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putnam_1999_b3 | abbrev putnam_1999_b3_solution : β := sorry
-- 3
theorem putnam_1999_b3
(A : Set (β Γ β) := {xy : β Γ β | 0 β€ xy.1 β§ xy.1 < 1 β§ 0 β€ xy.2 β§ xy.2 < 1})
(S : β β β β β := fun x y => β' m : β, β' n : β, if (m > 0 β§ n > 0 β§ 1/2 β€ m/n β§ m/n β€ 2) then x^m * y^n else 0)
: Tendsto (fun xy : (β Γ β) => (1 - xy.1 * xy.2^2) * (1 - xy.1^2 * xy.2) * (S xy.1 xy.2)) (π[A] β¨1,1β©) (π putnam_1999_b3_solution) :=
sorry | Let $A=\{(x,y):0\leq x,y<1\}$. For $(x,y)\in A$, let \[S(x,y) = \sum_{\frac{1}{2}\leq \frac{m}{n}\leq 2} x^m y^n,\] where the sum ranges over all pairs $(m,n)$ of positive integers satisfying the indicated inequalities. Evaluate \[\lim_{(x,y)\rightarrow (1,1), (x,y)\in A} (1-xy^2)(1-x^2y)S(x,y).\] | Show that the answer is 3. | ['algebra'] | null | theory putnam_1999_b3 imports
Complex_Main
begin
definition putnam_1999_b3_solution :: real where "putnam_1999_b3_solution \<equiv> undefined"
(* 3 *)
theorem putnam_1999_b3:
fixes A :: "(real \<times> real) set"
and S :: "real \<Rightarrow> real \<Rightarrow> real"
defines "A \<equiv> {xy :: real \<times> real. 0 \<le> (fst xy) \<and> (fst xy) < 1 \<and> 0 \<le> (snd xy) \<and> (snd xy) < 1}"
and "S \<equiv> \<lambda> x y. \<Sum> m :: nat. \<Sum> n :: nat. (if (m > 0 \<and> n > 0 \<and> 1 / 2 \<le> m / n \<and> m / n \<le> 2) then x ^ m * y ^ n else 0)"
shows "filterlim (\<lambda> xy :: real \<times> real. (1 - (fst xy) * (snd xy) ^ 2) * (1 - (fst xy) ^ 2 * (snd xy)) * (S (fst xy) (snd xy))) (nhds putnam_1999_b3_solution) (at (1, 1) within A)"
sorry
end | null |
putnam_1999_b4 | theorem putnam_1999_b4
(f : β β β)
(hf : ContDiff β 3 f)
(hpos: β n β€ 3, β x : β, iteratedDeriv n f x > 0)
(hle : β x : β, iteratedDeriv 3 f x β€ f x)
: β x : β, deriv f x < 2 * (f x) :=
sorry | Let $f$ be a real function with a continuous third derivative such that $f(x), f'(x), f''(x), f'''(x)$ are positive for all $x$. Suppose that $f'''(x)\leq f(x)$ for all $x$. Show that $f'(x)<2f(x)$ for all $x$. | null | ['analysis'] | null | theory putnam_1999_b4 imports Complex_Main
"HOL-Analysis.Derivative"
begin
theorem putnam_1999_b4:
fixes f :: "real\<Rightarrow>real"
assumes f_cont : "continuous_on UNIV ((deriv^^3) f)"
and f_pos : "\<forall>x. f x > 0"
and f'_pos : "\<forall>x. deriv f x > 0"
and f''_pos : "\<forall>x. (deriv^^2) f x > 0"
and f'''_pos : "\<forall>x. (deriv^^3) f x > 0"
and hf : "\<forall>x. (deriv^^3) f x \<le> f x"
shows "\<forall>x. deriv f x < 2 * f x"
sorry
end
| null |
putnam_1999_b5 | abbrev putnam_1999_b5_solution : β β β := sorry
-- fun n => 1 - n^2/4
theorem putnam_1999_b5
(n : β)
(hn : n β₯ 3)
(theta : β := 2 * Real.pi / n)
(A : Matrix (Fin n) (Fin n) β := fun j k => Real.cos ((j.1 + 1) * theta + (k.1 + 1) * theta))
: ((1 : Matrix (Fin n) (Fin n) β) + A).det = putnam_1999_b5_solution n :=
sorry | For an integer $n\geq 3$, let $\theta=2\pi/n$. Evaluate the determinant of the $n\times n$ matrix $I+A$, where $I$ is the $n\times n$ identity matrix and $A=(a_{jk})$ has entries $a_{jk}=\cos(j\theta+k\theta)$ for all $j,k$. | Show that the answer is $(1 - n^2)/4$. | ['linear_algebra'] | null | theory putnam_1999_b5 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Analysis.Determinants"
begin
definition putnam_1999_b5_solution :: "nat \<Rightarrow> real" where "putnam_1999_b5_solution \<equiv> undefined"
(* \<lambda> n. 1 - n ^ 2 / 4 *)
theorem putnam_1999_b5:
fixes n :: nat
and theta :: real
and A :: "real^'n^'n"
and idx :: "'n \<Rightarrow> nat"
defines "theta \<equiv> 2 * pi / n"
and "A \<equiv> \<chi> j k. cos (idx j * theta + idx k * theta)"
assumes hn: "n \<ge> 3"
and hncard: "CARD('n) = n"
and idxbij: "\<forall> i \<in> {1..n}. \<exists>! m :: 'n. idx m = i"
shows "det (mat 1 + A) = putnam_1999_b5_solution n"
sorry
end | null |
putnam_1999_b6 | theorem putnam_1999_b6
(S : Finset β€)
(hSgt : β s : β€, s β S β s > 1)
(hSgcd : β n : β€, β s : S, Int.gcd s n = 1 β¨ Int.gcd s n = (s : β€))
: β s t : S, Prime (Int.gcd s t) :=
sorry | Let $S$ be a finite set of integers, each greater than 1. Suppose that for each integer $n$ there is some $s\in S$ such that $\gcd(s,n)=1$ or $\gcd(s,n)=s$. Show that there exist $s,t\in S$ such that $\gcd(s,t)$ is prime. | null | ['number_theory'] | null | theory putnam_1999_b6 imports Complex_Main
"HOL-Computational_Algebra.Primes"
begin
theorem putnam_1999_b6:
fixes S::"int set"
assumes S_fin: "finite S"
and Sge1: "\<forall>s \<in> S. s > 1"
and hgcd: "\<forall>n::int. \<exists>s \<in> S. (gcd s n) = 1 \<or> (gcd s n) = s"
shows "\<exists> s \<in> S. \<exists> t \<in> S. prime(gcd s t)"
sorry
end
| null |
putnam_1974_a1 | abbrev putnam_1974_a1_solution : β := sorry
-- 11
theorem putnam_1974_a1
(conspiratorial : Set β€ β Prop := fun S => β a β S, β b β S, β c β S, (a > 0 β§ b > 0 β§ c > 0) β§ ((a β b β§ b β c β§ a β c) β (Int.gcd a b > 1 β¨ Int.gcd b c > 1 β¨ Int.gcd a c > 1)))
: (β S : Set β€, S β Icc 1 16 β conspiratorial S β S.encard β€ putnam_1974_a1_solution) β§ (β S : Set β€, S β Icc 1 16 β§ conspiratorial S β§ S.encard = putnam_1974_a1_solution) := sorry | Call a set of positive integers 'conspiratorial' if no three of them are pairwise relatively prime. What is the largest number of elements in any conspiratorial subset of the integers 1 through 16? | Show that the answer is 11. | ['number_theory'] | Section putnam_1974_a1.
Require Import Nat List Ensembles Finite_sets. From mathcomp Require Import div fintype perm ssrbool.
Definition putnam_1974_a1_solution := 11.
Theorem putnam_1974_a1
(s : list nat := seq 1 17)
: exists (m: nat),
((forall (p : nat -> Prop),
let E: Ensemble nat := fun x : nat => List.In x s /\ p x in
~ exists (p q r : nat), p <> q /\ q <> r /\ coprime p q /\ coprime q r /\ coprime p r /\ exists (n: nat), cardinal nat E n /\ n <= m) /\
(exists (p : nat -> Prop),
let E: Ensemble nat := fun x => List.In x s /\ p x in
~ exists (p q r : nat), p <> q /\ q <> r /\ coprime p q /\ coprime q r /\ coprime p r /\ exists (n: nat), cardinal nat E m)) <->
m = putnam_1974_a1_solution.
Proof. Admitted.
End putnam_1974_a1. | theory putnam_1974_a1 imports Complex_Main
begin
definition putnam_1974_a1_solution :: nat where "putnam_1974_a1_solution \<equiv> undefined"
(* 11 *)
theorem putnam_1974_a1:
fixes conspiratorial :: "nat set \<Rightarrow> bool"
defines "conspiratorial \<equiv> (\<lambda>S::nat set. (\<forall>a\<in>S. \<forall>b\<in>S. \<forall>c\<in>S. a \<noteq> 0 \<and> ((a \<noteq> b \<and> b \<noteq> c \<and> a \<noteq> c) \<longrightarrow> (gcd a b > 1 \<or> gcd b c > 1 \<or> gcd a c > 1))))"
shows "(GREATEST Scard::nat. (\<exists>S::nat set. S \<subseteq> {1..16} \<and> conspiratorial S \<and> Scard = card S)) = putnam_1974_a1_solution"
sorry
end
| null |
putnam_1974_a3 | abbrev putnam_1974_a3_solution : (Set β) Γ (Set β) := sorry
-- ({p : β | p.Prime β§ p β‘ 1 [MOD 8]}, {p : β | p.Prime β§ p β‘ 5 [MOD 8]})
theorem putnam_1974_a3
(assmption : β p : β, p.Prime β§ p > 2 β ((β m n : β€, p = m^2 + n^2) β p β‘ 1 [MOD 4]))
: β p : β, ((p.Prime β§ p > 2 β§ (β x y : β€, p = x^2 + 16*y^2)) β p β putnam_1974_a3_solution.1) β§ ((p.Prime β§ p > 2 β§ (β x y : β€, p = 4*x^2 + 4*x*y + 5*y^2)) β p β putnam_1974_a3_solution.2) :=
sorry | A well-known theorem asserts that a prime $p > 2$ can be written as the sum of two perfect squres if and only if $p \equiv 1 \bmod 4$. Find which primes $p > 2$ can be written in each of the following forms, using (not necessarily positive) integers $x$ and $y$: (a) $x^2 + 16y^2$, (b) $4x^2 + 4xy + 5y^2$. | Show that that the answer to (a) is the set of primes which are $1 \bmod 8$, and the solution to (b) is the set of primes which are $5 \bmod 8$. | ['number_theory'] | Section putnam_1974_a3.
Require Import Nat ZArith Znumtheory.
