Proof. Admitted.
End putnam_1980_b6. | theory putnam_1980_b6 imports
Complex_Main
"HOL-Computational_Algebra.Primes"
begin
theorem putnam_1980_b6:
fixes G :: "nat \<times> nat \<Rightarrow> rat"
assumes hG: "\<forall> d n :: nat. d \<le> n \<longrightarrow> ((d = 1 \<longrightarrow> G (d, n) = 1 / n) \<and> (d > 1 \<longrightarrow> G (d, n) = (d / n) * (\<Sum> i = d..n. G (d - 1, i - 1))))"
shows "\<forall> d p :: nat. (1 < d \<and> d \<le> p \<and> prime p) \<longrightarrow> \<not>(p dvd (snd (quotient_of (G (d, p)))))"
sorry
end
|
putnam_1984_a2 | abbrev putnam_1984_a2_solution : β := sorry
-- 2
theorem putnam_1984_a2
: β' k : Set.Ici 1, (6 ^ (k : β) / ((3 ^ ((k : β) + 1) - 2 ^ ((k : β) + 1)) * (3 ^ (k : β) - 2 ^ (k : β)))) = putnam_1984_a2_solution :=
sorry | Express $\sum_{k=1}^\infty (6^k/(3^{k+1}-2^{k+1})(3^k-2^k))$ as a rational number. | Show that the sum converges to $2$. | ['analysis'] | Section putnam_1984_a2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope nat_scope.
Definition putnam_1984_a2_solution := 2%R.
Theorem putnam_1984_a2:
let f (n: nat) := Rdiv (pow 6 n) ((pow 3 (n+1) - pow 2 (n+1)) * pow 3 n - pow 2 n) in
Series f = putnam_1984_a2_solution.
Proof. Admitted.
End putnam_1984_a2. | theory putnam_1984_a2 imports Complex_Main
begin
definition putnam_1984_a2_solution :: "rat" where "putnam_1984_a2_solution \<equiv> undefined"
(* 2 *)
theorem putnam_1984_a2:
shows "(\<Sum> (k :: nat) \<in> {1..}. 6^k/((3^(k+1) - 2^(k+1))*(3^k - 2^k))) = putnam_1984_a2_solution"
sorry
end |
putnam_1984_a3 | abbrev putnam_1984_a3_solution : MvPolynomial (Fin 3) β := sorry
-- (MvPolynomial.X 2) ^ 2 * ((MvPolynomial.X 0) ^ 2 - (MvPolynomial.X 1) ^ 2)
theorem putnam_1984_a3
(n : β)
(a b : β)
(Mn : β β Matrix (Fin (2 * n)) (Fin (2 * n)) β)
(polyabn : Fin 3 β β)
(npos : n > 0)
(aneb : a β b)
(hMnx : β x : β, β i : Fin (2 * n), (Mn x) i i = x)
(hMna : β x : β, β i j : Fin (2 * n), (i β j β§ Even ((i : β) + j)) β (Mn x) i j = a)
(hMnb : β x : β, β i j : Fin (2 * n), (i β j β§ Odd ((i : β) + j)) β (Mn x) i j = b)
(hpolyabn : polyabn 0 = a β§ polyabn 1 = b β§ polyabn 2 = n)
: Tendsto (fun x : β => (Mn x).det / (x - a) ^ (2 * n - 2)) (π[β ] a) (π (MvPolynomial.eval polyabn putnam_1984_a3_solution)) :=
sorry | Let $n$ be a positive integer. Let $a,b,x$ be real numbers, with $a \neq b$, and let $M_n$ denote the $2n \times 2n$ matrix whose $(i,j)$ entry $m_{ij}$ is given by
\[
m_{ij}=\begin{cases}
x & \text{if }i=j, \\
a & \text{if }i \neq j\text{ and }i+j\text{ is even}, \\
b & \text{if }i \neq j\text{ and }i+j\text{ is odd}.
\end{cases}
\]
Thus, for example, $M_2=\begin{pmatrix} x & b & a & b \\ b & x & b & a \\ a & b & x & b \\ b & a & b & x \end{pmatrix}$. Express $\lim_{x \to a} \det M_n/(x-a)^{2n-2}$ as a polynomial in $a$, $b$, and $n$, where $\det M_n$ denotes the determinant of $M_n$. | Show that $\lim_{x \to a} \frac{\det M_n}{(x-a)^{2n-2}}=n^2(a^2-b^2)$. | ['linear_algebra', 'analysis'] | null | theory putnam_1984_a3 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Analysis.Determinants"
begin
definition putnam_1984_a3_solution :: "real poly poly poly" where "putnam_1984_a3_solution \<equiv> undefined"
(* (monom 1 2) * ([:[:monom 1 2:]:] - [:monom 1 2:]) *)
theorem putnam_1984_a3:
fixes n :: "nat"
and a b :: "real"
and Mn :: "real \<Rightarrow> (real^'i^'i)"
and ni :: "'i \<Rightarrow> nat"
assumes hij : "CARD('i) = 2*n"
and hn : "bij_betw ni (UNIV :: 'i set) {1..2*n}"
and npos : "n > 0"
and aneb : "a \<noteq> b"
and hMnx : "\<forall> x :: real. \<forall> i :: 'i. (Mn x)$i$i = x"
and hMna : "\<forall> x :: real. \<forall> i j :: 'i. (ni i \<noteq> nj j \<and> even (ni i + nj j)) \<longrightarrow> (Mn x)$i$j = a"
and hMnb : "\<forall> x :: real. \<forall> i j :: 'i. (ni i \<noteq> nj j \<and> odd (ni i + nj j)) \<longrightarrow> (Mn x)$i$j = b"
shows "filterlim (\<lambda> x :: real. det (Mn x) / (x - a) ^ (2 * n - 2)) (nhds (poly (poly (poly putnam_1984_a3_solution [:[:(n::real):]:]) [:b:]) a)) (at a)"
sorry
end |
putnam_1984_a5 | abbrev putnam_1984_a5_solution : β Γ β Γ β Γ β Γ β := sorry
-- (1, 9, 8, 4, 25)
theorem putnam_1984_a5
(R : Set (Fin 3 β β))
(w : (Fin 3 β β) β β)
(hR : R = {p : Fin 3 β β | (β i : Fin 3, p i β₯ 0) β§ p 0 + p 1 + p 2 β€ 1})
(hw : β p : Fin 3 β β, w p = 1 - p 0 - p 1 - p 2)
: let (a, b, c, d, n) := putnam_1984_a5_solution; a > 0 β§ b > 0 β§ c > 0 β§ d > 0 β§ n > 0 β§ (β« p in R, (p 0) ^ 1 * (p 1) ^ 9 * (p 2) ^ 8 * (w p) ^ 4 = ((a)! * (b)! * (c)! * (d)! : β) / (n)!) :=
sorry | Let $R$ be the region consisting of all triples $(x,y,z)$ of nonnegative real numbers satisfying $x+y+z \leq 1$. Let $w=1-x-y-z$. Express the value of the triple integral $\iiint_R x^1y^9z^8w^4\,dx\,dy\,dz$ in the form $a!b!c!d!/n!$, where $a$, $b$, $c$, $d$, and $n$ are positive integers. | Show that the integral we desire is $1!9!8!4!/25!$. | ['analysis'] | Section putnam_1984_a5.
Require Import Reals Factorial Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1984_a5_solution := INR (fact 9 * fact 8 * fact 4 / fact 25).
Theorem putnam_1984_a5:
RInt (fun z => RInt (fun y => RInt (fun x => x * pow y 9 * pow z 8 * pow (1 - x - y - z) 4) 0 (1 - y - z)) 0 (1 - z)) 0 1 = putnam_1984_a5_solution.
Proof. Admitted.
End putnam_1984_a5. | theory putnam_1984_a5 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Analysis.Set_Integral"
begin
definition putnam_1984_a5_solution :: "nat \<times> nat \<times> nat \<times> nat \<times> nat" where "putnam_1984_a5_solution \<equiv> undefined"
(* (1, 9, 8, 4, 25) *)
theorem putnam_1984_a5:
fixes R :: "(real \<times> real \<times> real) set"
and w :: "(real \<times> real \<times> real) \<Rightarrow> real"
defines "R \<equiv> {(x :: real, y :: real, z :: real). x \<ge> 0 \<and> y \<ge> 0 \<and> z \<ge> 0 \<and> x + y + z \<le> 1}"
and "w \<equiv> \<lambda> (x :: real, y :: real, z :: real). 1 - x - y - z"
shows "let (a, b, c, d, n) = putnam_1984_a5_solution in a > 0 \<and> b > 0 \<and> c > 0 \<and> d > 0 \<and> n > 0 \<and>
set_lebesgue_integral lebesgue R (\<lambda> (x, y, z). x^1 * y^9 * z^8 * (w (x, y, z))^4) = (real_of_nat ((fact a) * (fact b) * (fact c) * (fact d)))/(fact n)"
sorry
end |
putnam_1984_a6 | abbrev putnam_1984_a6_solution : Prop Γ β := sorry
-- (True, 4)
theorem putnam_1984_a6
(lastnzdig : List β β β)
(f : β β β)
(kadistinct : β β (β β β) β Prop := fun k : β => fun a : β β β => (k β₯ 1 β§ β i j : Fin k, i β j β a i β a j))
(gpeq : (β β β) β β β Prop := fun g : β β β => fun p : β => (p > 0 β§ β s β₯ 1, g s = g (s + p)))
(hlastnzdig : β digs : List β, (β i : Fin digs.length, digs[i] β 0) β lastnzdig digs β 0 β§ (β i : Fin digs.length, digs[i] = lastnzdig digs β§ β j < i, digs[j] = 0))
(hf : β n > 0, f n = lastnzdig (Nat.digits 10 (n)!))
: β g : β β β, (β (k : β) (a : β β β), kadistinct k a β g (β i : Fin k, a i) = f (β i : Fin k, 5 ^ (a i))) β§
(if putnam_1984_a6_solution.1 = True then (gpeq g putnam_1984_a6_solution.2 β§ β p : β, gpeq g p β p β₯ putnam_1984_a6_solution.2) else (Β¬β p : β, gpeq g p)) :=
sorry | Let $n$ be a positive integer, and let $f(n)$ denote the last nonzero digit in the decimal expansion of $n!$. For instance, $f(5)=2$.
\begin{enumerate}
\item[(a)] Show that if $a_1,a_2,\dots,a_k$ are \emph{distinct} nonnegative integers, then $f(5^{a_1}+5^{a_2}+\dots+5^{a_k})$ depends only on the sum $a_1+a_2+\dots+a_k$.
\item[(b)] Assuming part (a), we can define $g(s)=f(5^{a_1}+5^{a_2}+\dots+5^{a_k})$, where $s=a_1+a_2+\dots+a_k$. Find the least positive integer $p$ for which $g(s)=g(s + p)$, for all $s \geq 1$, or else show that no such $p$ exists.
\end{enumerate} | Show that the least such $p$ is $p=4$. | ['algebra', 'number_theory'] | null | theory putnam_1984_a6 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
definition putnam_1984_a6_solution :: "bool \<times> nat" where "putnam_1984_a6_solution \<equiv> undefined"
(* (True, 4) *)
theorem putnam_1984_a6:
fixes f :: "nat \<Rightarrow> nat"
and kadistinct :: "nat \<Rightarrow> (nat \<Rightarrow> nat) \<Rightarrow> bool"
and gpeq :: "(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> bool"
defines "kadistinct \<equiv> \<lambda> (k :: nat) (a :: nat \<Rightarrow> nat). k \<ge> 1 \<and> (\<forall> i j :: nat. (i < k \<and> j < k \<and> i \<noteq> j) \<longrightarrow> a i \<noteq> a j)"
and "gpeq \<equiv> \<lambda> (g :: nat \<Rightarrow> nat) (p :: nat). p > 0 \<and> (\<forall> (s :: nat) \<ge> 1. g s = g (s + p))"
assumes hf : "\<forall> n > 0. f n = (if [n \<noteq> 0] (mod 10) then (n mod 10) else f (n div 10))"
shows "let (b, n) = putnam_1984_a6_solution in \<exists> g :: nat \<Rightarrow> nat.
(\<forall> (k :: nat) (a :: nat \<Rightarrow> nat). kadistinct k a \<longrightarrow> g (\<Sum> i=0..(k-1). a i) = f (\<Sum> i=0..(k-1). 5^(a i)))
\<and> (if b then gpeq g n \<and> (\<forall> p :: nat. gpeq g p \<longrightarrow> p \<ge> n) else \<not>(\<exists> p :: nat. gpeq g p))"
sorry
end |
putnam_1984_b1 | abbrev putnam_1984_b1_solution : Polynomial β Γ Polynomial β := sorry
-- (Polynomial.X + 3, -Polynomial.X - 2)
theorem putnam_1984_b1
(f : β β β€)
(hf : β n > 0, f n = β i : Set.Icc 1 n, ((i)! : β€))
: let (P, Q) := putnam_1984_b1_solution; β n β₯ 1, f (n + 2) = P.eval (n : β) * f (n + 1) + Q.eval (n : β) * f n :=
sorry | Let $n$ be a positive integer, and define $f(n)=1!+2!+\dots+n!$. Find polynomials $P(x)$ and $Q(x)$ such that $f(n+2)=P(n)f(n+1)+Q(n)f(n)$ for all $n \geq 1$. | Show that we can take $P(x)=x+3$ and $Q(x)=-x-2$. | ['algebra'] | Section putnam_1984_b1.
Require Import Factorial ZArith Reals Coquelicot.Coquelicot.
Open Scope Z.
Definition putnam_1984_b1_solution (coeff1 coeff2 : nat -> Z) (n1 n2 : Z) := (coeff1 = fun x => match x with | O => 3 | S O => 1 | _ => 0 end) /\ (coeff2 = fun x => match x with | O => -2 | S O => -1 | _ => 0 end) /\ n1 = 1 /\ n2 = 1.
Theorem putnam_1984_b1
(f : nat -> Z := fun n => (floor (sum_n (fun i => INR (fact (i + 1))) n)))
(p: (nat -> Z) -> Z -> (nat -> Z) := fun coeff n => (fun x => (floor (sum_n (fun i => IZR ((coeff i) * (Z.of_nat (x ^ i)))) (Z.to_nat n + 1)))))
: exists (coeff1 coeff2 : nat -> Z) (n1 n2 : Z),
let fix F (n: nat) :=
match n with
| O => f 0%nat
| S O => f 1%nat
| S ((S n'') as n') => (p coeff1 n1) n' * F n' + (p coeff2 n2) n'' * F n''
end in
f = F -> putnam_1984_b1_solution coeff1 coeff2 n1 n2.
Proof. Admitted.
End putnam_1984_b1. | theory putnam_1984_b1 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1984_b1_solution :: "(real poly) \<times> (real poly)" where "putnam_1984_b1_solution \<equiv> undefined"
(* (monom 1 1 + monom 3 0, monom (-1) 1 + monom (-2) 0) *)
theorem putnam_1984_b1:
fixes f :: "nat \<Rightarrow> nat"
assumes hf : "\<forall> n > 0. f n = (\<Sum> i=1..n. (fact i))"
shows "let (P, Q) = putnam_1984_b1_solution in \<forall> n \<ge> 1. f (n + 2) = (poly P n) * (f (n + 1)) + (poly Q n) * (f n)"
sorry
end |
putnam_1984_b2 | abbrev putnam_1984_b2_solution : β := sorry
-- 8
theorem putnam_1984_b2
(f : β β β β β)
(hf : β u v : β, f u v = (u - v) ^ 2 + (Real.sqrt (2 - u ^ 2) - 9 / v) ^ 2)
: (β u : Set.Ioo 0 (Real.sqrt 2), β v > 0, f u v = putnam_1984_b2_solution) β§ (β u : Set.Ioo 0 (Real.sqrt 2), β v > 0, f u v β₯ putnam_1984_b2_solution) :=
sorry | Find the minimum value of $(u-v)^2+(\sqrt{2-u^2}-\frac{9}{v})^2$ for $0<u<\sqrt{2}$ and $v>0$. | Show that the minimum value is $8$. | ['geometry', 'analysis'] | Section putnam_1984_b2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1984_b2_solution := 8.
Theorem putnam_1984_b2:
let f (x y: R) := pow (x - y) 2 + pow (sqrt (2 - pow x 2) - 9 / y) 2 in
exists (m: R),
(forall (x y: R), 0 < x < sqrt 2 /\ y > 0 -> f x y >= m) /\
(exists (x y: R), 0 < x < sqrt 2 /\ y > 0 -> f x y = m) ->
m = putnam_1984_b2_solution.
Proof. Admitted.
End putnam_1984_b2. | theory putnam_1984_b2 imports Complex_Main
begin
definition putnam_1984_b2_solution :: "real" where "putnam_1984_b2_solution \<equiv> undefined"
(* 8 *)
theorem putnam_1984_b2:
fixes f :: "real \<Rightarrow> real \<Rightarrow> real"
defines "f \<equiv> \<lambda> u v :: real. (u - v)^2 + (sqrt (2 - u^2) - 9/v)^2"
shows "(LEAST (r :: real) . (\<exists> u \<in> {0<..<sqrt 2}. \<exists> v > 0. f u v = r)) = putnam_1984_b2_solution"
sorry
end |
putnam_1984_b3 | abbrev putnam_1984_b3_solution : Prop := sorry
-- True
theorem putnam_1984_b3
: (β (F : Type*) (_ : Fintype F), Fintype.card F β₯ 2 β (β mul : F β F β F, β x y z : F, (mul x z = mul y z β x = y) β§ (mul x (mul y z) β mul (mul x y) z))) β putnam_1984_b3_solution :=
sorry | Prove or disprove the following statement: If $F$ is a finite set with two or more elements, then there exists a binary operation $*$ on F such that for all $x,y,z$ in $F$,
\begin{enumerate}
\item[(i)] $x*z=y*z$ implies $x=y$ (right cancellation holds), and
\item[(ii)] $x*(y*z) \neq (x*y)*z$ (\emph{no} case of associativity holds).
\end{enumerate} | Show that the statement is true. | ['abstract_algebra'] | null | theory putnam_1984_b3 imports Complex_Main
"HOL-Library.Cardinality"
begin
definition putnam_1984_b3_solution :: "bool" where "putnam_1984_b3_solution \<equiv> undefined"
(* True *)
theorem putnam_1984_b3:
shows "((CARD('a) \<ge> 2 \<and> (\<exists> n :: nat. CARD('a) = n)) \<longrightarrow> (\<exists> mul :: 'a \<Rightarrow> 'a \<Rightarrow> 'a. \<forall> x y z :: 'a. (mul x z = mul y z \<longrightarrow> x = y) \<and> (mul x (mul y z) \<noteq> mul (mul x y) z))) \<longleftrightarrow> putnam_1984_b3_solution"
sorry
end |
putnam_1984_b5 | abbrev putnam_1984_b5_solution : β€ Γ Polynomial β Γ Polynomial β := sorry
-- (2, (Polynomial.X * (Polynomial.X - 1)) / 2, Polynomial.X)
theorem putnam_1984_b5
(sumbits : List β β β)
(d : β β β)
(m : β)
(hsumbits : β bits : List β, sumbits bits = β i : Fin bits.length, bits[i])
(hd : β k : β, d k = sumbits (Nat.digits 2 k))
(mpos : m > 0)
: let (a, f, g) := putnam_1984_b5_solution; β k : Set.Icc 0 (2 ^ m - 1), (-(1 : β€)) ^ (d k) * (k : β) ^ m = (-1) ^ m * (a : β) ^ (f.eval (m : β)) * (g.eval m)! :=
sorry | For each nonnegative integer $k$, let $d(k)$ denote the number of $1$'s in the binary expansion of $k$ (for example, $d(0)=0$ and $d(5)=2$). Let $m$ be a positive integer. Express $\sum_{k=0}^{2^m-1} (-1)^{d(k)}k^m$ in the form $(-1)^ma^{f(m)}(g(m))!$, where $a$ is an integer and $f$ and $g$ are polynomials. | Show that $\sum_{k=0}^{2^m-1} (-1)^{d(k)}k^m=(-1)^m2^{m(m-1)/2}m!$. | ['algebra', 'analysis'] | Section putnam_1984_b5.
Require Import Nat Reals Coquelicot.Coquelicot.
Definition putnam_1984_b5_solution (coeff1 coeff2 : nat -> R) (n1 n2: nat) (a: Z) := a = 2%Z /\ (coeff1 = fun x => match x with | O => 0 | S O => -1 / 2 | S (S O) => 1 / 2 | _ => 0 end) /\ (coeff2 = fun x => match x with | O => 0 | S O => INR x | _ => 0 end) /\ n1 = 2%nat /\ n2 = 1%nat.
Theorem putnam_1984_b5
(f : positive -> nat := fix f (n : positive) : nat :=
match n with
| xH => 1%nat
| xO n' => f n'
| xI n' => add 1 (f n')
end)
(g : nat -> R := fun m => sum_n (fun k => (-1) ^ (f (Pos.of_nat k)) * INR k ^ m) (2 ^ m - 1))
(p: (nat -> R) -> nat -> (R -> R) := fun coeff n => (fun x => sum_n (fun i => coeff i * x ^ i) (n + 1)))
: forall (m : nat), exists (a : Z) (coeff1 coeff2 : nat -> R) (n1 n2: nat), g m = (-1) ^ m * Rpower (IZR a) ((p coeff1 n1) (INR m)) * INR (fact (Z.to_nat (floor ((p coeff2 n2) (INR m))))) <->
putnam_1984_b5_solution coeff1 coeff2 n1 n2 a.
Proof. Admitted.
End putnam_1984_b5. | theory putnam_1984_b5 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1984_b5_solution :: "int \<times> (real poly) \<times> (nat poly)" where "putnam_1984_b5_solution \<equiv> undefined"
(* (2, ((monom 1 1) * ((monom 1 1) - 1)) * (monom (1/2) 0), monom 1 1) *)
theorem putnam_1984_b5:
fixes d :: "nat \<Rightarrow> nat"
and m :: "nat"
assumes mpos : "m > 0"
and hd : "\<forall> n::nat. d n = (if n < 2 then n else ((n mod 2::nat) + d (n div 2::nat)))"
shows "let (a, f, g) = putnam_1984_b5_solution in (\<Sum> k=0..(2^m-1). (-1::int)^(d k) * k^m) = (-1::int)^m * (a powr (poly f m)) * (fact (poly g m))"
sorry
end |
putnam_2009_a1 | abbrev putnam_2009_a1_solution : Prop := sorry
-- True
theorem putnam_2009_a1
: ((β f : (β Γ β) β β, (β O v : β Γ β, v β (0, 0) β f (O.1, O.2) + f (O.1 + v.1, O.2 + v.2) + f (O.1 + v.1 - v.2, O.2 + v.2 + v.1) + f (O.1 - v.2, O.2 + v.1) = 0) β β P : β Γ β, f P = 0) β putnam_2009_a1_solution) :=
sorry | Let $f$ be a real-valued function on the plane such that for every square $ABCD$ in the plane, $f(A)+f(B)+f(C)+f(D)=0$. Does it follow that $f(P)=0$ for all points $P$ in the plane? | Prove that $f$ is identically $0$. | ['geometry', 'algebra'] | Section putnam_2009_a1.
Require Import Reals GeoCoq.Main.Tarski_dev.Ch16_coordinates_with_functions.
Context `{T2D:Tarski_2D} `{TE:@Tarski_euclidean Tn TnEQD}.
Open Scope R.
Definition putnam_2009_a1_solution := True.
Theorem putnam_2009_a1
(f: Tpoint -> R)
: ((forall (A B C D: Tpoint), Square A B C D ->
f A + f B + f C + f D = 0) ->
forall (P : Tpoint), f P = 0) <->
putnam_2009_a1_solution.
Proof. Admitted.
End putnam_2009_a1. | theory putnam_2009_a1 imports Complex_Main
begin
definition putnam_2009_a1_solution :: bool where "putnam_2009_a1_solution \<equiv> undefined"
(* True *)
theorem putnam_2009_a1:
shows "(\<forall>f::(real\<times>real)\<Rightarrow>real. (\<forall>A v::real\<times>real. v \<noteq> (0, 0) \<longrightarrow> f (fst A, snd A) + f (fst A+fst v, snd A+snd v) + f (fst A+fst v-snd v, snd A+snd v+fst v) + f (fst A-snd v, snd A+fst v) = 0) \<longrightarrow> (\<forall>P::real\<times>real. f P = 0)) \<longleftrightarrow> putnam_2009_a1_solution"
sorry
end
|
putnam_2009_a3 | abbrev putnam_2009_a3_solution : β := sorry
-- 0
theorem putnam_2009_a3
(cos_matrix : (n : β) β Matrix (Fin n) (Fin n) β)
(hM : β n : β, β i j : Fin n, (cos_matrix n) i j = Real.cos (1 + n * i + j))
: Tendsto (fun n => (cos_matrix n).det) atTop (π putnam_2009_a3_solution) :=
sorry | Let $d_n$ be the determinant of the $n \times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos 1, \cos 2, \dots, \cos n^2$. (For example,\[ d_3 = \left|\begin{matrix} \cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{matrix} \right|. \]The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty} d_n$. | Show that the limit is 0. | ['linear_algebra', 'analysis'] | null | theory putnam_2009_a3 imports Complex_Main
"HOL-Combinatorics.Permutations"
begin
definition putnam_2009_a3_solution :: real where "putnam_2009_a3_solution \<equiv> undefined"
(* 0 *)
theorem putnam_2009_a3:
fixes cos_matrix :: "nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real)"
and ndet :: "nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> real"
assumes hM: "\<forall>n::nat. \<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. (cos_matrix n) i j = cos (1 + n * i + j)"
defines "ndet \<equiv> (\<lambda>(n::nat)(A::nat\<Rightarrow>nat\<Rightarrow>real). (\<Sum>p\<in>{p'::nat\<Rightarrow>nat. p' permutes {0..(n-1)} \<and> (\<forall>i::nat\<ge>n. p' i = i)}. (sign p * (\<Prod>i::nat=0..(n-1). A i (p i)))))"
shows "filterlim (\<lambda>n::nat. ndet n (cos_matrix n)) (nhds putnam_2009_a3_solution) at_top"
sorry
end
|
putnam_2009_a4 | abbrev putnam_2009_a4_solution : Prop := sorry
-- False
theorem putnam_2009_a4
: ((β S : Set β, 0 β S β (β x β S, x + 1 β S β§ x - 1 β S) β (β x β S, x β ({0, 1} : Set β) β 1 / (x * (x - 1)) β S) β β r : β, r β S) β putnam_2009_a4_solution) :=
sorry | Let $S$ be a set of rational numbers such that
\begin{enumerate}
\item[(a)] $0 \in S$;
\item[(b)] If $x \in S$ then $x+1\in S$ and $x-1\in S$; and
\item[(c)] If $x\in S$ and $x\not\in\{0,1\}$, then $\frac{1}{x(x-1)}\in S$.
\end{enumerate}
Must $S$ contain all rational numbers? | Prove that $S$ need not contain all rationals. | ['number_theory'] | Section putnam_2009_a4.
Require Import Ensembles QArith.
Definition putnam_2009_a4_solution := False.
Theorem putnam_2009_a4:
forall (E: Ensemble Q),
(forall (q: Q),
E q <-> q = 0 /\
E q -> E (q + 1) /\ E (q - 1) /\
E q /\ q <> 0 /\ q <> 1 -> E (1 / (q * (q - 1)))) ->
forall (q: Q), E q <-> putnam_2009_a4_solution.
Proof. Admitted.
End putnam_2009_a4. | theory putnam_2009_a4 imports Complex_Main
begin
definition putnam_2009_a4_solution :: bool where "putnam_2009_a4_solution \<equiv> undefined"
(* False *)
theorem putnam_2009_a4:
shows "(\<forall>S::rat set. 0 \<in> S \<longrightarrow> (\<forall>x\<in>S. x+1 \<in> S \<and> x-1 \<in> S) \<longrightarrow> (\<forall>x\<in>S. x \<notin> {0, 1} \<longrightarrow> 1 / (x*(x-1)) \<in> S) \<longrightarrow> (\<forall>r::rat. r \<in> S)) \<longleftrightarrow> putnam_2009_a4_solution"
sorry
end
|
putnam_2009_a5 | abbrev putnam_2009_a5_solution : Prop := sorry
-- False
theorem putnam_2009_a5
: (β (G : Type*) (_ : CommGroup G) (_ : Fintype G), β g : G, orderOf g = 2^2009) β putnam_2009_a5_solution :=
sorry | Is there a finite abelian group $G$ such that the product of the orders of all its elements is 2^{2009}? | Show that the answer is no such finite abelian group exists. | ['abstract_algebra'] | null | theory putnam_2009_a5 imports Complex_Main
"HOL-Algebra.Multiplicative_Group"
begin
definition putnam_2009_a5_solution :: bool where "putnam_2009_a5_solution \<equiv> undefined"
(* False *)
theorem putnam_2009_a5:
assumes pacount: "\<exists>pamap::'a\<Rightarrow>nat. surj pamap"
shows "(\<exists>G::'a monoid. finite (carrier G) \<and> comm_group G \<and> (\<Prod>g\<in>(carrier G). (group.ord G) g) = 2^2009) \<longleftrightarrow> putnam_2009_a5_solution"
sorry
end
|
putnam_2009_b1 | theorem putnam_2009_b1
(isquotprodprimefact : β β Prop :=
fun q => (β (k m : β) (a : Fin k β β) (b : Fin m β β),
(β i : Fin k, Nat.Prime (a i)) β§ (β j : Fin m, Nat.Prime (b j))
β§ (q = (β i : Fin k, Nat.factorial (a i))/(β j : Fin m, Nat.factorial (b j)))
))
: β q : β, q > 0 β isquotprodprimefact q :=
sorry | Show that every positive rational number can be written as a quotient of products of factorails of (not necessarily distinct) primes. For example, 10/9 = (2! * 5!)/(3! * 3! * 3!). | null | ['number_theory'] | Section putnam_2009_b1.
Require Import List QArith Znumtheory Reals.
Open Scope Q.
Theorem putnam_2009_b1:
let fix factl (l : list nat) : list nat :=
match l with
| nil => nil
| h :: t => fact h :: t end in
forall (q: Q), q > 0 ->
exists (n d: list nat), (forall x, (In x n \/ In x d)-> prime (Z.of_nat x)) /\
inject_Z (Z.of_nat (fold_left Nat.mul (factl n) 1%nat)) / inject_Z (Z.of_nat (fold_left Nat.mul (factl d) 1%nat)) = q.
Proof. Admitted.
End putnam_2009_b1. | theory putnam_2009_b1 imports Complex_Main
"HOL-Computational_Algebra.Primes"
begin
(* uses (nat \<Rightarrow> nat) instead of (Fin k \<Rightarrow> nat) and (Fin m \<Rightarrow> nat) *)
theorem putnam_2009_b1:
fixes isquotprodprimefact :: "rat \<Rightarrow> bool"
defines "isquotprodprimefact \<equiv> (\<lambda>q::rat. (\<exists>(k::nat)(m::nat)(a::nat\<Rightarrow>nat)(b::nat\<Rightarrow>nat).
(\<forall>i::nat\<in>{0..(k-1)}. prime (a i)) \<and> (\<forall>j::nat\<in>{0..(m-1)}. prime (b j))
\<and> q = (\<Prod>i::nat=0..(k-1). fact (a i)) / (\<Prod>j::nat=0..(m-1). fact (b j))))"
shows "\<forall>q::rat. (q > 0 \<longrightarrow> isquotprodprimefact q)"
sorry
end
|
putnam_2009_b3 | abbrev putnam_2009_b3_solution : Set β€ := sorry
-- {n : β€ | β k β₯ 1, n = 2 ^ k - 1}
theorem putnam_2009_b3
(mediocre : β€ β Set β€ β Prop := fun n S β¦ (S β Icc 1 n) β§ β a β S, β b β S, 2 β£ a + b β (a + b) / 2 β S)
(A : β€ β β€ := fun n β¦ {S : Set β€ | mediocre n S}.ncard)
: ({n : β€ | n > 0 β§ A (n + 2) - 2 * A (n + 1) + A n = 1} = putnam_2009_b3_solution) :=
sorry | Call a subset $S$ of $\{1, 2, \dots, n\}$ \emph{mediocre} if it has the following property: Whenever $a$ and $b$ are elements of $S$ whose average is an integer, that average is also an element of $S$. Let $A(n)$ be the number of mediocre subsets of $\{1,2,\dots,n\}$. [For instance, every subset of $\{1,2,3\}$ except $\{1,3\}$ is mediocre, so $A(3) = 7$.] Find all positive integers $n$ such that $A(n+2) - 2A(n+1) + A(n) = 1$. | Show that the answer is $n = 2^k - 1$ for some integer $k$. | ['number_theory'] | null | theory putnam_2009_b3 imports Complex_Main
begin
definition putnam_2009_b3_solution :: "nat set" where "putnam_2009_b3_solution \<equiv> undefined"
(* {n::nat. (\<exists>k::nat\<ge>1. n = 2^k - 1)} *)
theorem putnam_2009_b3:
fixes mediocre :: "nat \<Rightarrow> nat set \<Rightarrow> bool"
and A :: "nat \<Rightarrow> int"
defines "mediocre \<equiv> (\<lambda>(n::nat)(S::nat set). S \<subseteq> {1..n} \<and> (\<forall>a\<in>S. \<forall>b\<in>S. 2 dvd (a + b) \<longrightarrow> (nat (round ((a + b)/2))) \<in> S))"
and "A \<equiv> (\<lambda>n::nat. card {S::nat set. mediocre n S})"
shows "{n::nat. n > 0 \<and> A (n+2) - 2*A (n+1) + A n = 1} = putnam_2009_b3_solution"
sorry
end
|
putnam_2009_b4 | abbrev putnam_2009_b4_solution : β := sorry
-- 2020050
theorem putnam_2009_b4
(balanced : MvPolynomial (Fin 2) β β Prop := fun P β¦ β r > 0, (β« x in Metric.sphere 0 r, MvPolynomial.eval x P) / (2 * Real.pi * r) = 0)
(V : Set (MvPolynomial (Fin 2) β)) [AddCommGroup V] [Module β V]
(hV : β P : MvPolynomial (Fin 2) β, P β V β balanced P β§ P.totalDegree β€ 2009)
: (Module.rank V = putnam_2009_b4_solution) :=
sorry | Say that a polynomial with real coefficients in two variables, $x,y$, is \emph{balanced} if the average value of the polynomial on each circle centered at the origin is $0$. The balanced polynomials of degree at most $2009$ form a vector space $V$ over $\mathbb{R}$. Find the dimension of $V$. | Prove that the dimension of $V$ is $2020050$. | ['algebra', 'linear_algebra'] | null | theory putnam_2009_b4 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_2009_b4_solution :: nat where "putnam_2009_b4_solution \<equiv> undefined"
(* 2020050 *)
definition mvpolyscale :: "real \<Rightarrow> (real poly poly) \<Rightarrow> (real poly poly)" where "mvpolyscale \<equiv> (\<lambda>(c::real)(P::real poly poly). smult (monom c 0) P)"
interpretation mvpolyspace: vector_space "mvpolyscale" sorry
theorem putnam_2009_b4:
fixes balanced :: "(real poly poly) \<Rightarrow> bool"
and V :: "(real poly poly) set"
defines "balanced \<equiv> (\<lambda>P::real poly poly. (\<forall>r::real>0. (set_lebesgue_integral lebesgue (sphere 0 r) (\<lambda>x::real^2. poly (poly P (monom (x$2) 0)) (x$1))) / (2*pi*r) = 0))"
assumes hV: "\<forall>P::real poly poly. (P \<in> V \<longleftrightarrow> (balanced P \<and> degree P \<le> 2009 \<and> (\<forall>i::nat. degree (coeff P i) \<le> 2009)))"
shows "mvpolyspace.dim V = putnam_2009_b4_solution"
sorry
end
|
putnam_2009_b5 | theorem putnam_2009_b5
(f : β β β)
(hfdiff : DifferentiableOn β f (Ioi 1))
(hf : β x > 1, deriv f x = (x ^ 2 - (f x) ^ 2) / ((x ^ 2) * ((f x) ^ 2 + 1)))
: (Tendsto f β€ β€) :=
sorry | Let $f: (1, \infty) \to \mathbb{R}$ be a differentiable function such that
\[
f'(x) = \frac{x^2 - f(x)^2}{x^2 (f(x)^2 + 1)}
\qquad \mbox{for all $x>1$.}
\]
Prove that $\lim_{x \to \infty} f(x) = \infty$. | null | ['analysis'] | Section putnam_2009_b5.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2009_b5:
forall (f: R -> R) (x: R), (1 < x /\ ex_derive f x /\
Derive f x = (x ^ 2 - (f x) ^ 2) / (x ^ 2 * ((f x) ^ 2 + 1))) ->
~ ex_lim_seq (fun n => f (INR n)).