Definition putnam_1974_a3_solution1 := fun x => x mod 8 = 1.
Definition putnam_1974_a3_solution2 := fun x => x mod 8 = 5.
Theorem putnam_1974_a3:
forall (p: nat), prime (Z.of_nat p) /\ odd p = true ->
(exists (m n: Z), (Z.of_nat p) = m*m + 16*n*n <-> putnam_1974_a3_solution1 (Z.of_nat p)) /\
(exists (m n: Z), (Z.of_nat p) = 4*m*m + 4*m*n + 5*n*n <-> putnam_1974_a3_solution2 (Z.of_nat p)).
Proof. Admitted.
End putnam_1974_a3. | theory putnam_1974_a3 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
definition putnam_1974_a3_solution :: "(nat set) \<times> (nat set)" where "putnam_1974_a3_solution \<equiv> undefined"
(* ({p::nat. prime p \<and> [p = 1] (mod 8)}, {p::nat. prime p \<and> [p = 5] (mod 8)}) *)
theorem putnam_1974_a3:
assumes assmption: "\<forall>p::nat. (prime p \<and> p > 2) \<longrightarrow> ((\<exists>m n::nat. p = m^2 + n^2) \<longleftrightarrow> [p = 1] (mod 4))"
shows "\<forall>p::nat. ((prime p \<and> p > 2 \<and> (\<exists>x y::int. p = x^2 + 16*y^2)) \<longleftrightarrow> p \<in> fst putnam_1974_a3_solution) \<and> ((prime p \<and> p > 2 \<and> (\<exists>x y::int. p = 4*x^2 + 4*x*y + 5*y^2)) \<longleftrightarrow> p \<in> snd putnam_1974_a3_solution)"
sorry
end
| null |
putnam_1974_a4 | abbrev putnam_1974_a4_solution : β β β := sorry
-- (fun n => (n / 2^(n-1)) * (n-1).choose (floor ((n-1)/2)))
theorem putnam_1974_a4
(n : β)
(hn : n > 0)
: (1 : β)/(2^(n-1)) * β k in Finset.Icc 0 ((ceil (n/2)) - 1), (n - 2*k)*(n.choose k) = putnam_1974_a4_solution n :=
sorry | Evaluate in closed form: $\frac{1}{2^{n-1}} \sum_{k < n/2} (n-2k)*{n \choose k}$. | Show that the solution is $\frac{n}{2^{n-1}} * {(n-1) \choose \left[ (n-1)/2 \right]}$. | ['algebra'] | Section putnam_1974_a4.
Require Import Nat Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1974_a4_solution (n: nat) := INR n * (Binomial.C (n - 1) (n / 2) - 1) / (2 ^ (n - 1)).
Theorem putnam_1974_a4
: forall (n: nat), 1 / (2 ^ (pred n)) * sum_n (fun i => (INR n - 2 * INR (S i)) * Binomial.C n (i + 1)) (n / 2) = putnam_1974_a4_solution n.
Proof. Admitted.
End putnam_1974_a4. | theory putnam_1974_a4 imports Complex_Main
begin
definition putnam_1974_a4_solution :: "nat \<Rightarrow> real" where "putnam_1974_a4_solution \<equiv> undefined"
(* (\<lambda>n::nat. (n / 2^(n-1)) * ((n-1) choose (nat \<lfloor>(n-1)/2\<rfloor>))) *)
theorem putnam_1974_a4:
fixes n :: nat
assumes hn: "n > 0"
shows "1/(2^(n-1)) * (\<Sum>k::nat=0..((nat \<lceil>n/2\<rceil>)-1). (n - 2*k) * (n choose k)) = putnam_1974_a4_solution n"
sorry
end
| null |
putnam_1974_a6 | abbrev putnam_1974_a6_solution : β := sorry
-- 25
theorem putnam_1974_a6
(n : β€ := 10^6)
(hdivnallx : Polynomial β€ β Prop := fun f => Monic f β§ (β x : β€, (n : β€) β£ f.eval x))
: sInf {d : β | β f : Polynomial β€, hdivnallx f β§ d = f.natDegree} = putnam_1974_a6_solution :=
sorry | Given $n$, let $k(n)$ be the minimal degree of any monic integral polynomial $f$ such that the value of $f(x)$ is divisible by $n$ for every integer $x$. Find the value of $k(1000000)$. | Show that the answer is 25. | ['algebra'] | Section putnam_1974_a6.
Require Import Factorial Reals Coquelicot.Coquelicot.
Definition putnam_1974_a6_solution1 (n N : nat) := fact N mod n = 0%nat /\ forall (N': nat), fact N' mod n = 0%nat -> ge N' N.
Definition putnam_1974_a6_solution2 : nat := 25.
Theorem putnam_1974_a6
(f : nat -> nat)
(cond : nat -> (nat -> Z) -> Prop := fun n coeff => coeff n = Z.of_nat 1 /\ let p : nat -> R := fun x => sum_n (fun i => IZR (coeff i) * INR x ^ i) (n + 1) in (forall (m: Z), Z.to_nat (floor (p (Z.to_nat m))) mod n = 0%nat))
(hf : forall (n: nat), exists (coeff: nat -> Z), cond (f n) coeff)
(hflb : forall (n: nat) (coeff: nat -> Z), cond n coeff -> ge n (f n))
: forall (n: nat), exists N, f n = N /\ putnam_1974_a6_solution1 n N /\ f 1000000%nat = putnam_1974_a6_solution2.
Proof. Admitted.
End putnam_1974_a6. | theory putnam_1974_a6 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1974_a6_solution :: nat where "putnam_1974_a6_solution \<equiv> undefined"
(* 25 *)
theorem putnam_1974_a6:
fixes n :: nat
and hdivnallx :: "(int poly) \<Rightarrow> bool"
defines "n \<equiv> 10^6"
and "hdivnallx \<equiv> (\<lambda>f::int poly. lead_coeff f = 1 \<and> (\<forall>x::int. n dvd (poly f x)))"
shows "(LEAST d::nat. (\<exists>f::int poly. hdivnallx f \<and> d = degree f)) = putnam_1974_a6_solution"
sorry
end
| null |
putnam_1974_b3 | theorem putnam_1974_b3
(Ξ± : β)
(ha : Real.cos (Real.pi * Ξ±) = (1 : β)/3)
: Irrational Ξ± :=
sorry | Prove that if $\alpha$ is a real number such that $\cos (\pi \alpha) = 1/3$, the $\alpha$ is irrational. | null | ['number_theory'] | Section putnam_1974_b3.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1974_b3_solution := 1.
Theorem putnam_1974_b3
: ~ exists (p q: Z), acos (1 / 3) / PI = IZR p / IZR q.
Proof. Admitted.
End putnam_1974_b3. | theory putnam_1974_b3 imports Complex_Main
begin
theorem putnam_1974_b3:
fixes \<alpha> :: real
assumes ha: "cos (pi*\<alpha>) = 1/3"
shows "\<alpha> \<notin> \<rat>"
sorry
end
| null |
putnam_1974_b4 | theorem putnam_1974_b4
(f : β Γ β β β)
(hfcontx : β y : β, Continuous (fun x => f β¨x, yβ©))
(hfconty : β x : β, Continuous (fun y => f β¨x, yβ©))
: β g : β β (β Γ β β β), (β n : β, Continuous (g n)) β§ (β x y : β, Tendsto (fun n => g n β¨x, yβ©) atTop (π (f β¨x, yβ©))) :=
sorry | Let $f : \mathbb{R} \to \mathbb{R}$ be continuous in each variable seperately. Show that there exists a sequence of continuous functions $g_n : \mathbb{R}^2 \to \mathbb{R}$ such that $f(x,y) = \lim_{n \to \infty} g_n(x,y)$ for all $(x,y) \in \mathbb{R}^2$. | null | ['analysis'] | null | theory putnam_1974_b4 imports Complex_Main
begin
theorem putnam_1974_b4:
fixes f :: "(real \<times> real) \<Rightarrow> real"
assumes hfcontx: "\<forall>y::real. continuous_on UNIV (\<lambda>x::real. f (x,y))"
assumes hfconty: "\<forall>x::real. continuous_on UNIV (\<lambda>y::real. f (x,y))"
shows "\<exists>g::nat\<Rightarrow>((real\<times>real)\<Rightarrow>real). (\<forall>n::nat. continuous_on UNIV (g n)) \<and> (\<forall>x y::real. filterlim (\<lambda>n::nat. g n (x,y)) (nhds (f (x,y))) at_top)"
sorry
end
| null |
putnam_1974_b5 | theorem putnam_1974_b5
: β n β₯ 0, β i in Finset.Icc (0 : β) n, (n^i : β)/(Nat.factorial i) > (Real.exp n)/2 :=
sorry | Show that $1 + (n/1!) + (n^2/2!) + \dots + (n^n/n!) > e^n/2$ for every integer $n \geq 0$. | null | ['analysis'] | Section putnam_1974_b5.
Require Import Factorial Reals Coquelicot.Coquelicot.
Definition putnam_1974_b5_solution := 1.
Theorem putnam_1974_b5
(f : nat -> R -> R := fun n x => sum_n (fun i => x ^ i / INR (fact i)) (n + 1))
: forall (n : nat), f n (INR n) > exp PI / 2.
Proof. Admitted.
End putnam_1974_b5. | theory putnam_1974_b5 imports Complex_Main
begin
theorem putnam_1974_b5:
shows "\<forall>n::nat\<ge>0. (\<Sum>i::nat=0..n. (n^i)/(fact i)) > (exp n)/2"
sorry
end
| null |
putnam_1974_b6 | abbrev putnam_1974_b6_solution : (β Γ β Γ β) := sorry
-- ((2^1000 - 1)/3, (2^1000 - 1)/3, 1 + (2^1000 - 1)/3)
theorem putnam_1974_b6
(n : β€ := 1000)
(count0 : β := {S | S β Finset.Icc 1 n β§ S.card β‘ 0 [MOD 3]}.ncard)
(count1 : β := {S | S β Finset.Icc 1 n β§ S.card β‘ 1 [MOD 3]}.ncard)
(count2 : β := {S | S β Finset.Icc 1 n β§ S.card β‘ 2 [MOD 3]}.ncard)
: (count0, count1, count2) = putnam_1974_b6_solution :=
sorry | For a set with $1000$ elements, how many subsets are there whose candinality is respectively $\equiv 0 \bmod 3, \equiv 1 \bmod 3, \equiv 2 \bmod 3$? | Show that there answer is that there are $(2^1000-1)/3$ subsets of cardinality $\equiv 0 \bmod 3$ and $\equiv 1 \bmod 3$, and $1 + (2^1000-1)/3$ subsets of cardinality $\equiv 2 \bmod 3$. | ['set_theory'] | Section putnam_1974_b6.
Require Import Ensembles Finite_sets Reals Coquelicot.Coquelicot.