Proof. Admitted.
End putnam_2009_b5. | theory putnam_2009_b5 imports Complex_Main
"HOL-Analysis.Derivative"
begin
(* uses (real \<Rightarrow> real) instead of ({1<..} \<Rightarrow> real) *)
theorem putnam_2009_b5:
fixes f :: "real \<Rightarrow> real"
assumes hfdiff: "f differentiable_on {1<..}"
and f: "\<forall>x::real>1. deriv f x = (x^2 - (f x)^2) / (x^2 * ((f x)^2 + 1))"
shows "filterlim f at_top at_top"
sorry
end
|
putnam_2009_b6 | theorem putnam_2009_b6
(n : β)
(npos : n > 0)
: (β a : β β β€, a 0 = 0 β§ a 2009 = n β§ β i : Icc 1 2009, ((β j k : β, j < i β§ a i = a j + 2 ^ k) β¨ β b c : β, b < i β§ c < i β§ a b > 0 β§ a c > 0 β§ a i = (a b) % (a c))) :=
sorry | Prove that for every positive integer $n$, there is a sequence of integers $a_0, a_1, \dots, a_{2009}$ with $a_0 = 0$ and $a_{2009} = n$ such that each term after $a_0$ is either an earlier term plus $2^k$ for some nonnegative integer $k$, or of the form $b\,\mathrm{mod}\,c$ for some earlier positive terms $b$ and $c$. [Here $b\,\mathrm{mod}\,c$ denotes the remainder when $b$ is divided by $c$, so $0 \leq (b\,\mathrm{mod}\,c) < c$.] | null | ['number_theory'] | Section putnam_2009_b6.
Require Import List ZArith Coquelicot.Coquelicot.
Open Scope Z.
Theorem putnam_2009_b6:
forall (n: Z), n > 0 ->
exists (a: list Z), length a = 2009%nat /\ nth 0 a 0 = 0 /\ nth 2008 a 0 = n /\
forall (i: nat), and (le 1 i) (lt i 2009) -> exists (j: nat), and (le 0 j) (lt j i) /\ ((exists (k: Z), k > 0 /\ nth i a 0 = nth j a 0 + 2 ^ k) \/ exists (b c: Z), b > 0 /\ c > 0 /\ nth i a 0 = b mod c).
Proof. Admitted.
End putnam_2009_b6. | theory putnam_2009_b6 imports Complex_Main
begin
(* uses (nat \<Rightarrow> int) instead of ({0..2009} \<Rightarrow> int) *)
theorem putnam_2009_b6:
fixes n :: nat
assumes npos: "n > 0"
shows "\<exists>a::nat\<Rightarrow>int. a 0 = 0 \<and> a 2009 = n \<and> (\<forall>i::nat\<in>{1..2009}. (\<exists>j::nat<i. \<exists>k::nat. a i = a j + 2^k) \<or> (\<exists>b::nat<i. \<exists>c::nat<i. a b > 0 \<and> a c > 0 \<and> a i = (a b) mod (a c)))"
sorry
end
|
putnam_2020_a1 | abbrev putnam_2020_a1_solution : β := sorry
-- 508536
theorem putnam_2020_a1
: Set.ncard {x : β | (2020 β£ x) β§ (Nat.log 10 x) + 1 β€ 2020 β§ (β k l, k β₯ l β§ x = β i in Finset.range (k-l+1), 10 ^ (i+l))} = putnam_2020_a1_solution :=
sorry | Find the number of positive integers $N$ satisfying: (i) $N$ is divisible by $2020$, (ii) $N$ has at most $2020$ decimal digits, (iii) The decimal digits of $N$ are a string of consecutive ones followed by a string of consecutive zeros. | Show that the solution is $508536$. | ['number_theory', 'algebra'] | Section putnam_2020_a1.
Require Import Ensembles Finite_sets Rdefinitions Reals Rpower. From mathcomp Require Import bigop div fintype ssralg ssrnat ssrnum.
Definition putnam_2020_a1_solution := 508536.
Theorem putnam_2020_a1:
exists (A: Ensemble nat),
forall (n: nat),
(
(2020 %| n = true
/\ (Rle (Rlog (INR 10) (INR n) + R1) (INR 2020))
/\ exists (k l: nat), k >= l = true /\ n = \sum_(i < k-l+1) 10^(i+l))
<-> A n
)
-> cardinal nat A putnam_2020_a1_solution.
Proof. Admitted.
End putnam_2020_a1. | theory putnam_2020_a1 imports Complex_Main
begin
definition putnam_2020_a1_solution::nat where "putnam_2020_a1_solution \<equiv> undefined"
(* 508536 *)
theorem putnam_2020_a1:
fixes S::"nat set"
defines "S \<equiv> {x. (2020 dvd x) \<and> \<lfloor>log 10 x\<rfloor> + 1 \<le> 2020 \<and>
(\<exists>k l::nat. k \<ge> l \<and> x = (\<Sum>i=0..(k-l). 10^(i+l)))}"
shows "card S = putnam_2020_a1_solution"
sorry
end
|
putnam_2020_a2 | abbrev putnam_2020_a2_solution : β β β := sorry
-- fun k β¦ 4 ^ k
theorem putnam_2020_a2
(k : β)
: (β j in Finset.Icc 0 k, 2 ^ (k - j) * Nat.choose (k + j) j = putnam_2020_a2_solution k) :=
sorry | Let $k$ be a nonnegative integer. Evaluate
\[
\sum_{j=0}^k 2^{k-j} \binom{k+j}{j}.
\]
| Show that the answer is $4^k$. | ['algebra'] | Section putnam_2020_a2.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2020_a2_solution := fun k => 4 ^ k.
Theorem putnam_2020_a2
: (fun k => sum_n (fun j => 2 ^ (k - j) * Binomial.C (k + j) j) (k + 1)) = putnam_2020_a2_solution.
Proof. Admitted.
End putnam_2020_a2. | theory putnam_2020_a2 imports
Complex_Main
begin
definition putnam_2020_a2_solution :: "nat \<Rightarrow> nat" where "putnam_2020_a2_solution \<equiv> undefined"
(* \<lambda> k. 4 ^ k *)
theorem putnam_2020_a2:
fixes k :: nat
shows "(\<Sum> j = 0..k. 2 ^ (k - j) * (k + j choose j)) = putnam_2020_a2_solution k"
sorry
end |
putnam_2020_a3 | abbrev putnam_2020_a3_solution : Prop := sorry
-- False
theorem putnam_2020_a3
(a : β β β)
(ha0 : a 0 = Real.pi / 2)
(ha : β n : β, n β₯ 1 β a n = Real.sin (a (n - 1)))
: (β L : β, Tendsto (fun m : β => β n : Icc 1 m, (a n)^2) atTop (π L)) β putnam_2020_a3_solution :=
sorry | Let $a_0 = \pi/2$, and let $a_n = \sin(a_{n-1})$ for $n \geq 1$. Determine whether
\[
\sum_{n=1}^\infty a_n^2
\]
converges. | The series diverges. | ['analysis'] | Section putnam_2020_a3.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2020_a3_solution := False.
Theorem putnam_2020_a3
(a : nat -> R := fix a (n: nat) :=
match n with
| O => PI / 2
| S n' => sin (a n')
end)
: ex_lim_seq (fun n => (a n) ^ 2) <-> putnam_2020_a3_solution.
Proof. Admitted.
End putnam_2020_a3. | theory putnam_2020_a3 imports Complex_Main
begin
definition putnam_2020_a3_solution :: bool where "putnam_2020_a3_solution \<equiv> undefined"
(* False *)
theorem putnam_2020_a3:
fixes a :: "nat \<Rightarrow> real"
assumes ha0: "a 0 = pi/2"
and ha: "\<forall>n::nat\<ge>1. a n = sin (a (n-1))"
shows "(\<exists>L::real. filterlim (\<lambda>m::nat. (\<Sum>n::nat=1..m. (a n)^2)) (nhds L) at_top) \<longleftrightarrow> putnam_2020_a3_solution"
sorry
end
|
putnam_2020_a5 | abbrev putnam_2020_a5_solution : β€ := sorry
-- (Nat.fib 4040) - 1
theorem putnam_2020_a5
(a : β€ β β := fun n : β€ => {S : Finset β | (β k β S, k > 0) β§ β k : S, Nat.fib k = n}.ncard)
: a putnam_2020_a5_solution = 2020 β§ β n : β€, a n = 2020 β n β€ putnam_2020_a5_solution :=
sorry | Let $a_n$ be the number of sets $S$ of positive integers for which
\[
\sum_{k \in S} F_k = n,
\]
where the Fibonacci sequence $(F_k)_{k \geq 1}$ satisfies $F_{k+2} = F_{k+1} + F_k$ and begins $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$. Find the largest integer $n$ such that $a_n = 2020$. | The answer is $n=F_{4040}-1$. | ['number_theory', 'combinatorics'] | null | theory putnam_2020_a5 imports Complex_Main
"HOL-Number_Theory.Fib"
begin
definition putnam_2020_a5_solution :: int where "putnam_2020_a5_solution \<equiv> undefined"
(* (fib 4040) - 1 *)
theorem putnam_2020_a5:
fixes a :: "int \<Rightarrow> nat"
defines "a \<equiv> (\<lambda>n::int. card {S::nat set. finite S \<and> (\<forall>k\<in>S. k > 0) \<and> (\<Sum>k\<in>S. fib k) = n})"
shows "(GREATEST n::int. a n = 2020) = putnam_2020_a5_solution"
sorry
end
|
putnam_2020_a6 | abbrev putnam_2020_a6_solution : β := sorry
-- Real.pi / 4
theorem putnam_2020_a6
(f : β€ β (β β β) := fun N : β€ => fun x : β =>
β n in Finset.Icc 0 N, (N + 1/2 - n)/((N + 1)*(2*n + 1)) * Real.sin ((2*n + 1)*x))
: (β N > 0, β x : β, f N x β€ putnam_2020_a6_solution) β§
β M : β, (β N > 0, β x : β, f N x β€ M) β M β₯ putnam_2020_a6_solution :=
sorry | For a positive integer $N$, let $f_N$ be the function defined by
\[
f_N(x) = \sum_{n=0}^N \frac{N+1/2-n}{(N+1)(2n+1)} \sin((2n+1)x).
\]
Determine the smallest constant $M$ such that $f_N(x) \leq M$ for all $N$ and all real $x$. | The smallest constant $M$ is $\pi/4$. | ['algebra'] | Section putnam_2020_a6.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2020_a6_solution := PI / 4.
Theorem putnam_2020_a6
(f : Z -> (R -> R) := fun Nz : Z => fun x : R =>
sum_n (fun n => let N := IZR Nz in let n := INR n in (N + 1 / 2 - n) / ((N + 1) * (2 * n + 1)) * sin ((2 * n + 1) * x)) (Z.to_nat Nz + 1))
(M : R)
(hM : forall (N: Z), Z.gt N 0 -> forall (x: R), f N x <= M)
(hMlb : forall (n: R), (forall (N: Z), Z.gt N 0 -> forall (x: R), f N x <= n) -> n >= M)
: M = putnam_2020_a6_solution.
Proof. Admitted.
End putnam_2020_a6. | theory putnam_2020_a6 imports Complex_Main
begin
definition putnam_2020_a6_solution :: real where "putnam_2020_a6_solution \<equiv> undefined"
(* pi/4 *)
theorem putnam_2020_a6:
fixes f :: "int \<Rightarrow> (real \<Rightarrow> real)"
assumes "f \<equiv> (\<lambda>N::int. (\<lambda>x::real. (\<Sum>n::int=0..N. (N + 1/2 - n) / ((N + 1)*(2*n + 1)) * sin ((2*n + 1)*x))))"
shows "(LEAST M::real. (\<forall>N::int>0. \<forall>x::real. f N x \<le> M)) = putnam_2020_a6_solution"
sorry
end
|
putnam_2020_b1 | abbrev putnam_2020_b1_solution : β := sorry
-- 1990
theorem putnam_2020_b1
(d : β β β := fun n : β => β i : Fin (Nat.digits 2 n).length, (Nat.digits 2 n)[i]!)
(S : β€ := β k : Icc 1 2020, ((-1)^(d k))*(k : β)^3)
: S % 2020 = putnam_2020_b1_solution :=
sorry | For a positive integer $n$, define $d(n)$ to be the sum of the digits of $n$ when written in binary (for example, $d(13) = 1+1+0+1=3)$. Let
\[
S = \sum_{k=1}^{2020} (-1)^{d(k)} k^3.
\]
Determine $S$ modulo 2020. | The answer is $1990$. | ['algebra'] | Section putnam_2020_b1.
Require Import ZArith Reals Coquelicot.Coquelicot.
Open Scope Z.
Definition putnam_2020_b1_solution := 1990.
Theorem putnam_2020_b1
(d : positive -> Z := fix d (n : positive) :=
match n with
| xH => 1
| xO n' => d n'%positive
| xI n' => 1 + d n'%positive
end)
(A := sum_n (fun k => IZR ((-1) ^ (d (Pos.of_nat (S k))) * (Z.of_nat k) ^ 3)) 2020)
: (floor A) mod 2020 = putnam_2020_b1_solution.
Proof. Admitted.
End putnam_2020_b1. | theory putnam_2020_b1 imports Complex_Main
begin
definition putnam_2020_b1_solution :: nat where "putnam_2020_b1_solution \<equiv> undefined"
(* 1990 *)
theorem putnam_2020_b1:
fixes d :: "int \<Rightarrow> nat"
and S :: "int"
assumes "d \<equiv> (\<lambda>n::int. if n = 0 then 0 else ((n mod 2) + d \<lfloor>n / 2\<rfloor>))"
and "S \<equiv> \<Sum>k::int=1..2020. (-1)^(d k) * k^3"
shows "S mod 2020 = putnam_2020_b1_solution"
sorry
end
|
putnam_2020_b4 | abbrev putnam_2020_b4_solution : β := sorry
-- 1 / 4040
theorem putnam_2020_b4
(V : β β Set (β β β€) := fun n β¦ ({s : β β β€ | s 0 = 0 β§ (β j β₯ 2 * n, s j = 0) β§ (β j β Icc 1 (2 * n), |s j - s (j - 1)| = 1)}))
(q : β β (β β β€) β β := fun n s β¦ 1 + β j in Finset.Icc 1 (2 * n - 1), 3 ^ (s j))
(M : β β β := fun n β¦ (β' v : V n, 1 / (q n v)) / (V n).ncard)
: (M 2020 = putnam_2020_b4_solution) :=
sorry | Let $n$ be a positive integer, and let $V_n$ be the set of integer $(2n+1)$-tuples $\mathbf{v} = (s_0, s_1, \cdots, s_{2n-1}, s_{2n})$ for which $s_0 = s_{2n} = 0$ and $|s_j - s_{j-1}| = 1$ for $j=1,2,\cdots,2n$. Define \[ q(\mathbf{v}) = 1 + \sum_{j=1}^{2n-1} 3^{s_j}, \] and let $M(n)$ be the average of $\frac{1}{q(\mathbf{v})}$ over all $\mathbf{v} \in V_n$. Evaluate $M(2020)$. | Show that the answer is $\frac{1}{4040}$. | ['algebra'] | null | theory putnam_2020_b4 imports
Complex_Main
begin
definition putnam_2020_b4_solution :: real where "putnam_2020_b4_solution \<equiv> undefined"
(* 1 / 4040 *)
theorem putnam_2020_b4:
fixes V :: "nat \<Rightarrow> (nat \<Rightarrow> int) set"
and q :: "nat \<Rightarrow> (nat \<Rightarrow> int) \<Rightarrow> real"
and M :: "nat \<Rightarrow> real"
defines "V \<equiv> \<lambda> n. {s. s 0 = 0 \<and> (\<forall> j \<ge> 2 * n. s j = 0) \<and> (\<forall> j \<in> {1 .. (2 * n)}. \<bar>s j - s (j - 1)\<bar> = 1)}"
and "q \<equiv> \<lambda> n s. 1 + (\<Sum> j = 1 .. 2 * n - 1. 3 powr (s j))"
and "M \<equiv> \<lambda> n. (\<Sum> v \<in> V n. 1 / (q n v)) / card (V n)"
shows "M 2020 = putnam_2020_b4_solution"
sorry
end |
putnam_2020_b5 | theorem putnam_2020_b5
(z : Fin 4 β β)
(hzle1 : β n, βz nβ < 1)
(hzne1 : β n, z n β 1)
: 3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) β 0:=
sorry | For $j \in \{1, 2, 3, 4\}$, let $z_j$ be a complex number with $|z_j| = 1$ and $z_j \neq 1$. Prove that \[ 3 - z_1 - z_2 - z_3 - z_4 + z_1 z_2 z_3 z_4 \neq 0. \] | null | ['algebra'] | null | theory putnam_2020_b5 imports Complex_Main
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_2020_b5:
fixes z :: "nat \<Rightarrow> complex"
assumes hz: "\<forall>n \<in> {1..4}. norm (z n) = 1"
and hzne1: "\<forall>n \<in> {1..4}. z n \<noteq> 1"
shows "3 - z 1 - z 2 - z 3 - z 4 + z 1 * z 2 * z 3 * z 4 \<noteq> 0"
sorry
end
|
putnam_2020_b6 | theorem putnam_2020_b6
(n : β)
(npos : n > 0)
: β k : Fin n, ((-1) ^ Int.floor ((k.1 + 1) * (Real.sqrt 2 - 1)) : β) β₯ 0 :=
sorry | Let $n$ be a positive integer. Prove that $\sum_{k=1}^n(-1)^{\lfloor k(\sqrt{2}-1) \rfloor} \geq 0$. | null | ['algebra'] | Section putnam_2020_b6.
Require Import Reals. From Coquelicot Require Import Coquelicot Hierarchy Rcomplements.
Local Open Scope R.
Theorem putnam_2020_b6:
let A (k: nat) := (-1)^(Z.to_nat (floor (INR k * (sqrt 2 - 1)))) in
let B (n: nat) := sum_n A n in
forall (n: nat), B n >= 0.
Proof. Admitted.
End putnam_2020_b6. | theory putnam_2020_b6 imports Complex_Main
begin
theorem putnam_2020_b6:
fixes n :: nat
assumes npos: "n > 0"
shows "(\<Sum>k::nat=1..n. (-1) ^ (nat \<lfloor>k * (sqrt 2 - 1)\<rfloor>)) \<ge> 0"
sorry
end
|
putnam_1970_a1 | theorem putnam_1970_a1
(a b : β)
(ha : a > 0)
(hb : b > 0)
(f : β β β := fun x : β => Real.exp (a*x) * Real.cos (b*x))
(p : β β β)
(hp : β c : β, c > 0 β§ β x β ball 0 c, β' n : β, (p n)*x^n = f x)
(S : Set β := {n : β | p n = 0})
: S = β
β¨ Β¬Finite S :=
sorry | Prove that, for all $a > 0$ and $b > 0$, the power series of $e^{ax} \cos (bx)$ with respect to $x$ has either zero or infinitely many zero coefficients. | null | ['analysis'] | null | theory putnam_1970_a1 imports Complex_Main
"HOL-Analysis.Elementary_Metric_Spaces"
begin
theorem putnam_1970_a1:
fixes a b :: real
and f :: "real \<Rightarrow> real"
and p :: "nat \<Rightarrow> real"
and S :: "nat set"
defines "f \<equiv> (\<lambda>x::real. exp (a*x) * cos (b*x))"
assumes hp: "\<exists>a::real>0. (\<forall>x::real\<in>(ball 0 a). (\<Sum>n::nat. (p n)*x^n) = f x)"
defines "S \<equiv> {n::nat. p n = 0}"
shows "S = {} \<or> infinite S"
sorry
end
|
putnam_1970_a3 | abbrev putnam_1970_a3_solution : β Γ β := sorry
-- (3, 1444)
theorem putnam_1970_a3
(L : β β β)
(hL : β n : β, L n β€ (Nat.digits 10 n).length β§
(β k : β, k < L n β (Nat.digits 10 n)[k]! = (Nat.digits 10 n)[0]!) β§
(L n β (Nat.digits 10 n).length β (Nat.digits 10 n)[L n]! β (Nat.digits 10 n)[0]!))
: (β n : β, (Nat.digits 10 (n^2))[0]! β 0 β§ L (n^2) = putnam_1970_a3_solution.1) β§
(β n : β, (Nat.digits 10 (n^2))[0]! β 0 β L (n^2) β€ putnam_1970_a3_solution.1) β§
(β m : β, m^2 = putnam_1970_a3_solution.2) β§
L (putnam_1970_a3_solution.2) = putnam_1970_a3_solution.1 β§
(Nat.digits 10 putnam_1970_a3_solution.2)[0]! β 0 β§
β n : β, (Nat.digits 10 (n^2))[0]! β 0 β§ L (n^2) = putnam_1970_a3_solution.1 β n^2 β₯ putnam_1970_a3_solution.2 :=
sorry | Find the length of the longest possible sequence of equal nonzero digits (in base 10) in which a perfect square can terminate. Also, find the smallest square that attains this length. | The maximum attainable length is $3$; the smallest such square is $38^2 = 1444$. | ['number_theory'] | null | theory putnam_1970_a3 imports Complex_Main
begin
fun digits :: "nat \<Rightarrow> (nat list)" where
"digits n = (if n < 10 then [n] else ([n mod 10::nat] @ digits (n div 10::nat)))"
definition putnam_1970_a3_solution :: "nat \<times> nat" where "putnam_1970_a3_solution \<equiv> undefined"
(* (3, 1444) *)
theorem putnam_1970_a3:
fixes L :: "nat \<Rightarrow> nat"
assumes hL: "\<forall>n::nat. L n \<le> length (digits n) \<and>
(\<forall>k::nat<(L n). (digits n)!k = (digits n)!0) \<and>
(L n \<noteq> length (digits n) \<longrightarrow> (digits n)!(L n) \<noteq> (digits n)!0)"
shows "(GREATEST d::nat. (\<exists>n::nat. (digits (n^2))!0 \<noteq> 0 \<and> d = L (n^2))) = fst putnam_1970_a3_solution \<and>
(LEAST m::nat. (\<exists>n::nat. n^2 = m) \<and> (digits m)!0 \<noteq> 0 \<and> L m = fst putnam_1970_a3_solution) = snd putnam_1970_a3_solution"
sorry
end
|
putnam_1970_a4 | theorem putnam_1970_a4
(x : β β β)
(hxlim : Tendsto (fun n => x n - x (n-2)) atTop (π 0))
: Tendsto (fun n => (x n - x (n-1))/n) atTop (π 0) :=
sorry | Suppose $(x_n)$ is a sequence such that $\lim_{n \to \infty} (x_n - x_{n-2} = 0$. Prove that $\lim_{n \to \infty} \frac{x_n - x_{n-1}}{n} = 0$. | null | ['analysis'] | null | theory putnam_1970_a4 imports Complex_Main
begin
theorem putnam_1970_a4:
fixes x :: "nat \<Rightarrow> real"
assumes hxlim: "filterlim (\<lambda>n::nat. x n - x (n-2)) (nhds 0) at_top"
shows "filterlim (\<lambda>n::nat. (x n - x (n-1)) / n) (nhds 0) at_top"
sorry
end
|
putnam_1970_b1 | abbrev putnam_1970_b1_solution : β := sorry
-- Real.exp (2 * Real.log 5 - 4 + 2 * Real.arctan 2)
theorem putnam_1970_b1
: Tendsto (fun n => 1/(n^4) * β i in Finset.Icc (1 : β€) (2*n), ((n^2 + i^2) : β)^((1 : β)/n)) atTop (π putnam_1970_b1_solution) :=
sorry | Evaluate the infinite product $\lim_{n \to \infty} \frac{1}{n^4} \prod_{i = 1}^{2n} (n^2 + i^2)^{1/n}$. | Show that the solution is $e^{2 \log(5) - 4 + 2 arctan(2)}$. | ['analysis'] | null | theory putnam_1970_b1 imports Complex_Main
begin
definition putnam_1970_b1_solution :: real where "putnam_1970_b1_solution \<equiv> undefined"
(* exp (2*ln 5 - 4 + 2*arctan 2) *)
theorem putnam_1970_b1:
shows "filterlim (\<lambda>n::nat. 1/(n^4) * (\<Prod>i::nat=1..(2*n). (n^2 + i^2) powr (1/n))) (nhds putnam_1970_b1_solution) at_top"
sorry
end
|
putnam_1970_b3 | theorem putnam_1970_b3
(S : Set (β Γ β))
(a b : β)
(hab : a < b)
(hS : β s β S, s.1 β Ioo a b)
(hSclosed : IsClosed S)
: IsClosed {y | β x : β, β¨x,yβ© β S} :=
sorry | A closed subset $S$ of $\mathbb{R}^2$ lies in $a < x < b$. Show that its projection on the y-axis is closed. | null | ['analysis'] | null | theory putnam_1970_b3 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1970_b3:
fixes S :: "(real^2) set"
and a b :: real
assumes hab: "a < b"
and hS: "\<forall>s\<in>S. s$1 \<in> {a<..<b}"
and hSclosed: "closed S"
shows "closed {y::real. (\<exists>s\<in>S. y = s$2)}"
sorry
end
|
putnam_1970_b5 | theorem putnam_1970_b5
(ramp : β€ β (β β β) := fun n => (fun x => if x β€ -n then -n else (if -n < x β§ x β€ n then x else n)))
(F : β β β)
: Continuous F β (β n : β, Continuous ((ramp n) β F)) :=
sorry | Let $u_n$ denote the function $u_n(x) = -n$ if $x \leq -n$, $x$ if $-n < x \leq n$, and $n$ otherwise. Let $F$ be a function on the reals. Show that $F$ is continuous if and only if $u_n \circ F$ is continuous for all natural numbers $n$. | null | ['analysis'] | null | theory putnam_1970_b5 imports Complex_Main
begin
theorem putnam_1970_b5:
fixes ramp :: "int \<Rightarrow> (real \<Rightarrow> real)"
and F :: "real \<Rightarrow> real"
defines "ramp \<equiv> (\<lambda>n::int. (\<lambda>x::real. if x \<le> -n then -n else if (-n < x \<and> x \<le> n) then x else n))"
shows "continuous_on UNIV F \<longleftrightarrow> (\<forall>n::nat. continuous_on UNIV ((ramp n) \<circ> F))"
sorry
end
|
putnam_1986_a1 | abbrev putnam_1986_a1_solution : β := sorry
-- 18
theorem putnam_1986_a1
(S : Set β := {x : β | x ^ 4 + 36 β€ 13 * x ^ 2})
(f : β β β := fun x β¦ x ^ 3 - 3 * x)
: (β x β S, f x β€ putnam_1986_a1_solution β§ β x β S, f x = putnam_1986_a1_solution) :=
sorry | Find, with explanation, the maximum value of $f(x)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36 \leq 13x^2$. | Show that the maximum value is $18$. | ['algebra', 'analysis'] | Section putnam_1986_a1.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1986_a1_solution := 18.
Theorem putnam_1986_a1:
let f (x: R) := pow x 3 in
let on_S (x: R) := pow x 4 - 13 * pow x 2 + 36 <= 0 in
exists (m: R),
(forall (x: R), on_S x -> m >= f x) /\
(exists (x: R), on_S x -> m = f x)
<-> m = putnam_1986_a1_solution.
Proof. Admitted.
End putnam_1986_a1. | theory putnam_1986_a1 imports Complex_Main
begin
definition putnam_1986_a1_solution::real where "putnam_1986_a1_solution \<equiv> undefined"
(* 18 *)
theorem putnam_1986_a1:
fixes f::"real\<Rightarrow>real"
defines "f \<equiv> \<lambda>x. x^3 - 3 * x"
shows "putnam_1986_a1_solution = (GREATEST y. (\<exists>x. y = f x \<and> x^4 + 36 \<le> 13 * x^2))"
sorry
end
|
putnam_1986_a2 | abbrev putnam_1986_a2_solution : β := sorry
-- 3
theorem putnam_1986_a2
: (Nat.floor ((10 ^ 20000 : β) / (10 ^ 100 + 3)) % 10 = putnam_1986_a2_solution) :=
sorry | What is the units (i.e., rightmost) digit of
\[
\left\lfloor \frac{10^{20000}}{10^{100}+3}\right\rfloor ?
\] | Show that the answer is $3$. | ['algebra'] | Section putnam_1986_a2.
Require Import Nat.
Definition putnam_1986_a2_solution := 3.
Theorem putnam_1986_a2:
10 ^ (20000) / (10 ^ (100) + 3) mod 10 = putnam_1986_a2_solution.
Proof. Admitted.
End putnam_1986_a2. | theory putnam_1986_a2 imports Complex_Main
begin
definition putnam_1986_a2_solution::nat where "putnam_1986_a2_solution \<equiv> undefined"
(* 3 *)
theorem putnam_1986_a2:
shows "putnam_1986_a2_solution = \<lfloor>(10^20000) / (10^100 + 3)\<rfloor> mod 10"
sorry
end
|
putnam_1986_a3 | abbrev putnam_1986_a3_solution : β := sorry
-- Real.pi / 2
theorem putnam_1986_a3
(cot : β β β := fun ΞΈ β¦ cos ΞΈ / sin ΞΈ)
(arccot : β β β)
(harccot : β t : β, t β₯ 0 β arccot t β Set.Ioc 0 (Real.pi / 2) β§ cot (arccot t) = t)
: (β' n : β, arccot (n ^ 2 + n + 1) = putnam_1986_a3_solution) :=
sorry | Evaluate $\sum_{n=0}^\infty \mathrm{Arccot}(n^2+n+1)$, where $\mathrm{Arccot}\,t$ for $t \geq 0$ denotes the number $\theta$ in the interval $0 < \theta \leq \pi/2$ with $\cot \theta = t$. | Show that the sum equals $\pi/2$. | ['analysis'] | Section putnam_1986_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1986_a3_solution := PI / 2.
Theorem putnam_1986_a3:
Series (fun n => 1/ atan (pow (INR n) 2 + INR n + 1)) = putnam_1986_a3_solution.
Proof. Admitted.
End putnam_1986_a3. | theory putnam_1986_a3 imports Complex_Main
begin
definition putnam_1986_a3_solution::real where "putnam_1986_a3_solution \<equiv> undefined"
(* pi / 2 *)
theorem putnam_1986_a3:
fixes arccot::"real\<Rightarrow>real"
defines "arccot \<equiv> \<lambda>y. (THE x. 0 < x \<and> x \<le> pi/2 \<and> cot x = y)"
shows "(\<Sum>n::nat. arccot (n^2 + n + 1)) = putnam_1986_a3_solution"
sorry
end
|
putnam_1986_a4 | abbrev putnam_1986_a4_solution : β Γ β Γ β Γ β Γ β Γ β Γ β := sorry
-- (1, 4, 2, 3, -4, 2, 1)
theorem putnam_1986_a4
(f : β β β := fun n β¦ {A : Matrix (Fin n) (Fin n) β€ | (β i j : Fin n, A i j β ({-1, 0, 1} : Set β€)) β§ β S : β€, β Ο : Perm (Fin n), β i : Fin n, A i (Ο i) = S}.ncard)
: let (a1, b1, a2, b2, a3, b3, a4) := putnam_1986_a4_solution;
(β n > 0, f n = a1 * b1 ^ n + a2 * b2 ^ n + a3 * b3 ^ n + a4) :=
sorry | A \emph{transversal} of an $n\times n$ matrix $A$ consists of $n$ entries of $A$, no two in the same row or column. Let $f(n)$ be the number of $n \times n$ matrices $A$ satisfying the following two conditions:
\begin{enumerate}
\item[(a)] Each entry $\alpha_{i,j}$ of $A$ is in the set
$\{-1,0,1\}$.
\item[(b)] The sum of the $n$ entries of a transversal is the same for all transversals of $A$.
\end{enumerate}
An example of such a matrix $A$ is
\[
A = \left( \begin{array}{ccc} -1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 0
\end{array}
\right).
\]
Determine with proof a formula for $f(n)$ of the form
\[
f(n) = a_1 b_1^n + a_2 b_2^n + a_3 b_3^n + a_4,
\]
where the $a_i$'s and $b_i$'s are rational numbers. | Prove that $f(n) = 4^n + 2 \cdot 3^n - 4 \cdot 2^n + 1$. | ['linear_algebra'] | null | theory putnam_1986_a4 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Combinatorics.Permutations"
begin
definition putnam_1986_a4_solution::"rat\<times>rat\<times>rat\<times>rat\<times>rat\<times>rat\<times>rat" where "putnam_1986_a4_solution \<equiv> undefined"
(* (1, 4, 2, 3, -4, 2, 1) *)
theorem putnam_1986_a4:
fixes n::nat and f::nat
defines "f \<equiv> card {A::int^'a^'a. CARD('a) = n \<and> (\<forall>i::'a. \<forall>j::'a. (A$i$j) \<in> {-1, 0, 1}) \<and>
(\<exists>S::int. \<forall>f::'a\<Rightarrow>'a. f permutes UNIV \<longrightarrow> S = (\<Sum>i::'a \<in> UNIV. A$i$(f i)))}"
assumes npos : "n > 0"
shows "let (a1, b1, a2, b2, a3, b3, a4) = putnam_1986_a4_solution in (f = a1 * b1^n + a2 * b2^n + a3 * b3^n + a4)"
sorry
end
|
putnam_1986_a6 | abbrev putnam_1986_a6_solution : (β β β) β β β β := sorry
-- fun b n β¦ (β i : Finset.Icc 1 n, b i) / Nat.factorial n
theorem putnam_1986_a6
(n : β)
(npos : n > 0)
(a : β β β)
(b : β β β)
(bpos : β i β Finset.Icc 1 n, b i > 0)
(binj : β i β Finset.Icc 1 n, β j β Finset.Icc 1 n, b i = b j β i = j)
(f : Polynomial β)
(hf : β x : β, (1 - x) ^ n * f.eval x = 1 + β i : Finset.Icc 1 n, (a i) * x ^ (b i))
: (f.eval 1 = putnam_1986_a6_solution b n) :=
sorry | Let $a_1, a_2, \dots, a_n$ be real numbers, and let $b_1, b_2, \dots, b_n$ be distinct positive integers. Suppose that there is a polynomial $f(x)$ satisfying the identity
\[
(1-x)^n f(x) = 1 + \sum_{i=1}^n a_i x^{b_i}.
\]
Find a simple expression (not involving any sums) for $f(1)$ in terms of $b_1, b_2, \dots, b_n$ and $n$ (but independent of $a_1, a_2, \dots, a_n$). | Show that $f(1) = b_1 b_2 \dots b_n / n!$. | ['algebra'] | Section putnam_1986_a6.
Require Import Reals Factorial Coquelicot.Coquelicot.
Definition putnam_1986_a6_solution (m: nat -> R) (n: nat) :=
let fix prod_n (m: nat -> R) (n : nat) : R :=
match n with
| O => m 0%nat
| S n' => m n' * prod_n m n'
end in
prod_n m n / INR (fact n).
Theorem putnam_1986_a6:
forall (n: nat),
forall (a m: nat -> R) (i j: nat), Nat.lt i j -> 0 < m i < m j ->
let p (x: R) := sum_n (fun n => a n * Rpower x (m n)) n in
exists (q: R -> R), forall (x: R),
p x = (1 - x) ^ n * (q x) ->
q 1 = putnam_1986_a6_solution m n.
Proof. Admitted.