Definition putnam_1974_b6_solution1 : nat := 21000/3.
Definition putnam_1974_b6_solution2 : nat := 21000/3.
Definition putnam_1974_b6_solution3 : nat := 21000/3 + 1.
Definition putnam_1974_b6_solution4 : nat := 21001/3 + 1.
Definition putnam_1974_b6_solution5 : nat := 21001/3.
Definition putnam_1974_b6_solution6 : nat := 21001/3 + 1.
Theorem putnam_1974_b6
(E : Ensemble R)
(E' : Ensemble R)
(pres : R -> Prop)
(hE : cardinal R E 1000)
(hE' : cardinal R E 1001)
(habc : nat -> Ensemble (Ensemble R) := fun val => fun (subE : Ensemble R) => exists (subEsz: nat), (forall x: R, subE x -> E x /\ pres x) /\ cardinal R subE subEsz /\ subEsz mod 3 = val)
(habc' : nat -> Ensemble (Ensemble R) := fun val => fun (subE' : Ensemble R) => exists (subEsz': nat), (forall x: R, subE' x -> E' x /\ pres x) /\ cardinal R subE' subEsz' /\ subEsz' mod 3 = val)
: cardinal (Ensemble R) (habc 0%nat) putnam_1974_b6_solution1 /\
cardinal (Ensemble R) (habc 1%nat) putnam_1974_b6_solution2 /\
cardinal (Ensemble R) (habc 2%nat) putnam_1974_b6_solution3 /\
cardinal (Ensemble R) (habc' 0%nat) putnam_1974_b6_solution4 /\
cardinal (Ensemble R) (habc' 1%nat) putnam_1974_b6_solution5 /\
cardinal (Ensemble R) (habc' 2%nat) putnam_1974_b6_solution6.
Proof. Admitted.
End putnam_1974_b6. | theory putnam_1974_b6 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
definition putnam_1974_b6_solution :: "nat \<times> nat \<times> nat" where "putnam_1974_b6_solution \<equiv> undefined"
(* (nat (round ((2^1000 - 1)/3)), nat (round ((2^1000 - 1)/3)), 1 + nat (round ((2^1000 - 1)/3))) *)
theorem putnam_1974_b6:
fixes n :: nat
and count0 :: nat
and count1 :: nat
and count2 :: nat
defines "n \<equiv> 1000"
and "count0 \<equiv> card {S::nat set. S \<subseteq> {1..n} \<and> [card S = 0] (mod 3)}"
and "count1 \<equiv> card {S::nat set. S \<subseteq> {1..n} \<and> [card S = 1] (mod 3)}"
and "count2 \<equiv> card {S::nat set. S \<subseteq> {1..n} \<and> [card S = 2] (mod 3)}"
shows "(count0, count1, count2) = putnam_1974_b6_solution"
sorry
end
| null |
putnam_1998_a3 | theorem putnam_1998_a3
(f : β β β)
(hf : ContDiff β 3 f)
: β a : β, (f a) * (deriv f a) * (iteratedDeriv 2 f a) * (iteratedDeriv 3 f a) β₯ 0 :=
sorry | Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that \[f(a)\cdot f'(a) \cdot f''(a) \cdot f'''(a)\geq 0 .\] | null | ['analysis'] | null | theory putnam_1998_a3 imports Complex_Main
"HOL-Analysis.Derivative"
begin
theorem putnam_1998_a3:
fixes f :: "real \<Rightarrow> real"
assumes hf : "continuous_on UNIV ((deriv^^3) f)"
shows "\<exists> a :: real. f a * deriv f a * (deriv^^2) f a * (deriv^^3) f a \<ge> 0"
sorry
end
| null |
putnam_1998_a4 | abbrev putnam_1998_a4_solution : Set β := sorry
-- {n | n β‘ 1 [MOD 6]}
theorem putnam_1998_a4
(A : β β β)
(hA1 : A 1 = 0)
(hA2 : A 2 = 1)
(hA : β n : β, n > 2 β A n = Nat.ofDigits 10 (Nat.digits 10 (A (n - 2)) ++ Nat.digits 10 (A (n - 1))))
: putnam_1998_a4_solution = {n | 11 β£ A n} :=
sorry | Let $A_1=0$ and $A_2=1$. For $n>2$, the number $A_n$ is defined by concatenating the decimal expansions of $A_{n-1}$ and $A_{n-2}$ from left to right. For example $A_3=A_2 A_1=10$, $A_4=A_3 A_2 = 101$, $A_5=A_4 A_3 = 10110$, and so forth. Determine all $n$ such that $11$ divides $A_n$. | Show that the solution is those n for which n can be written as 6k+1 for some integer k. | ['algebra'] | Section putnam_1998_a4.
Require Import Nat ZArith Reals Coquelicot.Coquelicot.
Open Scope nat_scope.
Definition putnam_1998_a4_solution (n: nat) := exists (k: nat), n = 6 * k + 1.
Theorem putnam_1998_a4:
let concatenate (x y : nat) : nat := Nat.pow 10 (Z.to_nat (floor (Rdiv (ln (INR y)) (ln 10))) + 1) * x + y in
let fix A (n: nat) :=
match n with
| O => O
| S O => 1
| S ((S n'') as n') => concatenate (A n') (A n'')
end in
forall (n: nat), A (n+1) mod 11 = 0 <->
putnam_1998_a4_solution n.
Proof. Admitted.
End putnam_1998_a4. | theory putnam_1998_a4 imports Complex_Main "HOL-Number_Theory.Cong"
begin
definition putnam_1998_a4_solution::"nat set" where "putnam_1998_a4_solution \<equiv> undefined"
(* {n::nat. [n = 1] (mod 6)} *)
fun digits::"nat \<Rightarrow> nat list" where
"digits n = (if n < 10 then [n] else ([n mod 10::nat] @ digits (n div 10::nat)))"
fun from_digits::"nat list \<Rightarrow> nat" where
"from_digits L = foldr (\<lambda>a. \<lambda>b. a + 10 * b) L 0"
theorem putnam_1998_a4:
fixes A::"nat\<Rightarrow>nat"
assumes hA1 : "A 1 = 0"
and hA2 : "A 2 = 1"
and hA : "\<forall>n::nat > 2. A n = from_digits (digits (A (n-2)) @ digits (A (n-1)))"
shows "putnam_1998_a4_solution = {n::nat. 11 dvd (A n)}"
sorry
end | null |
putnam_1998_a5 | theorem putnam_1998_a5
(k : β)
(c : Fin k β (β Γ β))
(r : Fin k β β)
(hr : β i : Fin k, r i > 0)
(E : Set (β Γ β))
(hE : E β β i : Fin k, ball (c i) (r i))
: β (n : β) (t : Fin n β Fin k), (β i j : Fin n, i β j β (ball (c (t i)) (r (t i)) β© ball (c (t j)) (r (t j)) = β
)) β§ E β β i : Fin n, ball (c (t i)) (3 * (r (t i))) :=
sorry | Let $\mathcal F$ be a finite collection of open discs in $\mathbb R^2$ whose union contains a set $E\subseteq \mathbb R^2$. Show that there is a pairwise disjoint subcollection $D_1,\ldots, D_n$ in $\mathcal F$ such that \[E\subseteq \cup_{j=1}^n 3D_j.\] Here, if $D$ is the disc of radius $r$ and center $P$, then $3D$ is the disc of radius $3r$ and center $P$. | null | ['analysis'] | null | theory putnam_1998_a5 imports Complex_Main "HOL-Analysis.Elementary_Metric_Spaces"
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_1998_a5:
fixes k::nat and c::"nat \<Rightarrow> (real \<times> real)" and r::"nat \<Rightarrow> real" and E::"(real \<times> real) set"
assumes hr : "\<forall>i \<in> {0..<k}. r i > 0"
and hE : "E \<subseteq> (\<Union> i \<in> {0..<k}. ball (c i) (r i))"
shows "\<exists>n::nat. \<exists>t::nat\<Rightarrow>nat. (\<forall>i \<in> {0..<n}. \<forall>j \<in> {0..<n}. i \<noteq> j \<longrightarrow> ((ball (c (t i)) (r (t i)) \<inter> (ball (c (t j)) (r (t j))) = {})))
\<and> E \<subseteq> (\<Union> i \<in> {0..<n}. ball (c (t i)) (3 * (r (t i))))"
sorry
end | null |
putnam_1998_b1 | abbrev putnam_1998_b1_solution : β := sorry
-- 6
theorem putnam_1998_b1
: sInf {((x + 1/x)^6 - (x^6 + 1/x^6) - 2)/((x + 1/x)^3 + (x^3 + 1/x^3)) | x > (0 : β)} = putnam_1998_b1_solution :=
sorry | Find the minimum value of \[\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}\] for $x>0$. | Show that the minimum value is 6. | ['algebra'] | Section putnam_1998_b1.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1998_b1_solution := 1.
Theorem putnam_1998_b1:
let f (x: R) := ((x + 1 / x) ^ 6 - (x ^ 6 + 1 / (x ^ 6)) - 2) / (x + 1 / x) ^ 3 + (x ^ 3 + 1 / (x ^ 3)) in
exists (minval: R),
(forall (x: R), x > 0 -> f x >= minval) /\
(exists (x: R), x > 0 -> f x = minval) ->
minval = putnam_1998_b1_solution.
Proof. Admitted.