End putnam_1986_a6. | theory putnam_1986_a6 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1986_a6_solution::"(nat\<Rightarrow>nat) \<Rightarrow> nat \<Rightarrow> real" where "putnam_1986_a6_solution \<equiv> undefined"
(* \<lambda>b. \<lambda>n. (\<Prod>i=1..n. b i) / fact n *)
theorem putnam_1986_a6:
fixes n::nat and a::"nat\<Rightarrow>real" and b::"nat\<Rightarrow>nat" and f::"real poly"
assumes npos : "n > 0"
and bpos : "\<forall>i \<in> {1..n::nat}. b i > 0"
and binj : "\<forall>i \<in> {1..n::nat}. \<forall>j \<in> {1..n::nat}. b i = b j \<longrightarrow> i = j"
and hf : "\<forall>x::real. (1 - x) ^ n * (poly f x) = 1 + (\<Sum>i=1..n. (a i) * x ^ (b i))"
shows "poly f 1 = putnam_1986_a6_solution b n"
sorry
end
|
putnam_1986_b2 | abbrev putnam_1986_b2_solution : Finset (β Γ β Γ β) := sorry
-- {(0, 0, 0), (0, -1, 1), (1, 0, -1), (-1, 1, 0)}
theorem putnam_1986_b2
: ({T : β Γ β Γ β | β x y z : β, T = (x - y, y - z, z - x) β§ x * (x - 1) + 2 * y * z = y * (y - 1) + 2 * z * x β§ y * (y - 1) + 2 * z * x = z * (z - 1) + 2 * x * y} = putnam_1986_b2_solution) :=
sorry | Prove that there are only a finite number of possibilities for the ordered triple $T=(x-y,y-z,z-x)$, where $x,y,z$ are complex numbers satisfying the simultaneous equations
\[
x(x-1)+2yz = y(y-1)+2zx = z(z-1)+2xy,
\]
and list all such triples $T$. | Show that the possibilities for $T$ are $(0, 0, 0), \, (0, -1, 1), \, (1, 0, -1), \, (-1, 1, 0)$. | ['algebra'] | Section putnam_1986_b2.
Require Import Reals Ensembles Finite_sets Coquelicot.Coquelicot.
Open Scope C.
Definition putnam_1986_b2_solution (xyz : C*C*C) := xyz = (RtoC 0, RtoC 0, RtoC 0) \/ xyz = (RtoC 1, RtoC 0, RtoC (-1)) \/ xyz = (RtoC (-1), RtoC 1, RtoC 0) \/ xyz = (RtoC 0, RtoC (-1), RtoC 1).
Theorem putnam_1986_b2:
exists (n: nat),
forall (E: Ensemble (C*C*C)) (xyz: C*C*C),
let x := fst (fst xyz) in
let y := snd (fst xyz) in
let z := snd xyz in
(E (x-y,y-z,z-x) <->
x * (x - 1) * 2 * y * z = y * (y - 1) * 2 * z * x /\ y * (y - 1) * 2 * z * x = z * (z - 1) + 2 * x * y) ->
exists (n: nat), cardinal (C*C*C) E n /\ putnam_1986_b2_solution xyz.
Proof. Admitted.
End putnam_1986_b2. | theory putnam_1986_b2 imports Complex_Main
begin
definition putnam_1986_b2_solution::"(complex\<times>complex\<times>complex) set" where "putnam_1986_b2_solution \<equiv> undefined"
(* {(0, 0, 0), (0, -1, 1), (1, 0, -1), (-1, 1, 0)} *)
theorem putnam_1986_b2:
shows "putnam_1986_b2_solution = {T. \<exists>x y z::complex. T = (x - y, y - z, z - x) \<and> x*(x-1) + 2*y*z = y*(y-1) + 2*z*x
\<and> y*(y-1) + 2*z*x = z*(z-1) + 2*x*y}"
sorry
end
|
putnam_1986_b3 | theorem putnam_1986_b3
(cong : Polynomial β€ β Polynomial β€ β β€ β Prop := fun f g m β¦ β i : β, m β£ (f - g).coeff i)
(n p : β)
(nppos : n > 0 β§ p > 0)
(pprime : Nat.Prime p)
(f g h r s : Polynomial β€)
(hcoprime : cong (r * f + s * g) 1 p)
(hprod : cong (f * g) h p)
: (β F G : Polynomial β€, cong F f p β§ cong G g p β§ cong (F * G) h (p ^ n)) :=
sorry | Let $\Gamma$ consist of all polynomials in $x$ with integer coefficients. For $f$ and $g$ in $\Gamma$ and $m$ a positive integer, let $f \equiv g \pmod{m}$ mean that every coefficient of $f-g$ is an integral multiple of $m$. Let $n$ and $p$ be positive integers with $p$ prime. Given that $f,g,h,r$ and $s$ are in $\Gamma$ with $rf+sg\equiv 1 \pmod{p}$ and $fg \equiv h \pmod{p}$, prove that there exist $F$ and $G$ in $\Gamma$ with $F \equiv f \pmod{p}$, $G \equiv g \pmod{p}$, and $FG \equiv h \pmod{p^n}$. | null | ['number_theory', 'algebra'] | null | theory putnam_1986_b3 imports Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Computational_Algebra.Primes"
begin
theorem putnam_1986_b3:
fixes cong::"(int poly) \<Rightarrow> (int poly) \<Rightarrow> int \<Rightarrow> bool" and n p::nat and f g h r s::"int poly"
defines "cong \<equiv> \<lambda>f. \<lambda>g. \<lambda>m. \<forall>i::nat. m dvd (coeff (f - g) i)"
assumes nppos : "n > 0 \<and> p > 0"
and pprime : "prime p"
and hcoprime : "cong (r * f + s * g) [:1:] p"
and hprod : "cong (f * g) h p"
shows "\<exists>F G:: int poly. cong F f p \<and> cong G g p \<and> cong (F * G) h (p ^ n)"
sorry
end
|
putnam_1986_b4 | abbrev putnam_1986_b4_solution : Prop := sorry
-- True
theorem putnam_1986_b4
(G : β β β)
(hGeq : β r : β, β m n : β€, G r = |r - sqrt (m ^ 2 + 2 * n ^ 2)|)
(hGlb : β r : β, β m n : β€, G r β€ |r - sqrt (m ^ 2 + 2 * n ^ 2)|)
: (Tendsto G β€ (π 0) β putnam_1986_b4_solution) :=
sorry | For a positive real number $r$, let $G(r)$ be the minimum value of $|r - \sqrt{m^2+2n^2}|$ for all integers $m$ and $n$. Prove or disprove the assertion that $\lim_{r\to \infty}G(r)$ exists and equals $0$. | Show that the limit exists and equals $0$. | ['analysis'] | Section putnam_1986_b4.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1986_b4_solution := True.
Theorem putnam_1986_b4
(G : R -> R)
(hGeq : forall (r: R), exists (m n: Z), G r = Rabs (r - sqrt (IZR (m ^ 2 + 2 * n ^ 2))))
(hGlb : forall (r: R), forall (m n: Z), G r <= Rabs (r - sqrt (IZR (m ^ 2 + 2 * n ^ 2))))
: Lim_seq (fun n => G (INR n)) = 0 <-> putnam_1986_b4_solution.
Proof. Admitted.
End putnam_1986_b4. | theory putnam_1986_b4 imports Complex_Main
begin
definition putnam_1986_b4_solution::bool where "putnam_1986_b4_solution \<equiv> undefined"
(* True *)
theorem putnam_1986_b4:
fixes G::"real\<Rightarrow>real"
defines "G \<equiv> \<lambda>r. (LEAST y. \<exists> m n::int. y = \<bar>r - sqrt (m^2 + 2*n^2)\<bar>)"
shows "(G \<longlonglongrightarrow> 0) \<longleftrightarrow> putnam_1986_b4_solution"
sorry
end
|
putnam_1986_b5 | abbrev putnam_1986_b5_solution : Prop := sorry
-- False
theorem putnam_1986_b5
(f : MvPolynomial (Fin 3) β := (X 0) ^ 2 + (X 1) ^ 2 + (X 2) ^ 2 + (X 0) * (X 1) * (X 2))
(perms : Set (Set (MvPolynomial (Fin 3) β)) := {{X 0, X 1, X 2}, {X 0, -X 1, -X 2}, {-X 0, X 1, -X 2}, {-X 0, -X 1, X 2}})
: ((β pqr : Fin 3 β MvPolynomial (Fin 3) β, (β xyz : Fin 3 β β, MvPolynomial.eval (fun i β¦ MvPolynomial.eval xyz (pqr i)) f = MvPolynomial.eval xyz f) β ({pqr 0, pqr 1, pqr 2} β perms)) β putnam_1986_b5_solution) :=
sorry | Let $f(x,y,z) = x^2+y^2+z^2+xyz$. Let $p(x,y,z), q(x,y,z)$, $r(x,y,z)$ be polynomials with real coefficients satisfying
\[
f(p(x,y,z), q(x,y,z), r(x,y,z)) = f(x,y,z).
\]
Prove or disprove the assertion that the sequence $p,q,r$ consists of some permutation of $\pm x, \pm y, \pm z$, where the number of minus signs is $0$ or $2$. | Prove that the assertion is false. | ['algebra'] | null | theory putnam_1986_b5 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1986_b5_solution::bool where "putnam_1986_b5_solution \<equiv> undefined"
(* False *)
theorem putnam_1986_b5:
fixes f::"real poly poly poly" and perms::"((real poly poly poly) set) set" and eval::"(real poly poly poly) \<Rightarrow>real\<Rightarrow>real\<Rightarrow>real\<Rightarrow>real"
defines "f \<equiv> [:[:[: 0, 0, 1:], 0, 1:], [:0, [:0, 1:]:], 1:]" (* x^2 + y^2 + z^2 + xyz = ((0 + 0z + z^2) + 0y + y^2) + (0 + (0 + z)y) x + x^2 *)
and "perms \<equiv> {
{[:0, 1:], [:[:0, 1:]:], [:[:[:0, 1:]:]:]},
{[:0, 1:], [:[:0, -1:]:], [:[:[:0, -1:]:]:]},
{[:0, -1:], [:[:0, 1:]:], [:[:[:0, -1:]:]:]},
{[:0, -1:], [:[:0, -1:]:], [:[:[:0, 1:]:]:]}
}"
and "eval \<equiv> \<lambda>P. \<lambda>x. \<lambda>y. \<lambda>z. poly (poly (poly P [:[:z:]:]) [:y:]) x"
shows "(\<forall>p q r::real poly poly poly. (\<forall>x y z::real. eval f x y z = eval f (eval p x y z) (eval q x y z) (eval r x y z)) \<longrightarrow> ({p, q, r} \<in> perms)) \<longleftrightarrow> putnam_1986_b5_solution"
sorry
end
|
putnam_1986_b6 | theorem putnam_1986_b6
(n : β)
(npos : n > 0)
(F : Type*) [Field F]
(A B C D : Matrix (Fin n) (Fin n) F)
(hsymm : IsSymm (A * Bα΅) β§ IsSymm (C * Dα΅))
(hid : A * Dα΅ - B * Cα΅ = 1)
: (Aα΅ * D - Cα΅ * B = 1) :=
sorry | Suppose $A,B,C,D$ are $n \times n$ matrices with entries in a field $F$, satisfying the conditions that $AB^T$ and $CD^T$ are symmetric and $AD^T - BC^T = I$. Here $I$ is the $n \times n$ identity matrix, and if $M$ is an $n \times n$ matrix, $M^T$ is its transpose. Prove that $A^T D - C^T B = I$. | null | ['linear_algebra'] | null | theory putnam_1986_b6 imports Complex_Main "HOL-Algebra.Ring" "HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1986_b6:
fixes n::nat and F::"('a::{semiring_1, minus}, 'm) ring_scheme" (structure) and A B C D::"'a^'b^'b"
assumes npos : "n > 0"
and Ffield : "field F"
and matdim : "CARD('b) = n"
and hsymm1 : "A ** transpose B = transpose (A ** transpose B)"
and hsymm2 : "C ** transpose D = transpose (C ** transpose D)"
and hid : "A ** transpose D - B ** transpose C = 1"
shows "transpose A ** D - transpose C ** B = 1"
sorry
end
|
putnam_2000_a1 | abbrev putnam_2000_a1_solution : β β Set β := sorry
-- (fun A : β => Set.Ioo 0 (A ^ 2))
theorem putnam_2000_a1
(A : β)
(Apos : A > 0)
: ({S : β | β x : β β β, (β j : β, x j > 0) β§ (β' j : β, x j) = A β§ (β' j : β, (x j) ^ 2) = S} = putnam_2000_a1_solution A) :=
sorry | Let $A$ be a positive real number. What are the possible values of $\sum_{j=0}^\infty x_j^2$, given that $x_0,x_1,\ldots$ are positive numbers for which $\sum_{j=0}^\infty x_j=A$? | Show that the possible values comprise the interval $(0,A^2)$. | ['analysis'] | Section putnam_2000_a1.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2000_a1_solution (x A: R) := 0 < x < A ^ 2.
Theorem putnam_2000_a1:
forall (A: R), A > 0 ->
forall (x: nat -> R), Series x = A ->
putnam_2000_a1_solution (Series (fun j => x j ^ 2)) A.
Proof. Admitted.
End putnam_2000_a1. | theory putnam_2000_a1 imports
Complex_Main
begin
definition putnam_2000_a1_solution :: "real \<Rightarrow> real set" where "putnam_2000_a1_solution \<equiv> undefined"
(* \<lambda> A. {0 <..< A ^ 2} *)
theorem putnam_2000_a1:
fixes A :: real
assumes Apos: "A > 0"
shows "{S :: real. \<exists> x :: nat \<Rightarrow> real. (\<forall> j :: nat. x j > 0) \<and> (\<Sum> j :: nat. x j) = A \<and> (\<Sum> j :: nat. (x j) ^ 2) = S} = putnam_2000_a1_solution A"
sorry
end |
putnam_2000_a2 | theorem putnam_2000_a2
: β n : β, β N : β€, β i : Fin 6 β β, N > n β§ N = (i 0)^2 + (i 1)^2 β§ N + 1 = (i 2)^2 + (i 3)^2 β§ N + 2 = (i 4)^2 + (i 5)^2 :=
sorry | Prove that there exist infinitely many integers $n$ such that $n,n+1,n+2$ are each the sum of the squares of two integers. | null | ['number_theory'] | null | theory putnam_2000_a2 imports Complex_Main
begin
definition sum_of_squares:: "int \<Rightarrow> bool" where
"sum_of_squares n \<equiv> \<exists>a b::int. n = a^2 + b^2"
theorem putnam_2000_a2:
shows "\<forall>n :: int. \<exists>m::int. m > n \<and>
sum_of_squares m \<and> sum_of_squares (m+1) \<and> sum_of_squares (m+2)"
sorry
end
|
putnam_2000_a4 | theorem putnam_2000_a4
: β y : β, Tendsto (fun B : β => β« x in Set.Ioo 0 B, Real.sin x * Real.sin (x ^ 2)) atTop (π y) :=
sorry | Show that the improper integral $\lim_{B \to \infty} \int_0^B \sin(x)\sin(x^2)\,dx$ converges. | null | ['analysis'] | Section putnam_2000_a4.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2000_a4:
ex_lim_seq (fun n => sum_n (fun x => sin (INR x) * sin ((INR x) ^ 2)) n).
Proof. Admitted.
End putnam_2000_a4. | theory putnam_2000_a4 imports
Complex_Main
"HOL-Analysis.Interval_Integral"
begin
theorem putnam_2000_a4:
shows "\<exists> y :: real. ((\<lambda> B. interval_lebesgue_integral lebesgue 0 B (\<lambda> x. sin x * sin (x ^ 2))) \<longlongrightarrow> y) at_top"
sorry
end |
putnam_2000_a5 | theorem putnam_2000_a5
(r : β)
(z : Fin 2 β β)
(p : Fin 3 β (Fin 2 β β))
(rpos : r > 0)
(pdiff : β n m : Fin 3, (n β m) β (p n β p m))
(pint : β (n : Fin 3) (i : Fin 2), p n i = round (p n i))
(pcirc : β n : Fin 3, p n β Metric.sphere z r)
: β n m : Fin 3, (n β m) β§ (dist (p n) (p m) β₯ r ^ ((1 : β) / 3)) :=
sorry | Three distinct points with integer coordinates lie in the plane on a circle of radius $r>0$. Show that two of these points are separated by a distance of at least $r^{1/3}$. | null | ['algebra'] | null | theory putnam_2000_a5 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_2000_a5:
fixes r :: real
and z :: "real^2"
and S :: "(real^2) set"
assumes rpos: "r > 0"
and Scard: "finite S \<and> card S = 3"
and pint: "\<forall> p \<in> S. p$1 = round (p$1) \<and> p$2 = round (p$2)"
and pcirc: "\<forall> p \<in> S. p \<in> sphere z r"
shows "\<exists> p \<in> S. \<exists> q \<in> S. dist p q \<ge> r powr (1 / 3)"
sorry
end |
putnam_2000_a6 | theorem putnam_2000_a6
(f : Polynomial β€)
(a : β β β€)
(ha0 : a 0 = 0)
(ha : β n : β, a (n + 1) = f.eval (a n))
: ((β m > 0, a m = 0) β (a 1 = 0 β¨ a 2 = 0)) :=
sorry | Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0,a_1,\ldots$ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n\geq 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$. | null | ['algebra'] | null | theory putnam_2000_a6 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_2000_a6:
fixes f :: "int poly"
and a :: "nat \<Rightarrow> int"
assumes ha0: "a 0 = 0"
and ha: "\<forall>n::nat. a (n + 1) = poly f (a n)"
shows "(\<exists>m::nat>0. a m = 0) \<longrightarrow> (a 1 = 0 \<or> a 2 = 0)"
sorry
end
|
putnam_2000_b1 | theorem putnam_2000_b1
(N : β)
(a b c : Fin N β β€)
(Nge1 : N β₯ 1)
(hodd : β j : Fin N, Odd (a j) β¨ Odd (b j) β¨ Odd (c j))
: (β r s t : β€, {j : Fin N | Odd (r * a j + s * b j + t * c j)}.ncard β₯ (4 * N : β) / 7) :=
sorry | Let $a_j,b_j,c_j$ be integers for $1\leq j\leq N$. Assume for each $j$, at least one of $a_j,b_j,c_j$ is odd. Show that there exist integers $r$, $s$, $t$ such that $ra_j+sb_j+tc_j$ is odd for at least $4N/7$ values of $j$, $1\leq j\leq N$. | null | ['algebra'] | Section putnam_2000_b1.
Require Import List Nat Reals ZArith.
Open Scope Z.
Theorem putnam_2000_b1:
forall (a b c: nat -> Z) (n: nat),
(forall (j: nat), and (le 1 j) (le j n) ->
Z.odd (a j) =true \/ Z.odd (b j) = true \/ Z.odd (c j) = true) ->
exists (l: list nat), ge (length l) (4 * n / 7) /\ forall (j: nat), In j l -> and (le 1 j) (le j n) /\
exists (r s t: Z),
Z.odd (Z.add (Z.add (Z.mul r (a j)) (Z.mul s (b j))) (Z.mul t (c j))) = true.
Proof. Admitted.
End putnam_2000_b1. | theory putnam_2000_b1 imports Complex_Main
begin
(* uses (nat \<Rightarrow> int) instead of (Fin N \<Rightarrow> int) *)
theorem putnam_2000_b1:
fixes n :: nat
and a b c :: "nat \<Rightarrow> int"
assumes Nge1: "N \<ge> 1"
and hodd: "\<forall>j::nat\<in>{0..(N-1)}. odd (a j) \<or> odd (b j) \<or> odd (c j)"
shows "\<exists>r s t::int. card {j::nat\<in>{0..(N-1)}. odd (r * a j + s * b j + t * c j)} \<ge> 4*N/7"
sorry
end
|
putnam_2000_b2 | theorem putnam_2000_b2
: (β m n : β, m β₯ 1 β n β₯ m β n β£ Nat.gcd m n * Nat.choose n m) :=
sorry | Prove that the expression
\[
\frac{gcd(m,n)}{n}\binom{n}{m}
\]
is an integer for all pairs of integers $n\geq m\geq 1$. | null | ['number_theory', 'algebra'] | Section putnam_2000_b2.
Require Import Nat Reals.
Open Scope R.
Theorem putnam_2000_b2:
forall (n m: nat), and (ge n m) (ge m 1) ->
exists (c: Z),
INR (gcd m n) / INR n * Binomial.C n m = IZR c.
Proof. Admitted.
End putnam_2000_b2. | theory putnam_2000_b2 imports Complex_Main
begin
theorem putnam_2000_b2:
shows "\<forall>m n::nat. (m \<ge> 1 \<longrightarrow> n \<ge> m \<longrightarrow> n dvd ((gcd m n) * (n choose m)))"
sorry
end
|
putnam_2000_b3 | theorem putnam_2000_b3
(N : β)
(Npos : N > 0)
(a : Fin (N + 1) β β)
(haN : a N β 0)
(f : β β β := fun t β¦ β j : Icc 1 N, a j * Real.sin (2 * Real.pi * j * t))
(mult : (β β β) β β β β)
(hmult : β g : β β β, β t : β, (β c : β, iteratedDeriv c g t β 0) β (iteratedDeriv (mult g t) g t β 0 β§ β k < (mult g t), iteratedDeriv k g t = 0))
(M : β β β := fun k β¦ β' t : Ico (0 : β) 1, mult (iteratedDeriv k f) t)
: ((β i j : β, i β€ j β M i β€ M j) β§ Tendsto M β€ (π (2 * N))) :=
sorry | Let $f(t)=\sum_{j=1}^N a_j \sin(2\pi jt)$, where each $a_j$ is real and $a_N$ is not equal to $0$. Let $N_k$ denote the number of zeroes (including multiplicities) of $\frac{d^k f}{dt^k}$. Prove that
\[
N_0\leq N_1\leq N_2\leq \cdots \mbox{ and } \lim_{k\to\infty} N_k = 2N.
\] | null | ['analysis'] | null | theory putnam_2000_b3 imports Complex_Main
"HOL-Analysis.Derivative"
begin
(* uses (nat \<Rightarrow> real) instead of (Fin (N + 1) \<Rightarrow> real) *)
theorem putnam_2000_b3:
fixes N :: nat
and a :: "nat \<Rightarrow> real"
and f :: "real \<Rightarrow> real"
and mult :: "(real \<Rightarrow> real) \<Rightarrow> real \<Rightarrow> nat"
and M :: "nat \<Rightarrow> nat"
assumes Npos: "N > 0"
and haN: "a N \<noteq> 0"
defines "f \<equiv> (\<lambda>t::real. (\<Sum>j::nat=1..N. a j * sin (2*pi*j*t)))"
assumes hmult: "\<forall>(g::real\<Rightarrow>real)(t::real). ((\<exists>c::nat. (deriv^^c) g t \<noteq> 0) \<longrightarrow> mult g t = (LEAST k::nat. (deriv^^k) g t \<noteq> 0))"
defines "M \<equiv> (\<lambda>k::nat. (\<Sum>\<^sub>\<infinity>t::real\<in>{0..<1}. mult ((deriv^^k) f) t))"
shows "(\<forall>i j::nat. (i \<le> j \<longrightarrow> M i \<le> M j)) \<and> filterlim M (nhds (2*N)) at_top"
sorry
end
|
putnam_2000_b4 | theorem putnam_2000_b4
(f : β β β)
(hfcont : Continuous f)
(hf : β x : β, f (2 * x ^ 2 - 1) = 2 * x * f x)
: β x : β, x β Icc (-1) 1 β f x = 0 :=
sorry | Let $f(x)$ be a continuous function such that $f(2x^2-1)=2xf(x)$ for all $x$. Show that $f(x)=0$ for $-1\leq x\leq 1$. | null | ['analysis'] | null | theory putnam_2000_b4 imports Complex_Main
begin
theorem putnam_2000_b4:
fixes f :: "real \<Rightarrow> real"
assumes hf : "\<forall>x. f (2 * x^2 - 1) = 2 * x * f x"
and f_cont : "continuous_on UNIV f"
shows "\<forall>x. x \<ge> -1 \<and> x \<le> 1 \<longrightarrow> f x = 0"
sorry
end
|
putnam_2000_b5 | theorem putnam_2000_b5
(S : β β Set β€)
(hSfin : β n : β, Set.Finite (S n))
(hSpos : β n : β, β s β S n, s > 0)
(hSdef : β n : β, β a : β€, a β S (n + 1) β Xor' (a - 1 β S n) (a β S n))
: (β n : β, β N β₯ n, S N = S 0 βͺ {M : β€ | M - N β S 0}) :=
sorry | Let $S_0$ be a finite set of positive integers. We define finite sets $S_1,S_2,\ldots$ of positive integers as follows: the integer $a$ is in $S_{n+1}$ if and only if exactly one of $a-1$ or $a$ is in $S_n$. Show that there exist infinitely many integers $N$ for which $S_N=S_0\cup\{N+a: a\in S_0\}$. | null | ['algebra'] | null | theory putnam_2000_b5 imports Complex_Main
begin
theorem putnam_2000_b5:
fixes S :: "nat \<Rightarrow> int set"
assumes hSfin: "\<forall>n::nat. finite (S n)"
and hSpos: "\<forall>n::nat. (\<forall>s\<in>(S n). s > 0)"
and hSdef: "\<forall>n::nat. (\<forall>a::int. (a \<in> S (n + 1) \<longleftrightarrow> (((a - 1) \<in> S n) \<noteq> (a \<in> S n))))"
shows "\<forall>n::nat. \<exists>N::nat\<ge>n. S N = S 0 \<union> {M::int. M - N \<in> S 0}"
sorry
end
|
putnam_1995_a1 | theorem putnam_1995_a1
(S : Set β)
(hS : β a β S, β b β S, a * b β S)
(T U : Set β)
(hsub : T β S β§ U β S)
(hunion : T βͺ U = S)
(hdisj : T β© U = β
)
(hT3 : β a β T, β b β T, β c β T, a * b * c β T)
(hU3 : β a β U, β b β U, β c β U, a * b * c β U)
: (β a β T, β b β T, a * b β T) β¨ (β a β U, β b β U, a * b β U) :=
sorry | Let $S$ be a set of real numbers which is closed under multiplication (that is, if $a$ and $b$ are in $S$, then so is $ab$). Let $T$ and $U$ be disjoint subsets of $S$ whose union is $S$. Given that the product of any {\em three} (not necessarily distinct) elements of $T$ is in $T$ and that the product of any three elements of $U$ is in $U$, show that at least one of the two subsets $T,U$ is closed under multiplication. | null | ['algebra'] | null | theory putnam_1995_a1 imports Complex_Main
begin
theorem putnam_1995_a1:
fixes S :: "real set"
and T U :: "real set"
assumes hS: "\<forall>a\<in>S. \<forall>b\<in>S. a * b \<in> S"
and hsub: "T \<subseteq> S \<and> U \<subseteq> S"
and hunion: "T \<union> U = S"
and hdisj: "T \<inter> U = {}"
and hT3: "\<forall>a\<in>T. \<forall>b\<in>T. \<forall>c\<in>T. a * b * c \<in> T"
and hU3: "\<forall>a\<in>U. \<forall>b\<in>U. \<forall>c\<in>U. a * b * c \<in> U"
shows "(\<forall>a\<in>T. \<forall>b\<in>T. a * b \<in> T) \<or> (\<forall>a\<in>U. \<forall>b\<in>U. a * b \<in> U)"
sorry
end
|
putnam_1995_a2 | abbrev putnam_1995_a2_solution : Set (β Γ β) := sorry
-- {x | let β¨a,bβ© := x; a = b}
theorem putnam_1995_a2
(habconv : (β Γ β) β Prop := fun β¨a,bβ© =>
β limit : β, Tendsto (fun t : β => β« x in (Set.Icc b t), (sqrt (sqrt (x + a) - sqrt x) - sqrt (sqrt x - sqrt (x - b)))) atTop (π limit))
: β ab : β Γ β, habconv ab β ab β putnam_1995_a2_solution :=
sorry | For what pairs $(a,b)$ of positive real numbers does the improper integral \[ \int_{b}^{\infty} \left( \sqrt{\sqrt{x+a}-\sqrt{x}} - \sqrt{\sqrt{x}-\sqrt{x-b}} \right)\,dx \] converge? | Show that the solution is those pairs $(a,b)$ where $a = b$. | ['analysis'] | Section putnam_1995_a2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1995_a2_solution (a b: R) := a = b.
Theorem putnam_1995_a2:
forall (a b: R), a > 0 /\ b > 0 /\
ex_lim_seq (fun n => RInt (fun x => sqrt (sqrt (x + a) - sqrt x) - sqrt (sqrt x - (x - b))) b (INR n)) <->
putnam_1995_a2_solution a b.
Proof. Admitted.
End putnam_1995_a2. | theory putnam_1995_a2 imports Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1995_a2_solution :: "(real \<times> real) set" where "putnam_1995_a2_solution \<equiv> undefined"
(* {(a::real,b::real). a = b} *)
theorem putnam_1995_a2:
fixes habconv :: "(real \<times> real) \<Rightarrow> bool"
defines "habconv \<equiv> (\<lambda>(a::real,b::real). (\<exists>limit::real. filterlim (\<lambda>t::real. interval_lebesgue_integral lebesgue b t (\<lambda>x::real. sqrt (sqrt (x+a) - sqrt x) - sqrt (sqrt x - sqrt (x-b)))) (nhds limit) at_top))"
shows "\<forall>ab::real\<times>real. habconv ab \<longleftrightarrow> ab \<in> putnam_1995_a2_solution"
sorry
end
|
putnam_1995_a3 | theorem putnam_1995_a3
(relation : (Fin 9 β β€) β (Fin 9 β β€) β Prop)
(digits_to_num : (Fin 9 β β€) β β€ := fun dig => β i : Fin 9, (dig i) * 10^i.1)
(hrelation : β d e : (Fin 9 β β€), relation d e β (β i : Fin 9, d i < 10 β§ d i β₯ 0 β§ e i < 10 β§ e i β₯ 0) β§ (β i : Fin 9, 7 β£ (digits_to_num (fun j : Fin 9 => if j = i then e j else d j))))
: β d e f : (Fin 9 β β€), ((relation d e) β§ (relation e f)) β (β i : Fin 9, 7 β£ 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.] | null | ['number_theory'] | null | theory putnam_1995_a3 imports Complex_Main
begin
(* uses (nat \<Rightarrow> int) instead of (Fin 9 \<Rightarrow> int) *)
theorem putnam_1995_a3:
fixes relation :: "(nat \<Rightarrow> int) \<Rightarrow> (nat \<Rightarrow> int) \<Rightarrow> bool"
and digits_to_num :: "(nat \<Rightarrow> int) \<Rightarrow> int"
defines "digits_to_num \<equiv> (\<lambda>dig::nat\<Rightarrow>int. (\<Sum>i::nat=0..8. (dig i) * 10^i))"
assumes hrelation: "\<forall>d e::nat\<Rightarrow>int. (relation d e \<longleftrightarrow> ((\<forall>i::nat\<in>{0..8}. d i < 10 \<and> d i \<ge> 0 \<and> e i < 10 \<and> e i \<ge> 0) \<and> (\<forall>i::nat\<in>{0..8}. 7 dvd (digits_to_num (\<lambda>j::nat. if j = i then e j else d j)))))"
shows "\<forall>d e f::nat\<Rightarrow>int. ((relation d e \<and> relation e f) \<longrightarrow> (\<forall>i::nat\<in>{0..8}. 7 dvd (d i - f i)))"
sorry
end
|
putnam_1995_a4 | theorem putnam_1995_a4
(n : β)
(hn : n > 0)
(necklace : Fin n β β€)
(hnecklacesum : β i : Fin n, necklace i = n - 1)
: β cut : Fin n, β k : Fin n, β i : {j : Fin n | j.1 β€ k}, necklace (cut + i) β€ k :=
sorry | Suppose we have a necklace of $n$ beads. Each bead is labeled with an integer and the sum of all these labels is $n-1$. Prove that we can cut the necklace to form a string whose consecutive labels $x_{1},x\_{2},\dots,x_{n}$ satisfy \[\sum_{i=1}^{k} x_{i} \leq k-1 \qquad \mbox{for} \quad k=1,2,\dots,n.\] | null | ['combinatorics'] | null | theory putnam_1995_a4 imports Complex_Main
begin
(* uses (nat \<Rightarrow> int) instead of (Fin n \<Rightarrow> int) *)
theorem putnam_1995_a4:
fixes n :: nat
and necklace :: "nat \<Rightarrow> int"
assumes hn: "n > 0"
and hnecklacesum: "(\<Sum>i::nat=0..(n-1). necklace i) = n - 1"
shows "\<exists>cut::nat\<in>{0..(n-1)}. \<forall>k::nat\<in>{0..(n-1)}. (\<Sum>i::nat\<le>k. necklace ((cut + i) mod n)) \<le> k"
sorry
end
|
putnam_1995_a5 | abbrev putnam_1995_a5_solution : Prop := sorry
-- True
theorem putnam_1995_a5
(hdiffx : (n : β) β (Fin n β (β β β)) β Prop := (fun (n : β) (x : Fin n β (β β β)) => β i : Fin n, Differentiable β (x i)))
(ha : (n : β) β (Fin n β Fin n β β) β Prop := (fun (n : β) (a : Fin n β Fin n β β) => β i j : Fin n, a i j > 0))
(hcomb : (n : β) β (Fin n β (β β β)) β (Fin n β Fin n β β) β Prop := (fun (n : β) (x : Fin n β (β β β)) (a : Fin n β Fin n β β) => β t : β, β i : Fin n, (deriv (x i)) t = β j : Fin n, (a i j) * ((x j) t)))
(hxlim : (n : β) β (Fin n β (β β β)) β Prop := (fun (n : β) (x : Fin n β (β β β)) => β i : Fin n, Tendsto (x i) atTop (π 0)))
: putnam_1995_a5_solution β (β (n : β) (x : Fin n β (β β β)) (a : Fin n β Fin n β β), (n > 0 β§ hdiffx n x β§ ha n a β§ hcomb n x a β§ hxlim n x) β Β¬(β b : Fin n β β, (β t : β, β i : Fin n, (b i) * ((x i) t) = 0) β (β i : Fin n, b i = 0))) :=
sorry | Let $x_{1},x_{2},\dots,x_{n}$ be differentiable (real-valued) functions of a single variable $f$ which satisfy \begin{align*} \frac{dx_{1}}{dt} &= a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} \ \frac{dx_{2}}{dt} &= a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} \ \vdots && \vdots \ \frac{dx_{n}}{dt} &= a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} \end{align*} for some constants $a_{ij}>0$. Suppose that for all $i$, $x_{i}(t) \to 0$ as $t \to \infty$. Are the functions $x_{1},x_{2},\dots,x_{n}$ necessarily linearly dependent? | Show that the answer is yes, the functions must be linearly dependent. | ['linear_algebra', 'analysis'] | null | theory putnam_1995_a5 imports Complex_Main
"HOL-Analysis.Derivative"
begin
(* uses (nat \<Rightarrow> (real \<Rightarrow> real)) instead of (Fin n \<Rightarrow> (real \<Rightarrow> real)) and (nat \<Rightarrow> nat \<Rightarrow> real) instead of (Fin n \<Rightarrow> Fin n \<Rightarrow> real) *)
definition putnam_1995_a5_solution :: bool where "putnam_1995_a5_solution \<equiv> undefined"
(* True *)
theorem putnam_1995_a5:
fixes hdiffx :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<Rightarrow> real)) \<Rightarrow> bool"
and ha :: "nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> bool"
and hcomb :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<Rightarrow> real)) \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> bool"
and hxlim :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<Rightarrow> real)) \<Rightarrow> bool"
defines "hdiffx \<equiv> (\<lambda>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real)). (\<forall>i::nat\<in>{0..(n-1)}. (x i) differentiable_on UNIV))"
and "ha \<equiv> (\<lambda>(n::nat)(a::nat\<Rightarrow>nat\<Rightarrow>real). (\<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. a i j > 0))"
and "hcomb \<equiv> (\<lambda>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real))(a::nat\<Rightarrow>nat\<Rightarrow>real). (\<forall>t::real. \<forall>i::nat\<in>{0..(n-1)}. deriv (x i) t = (\<Sum>j::nat=0..(n-1). (a i j) * ((x j) t))))"
and "hxlim \<equiv> (\<lambda>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real)). (\<forall>i::nat\<in>{0..(n-1)}. filterlim (x i) (nhds 0) at_top))"
shows "putnam_1995_a5_solution \<longleftrightarrow> (\<forall>(n::nat)(x::nat\<Rightarrow>(real\<Rightarrow>real))(a::nat\<Rightarrow>nat\<Rightarrow>real). (n > 0 \<and> hdiffx n x \<and> ha n a \<and> hcomb n x a \<and> hxlim n x) \<longrightarrow> \<not>(\<forall>b::nat\<Rightarrow>real. (\<forall>t::real. (\<Sum>i::nat=0..(n-1). (b i) * ((x i) t)) = 0) \<longrightarrow> (\<forall>i::nat\<in>{0..(n-1)}. b i = 0)))"
sorry
end
|
putnam_1995_b1 | theorem putnam_1995_b1
(part_ct : Finpartition (Finset.range 9) β (Finset.range 9) β β)
(hp : β partition k, part_ct partition k = (Exists.choose (Finpartition.exists_mem partition k.2)).card)
: β Pt1 Pt2 : Finpartition (Finset.range 9), β x y : Finset.range 9, x β y β§ part_ct Pt1 x = part_ct Pt1 y β§ part_ct Pt2 x = part_ct Pt2 y :=