End putnam_1998_b1. | theory putnam_1998_b1 imports Complex_Main
begin
definition putnam_1998_b1_solution::real where "putnam_1998_b1_solution \<equiv> undefined"
(* 6 *)
theorem putnam_1998_b1:
shows "putnam_1998_b1_solution = (LEAST y. (\<exists>x::real > 0. y = ((x+1/x)^6 - (x^6 + 1/x^6) - 2) / ((x+1/x)^3 + (x^3 +1/x^3))))"
sorry
end | null |
putnam_1998_b2 | abbrev putnam_1998_b2_solution : β β β β β := sorry
-- fun a b => if a > b then Real.sqrt (2*a^2 + 2*b^2) else 0
theorem putnam_1998_b2
(a b : β)
(hab : 0 < b β§ b < a)
: sInf {d : β | β (c : β) (x : β), d = Euclidean.dist (a, b) (c, 0) + Euclidean.dist (c, 0) (x, x) + Euclidean.dist (a, b) (x, x) β§
Euclidean.dist (a, b) (c, 0) + Euclidean.dist (c, 0) (x, x) > Euclidean.dist (a, b) (x, x) β§
Euclidean.dist (a, b) (c, 0) + Euclidean.dist (a, b) (x, x) > Euclidean.dist (c, 0) (x, x) β§
Euclidean.dist (c, 0) (x, x) + Euclidean.dist (a, b) (x, x) > Euclidean.dist (a, b) (c, 0)}
= putnam_1998_b2_solution a b :=
sorry | Given a point $(a,b)$ with $0<b<a$, determine the minimum perimeter of a triangle with one vertex at $(a,b)$, one on the $x$-axis, and one on the line $y=x$. You may assume that a triangle of minimum perimeter exists. | Show that the solution is $\sqrt{2a^2 + 2b^2}. | ['geometry', 'algebra'] | null | theory putnam_1998_b2 imports Complex_Main "HOL-Analysis.Topology_Euclidean_Space"
begin
definition putnam_1998_b2_solution::"real\<Rightarrow>real\<Rightarrow>real" where "putnam_1998_b2_solution \<equiv> undefined"
(* \<lambda>a. \<lambda>b. sqrt (2*a^2 + 2*b^2) *)
theorem putnam_1998_b2:
fixes a b::real
assumes hab : "0 < b \<and> b < a"
shows "putnam_1998_b2_solution a b = (LEAST p. \<exists>c x::real. p = dist (a, b) (c, 0) + dist (c, 0) (x, x) + dist (a, b) (x, x) \<and>
dist (a, b) (c, 0) + dist (c, 0) (x, x) > dist (a, b) (x, x) \<and>
dist (a, b) (c, 0) + dist (a, b) (x, x) > dist (c, 0) (x, x) \<and>
dist (c, 0) (x, x) + dist (a, b) (x, x) > dist (a, b) (c, 0))"
sorry
end | null |
putnam_1998_b4 | abbrev putnam_1998_b4_solution : Set (β Γ β) := sorry
-- {nm | let β¨n,mβ© := nm; multiplicity 2 n β multiplicity 2 m}
theorem putnam_1998_b4
(hsum : β β β β β€ := fun n m => β i in Finset.range (m * n), (-1)^(i/m + i/n))
: β n m : β, n > 0 β§ m > 0 β ((hsum n m) = 0 β β¨n, mβ© β putnam_1998_b4_solution) :=
sorry | Find necessary and sufficient conditions on positive integers $m$ and $n$ so that \[\sum_{i=0}^{mn-1} (-1)^{\lfloor i/m \rfloor +\lfloor i/n\rfloor}=0.\] | Show that the sum is 0 if and only if the largest powers of $2$ dividing $m$ and $n$ are different. | ['number_theory'] | Section putnam_1998_b4.
Require Import Nat ZArith Reals Coquelicot.Coquelicot.
Definition putnam_1998_b4_solution (m n: nat) := exists (m2 n2: nat), m mod (2 ^ m2) = 0%nat /\ m mod (2 ^ (m2 + 1)) <> 0%nat /\ n mod (2 ^ n2) = 0%nat /\ n mod (2 ^ n2 + 1) <> 0%nat.
Theorem putnam_1998_b4:
forall (m n: nat), sum_n (fun i => Rpower (-1) (IZR (floor (INR i / INR m)) + IZR (floor (INR i / INR n)))) (m * n - 1) = 0 <->
putnam_1998_b4_solution m n.
Proof. Admitted.
End putnam_1998_b4. | theory putnam_1998_b4 imports Complex_Main "HOL-Computational_Algebra.Factorial_Ring"
begin
definition putnam_1998_b4_solution::"nat\<Rightarrow>nat\<Rightarrow>bool" where "putnam_1998_b4_solution \<equiv> undefined"
(* \<lambda>n. \<lambda>m. multiplicity 2 n \<noteq> multiplicity 2 m *)
theorem putnam_1998_b4:
fixes hsum::"nat\<Rightarrow>nat\<Rightarrow>int"
defines "hsum \<equiv> \<lambda>n. \<lambda>m. (\<Sum>i=0..<m*n. (-1)^(i div m + i div n))"
shows "\<forall>n m::nat. n > 0 \<and> m > 0 \<longrightarrow> (hsum n m = 0 \<longleftrightarrow> putnam_1998_b4_solution n m)"
sorry
end | null |
putnam_1998_b5 | abbrev putnam_1998_b5_solution : β := sorry
-- 1
theorem putnam_1998_b5
(N : β := β i in Finset.range 1998, 10^i)
: putnam_1998_b5_solution = (Nat.floor (10^1000 * Real.sqrt N)) % 10 :=
sorry | Let $N$ be the positive integer with 1998 decimal digits, all of them 1; that is, \[N=1111\cdots 11.\] Find the thousandth digit after the decimal point of $\sqrt N$. | Show that the thousandth digit is 1. | ['number_theory'] | Section putnam_1998_b5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1998_b5_solution : nat := 1.
Theorem putnam_1998_b5:
let N := sum_n (fun i => 10^i) 1998 in
Z.to_nat (floor (10^1000 * sqrt N)) mod 10 = putnam_1998_b5_solution.
Proof. Admitted.
End putnam_1998_b5. | theory putnam_1998_b5 imports Complex_Main
begin
definition putnam_1998_b5_solution::nat where "putnam_1998_b5_solution \<equiv> undefined"
(* 1 *)
theorem putnam_1998_b5:
fixes N::nat
defines "N \<equiv> (\<Sum>i=0..<1998. 10^i)"
shows "putnam_1998_b5_solution = (\<lfloor>10^1000 * sqrt N\<rfloor> mod 10)"
sorry
end | null |
putnam_1998_b6 | theorem putnam_1998_b6
: β a b c : β€, β n : β€, n > 0 β§ Β¬(β k : β€, k = Real.sqrt (n^3 + a * n^2 + b * n + c)) :=
sorry | Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer. | null | ['number_theory'] | null | theory putnam_1998_b6 imports Complex_Main
begin
theorem putnam_1998_b6:
fixes a b c :: int
shows "\<exists> n:: int > 0 . sqrt (n^3 + a * n^2 + b * n + c) \<notin> \<int>"
sorry
end
| null |
putnam_2006_a3 | theorem putnam_2006_a3
(x : β€ β β€)
(hxlo : β k : β€, 0 β€ k β§ k β€ 2006 β x k = k)
(hxhi : β k : β€, k β₯ 2006 β x (k + 1) = x k + x (k - 2005))
: (β i : β, i > 0 β§ β j : Finset.range 2005, 2006 β£ x (i + j)) :=
sorry | Let $1, 2, 3, \dots, 2005, 2006, 2007, 2009, 2012, 2016, \dots$ be a sequence defined by $x_k = k$ for $k=1, 2, \dots, 2006$ and $x_{k+1} = x_k + x_{k-2005}$ for $k \geq 2006$. Show that the sequence has $2005$ consecutive terms each divisible by $2006$. | null | ['algebra'] | null | theory putnam_2006_a3 imports
Complex_Main
begin
theorem putnam_2006_a3:
fixes x :: "nat \<Rightarrow> nat"
assumes hxlo: "\<forall> k \<in> {1 .. 2006}. x k = k"
and hxhi: "\<forall> k \<ge> 2006. x (k + 1) = x k + x (k - 2005)"
shows "\<exists> i :: nat. i > 0 \<and> (\<forall> j \<in> {0 ..< 2005}. 2006 dvd x (i + j))"
sorry
end | null |
putnam_2006_a4 | abbrev putnam_2006_a4_solution : β β β := sorry
-- (fun n : β => (n + 1) / 3)
theorem putnam_2006_a4
(n : β)
(pnat : Equiv.Perm (Fin n) β (β β β))
(pcount : Equiv.Perm (Fin n) β β)
(ngt1 : n > 1)
(hpnat : β p : Equiv.Perm (Fin n), β k : Fin n, (pnat p) k = p k)
(hpcount : β p : Equiv.Perm (Fin n), pcount p = {k : Fin n | (k.1 = 0 β¨ (pnat p) (k - 1) < (pnat p) k) β§ (k = n - 1 β¨ (pnat p) k > (pnat p) (k + 1))}.encard)
: (β p : Equiv.Perm (Fin n), pcount p) / {p : Equiv.Perm (Fin n) | true}.ncard = putnam_2006_a4_solution n :=
sorry | Let $S=\{1,2,\dots,n\}$ for some integer $n>1$. Say a permutation $\pi$ of $S$ has a \emph{local maximum} at $k \in S$ if
\begin{enumerate}
\item[(i)] $\pi(k)>\pi(k+1)$ for $k=1$;
\item[(ii)] $\pi(k-1)<\pi(k)$ and $\pi(k)>\pi(k+1)$ for $1<k<n$;
\item[(iii)] $\pi(k-1)<\pi(k)$ for $k=n$.
\end{enumerate}
(For example, if $n=5$ and $\pi$ takes values at $1,2,3,4,5$ of $2,1,4,5,3$, then $\pi$ has a local maximum of 2 at $k=1$, and a local maximum of 5 at $k=4$.) What is the average number of local maxima of a permutation of $S$, averaging over all permutations of $S$? | Show that the average number of local maxima is $\frac{n+1}{3}$. | ['algebra'] | null | theory putnam_2006_a4 imports Complex_Main
"HOL-Combinatorics.Permutations"
begin
(* uses (nat \<Rightarrow> nat) instead of (Equiv.Perm (Fin n)) *)
definition putnam_2006_a4_solution :: "nat \<Rightarrow> real" where "putnam_2006_a4_solution \<equiv> undefined"
(* (\<lambda>n::nat. (n+1) / 3) *)
theorem putnam_2006_a4:
fixes n :: nat
and nperms :: "(nat \<Rightarrow> nat) set"
and pcount :: "(nat \<Rightarrow> nat) \<Rightarrow> nat"
assumes ngt1: "n > 1"
and hnperms: "nperms \<equiv> {p::nat\<Rightarrow>nat. p permutes {0..(n-1)} \<and> (\<forall>i::nat\<ge>n. p i = i)}"
and hpcount: "\<forall>p::nat\<Rightarrow>nat. pcount p = card {k::nat\<in>{0..(n-1)}. (k = 0 \<or> p (k-1) < p k) \<and> (k = n-1 \<or> p k > p (k+1))}"
shows "(\<Sum>p\<in>nperms. pcount p) / card nperms = putnam_2006_a4_solution n"
sorry
end
| null |
putnam_2006_a5 | abbrev putnam_2006_a5_solution : β β β€ := sorry
-- (fun n : β => if (n β‘ 1 [MOD 4]) then n else -n)
theorem putnam_2006_a5
(n : β)
(theta : β)
(a : Set.Icc 1 n β β)
(nodd : Odd n)
(thetairr : Irrational (theta / Real.pi))
(ha : β k : Set.Icc 1 n, a k = Real.tan (theta + (k * Real.pi) / n))
: (β k : Set.Icc 1 n, a k) / (β k : Set.Icc 1 n, a k) = putnam_2006_a5_solution n :=
sorry | Let $n$ be a positive odd integer and let $\theta$ be a real number such that $\theta/\pi$ is irrational. Set $a_k=\tan(\theta+k\pi/n)$, $k=1,2,\dots,n$. Prove that $\frac{a_1+a_2+\cdots+a_n}{a_1a_2 \cdots a_n}$ is an integer, and determine its value. | Show that $\frac{a_1+\cdots+a_n}{a_1 \cdots a_n}=\begin{cases} n & n \equiv 1 \pmod{4} \\ -n & n \equiv 3 \pmod{4}. \end{cases}$ | ['algebra'] | Section putnam_2006_a5.