sorry | For a partition $\pi$ of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, let $\pi(x)$ be the number of elements in the part containing $x$. Prove that for any two partitions $\pi$ and $\pi'$, there are two distinct numbers $x$ and $y$ in $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that $\pi(x) = \pi(y)$ and $\pi'(x) = \pi'(y)$. [A {\em partition} of a set $S$ is a collection of disjoint subsets (parts) whose union is $S$.] | null | ['combinatorics'] | null | theory putnam_1995_b1 imports Complex_Main
"HOL-Library.Disjoint_Sets"
begin
theorem putnam_1995_b1:
fixes part_ct :: "(nat set set) \<Rightarrow> nat \<Rightarrow> nat"
assumes hp: "\<forall>(partition::nat set set)(k::nat). (\<forall>part\<in>partition. (k \<in> part \<longrightarrow> part_ct partition k = card part))"
shows "\<forall>Pt1 Pt2::nat set set. ((partition_on {1..9} Pt1 \<and> partition_on {1..9} Pt2) \<longrightarrow> (\<exists>x::nat\<in>{1..9}. \<exists>y::nat\<in>{1..9}. x \<noteq> y \<and> part_ct Pt1 x = part_ct Pt1 y \<and> part_ct Pt2 x = part_ct Pt2 y))"
sorry
end
|
putnam_1995_b3 | abbrev putnam_1995_b3_solution : β β β€ := sorry
-- fun n => if n = 1 then 45 else if n = 2 then 10 * 45^2 else 0
theorem putnam_1995_b3
(n : β)
(hn : n > 0)
(digits_set := {f : β β β | f 0 β 0 β§ (β i : Fin (n ^ 2), f i β€ 9) β§ (β i β₯ n ^ 2, f i = 0)})
(digits_to_matrix : (β β β) β Matrix (Fin n) (Fin n) β€ := fun f => (fun i j => f (i.1 * n + j.1)))
: β' f : digits_set, (digits_to_matrix f).det = putnam_1995_b3_solution n :=
sorry | To each positive integer with $n^{2}$ decimal digits, we associate the determinant of the matrix obtained by writing the digits in order across the rows. For example, for $n=2$, to the integer 8617 we associate $\det \left( \begin{array}{cc} 8 & 6 \ 1 & 7 \end{array} \right) = 50$. Find, as a function of $n$, the sum of all the determinants associated with $n^{2}$-digit integers. (Leading digits are assumed to be nonzero; for example, for $n=2$, there are 9000 determinants.) | Show that the solution is $45$ if $n = 1$, $45^2*10$ if $n = 2$, and $0$ if $n$ is greater than 3. | ['linear_algebra'] | null | theory putnam_1995_b3 imports Complex_Main
"HOL-Analysis.Determinants"
begin
(* Boosted the domain/range of digits_set to nat *)
definition putnam_1995_b3_solution :: "nat \<Rightarrow> int" where "putnam_1995_b3_solution \<equiv> undefined"
(* (\<lambda>n::nat. if n = 1 then 45 else if n = 2 then (10 * 45^2) else 0) *)
theorem putnam_1995_b3:
fixes n :: nat
and digits_set :: "(nat \<Rightarrow> nat) set"
and pnind :: "'n::finite \<Rightarrow> nat"
and digits_to_matrix :: "(nat \<Rightarrow> nat) \<Rightarrow> (int^'n^'n)"
assumes hn: "n > 0"
defines "digits_set \<equiv> {f::nat\<Rightarrow>nat. f 0 \<noteq> 0 \<and> (\<forall>i::nat\<in>{0..(n^2-1)}. f i \<le> 9) \<and> (\<forall>i::nat\<ge>n^2. f i = 0)}"
assumes pncard: "CARD('n) = n"
and hpnind: "pnind ` UNIV = {0..(n-1)}"
defines "digits_to_matrix \<equiv> (\<lambda>f::nat\<Rightarrow>nat. (\<chi> i j::'n. f ((pnind i) * n + (pnind j))))"
shows "(\<Sum>f\<in>digits_set. det (digits_to_matrix f)) = putnam_1995_b3_solution n"
sorry
end
|
putnam_1995_b4 | abbrev putnam_1995_b4_solution : β€ Γ β€ Γ β€ Γ β€ := sorry
-- β¨3,1,5,2β©
theorem putnam_1995_b4
(contfrac : β)
(hcontfrac : contfrac = 2207 - 1/contfrac)
: let β¨a,b,c,dβ© := putnam_1995_b4_solution; contfrac^((1 : β)/8) = (a + b * sqrt c)/d :=
sorry | Evaluate \[ \sqrt[8]{2207 - \frac{1}{2207-\frac{1}{2207-\dots}}}. \] Express your answer in the form $\frac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are integers. | Show that the solution is $(3 + 1*\sqrt{5})/2. | ['algebra'] | Section putnam_1995_b4.
Require Import Reals ZArith Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1995_b4_solution (a b c d: Z) := (a, b, c, d) = (3%Z,1%Z,5%Z,2%Z).
Theorem putnam_1995_b4:
exists (a b c d: Z),
exists (contfrac: R),
contfrac = 2207 - 1 / contfrac ->
pow contfrac (1 / 8) = (IZR a + IZR b * sqrt (IZR c))/IZR d <->
putnam_1995_b4_solution a b c d.
Proof. Admitted.
End putnam_1995_b4. | theory putnam_1995_b4 imports Complex_Main
begin
definition putnam_1995_b4_solution :: "int \<times> int \<times> int \<times> int" where "putnam_1995_b4_solution \<equiv> undefined"
(* (3,1,5,2) *)
theorem putnam_1995_b4:
fixes contfrac :: real
assumes hcontfrac: "contfrac = 2207 - 1/contfrac"
shows "let (a,b,c,d) = putnam_1995_b4_solution in (contfrac powr (1/8) = (a + b * sqrt c) / d)"
sorry
end
|
putnam_1995_b6 | theorem putnam_1995_b6
(S : β β Set β := fun Ξ± => {x : β | β n : β, n β₯ 1 β§ x = floor (n * Ξ±)})
: Β¬ β Ξ± Ξ² Ξ³ : β, Ξ± > 0 β§ Ξ² > 0 β§ Ξ³ > 0 β§ (S Ξ±) β© (S Ξ²) = β
β§ (S Ξ²) β© (S Ξ³) = β
β§ (S Ξ±) β© (S Ξ³) = β
β§ β+ = (S Ξ±) βͺ (S Ξ²) βͺ (S Ξ³) :=
sorry | For a positive real number $\alpha$, define \[ S(\alpha) = \{ \lfloor n\alpha \rfloor : n = 1,2,3,\dots \}. \] Prove that $\{1,2,3,\dots\}$ cannot be expressed as the disjoint union of three sets $S(\alpha), S(\beta)$ and $S(\gamma)$. [As usual, $\lfloor x \rfloor$ is the greatest integer $\leq x$.] | null | ['algebra', 'number_theory'] | null | theory putnam_1995_b6 imports Complex_Main
begin
theorem putnam_1995_b6:
fixes S :: "real \<Rightarrow> (nat set)"
defines "S \<equiv> (\<lambda>\<alpha>::real. {x::nat. (\<exists>n::nat. n \<ge> 1 \<and> x = \<lfloor>n*\<alpha>\<rfloor>)})"
shows "\<not>(\<exists>\<alpha> \<beta> \<gamma>::real. \<alpha> > 0 \<and> \<beta> > 0 \<and> \<gamma> > 0 \<and> (S \<alpha>) \<inter> (S \<beta>) = {} \<and> (S \<beta>) \<inter> (S \<gamma>) = {} \<and> (S \<alpha>) \<inter> (S \<gamma>) = {} \<and> {1::nat..} = (S \<alpha>) \<union> (S \<beta>) \<union> (S \<gamma>))"
sorry
end
|
putnam_1989_a1 | abbrev putnam_1989_a1_solution : β := sorry
-- 1
theorem putnam_1989_a1
(pdigalt : List β β Prop)
(hpdigalt : β pdig : List β, pdigalt pdig = Odd pdig.length β§ (β i : Fin pdig.length, pdig.get i = if Even (i : β) then 1 else 0))
: {p : β | p > 0 β§ p.Prime β§ pdigalt (Nat.digits 10 p)}.encard = putnam_1989_a1_solution :=
sorry | How many primes among the positive integers, written as usual in base $10$, are alternating $1$'s and $0$'s, beginning and ending with $1$? | Show that there is only one such prime. | ['algebra', 'number_theory'] | Section putnam_1989_a1.
Require Import Nat Reals ZArith Znumtheory Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1989_a1_solution (x: R) := x = INR 101.
Theorem putnam_1989_a1:
let a (n: nat) : R := sum_n (fun n => if odd n then INR (10^(n-1)) else R0) (2*n+2) in
forall (n: nat), prime (floor (a n)) -> putnam_1989_a1_solution (a n).
Proof. Admitted.
End putnam_1989_a1. | theory putnam_1989_a1 imports Complex_Main "HOL-Computational_Algebra.Primes"
begin
definition putnam_1989_a1_solution::nat where "putnam_1989_a1_solution \<equiv> undefined"
(* 1 *)
theorem putnam_1989_a1:
fixes pdigalt::"(nat list) \<Rightarrow> bool"
defines "pdigalt \<equiv> \<lambda>pdig. odd (length pdig) \<and> (\<forall>i \<in> {0..<(length pdig)}. pdig!i = (if (even i) then 1 else 0))"
shows "putnam_1989_a1_solution = card {p::nat. p > 0 \<and> prime p \<and> (\<forall>dig. (foldr (\<lambda>a b. a + 10 * b) dig 0) = p \<longrightarrow> pdigalt dig)}"
sorry
end |
putnam_1989_a2 | abbrev putnam_1989_a2_solution : β β β β β := sorry
-- (fun a b : β => (Real.exp (a ^ 2 * b ^ 2) - 1) / (a * b))
theorem putnam_1989_a2
(a b : β)
(abpos : a > 0 β§ b > 0)
: β« x in Set.Ioo 0 a, β« y in Set.Ioo 0 b, Real.exp (max (b ^ 2 * x ^ 2) (a ^ 2 * y ^ 2)) = putnam_1989_a2_solution a b :=
sorry | Evaluate $\int_0^a \int_0^b e^{\max\{b^2x^2,a^2y^2\}}\,dy\,dx$ where $a$ and $b$ are positive. | Show that the value of the integral is $(e^{a^2b^2}-1)/(ab)$. | ['analysis'] | Section putnam_1989_a2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1989_a2_solution (a b: R) := (exp (pow (a*b) 2) - 1)/(a * b).
Theorem putnam_1989_a2:
forall (a b: R),
let f (x y: R) := Rmax (pow (b*x) 2) (pow (a*y) 2) in
RInt (fun x => (RInt (fun y => exp (f x y)) 0 b)) 0 a = putnam_1989_a2_solution a b.
Proof. Admitted.
End putnam_1989_a2. | theory putnam_1989_a2 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1989_a2_solution::"real\<Rightarrow>real\<Rightarrow>real" where "putnam_1989_a2_solution \<equiv> undefined"
(* \<lambda>a b::real. (exp (a^2 * b^2) - 1) / (a * b) *)
theorem putnam_1989_a2:
fixes a b::real
assumes abpos : "a > 0 \<and> b > 0"
shows "set_lebesgue_integral lebesgue {(x::real, y::real). x \<in> {0..a} \<and> y \<in> {0..b}} (\<lambda> (x, y). exp (max (b^2 * x^2) (a^2 * y^2)))
= putnam_1989_a2_solution a b"
sorry
end |
putnam_1989_a3 | theorem putnam_1989_a3
(z : β)
(hz : 11 * z ^ 10 + 10 * I * z ^ 9 + 10 * I * z - 11 = 0)
: (βzβ = 1) :=
sorry | Prove that if
\[
11z^{10}+10iz^9+10iz-11=0,
\]
then $|z|=1.$ (Here $z$ is a complex number and $i^2=-1$.) | null | ['algebra'] | Section putnam_1989_a3.
Require Import Reals Coquelicot.Coquelicot. From Coqtail Require Import Cpow.
Open Scope C.
Theorem putnam_1989_a3:
let f (z: C) := 11 * Cpow z 10 + 10 * Ci * Cpow z 9 + 10 * Ci * z - 11 in
forall (x: C), f x = 0 <-> Cmod x = R1.
Proof. Admitted.
End putnam_1989_a3. | theory putnam_1989_a3 imports
Complex_Main
begin
theorem putnam_1989_a3:
fixes z :: complex
assumes hz: "11 * z ^ 10 + 10 * \<i> * z ^ 9 + 10 * \<i> * z - 11 = 0"
shows "norm z = 1"
sorry
end |
putnam_1989_a6 | theorem putnam_1989_a6
(F : Type*) [Field F] [Fintype F]
(hF : Fintype.card F = 2)
(Ξ± : PowerSeries F)
(hΞ± : β n : β, let bin := [1] ++ (digits 2 n) ++ [1]; PowerSeries.coeff F n Ξ± = ite (β i j : Fin bin.length, i < j β bin.get i = 1 β bin.get j = 1 β (β k, i < k β k < j β bin.get k = 0) β Even ((j : β) - (i : β) - 1)) 1 0)
: (Ξ± ^ 3 + PowerSeries.X * Ξ± + 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 $\alpha^3+x\alpha+1=0.$ | null | ['algebra', 'abstract_algebra'] | null | theory putnam_1989_a6 imports
Complex_Main
"HOL-Algebra.Ring"
"HOL-Computational_Algebra.Formal_Power_Series"
"HOL-Library.Cardinality"
begin
fun digits :: "nat \<Rightarrow> nat list" where
"digits n = (if n = 0 then [] else ([n mod 2] @ digits (n div 2)))"
theorem putnam_1989_a6:
fixes \<alpha> :: "('a::field) fps"
and X :: "'a fps"
defines "\<alpha> \<equiv> Abs_fps (\<lambda> n :: nat. let bin = [1] @ (digits n) @ [1] in (if (\<forall> i \<in> {0 ..< length bin}. \<forall> j \<in> {0 ..< length bin}. i < j \<longrightarrow> bin!i = 1 \<longrightarrow> bin!j = 1 \<longrightarrow> (\<forall> k :: nat. i < k \<longrightarrow> k < j \<longrightarrow> bin!k = 0) \<longrightarrow> even (j - i - 1)) then 1 else 0))"
and "X \<equiv> Abs_fps (\<lambda> n :: nat. if n = 1 then 1 else 0)"
assumes hF: "CARD ('a) = 2"
shows "\<alpha> ^ 3 + X * \<alpha> + 1 = 0"
sorry
end |
putnam_1989_b2 | abbrev putnam_1989_b2_solution : Prop := sorry
-- True
theorem putnam_1989_b2
(pow : (S : Type) β β β S β S)
(hpow1 : β (S : Type) (_ : Semigroup S), β s : S, pow S 1 s = s)
(hpown : β (S : Type) (_ : Semigroup S), β s : S, β n > 0, pow S (n + 1) s = s * (pow S n s))
: ((β (S : Type) (_ : Nonempty S) (_ : Semigroup S) (_ : IsCancelMul S), (β a : S, Finite {x | β n : β, n > 0 β§ pow S n a = x}) β β (_ : Group S), True) β putnam_1989_b2_solution) :=
sorry | Let $S$ be a non-empty set with an associative operation that is left and right cancellative ($xy=xz$ implies $y=z$, and $yx=zx$ implies $y=z$). Assume that for every $a$ in $S$ the set $\{a^n:\,n=1, 2, 3, \ldots\}$ is finite. Must $S$ be a group? | Prove that $S$ must be a group. | ['abstract_algebra'] | null | theory putnam_1989_b2 imports
Complex_Main
"HOL-Algebra.Complete_Lattice"
"HOL-Library.FuncSet"
begin
record 'a semigroup = "'a partial_object" +
mult :: "['a, 'a] \<Rightarrow> 'a" (infixl "\<bullet>\<index>" 70)
locale semigroup =
fixes G (structure)
assumes m_closed [intro, simp]:
"\<lbrakk>x \<in> carrier G; y \<in> carrier G\<rbrakk> \<Longrightarrow> x \<bullet> y \<in> carrier G"
and m_assoc:
"\<lbrakk>x \<in> carrier G; y \<in> carrier G; z \<in> carrier G\<rbrakk>
\<Longrightarrow> (x \<bullet> y) \<bullet> z = x \<bullet> (y \<bullet> z)"
(* Note: this problem includes a quantifier over algebraic structures of a certain type. We therefore assume that this type has cardinality at least that of the reals.*)
definition putnam_1989_b2_solution :: bool where "putnam_1989_b2_solution \<equiv> undefined"
(* True *)
theorem putnam_1989_b2:
fixes cancel :: "'a semigroup \<Rightarrow> bool"
and group :: "'a semigroup \<Rightarrow> bool"
defines "cancel \<equiv> \<lambda> S. \<forall> x \<in> carrier S. \<forall> y \<in> carrier S. \<forall> z \<in> carrier S. (x \<bullet>\<^bsub>S\<^esub> y = x \<bullet>\<^bsub>S\<^esub> z \<longrightarrow> y = z) \<and> (y \<bullet>\<^bsub>S\<^esub> x = z \<bullet>\<^bsub>S\<^esub> x \<longrightarrow> y = z)"
and "group \<equiv> \<lambda> S. \<exists> e \<in> carrier S. \<forall> x \<in> carrier S. x \<bullet>\<^bsub>S\<^esub> e = x \<and> e \<bullet>\<^bsub>S\<^esub> x = x \<and> (\<exists> y \<in> carrier S. x \<bullet>\<^bsub>S\<^esub> y = e \<and> y \<bullet>\<^bsub>S\<^esub> x = e)"
assumes pacard: "\<exists> pamap :: 'a \<Rightarrow> real. surj pamap"
shows "(\<forall> S :: 'a semigroup. (carrier S \<noteq> {} \<and> cancel S \<and> (\<forall> a \<in> carrier S. finite {x \<in> carrier S. \<exists> n :: nat. x = foldr (\<bullet>\<^bsub>S\<^esub>) (replicate n a) a})) \<longrightarrow> group S) \<longleftrightarrow> putnam_1989_b2_solution"
sorry
end |
putnam_1989_b3 | abbrev putnam_1989_b3_solution : β β β β β := sorry
-- fun n c β¦ c * factorial n / (3 ^ n * β m in Finset.Icc (1 : β€) n, (1 - 2 ^ (-m)))
theorem putnam_1989_b3
(f : β β β)
(hfdiff : Differentiable β f)
(hfderiv : β x > 0, deriv f x = -3 * f x + 6 * f (2 * x))
(hdecay : β x β₯ 0, |f x| β€ Real.exp (-Real.sqrt x))
(ΞΌ : β β β := fun n β¦ β« x in Set.Ioi 0, x ^ n * (f x))
: ((β n : β, ΞΌ n = putnam_1989_b3_solution n (ΞΌ 0)) β§ (β L : β, Tendsto (fun n β¦ (ΞΌ n) * 3 ^ n / factorial n) β€ (π L)) β§ (Tendsto (fun n β¦ (ΞΌ n) * 3 ^ n / factorial n) β€ (π 0) β ΞΌ 0 = 0)) :=
sorry | Let $f$ be a function on $[0,\infty)$, differentiable and satisfying
\[
f'(x)=-3f(x)+6f(2x)
\]
for $x>0$. Assume that $|f(x)|\le e^{-\sqrt{x}}$ for $x\ge 0$ (so that $f(x)$ tends rapidly to $0$ as $x$ increases). For $n$ a non-negative integer, define
\[
\mu_n=\int_0^\infty x^n f(x)\,dx
\]
(sometimes called the $n$th moment of $f$).
\begin{enumerate}
\item[a)] Express $\mu_n$ in terms of $\mu_0$.