Require Import Nat Reals Coquelicot.Coquelicot.
Definition putnam_2006_a5_solution (n: nat) := if eqb (n mod 4) (1%nat) then (INR n) else (-1 * INR n).
Theorem putnam_2006_a5:
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
forall (n: nat) (th: R), odd n = true /\ ~ exists (p q: Z), th / PI = IZR (p / q) /\
let a (k: nat) := tan (th + INR k * PI / INR n) in
sum_n a n / prod_n a n = putnam_2006_a5_solution n.
Proof. Admitted.
End putnam_2006_a5. | theory putnam_2006_a5 imports Complex_Main
begin
(* uses (nat \<Rightarrow> real) instead of ({1..n} \<Rightarrow> real) *)
definition putnam_2006_a5_solution :: "nat \<Rightarrow> int" where "putnam_2006_a5_solution \<equiv> undefined"
(* (\<lambda>n::nat. if ([n = 1] (mod 4)) then n else -n) *)
theorem putnam_2006_a5:
fixes n :: nat
and theta :: real
and a :: "nat \<Rightarrow> real"
assumes nodd: "odd n"
and thetairr: "theta/pi \<notin> \<rat>"
and ha: "\<forall>k::nat\<in>{1..n}. a k = tan (theta + (k*pi)/n)"
shows "(\<Sum>k::nat=1..n. a k) / (\<Prod>k::nat=1..n. a k) = putnam_2006_a5_solution n"
sorry
end
| null |
putnam_2006_b2 | theorem putnam_2006_b2
(n : β)
(npos : n > 0)
(X : Finset β)
(hXcard : X.card = n)
: (β S β X, S β β
β§ β m : β€, |m + β s in S, s| β€ 1 / (n + 1)) :=
sorry | Prove that, for every set $X = \{x_1, x_2, \dots, x_n\}$ of $n$ real numbers, there exists a non-empty subset $S$ of $X$ and an integer $m$ such that
\[
\left| m + \sum_{s \in S} s \right| \leq \frac{1}{n+1}.
\] | null | ['analysis'] | Section putnam_2006_b2.
Require Import List Reals Coquelicot.Coquelicot.
Theorem putnam_2006_b2
(n : nat)
(npos : gt n 0)
(X : list R)
(hXcard : length X = n)
: exists (presS: R -> Prop) (m: Z) (S: list R),
(neq (length S) 0) /\ (forall (x: R), In x S <-> (In x X /\ presS x)) /\
(Rabs (IZR m + (fold_left Rplus S 0)) <= 1 / INR (n + 1)).
Proof. Admitted.
End putnam_2006_b2. | theory putnam_2006_b2 imports
Complex_Main
begin
theorem putnam_2006_b2:
fixes n :: nat
and X :: "real set"
assumes npos: "n > 0"
and hXcard: "finite X \<and> card X = n"
shows "\<exists> S \<subseteq> X. (S \<noteq> {}) \<and> (\<exists> m :: int. \<bar>m + (\<Sum> s \<in> S. s)\<bar> \<le> 1 / (n + 1))"
sorry
end | null |
putnam_2006_b3 | abbrev putnam_2006_b3_solution : β β β := sorry
-- (fun n : β => (Nat.choose n 2) + 1)
theorem putnam_2006_b3
(SABpart : Finset (Fin 2 β β) β Finset (Finset (Fin 2 β β)) β Prop)
(LS : Finset (Fin 2 β β) β β)
(n : β)
(hSABpart : β (S : Finset (Fin 2 β β)) (AB : Finset (Finset (Fin 2 β β))), SABpart S AB = (AB.card = 2 β§ β A β AB, β B β AB, (A βͺ B = S) β§ (A β© B = β
) β§ (β m b : β, (β p β A, p 1 > m * p 0 + b) β§ (β p β B, p 1 < m * p 0 + b))))
(hLS : β S : Finset (Fin 2 β β), LS S = {AB : Finset (Finset (Fin 2 β β)) | SABpart S AB}.encard)
(npos : n > 0)
: (β S : Finset (Fin 2 β β), S.card = n β§ LS S = putnam_2006_b3_solution n) β§ (β S : Finset (Fin 2 β β), S.card = n β LS S β€ putnam_2006_b3_solution n) :=
sorry | Let $S$ be a finite set of points in the plane. A linear partition of $S$ is an unordered pair $\{A,B\}$ of subsets of $S$ such that $A \cup B=S$, $A \cap B=\emptyset$, and $A$ and $B$ lie on opposite sides of some straight line disjoint from $S$ ($A$ or $B$ may be empty). Let $L_S$ be the number of linear partitions of $S$. For each positive integer $n$, find the maximum of $L_S$ over all sets $S$ of $n$ points. | Show that the maximum is $\binom{n}{2}+1$. | ['geometry'] | null | theory putnam_2006_b3 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
definition putnam_2006_b3_solution :: "nat \<Rightarrow> nat" where "putnam_2006_b3_solution \<equiv> undefined"
(* (\<lambda>n::nat. n choose 2 + 1) *)
theorem putnam_2006_b3:
fixes SABpart :: "((real^2) set) \<Rightarrow> (((real^2) set) set) \<Rightarrow> bool"
and LS :: "((real^2) set) \<Rightarrow> nat"
and n :: nat
assumes hSABpart: "\<forall>(S::(real^2) set)(AB::((real^2) set) set). SABpart S AB = (card AB = 2 \<and> (\<exists>A\<in>AB. \<exists>B\<in>AB. A \<union> B = S \<and> A \<inter> B = {} \<and> (\<exists>m b::real. (\<forall>p\<in>A. p$2 > m*p$1 + b) \<and> (\<forall>p\<in>B. p$2 < m*p$1 + b))))"
and hLS: "\<forall>S::(real^2) set. LS S = card {AB::((real^2) set) set. SABpart S AB}"
and npos: "n > 0"
shows "(GREATEST LSS::nat. (\<exists>S::(real^2) set. card S = n \<and> LS S = LSS)) = putnam_2006_b3_solution n"
sorry
end
| null |
putnam_2006_b4 | abbrev putnam_2006_b4_solution : β β β := sorry
-- fun k β¦ 2 ^ k
theorem putnam_2006_b4
(n : β)
(npos : n > 0)
(k : β)
(hk : k β€ n)
(Z : Set (Fin n β β) := {P : Fin n β β | β j : Fin n, P j = 0 β¨ P j = 1})
(max : β)
(hmaxeq : β V : Subspace β (Fin n β β), Module.rank V = k β§ (Z β© V).ncard = max)
(hmaxub : β V : Subspace β (Fin n β β), Module.rank V = k β (Z β© V).ncard β€ max)
: (max = putnam_2006_b4_solution k) :=
sorry | Let $Z$ denote the set of points in $\mathbb{R}^n$ whose coordinates are $0$ or $1$. (Thus $Z$ has $2^n$ elements, which are the vertices of a unit hypercube in $\mathbb{R}^n$.) Given a vector subspace $V$ of $\mathbb{R}^n$, let $Z(V)$ denote the number of members of $Z$ that lie in $V$. Let $k$ be given, $0 \leq k \leq n$. Find the maximum, over all vector subspaces $V \subseteq \mathbb{R}^n$ of dimension $k$, of the number of points in $V \cap Z$. | Prove that the maximum is $2^k$. | ['linear_algebra'] | null | theory putnam_2006_b4 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
definition putnam_2006_b4_solution :: "nat \<Rightarrow> nat" where "putnam_2006_b4_solution \<equiv> undefined"
(* (\<lambda>k::nat. 2^k) *)
theorem putnam_2006_b4:
fixes n :: nat
and k :: nat
and Z :: "(real^'n) set"
and max :: nat
assumes npos: "n > 0"
and pncard: "CARD('n) = n"
and hk: "k \<le> n"
defines "Z \<equiv> {P::real^'n. (\<forall>j::'n. P$j = 0 \<or> P$j = 1)}"
and "max \<equiv> GREATEST ZVcard::nat. (\<exists>V::(real^'n) set. subspace V \<and> dim V = k \<and> ZVcard = card (Z \<inter> V))"
shows "max = putnam_2006_b4_solution k"
sorry
end
| null |
putnam_2006_b5 | abbrev putnam_2006_b5_solution : β := sorry
-- 1 / 16
theorem putnam_2006_b5
(I : (β β β) β β := fun f β¦ β« x in (0)..1, x ^ 2 * (f x))
(J : (β β β) β β := fun f β¦ β« x in (0)..1, x * (f x) ^ 2)
(max : β)
(heqmax : β f : β β β, ContinuousOn f (Icc 0 1) β§ I f - J f = max)
(hmaxub : β f : β β β, ContinuousOn f (Icc 0 1) β I f - J f β€ max)
: (max = putnam_2006_b5_solution) :=
sorry | For each continuous function $f: [0,1] \to \mathbb{R}$, let $I(f) = \int_0^1 x^2 f(x)\,dx$ and $J(x) = \int_0^1 x \left(f(x)\right)^2\,dx$. Find the maximum value of $I(f) - J(f)$ over all such functions $f$. | Show that the answer is \frac{1}{16}. | ['analysis', 'algebra'] | Section putnam_2006_b5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_2006_b5_solution := 1 / 16.
Theorem putnam_2006_b5:
let I (f: R -> R) := RInt (fun x => x ^ 2 * f x) 0 1 in
let J (f: R -> R) := RInt (fun x => x * (f x) ^ 2) 0 1 in
exists (maxval: R),
(forall (f: R -> R) (x: R), 0 <= x <= 1 /\ continuity_pt f x /\ maxval >= I f - J f) /\
(exists (f: R -> R) (x: R), 0 <= x <= 1 /\ continuity_pt f x /\ maxval = I f - J f) ->
maxval = putnam_2006_b5_solution.
Proof. Admitted.
End putnam_2006_b5. | theory putnam_2006_b5 imports
Complex_Main
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_2006_b5_solution :: real where "putnam_2006_b5_solution \<equiv> undefined"
(* 1 / 16 *)
theorem putnam_2006_b5:
fixes I :: "(real \<Rightarrow> real) \<Rightarrow> real"
and J :: "(real \<Rightarrow> real) \<Rightarrow> real"
defines "I \<equiv> \<lambda> f. set_lebesgue_integral lebesgue {0..1} (\<lambda> x. x ^ 2 * (f x))"
and "J \<equiv> \<lambda> f. set_lebesgue_integral lebesgue {0..1} (\<lambda> x. x * (f x) ^ 2)"
shows "(GREATEST y. \<exists> f. continuous_on {0..1} f \<and> I f - J f = y) = putnam_2006_b5_solution"
sorry
end | null |
putnam_2006_b6 | abbrev putnam_2006_b6_solution : β β β := sorry
-- fun k => ((k+1)/k)^k
theorem putnam_2006_b6
(k : β)
(hk : k > 1)
(a : β β β)
(ha0 : a 0 > 0)
(ha : β n : β, a (n + 1) = a n + 1/((a n)^((1 : β)/k)))
: Tendsto (fun n => (a n)^(k+1)/(n ^ k)) atTop (π (putnam_2006_b6_solution k)) :=
sorry | Let $k$ be an integer greater than 1. Suppose $a_0 > 0$, and define \[ a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}} \] for $n > 0$. Evaluate \[\lim_{n \to \infty} \frac{a_n^{k+1}}{n^k}.\] | Show that the solution is $(\frac{k+1}{k})^k$. | ['analysis'] | Section putnam_2006_b6.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2006_b6_solution (k: nat) := ((INR k + 1) / INR k) ^ k.
Theorem putnam_2006_b6:
forall (a0: R) (k: nat), a0 > 0 /\ ge k 1 /\
let fix a (n: nat) : R :=
match n with
| O => a0
| S n' => a n' + 1 / (a n') ^ (1 / k)
end in
Lim_seq (fun n => (a n) ^ (k + 1) / INR n ^ k) = putnam_2006_b6_solution k.
Proof. Admitted.
End putnam_2006_b6. | theory putnam_2006_b6 imports Complex_Main
begin
definition putnam_2006_b6_solution :: "nat \<Rightarrow> real" where "putnam_2006_b6_solution \<equiv> undefined"
(* (\<lambda>k::nat. ((k+1)/k) ^ k) *)
theorem putnam_2006_b6:
fixes k :: nat
and a :: "nat \<Rightarrow> real"
assumes hk: "k > 1"
and ha0: "a 0 > 0"
and ha: "\<forall>n::nat. a (n+1) = a n + 1 / ((a n) powr (1/k))"
shows "filterlim (\<lambda>n::nat. (a n)^(k+1) / (n^k)) (nhds (putnam_2006_b6_solution k)) at_top"
sorry
end
| null |
putnam_1991_a2 | abbrev putnam_1991_a2_solution : Prop := sorry
-- False
theorem putnam_1991_a2
(n : β)
(npos : n β₯ 1)
: (β A B : Matrix (Fin n) (Fin n) β, A β B β§ A ^ 3 = B ^ 3 β§ A ^ 2 * B = B ^ 2 * A β§ Nonempty (Invertible (A ^ 2 + B ^ 2))) β putnam_1991_a2_solution :=
sorry | Let $\mathbf{A}$ and $\mathbf{B}$ be different $n \times n$ matrices with real entries. If $\mathbf{A}^3=\mathbf{B}^3$ and $\mathbf{A}^2\mathbf{B}=\mathbf{B}^2\mathbf{A}$, can $\mathbf{A}^2+\mathbf{B}^2$ be invertible? | Show that the answer is no. | ['linear_algebra'] | Section putnam_1991_a2.
From mathcomp Require Import matrix ssralg ssrbool.
Open Scope ring_scope.
Definition putnam_1991_a2_solution := False.
Theorem putnam_1991_a2:
forall (R: comUnitRingType) (n: nat) (A B: 'M[R]_n),
A <> B ->
mulmx (mulmx A A) A = mulmx (mulmx B B) B /\
mulmx (mulmx A A) B = mulmx (mulmx B B) A ->
(mulmx A A + mulmx B B) \in unitmx <->
putnam_1991_a2_solution.
Proof. Admitted.
End putnam_1991_a2. | theory putnam_1991_a2 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
definition putnam_1991_a2_solution :: bool where "putnam_1991_a2_solution \<equiv> undefined"
(* False *)
theorem putnam_1991_a2:
fixes n :: nat
assumes npos: "n > 0"
and cardn: "CARD('n) = n"
shows "(\<exists> A B :: real^'n^'n. A \<noteq> B \<and> A ** A ** A = B ** B ** B \<and> A ** A ** B = B ** B ** A \<and> invertible (A ** A + B ** B)) \<longleftrightarrow> putnam_1991_a2_solution"
sorry
end | null |
putnam_1991_a3 | abbrev putnam_1991_a3_solution : Set (Polynomial β) := sorry
-- {p : Polynomial β | p.degree = 2 β§ (β r1 r2 : β, r1 β r2 β§ p.eval r1 = 0 β§ p.eval r2 = 0)}
theorem putnam_1991_a3
(p : Polynomial β)
(n : β)
(pr : Prop)
(hn : n = p.degree)
(hpr : pr = β r : β β β, (β i : Fin (n - 1), r i < r (i + 1)) β§ (β i : Fin n, p.eval (r i) = 0) β§ (β i : Fin (n - 1), (Polynomial.derivative p).eval ((r i + r (i + 1)) / 2) = 0))
: (n β₯ 2 β§ pr) β p β putnam_1991_a3_solution :=
sorry | Find all real polynomials $p(x)$ of degree $n \geq 2$ for which there exist real numbers $r_1<r_2<\cdots<r_n$ such that
\begin{enumerate}
\item $p(r_i)=0, \qquad i=1,2,\dots,n$, and
\item $p'(\frac{r_i+r_{i+1}}{2})=0 \qquad i=1,2,\dots,n-1$,
\end{enumerate}
where $p'(x)$ denotes the derivative of $p(x)$. | Show that the real polynomials with the required property are exactly those that are of degree $2$ with $2$ distinct real zeros. | ['algebra', 'analysis'] | Section putnam_1991_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1991_a3_solution (coeff: nat -> R) (n: nat) : Prop := exists (A r1 r2: R), coeff = (fun x => match x with | O => A * r1 * r2 | S O => -A * (r1 + r2) | S (S O) => A | _ => 0 end) /\ n = 2%nat.
Theorem putnam_1991_a3
(p: (nat -> R) -> nat -> (R -> R) := fun coeff n => (fun x => sum_n (fun i => coeff i * x ^ i) (n + 1)))
: forall (coeff: nat -> R) (n: nat), ge n 2 ->
(exists (r: nat -> R),
(forall (i j: nat), lt i j -> r i < r j) /\
forall (i: nat), lt i n -> p coeff n (r i) = 0 /\ lt i (n - 1) -> (Derive (p coeff n)) ((r i + r (S i)) / 2) = 0) <->
putnam_1991_a3_solution coeff n.
Proof. Admitted.
End putnam_1991_a3. | theory putnam_1991_a3 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1991_a3_solution :: "real poly set" where "putnam_1991_a3_solution \<equiv> undefined"
(* {p :: real poly. degree p = 2 \<and> (\<exists> r1 r2 :: real. r1 \<noteq> r2 \<and> poly p r1 = 0 \<and> poly p r2 = 0)} *)
theorem putnam_1991_a3:
fixes p :: "real poly"
and n :: nat
and pr :: bool
defines "n \<equiv> degree p"
and "pr \<equiv> \<exists> r :: nat \<Rightarrow> real. (\<forall> i \<in> {1 .. n - 1}. r i < r (i + 1)) \<and> (\<forall> i \<in> {1 .. n}. poly p (r i) = 0) \<and> (\<forall> i \<in> {1 .. n - 1}. poly (pderiv p) ((r i + r (i + 1)) / 2) = 0)"
shows "(n \<ge> 2 \<and> pr) \<longleftrightarrow> p \<in> putnam_1991_a3_solution"
sorry
end | null |
putnam_1991_a4 | abbrev putnam_1991_a4_solution : Prop := sorry
-- True
theorem putnam_1991_a4
(climit : (β β (Fin 2 β β)) β Prop)
(rareas : (β β β) β Prop)
(crline : (β β (Fin 2 β β)) β (β β β) β Prop)
(hclimit : β c : β β (Fin 2 β β), climit c = Β¬β (p : Fin 2 β β), β Ξ΅ : β, Ξ΅ > 0 β β i : β, c i β Metric.ball p Ξ΅)
(hrareas : β r : β β β, rareas r = β A : β, Tendsto (fun n : β => β i : Fin n, Real.pi * (r i) ^ 2) atTop (π A))
(hcrline : β (c : β β (Fin 2 β β)) (r : β β β), crline c r = (β v w : Fin 2 β β, w β 0 β β i : β, {p : Fin 2 β β | β t : β, p = v + t β’ w} β© Metric.closedBall (c i) (r i) β β
))
: (β (c : β β (Fin 2 β β)) (r : β β β), (β i : β, r i β₯ 0) β§ climit c β§ rareas r β§ crline c r) β putnam_1991_a4_solution :=
sorry | Does there exist an infinite sequence of closed discs $D_1,D_2,D_3,\dots$ in the plane, with centers $c_1,c_2,c_3,\dots$, respectively, such that
\begin{enumerate}
\item the $c_i$ have no limit point in the finite plane,
\item the sum of the areas of the $D_i$ is finite, and
\item every line in the plane intersects at least one of the $D_i$?
\end{enumerate} | Show that the answer is yes, such a sequence of closed discs exists. | ['geometry', 'analysis'] | null | theory putnam_1991_a4 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Analysis.Elementary_Metric_Spaces"
begin
definition putnam_1991_a4_solution :: bool where "putnam_1991_a4_solution \<equiv> undefined"
(* True *)
theorem putnam_1991_a4:
fixes climit :: "(nat \<Rightarrow> real^2) \<Rightarrow> bool"
and rareas :: "(nat \<Rightarrow> real) \<Rightarrow> bool"
and crline :: "(nat \<Rightarrow> real^2) \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> bool"
defines "climit \<equiv> \<lambda> c. \<not>(\<exists> p :: real^2. \<forall> \<epsilon> :: real. \<epsilon> > 0 \<longrightarrow> (\<exists> i :: nat. c i \<in> ball p \<epsilon>))"
and "rareas \<equiv> \<lambda> r. convergent (\<lambda> N. \<Sum> i = 0 .. N. pi * (r i) ^ 2)"
and "crline \<equiv> \<lambda> c r. \<forall> v w :: real^2. w \<noteq> 0 \<longrightarrow> (\<exists> i :: nat. {p :: real^2. \<exists> t :: real. p = v + t *s w} \<inter> cball (c i) (r i) \<noteq> {})"
shows "(\<exists> c :: nat \<Rightarrow> real^2. \<exists> r :: nat \<Rightarrow> real. (\<forall> i :: nat. r i \<ge> 0) \<and> climit c \<and> rareas r \<and> crline c r) \<longleftrightarrow> putnam_1991_a4_solution"
sorry
end | null |
putnam_1991_a5 | abbrev putnam_1991_a5_solution : β := sorry
-- 1 / 3
theorem putnam_1991_a5
(f : Set.Icc (0 : β) 1 β β)
(hf : β y : Set.Icc 0 1, f y = β« x in Set.Ioo 0 y, Real.sqrt (x ^ 4 + (y - y ^ 2) ^ 2))
: (β y : Set.Icc 0 1, f y = putnam_1991_a5_solution) β§ (β y : Set.Icc 0 1, f y β€ putnam_1991_a5_solution) :=
sorry | Find the maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2}\,dx$ for $0 \leq y \leq 1$. | Show that the maximum value of the integral is $1/3$. | ['analysis'] | Section putnam_1991_a5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1991_a5_solution := 1/3.