\item[b)] Prove that the sequence $\{\mu_n \frac{3^n}{n!}\}$ always converges, and that the limit is $0$ only if $\mu_0=0$.
\end{enumerate} | Show that for each $n \geq 0$, $\mu_n = \frac{n!}{3^n} \left( \prod_{m=1}^{n}(1 - 2^{-m}) \right)^{-1} \mu_0$. | ['analysis'] | null | theory putnam_1989_b3 imports
Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1989_b3_solution :: "nat \<Rightarrow> real \<Rightarrow> real" where "putnam_1989_b3_solution \<equiv> undefined"
(* \<lambda> n c. c * fact n / (3 ^ n * (\<Prod> m = 1..n. 1 - 2 powi (-m))) *)
theorem putnam_1989_b3:
fixes f :: "real \<Rightarrow> real"
and \<mu> :: "nat \<Rightarrow> real"
defines "\<mu> \<equiv> \<lambda> n. interval_lebesgue_integral lebesgue 0 \<infinity> (\<lambda> x. x ^ n * f x)"
assumes hfdiff: "f differentiable_on {0..}"
and hfderiv: "\<forall> x > 0. deriv f x = -3 * f x + 6 * f (2 * x)"
and hdecay: "\<forall> x \<ge> 0. \<bar>f x\<bar> \<le> exp (-sqrt x)"
shows "(\<forall> n :: nat. \<mu> n = putnam_1989_b3_solution n (\<mu> 0)) \<and> convergent (\<lambda> n :: nat. \<mu> n * 3 ^ n / fact n) \<and> (((\<lambda> n :: nat. \<mu> n * 3 ^ n / fact n) \<longlonglongrightarrow> 0) \<longrightarrow> (\<mu> 0 = 0))"
sorry
end |
putnam_1989_b4 | abbrev putnam_1989_b4_solution : Prop := sorry
-- True
theorem putnam_1989_b4
: ((β S : Type, Countable S β§ Infinite S β§ β C : Set (Set S), Β¬Countable C β§ (β R β C, R β β
) β§ (β A β C, β B β C, A β B β (A β© B).Finite)) β putnam_1989_b4_solution) :=
sorry | Can a countably infinite set have an uncountable collection of non-empty subsets such that the intersection of any two of them is finite? | Prove that such a collection exists. | ['set_theory'] | null | theory putnam_1989_b4 imports
Complex_Main
"HOL-Library.Countable_Set"
begin
definition putnam_1989_b4_solution :: bool where "putnam_1989_b4_solution \<equiv> undefined"
(* True *)
theorem putnam_1989_b4:
shows "(\<exists> C :: nat set set. uncountable C \<and> (\<forall> R \<in> C. R \<noteq> {}) \<and> (\<forall> A \<in> C. \<forall> B \<in> C. A \<noteq> B \<longrightarrow> finite (A \<inter> B))) \<longleftrightarrow> putnam_1989_b4_solution"
sorry
end |
putnam_1997_a3 | abbrev putnam_1997_a3_solution : β := sorry
-- Real.sqrt (Real.exp 1)
theorem putnam_1997_a3
(series1 : β β β := fun x => β' n : β, (-1)^n * x^(2*n + 1)/(β i : Finset.range n, 2 * ((i : β) + 1)))
(series2 : β β β := fun x => β' n : β, x^(2*n)/(β i : Finset.range n, (2 * ((i : β) + 1))^2))
: Tendsto (fun t => β« x in Set.Icc 0 t, series1 x * series2 x) atTop (π (putnam_1997_a3_solution)) :=
sorry | Evaluate \begin{gather*} \int_0^\infty \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right) \\ \left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2 \cdot 6^2}+\cdots\right)\,dx. \end{gather*} | Show that the solution is $\sqrt{e}$. | ['analysis'] | Section putnam_1997_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1997_a3_solution := sqrt (exp 1).
Theorem putnam_1997_a3:
let fix even_fact (n : nat) : R :=
match n with
| O => 1
| S n' => (2 * INR n) * even_fact n'
end in
let fix even_fact_sqr (n : nat) : R :=
match n with
| O => 1
| S n' => pow (2 * INR n) 2 * even_fact n'
end in
let f (x: R) := Series (fun n => pow (-1) n * pow x (2 * n + 1) / even_fact n) in
let g (x: R) := Series (fun n => pow x (2 * n) / even_fact_sqr n) in
Lim_seq (fun n => sum_n (fun m => RInt (fun x => f x * g x) 0 (INR m)) n) = putnam_1997_a3_solution.
Proof. Admitted.
End putnam_1997_a3. | theory putnam_1997_a3 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1997_a3_solution::real where "putnam_1997_a3_solution \<equiv> undefined"
(* sqrt (exp 1) *)
theorem putnam_1997_a3:
fixes series1 series2::"real\<Rightarrow>real"
defines "series1 \<equiv> \<lambda>x. (\<Sum>n::nat. (-1)^n * x^(2*n+1) / (\<Prod>i=0..<n. 2 * (i + 1)))"
and "series2 \<equiv> \<lambda>x. (\<Sum>n::nat. x^(2*n) / (\<Prod>i=0..<n. (2 * (i+1))^2))"
shows "((\<lambda>t::real. interval_lebesgue_integral lebesgue 0 t (\<lambda>x. series1 x * series2 x)) \<longlongrightarrow> putnam_1997_a3_solution) at_top"
sorry
end |
putnam_1997_a4 | theorem putnam_1997_a4
(G : Type*)
[Group G]
(Ο : G β G)
(hΟ : β g1 g2 g3 h1 h2 h3 : G, (g1 * g2 * g3 = 1 β§ h1 * h2 * h3 = 1) β Ο g1 * Ο g2 * Ο g3 = Ο h1 * Ο h2 * Ο h3)
: β a : G, let Ο := fun g => a * Ο g; β x y : G, Ο (x * y) = Ο x * Ο y :=
sorry | Let $G$ be a group with identity $e$ and $\phi:G\rightarrow G$ a function such that \[\phi(g_1)\phi(g_2)\phi(g_3)=\phi(h_1)\phi(h_2)\phi(h_3)\] whenever $g_1g_2g_3=e=h_1h_2h_3$. Prove that there exists an element $a\in G$ such that $\psi(x)=a\phi(x)$ is a homomorphism (i.e. $\psi(xy)=\psi(x)\psi(y)$ for all $x,y\in G$). | null | ['abstract_algebra'] | null | theory putnam_1997_a4 imports Complex_Main
begin
theorem putnam_1997_a4:
fixes Gmul::"'g \<Rightarrow> 'g \<Rightarrow> 'g" (infixl "\<^bold>*" 70) and e::'g and inv::"'g\<Rightarrow>'g" and \<phi>::"'g\<Rightarrow>'g"
assumes hgroup : "group (\<^bold>*) e inv"
and hphi : "\<forall> g1 g2 g3 h1 h2 h3::'g. (g1 \<^bold>* g2 \<^bold>* g3 = e \<and> h1 \<^bold>* h2 \<^bold>* h3 = e) \<longrightarrow> \<phi> g1 \<^bold>* \<phi> g2 \<^bold>* \<phi> g3 = \<phi> h1 \<^bold>* \<phi> h2 \<^bold>* \<phi> h3"
shows "\<exists>a::'g. \<forall>x y::'g. a \<^bold>* \<phi> (x \<^bold>* y) = (a \<^bold>* \<phi> x) \<^bold>* (a \<^bold>* \<phi> y)"
sorry
end |
putnam_1997_a5 | abbrev putnam_1997_a5_solution : Prop := sorry
-- True
theorem putnam_1997_a5
(N := fun (n : β+) => {t : Fin n β β+ | (β i j : Fin n, i < j β t i <= t j) β§ (β i : Fin n, (1 : β)/(t i) = 1) })
: Odd (N 10).ncard β putnam_1997_a5_solution :=
sorry | Let $N_n$ denote the number of ordered $n$-tuples of positive integers $(a_1,a_2,\ldots,a_n)$ such that $1/a_1 + 1/a_2 +\ldots + 1/a_n=1$. Determine whether $N_{10}$ is even or odd. | Show that $N_{10}$ is odd. | ['number_theory'] | Section putnam_1997_a5.
Require Import Nat Ensembles Finite_sets List Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1997_a5_solution := True.
Theorem putnam_1997_a5:
forall (E: Ensemble (list nat)) (l: list nat),
length l = 10%nat /\ (E l <-> sum_n (fun i => 1/ INR (nth i l 0%nat)) 10 = 1) ->
exists (m: nat), cardinal (list nat) E m /\ odd m = true <->
putnam_1997_a5_solution.
Proof. Admitted.
End putnam_1997_a5. | theory putnam_1997_a5 imports Complex_Main
begin
definition putnam_1997_a5_solution where "putnam_1997_a5_solution \<equiv> undefined"
(* True *)
theorem putnam_1997_a5:
fixes N::"nat\<Rightarrow>nat"
defines "N \<equiv> \<lambda>n. card {t::nat list. (size t = n) \<and> (\<forall>i \<in> {0..<n}. t!i > 0) \<and> (\<forall>i \<in> {0..<n}. \<forall>j \<in> {0..<n}. i < j \<longrightarrow> t!i \<le> t!j)
\<and> (\<Sum>i=0..<n. 1 / (t!i)) = 1}"
shows "odd (N 10) \<longleftrightarrow> putnam_1997_a5_solution"
sorry
end |
putnam_1997_a6 | abbrev putnam_1997_a6_solution : β€ β β€ β β := sorry
-- fun n k => Nat.choose (n.toNat-1) (k.toNat-1)
theorem putnam_1997_a6
(n : β€)
(hn : n > 0)
(C : β)
(x : β β (β€ β β))
(hx0 : β c : β, x c 0 = 0)
(hx1 : β c : β, x c 1 = 1)
(hxk : β c : β, β k β₯ 0, x c (k + 2) = (c*(x c (k + 1)) - (n - k)*(x c k))/(k + 1))
(S : Set β := {c : β | x c (n + 1) = 0})
(hC : C = sSup S)
: β k : Set.Icc 1 n, x C k = putnam_1997_a6_solution n k :=
sorry | For a positive integer $n$ and any real number $c$, define $x_k$ recursively by $x_0=0$, $x_1=1$, and for $k\geq 0$, \[x_{k+2}= rac{cx_{k+1}-(n-k)x_k}{k+1}.\] Fix $n$ and then take $c$ to be the largest value for which $x_{n+1}=0$. Find $x_k$ in terms of $n$ and $k$, $1\leq k\leq n$. | Show that the solution is that $x_k = {n - 1 \choose k - 1}$. | ['algebra'] | Section putnam_1997_a6.
Require Import Binomial Nat Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1997_a6_solution (n k: nat) := Binomial.C (n - 1) (k - 1).
Theorem putnam_1997_a6:
let fix x (n: nat) (c: R) (k: nat) : R :=
match k with
| O => 0
| S O => 1
| S ((S k'') as k') => (c * x n c k' - INR (n - k) * x n c k'') / INR k'
end in
forall (n: nat), exists (maxc: R), forall (c: R),
x n c (S n) = 0 /\ x n maxc (S n) = 0 -> c <= maxc ->
forall (k: nat), and (le 1 k) (le k n) ->
x n c k = putnam_1997_a6_solution n k.
Proof. Admitted.
End putnam_1997_a6. | theory putnam_1997_a6 imports Complex_Main
begin
definition putnam_1997_a6_solution :: "int \<Rightarrow> int \<Rightarrow> real" where
"putnam_1997_a6_solution \<equiv> undefined"
(* \<lambda> n k. (nat n-1) choose (nat k-1) *)
theorem putnam_1997_a6:
fixes n :: "int"
and C :: "real"
and x :: "real \<Rightarrow> (int \<Rightarrow> real)"
and S :: "real set"
defines "S \<equiv> {c :: real. x c (n + 1) = 0}"
assumes hx0 : "\<forall> c :: real. x c 0 = 0"
and hx1 : "\<forall> c :: real. x c 1 = 1"
and hxk : "\<forall> (c :: real) (k :: nat). x c (k + 2) = (c * (x c (k+1)) - (n-k) * (x c k))/(k+1)"
and hC : "C = (GREATEST s. s \<in> S)"
and hn : "n > 0"
shows "\<forall> k \<in> {1..n}. x C k = putnam_1997_a6_solution n k"
sorry
end |
putnam_1997_b1 | abbrev putnam_1997_b1_solution : β β β := sorry
-- fun n => n
noncomputable def dist_to_int : β β β := fun r => |r - round r|
theorem putnam_1997_b1
(F : β β β := fun n => β m in Finset.Icc 1 (6 * n - 1), min (dist_to_int (m/(6*n)) ) (dist_to_int (m/(3*n))))
: β n, n > 0 β F n = putnam_1997_b1_solution n :=
sorry | Let $\{x\}$ denote the distance between the real number $x$ and the nearest integer. For each positive integer $n$, evaluate \[F_n=\sum_{m=1}^{6n-1} \min(\{\frac{m}{6n}\},\{\frac{m}{3n}\}).\] (Here $\min(a,b)$ denotes the minimum of $a$ and $b$.) | Show that the solution is $n$. | ['algebra'] | Section putnam_1997_b1.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1997_b1_solution (n: nat) := INR n.
Theorem putnam_1997_b1:
let rnd (x: R) := Rmin (Rabs (IZR (floor x) - x)) (Rabs (IZR (floor (x + 1)) - x)) in
forall (n: nat), gt n 0 ->
sum_n (fun m => Rmin (rnd ((INR m + 1) / (6 * INR n))) (rnd ((INR m + 1) / (3 * INR n)))) (6 * n - 1) = putnam_1997_b1_solution n.
Proof. Admitted.
End putnam_1997_b1. | theory putnam_1997_b1 imports Complex_Main
begin
definition putnam_1997_b1_solution :: "nat \<Rightarrow> real" where
"putnam_1997_b1_solution \<equiv> undefined"
(* \<lambda> n. n *)
definition dist_to_int :: "real \<Rightarrow> real" where "dist_to_int r = \<bar>r - round r\<bar>"
theorem putnam_1997_b1:
fixes F :: "nat \<Rightarrow> real"
defines "F \<equiv> \<lambda> n. (\<Sum> m \<in> {1..6*n-1}. min (dist_to_int (m/(6*n))) (dist_to_int (m/(3*n))))"
shows "\<forall> n :: nat > 0. F n = putnam_1997_b1_solution n"
sorry
end |
End of preview. Expand
in Dataset Viewer.
YAML Metadata
Warning:
empty or missing yaml metadata in repo card
(https://huggingface.co/docs/hub/datasets-cards)
Link to the repository on GitHub: https://github.com/trishullab/PUTNAM
PutnamBench
PutnamBench is a benchmark for the evaluation of theorem-proving algorithms on competition mathematics problems sourced from the William Lowell Putnam Mathematical Competition years 1965 - 2023. Our formalizations currently support three formal languages: Lean 4 $\land$ Isabelle $\land$ Coq. PutnamBench comprises over 1300 manual formalizations, aggregated over all languages.
PutnamBench aims to support research in automated mathematical reasoning by providing a multilingual benchmark for evaluating theorem-proving algorithms. It is released under permissive licenses (Apache 2.0 for Lean 4 and Isabelle, MIT for Coq), and we encourage community contributions (TODO: After initial release)
Author List
George Tsoukalas (corresponding author email: george.tsoukalas@utexas.edu) Jasper Lee John Jennings Jimmy Xin Michelle Ding Michael Jennings Amitayush Thakur Swarat Chaudhuri
Statistics
| Language | Count |
|---|---|
| Lean 4 | 514 |
| Isabelle | 514 |
| Coq | 309 |
We also report the number of problems in a certain category. Note that some problems fall under multiple categories.
| Category | Total Quantity |
|---|---|
| Algebra | 218 |
| Analysis | 176 |
| Number Theory | 97 |
| Linear Algebra | 43 |
| Abstract Algebra | 25 |
| Geometry | 22 |
| Combinatorics | 12 |
| Set Theory | 4 |
| Probability | 2 |
license: apache-2.0
- Downloads last month
- 5