Theorem putnam_1991_a5:
exists (m: R),
(forall (y: R), 0 <= y <= 1 ->
m >= RInt (fun x => sqrt (pow x 4 + pow (y - pow y 2) 2)) 0 y) /\
(exists (y: R), 0 <= y <= 1 ->
m >= RInt (fun x => sqrt (pow x 4 + pow (y - pow y 2) 2)) 0 y) ->
m = putnam_1991_a5_solution.
Proof. Admitted.
End putnam_1991_a5. | theory putnam_1991_a5 imports
Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1991_a5_solution :: real where "putnam_1991_a5_solution \<equiv> undefined"
(* 1 / 3 *)
theorem putnam_1991_a5:
fixes f :: "real \<Rightarrow> real"
defines "f \<equiv> \<lambda> y. interval_lebesgue_integral lebesgue 0 y (\<lambda> x. sqrt (x ^ 4 + (y - y ^ 2) ^ 2))"
shows "(GREATEST I. \<exists> y \<in> {0..1}. f y = I) = putnam_1991_a5_solution"
sorry
end | null |
putnam_1991_a6 | theorem putnam_1991_a6
(nabsum : β β β Γ (β β β) β Prop)
(agt : β Γ (β β β) β Prop)
(A : β β β)
(bge : β Γ (β β β) β Prop)
(g : β β β)
(bg1 : β Γ (β β β) β Prop)
(bg2 : β Γ (β β β) β Prop)
(B : β β β)
(hnabsum : β n β₯ 1, β ab : β Γ (β β β), nabsum n ab = (ab.1 β₯ 1 β§ (β i < ab.1, ab.2 i > 0) β§ (β i β₯ ab.1, ab.2 i = 0) β§ (β i : Fin ab.1, ab.2 i) = n))
(hagt : β a : β Γ (β β β), agt a = β i : Fin (a.1 - 1), a.2 i > a.2 (i + 1) + a.2 (i + 2))
(hA : β n β₯ 1, A n = {a : β Γ (β β β) | nabsum n a β§ agt a}.encard)
(hbge : β b : β Γ (β β β), bge b = β i : Fin (b.1 - 1), b.2 i β₯ b.2 (i + 1))
(hg : g 0 = 1 β§ g 1 = 2 β§ (β j β₯ 2, g j = g (j - 1) + g (j - 2) + 1))
(hbg1 : β b : β Γ (β β β), bg1 b = β i : Fin b.1, β j : β, b.2 i = g j)
(hbg2 : β b : β Γ (β β β), bg2 b = β k : β, b.2 0 = g k β§ (β j β€ k, β i : Fin b.1, b.2 i = g j))
(hB : β n β₯ 1, B n = {b : β Γ (β β β) | nabsum n b β§ bge b β§ bg1 b β§ bg2 b}.encard)
: β n β₯ 1, A n = B n :=
sorry | Let $A(n)$ denote the number of sums of positive integers $a_1+a_2+\cdots+a_r$ which add up to $n$ with $a_1>a_2+a_3,a_2>a_3+a_4,\dots,a_{r-2}>a_{r-1}+a_r,a_{r-1}>a_r$. Let $B(n)$ denote the number of $b_1+b_2+\cdots+b_s$ which add up to $n$, with
\begin{enumerate}
\item $b_1 \geq b_2 \geq \dots \geq b_s$,
\item each $b_i$ is in the sequence $1,2,4,\dots,g_j,\dots$ defined by $g_1=1$, $g_2=2$, and $g_j=g_{j-1}+g_{j-2}+1$, and
\item if $b_1=g_k$ then every element in $\{1,2,4,\dots,g_k\}$ appears at least once as a $b_i$.
\end{enumerate}
Prove that $A(n)=B(n)$ for each $n \geq 1$. (For example, $A(7)=5$ because the relevant sums are $7,6+1,5+2,4+3,4+2+1$, and $B(7)=5$ because the relevant sums are $4+2+1,2+2+2+1,2+2+1+1+1,2+1+1+1+1+1,1+1+1+1+1+1+1$.) | null | ['algebra'] | null | theory putnam_1991_a6 imports
Complex_Main
begin
(* uses (nat \<rightarrow> nat) instead of (Fin r \<rightarrow> nat) and (Fin s \<rightarrow> nat) *)
theorem putnam_1991_a6:
fixes nabsum :: "nat \<Rightarrow> nat \<times> (nat \<Rightarrow> nat) \<Rightarrow> bool"
and agt :: "nat \<times> (nat \<Rightarrow> nat) \<Rightarrow> bool"
and A :: "nat \<Rightarrow> nat"
and bge :: "nat \<times> (nat \<Rightarrow> nat) \<Rightarrow> bool"
and g :: "nat \<Rightarrow> nat"
and bg1 :: "nat \<times> (nat \<Rightarrow> nat) \<Rightarrow> bool"
and bg2 :: "nat \<times> (nat \<Rightarrow> nat) \<Rightarrow> bool"
and B :: "nat \<Rightarrow> nat"
defines "nabsum \<equiv> \<lambda> (n :: nat) (ab :: nat \<times> (nat \<Rightarrow> nat)). (fst ab) \<ge> 1 \<and> (\<forall> i < fst ab. (snd ab) i > 0) \<and> (\<forall> i \<ge> fst ab. (snd ab) i = 0) \<and> (\<Sum> i = 0 .. fst ab - 1. (snd ab) i) = n"
and "agt \<equiv> \<lambda> a :: nat \<times> (nat \<Rightarrow> nat). \<forall> i \<in> {0 .. fst a - 2}. (snd a) i > (snd a) (i + 1) + (snd a) (i + 2)"
and "A \<equiv> \<lambda> n :: nat. card {a :: nat \<times> (nat \<Rightarrow> nat). nabsum n a \<and> agt a}"
and "bge \<equiv> \<lambda> b :: nat \<times> (nat \<Rightarrow> nat). \<forall> i \<in> {0 .. fst b - 2}. (snd b) i \<ge> (snd b) (i + 1)"
and "bg1 \<equiv> \<lambda> b :: nat \<times> (nat \<Rightarrow> nat). \<forall> i \<in> {0 .. fst b - 1}. \<exists> j :: nat. (snd b) i = g j"
and "bg2 \<equiv> \<lambda> b :: nat \<times> (nat \<Rightarrow> nat). \<exists> k :: nat. (snd b) 0 = g k \<and> (\<forall> j \<le> k. \<exists> i \<in> {0 .. fst b - 1}. (snd b) i = g j)"
and "B \<equiv> \<lambda> n :: nat. card {b :: nat \<times> (nat \<Rightarrow> nat). nabsum n b \<and> bge b \<and> bg1 b \<and> bg2 b}"
assumes hg: "g 0 = 1 \<and> g 1 = 2 \<and> (\<forall> j \<ge> 2. g j = g (j - 1) + g (j - 2) + 1)"
shows "\<forall> n \<ge> 1. A n = B n"
sorry
end | null |
putnam_1991_b1 | abbrev putnam_1991_b1_solution : Set β€ := sorry
-- {A : β€ | β x > 0, A = x ^ 2}
theorem putnam_1991_b1
(m : β€ β β€)
(S : β€ β β€)
(A : β€)
(a : β β β€)
(hm : β n : β€, n β₯ 0 β (m n) ^ 2 β€ n β§ (β m' : β€, m' ^ 2 β€ n β m' β€ m n))
(hS : β n : β€, n β₯ 0 β S n = n - (m n) ^ 2)
(ha : a 0 = A β§ (β k : β, a (k + 1) = a k + S (a k)))
: (A > 0 β§ (β (K : β) (c : β), β k β₯ K, a k = c)) β A β putnam_1991_b1_solution :=
sorry | For each integer $n \geq 0$, let $S(n)=n-m^2$, where $m$ is the greatest integer with $m^2 \leq n$. Define a sequence $(a_k)_{k=0}^\infty$ by $a_0=A$ and $a_{k+1}=a_k+S(a_k)$ for $k \geq 0$. For what positive integers $A$ is this sequence eventually constant? | Show that this sequence is eventually constant if and only if $A$ is a perfect square. | ['algebra'] | Section putnam_1991_b1.
Require Import Nat Coquelicot.Coquelicot.
Open Scope nat_scope.
Definition putnam_1991_b1_solution (A: nat) := exists (m: nat), A = pow m 2.
Theorem putnam_1991_b1:
let eS (n: nat) := sub n (pow (sqrt n) 2) in
let a_seq :=
fix a (A k: nat) :=
match k with
| O => A
| S k' => a A k' + eS (a A k')
end in
forall (A: nat), A > 0 -> Lim_seq (fun k => Raxioms.INR (a_seq A k)) = Rdefinitions.R0 <->
putnam_1991_b1_solution A.
Proof. Admitted.
End putnam_1991_b1. | theory putnam_1991_b1 imports
Complex_Main
begin
definition putnam_1991_b1_solution :: "nat set" where "putnam_1991_b1_solution \<equiv> undefined"
(* {A :: nat. \<exists> x > 0. A = x ^ 2} *)
theorem putnam_1991_b1:
fixes m :: "nat \<Rightarrow> nat"
and S :: "nat \<Rightarrow> nat"
and A :: nat
and a :: "nat \<Rightarrow> nat"
defines "m \<equiv> \<lambda> n. GREATEST M :: nat. M ^ 2 \<le> n"
and "S \<equiv> \<lambda> n. n - (m n) ^ 2"
assumes ha: "a 0 = A \<and> (\<forall> k :: nat. a (k + 1) = a k + S (a k))"
shows "(A > 0 \<and> (\<exists> K c :: nat. \<forall> k \<ge> K. a k = c)) \<longleftrightarrow> A \<in> putnam_1991_b1_solution"
sorry
end | null |
putnam_1991_b2 | theorem putnam_1991_b2
(f g : β β β)
(fgnconst : Β¬β c : β, f = Function.const β c β¨ g = Function.const β c)
(fgdiff : Differentiable β f β§ Differentiable β g)
(fadd : β x y : β, f (x + y) = f x * f y - g x * g y)
(gadd : β x y : β, g (x + y) = f x * g y + g x * f y)
: (deriv f 0 = 0) β (β x : β, (f x) ^ 2 + (g x) ^ 2 = 1) :=
sorry | Suppose $f$ and $g$ are non-constant, differentiable, real-valued functions defined on $(-\infty,\infty)$. Furthermore, suppose that for each pair of real numbers $x$ and $y$,
\begin{align*}
f(x+y)&=f(x)f(y)-g(x)g(y), \\
g(x+y)&=f(x)g(y)+g(x)f(y).
\end{align*}
If $f'(0)=0$, prove that $(f(x))^2+(g(x))^2=1$ for all $x$. | null | ['analysis'] | Section putnam_1991_b2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1991_b2:
forall (f g: R -> R), (~ exists (c: R), f = (fun _ => c) \/ g = (fun _ => c)) /\ forall x, ex_derive_n f 1 x /\ forall x, ex_derive_n g 1 x ->
forall (x y: R), f (x + y) = f x * f y - g x * g y /\
g (x + y) = f x * g y - g x * f y ->
Derive f 0 = 0 ->
forall (x: R), pow (f x) 2 + pow (g x) 2 = 1.
Proof. Admitted.
End putnam_1991_b2. | theory putnam_1991_b2 imports
Complex_Main
"HOL-Analysis.Derivative"
begin
theorem putnam_1991_b2:
fixes f g :: "real \<Rightarrow> real"
assumes fgnconst: "\<not>(\<exists> c :: real. f = (\<lambda> x. c) \<or> g = (\<lambda> x. c))"
and fgdiff: "f differentiable_on UNIV \<and> g differentiable_on UNIV"
and fadd: "\<forall> x y :: real. f (x + y) = f x * f y - g x * g y"
and gadd: "\<forall> x y :: real. g (x + y) = f x * g y + g x * f y"
shows "(deriv f 0 = 0) \<longrightarrow> (\<forall> x :: real. (f x) ^ 2 + (g x) ^ 2 = 1)"
sorry
end | null |
putnam_1991_b4 | theorem putnam_1991_b4
(p : β)
(podd : Odd p)
(pprime : Prime p)
: (β j : Fin (p + 1), (p.choose j) * ((p + j).choose j)) β‘ (2 ^ p + 1) [MOD (p ^ 2)] :=
sorry | Suppose $p$ is an odd prime. Prove that $\sum_{j=0}^p \binom{p}{j}\binom{p+j}{j} \equiv 2^p+1 \pmod{p^2}$. | null | ['number_theory', 'algebra'] | Section putnam_1991_b4.
Require Import Nat Reals ZArith Znumtheory Binomial Coquelicot.Coquelicot.
Theorem putnam_1991_b4:
forall (p: nat), odd p = true /\ prime (Z.of_nat p) ->
let expr : R := sum_n (fun j => Binomial.C p j * Binomial.C (p + j) j) p in
(floor expr) mod (Z.pow (Z.of_nat p) 2) = Z.add (Z.pow 2 (Z.of_nat p)) 1.
Proof. Admitted.
End putnam_1991_b4. | theory putnam_1991_b4 imports
Complex_Main
"HOL-Computational_Algebra.Primes"
"HOL-Number_Theory.Cong"
begin
theorem putnam_1991_b4:
fixes p :: nat
assumes podd: "odd p"
and pprime: "prime p"
shows "[(\<Sum> j = 0 .. p. (p choose j) * ((p + j) choose j)) = 2 ^ p + 1] (mod p ^ 2)"
sorry
end | null |
putnam_1991_b5 | abbrev putnam_1991_b5_solution : β β β := sorry
-- (fun p : β => Nat.ceil ((p : β) / 4))
theorem putnam_1991_b5
(p : β)
(podd : Odd p)
(pprime : Prime p)
: ({z : ZMod p | β x : ZMod p, z = x ^ 2} β© {z : ZMod p | β y : ZMod p, z = y ^ 2 + 1}).encard = putnam_1991_b5_solution p :=
sorry | Let $p$ be an odd prime and let $\mathbb{Z}_p$ denote (the field of) integers modulo $p$. How many elements are in the set $\{x^2:x \in \mathbb{Z}_p\} \cap \{y^2+1:y \in \mathbb{Z}_p\}$? | Show that the number of elements in the intersection is $\lceil p/4 \rceil$. | ['number_theory'] | Section putnam_1991_b5.
Require Import Reals Nat ZArith Znumtheory Ensembles Finite_sets Coquelicot.Coquelicot. From mathcomp Require Import fintype seq ssrbool.
Open Scope R.
Definition putnam_1991_b5_solution (p: nat) : nat := p / 4 + 1.
Variable A: Ensemble Z.
Theorem putnam_1991_b5:
forall (p: nat), odd p = true /\ prime (Z.of_nat p) ->
exists (A B: Ensemble Z), forall (z: Z),
(A z <-> exists (m: 'I_p), z = Z.of_nat (pow (nat_of_ord m) 2)) /\
(B z <-> exists (m: 'I_p), z = Z.of_nat (pow (nat_of_ord m) 2 + 1)) ->
let C : Ensemble Z := Intersection Z A B in
cardinal Z C (putnam_1991_b5_solution p).
Proof. Admitted.
End putnam_1991_b5. | theory putnam_1991_b5 imports
Complex_Main
"HOL-Computational_Algebra.Primes"
"HOL-Number_Theory.Cong"
begin
(* Note: Uses modular congruence instead of the field Zmod p as in Lean *)
definition putnam_1991_b5_solution :: "nat \<Rightarrow> nat" where "putnam_1991_b5_solution \<equiv> undefined"
(* \<lambda> p :: nat. nat (ceiling (p / 4)) *)
theorem putnam_1991_b5:
fixes p :: nat
assumes podd: "odd p"
and pprime: "prime p"
shows "card ({z \<in> {0 .. p - 1}. \<exists> x \<in> {0 .. p - 1}. [z = x ^ 2] (mod p)} \<inter> {z \<in> {0 .. p - 1}. \<exists> y \<in> {0 .. p - 1}. [z = y ^ 2 + 1] (mod p)}) = putnam_1991_b5_solution p"
sorry
end | null |
putnam_1991_b6 | abbrev putnam_1991_b6_solution : β β β β β := sorry
-- (fun a b : β => |Real.log (a / b)|)
theorem putnam_1991_b6
(a b : β)
(cle : β β Prop)
(abpos : a > 0 β§ b > 0)
(hcle : β c : β, cle c = β u : β, (0 < |u| β§ |u| β€ c) β (β x β Set.Ioo 0 1, a ^ x * b ^ (1 - x) β€ a * (Real.sinh (u * x) / Real.sinh u) + b * (Real.sinh (u * (1 - x)) / Real.sinh u)))
: cle (putnam_1991_b6_solution a b) β§ (β c : β, cle c β c β€ putnam_1991_b6_solution a b) :=
sorry | Let $a$ and $b$ be positive numbers. Find the largest number $c$, in terms of $a$ and $b$, such that $a^xb^{1-x} \leq a\frac{\sinh ux}{\sinh u}+b\frac{\sinh u(1-x)}{\sinh u}$ for all $u$ with $0<|u| \leq c$ and for all $x$, $0<x<1$. (Note: $\sinh u=(e^u-e^{-u})/2$.) | Show that the largest $c$ for which the inequality holds for $0<|u| \leq c$ is $c=|\ln(a/b)|$. | ['analysis'] | Section putnam_1991_b6.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1991_b6:
forall (a b: R),
let ineq_holds (c: R) := forall (u x: R), 0 < Rabs u <= c /\ 0 < x < 1 -> Rpower a x * Rpower b (1 - x) <= a * sinh (u * x) / sinh u + b * sinh (1 - x) / sinh u in
exists (mc: R), ineq_holds mc /\
forall (c: R), ineq_holds c -> c <= mc.
Proof. Admitted.
End putnam_1991_b6. | theory putnam_1991_b6 imports
Complex_Main
begin
definition putnam_1991_b6_solution :: "real \<Rightarrow> real \<Rightarrow> real" where "putnam_1991_b6_solution \<equiv> undefined"
(* \<lambda> a b :: real. \<bar>ln (a / b)\<bar> *)
theorem putnam_1991_b6:
fixes a b :: real
and cle :: "real \<Rightarrow> bool"
defines "cle \<equiv> \<lambda> c :: real. \<forall> u :: real. (0 < \<bar>u\<bar> \<and> \<bar>u\<bar> \<le> c) \<longrightarrow> (\<forall> x \<in> {0 <..< 1}. a powr x * b powr (1 - x) \<le> a * (sinh (u * x) / sinh u) + b * (sinh (u * (1 - x)) / sinh u))"
assumes abpos: "a > 0 \<and> b > 0"
shows "(GREATEST c. cle c) = putnam_1991_b6_solution a b"
sorry
end | null |
putnam_1985_a1 | abbrev putnam_1985_a1_solution : β Γ β Γ β Γ β := sorry
-- (10, 10, 0, 0)
theorem putnam_1985_a1
: (let (a, b, c, d) := putnam_1985_a1_solution; {(A1, A2, A3) : Set β€ Γ Set β€ Γ Set β€ | A1 βͺ A2 βͺ A3 = Icc 1 10 β§ A1 β© A2 β© A3 = β
}.ncard = 2 ^ a * 3 ^ b * 5 ^ c * 7 ^ d) :=
sorry | Determine, with proof, the number of ordered triples $(A_1, A_2, A_3)$ of sets which have the property that
\begin{enumerate}
\item[(i)] $A_1 \cup A_2 \cup A_3 = \{1,2,3,4,5,6,7,8,9,10\}$, and
\item[(ii)] $A_1 \cap A_2 \cap A_3 = \emptyset$.
\end{enumerate}
Express your answer in the form $2^a 3^b 5^c 7^d$, where $a,b,c,d$ are nonnegative integers. | Prove that the number of such triples is $2^{10}3^{10}$. | ['algebra'] | Section putnam_1985_a1.
Require Import Ensembles List Finite_sets Nat Coquelicot.Coquelicot.
Import ListNotations.
Definition putnam_1985_a1_solution := (10, 10, 0, 0).
Theorem putnam_1985_a1
: let E: Ensemble (list (Ensemble nat)) := fun A =>
match A with
| A1 :: A2 :: A3 :: _ =>
Union nat (Union nat A1 A2) A3 = fun x => 1 <= x <= 11 /\ Intersection nat (Intersection nat A1 A2) A3 = Empty_set nat
| _ => True
end in
exists (a b c d : nat), cardinal (list (Ensemble nat)) E (2 ^ a * 3 ^ b * 5 ^ c * 7 ^ d) <->
(a, b, c, d) = putnam_1985_a1_solution.
Proof. Admitted.
End putnam_1985_a1. | theory putnam_1985_a1 imports Complex_Main
begin
definition putnam_1985_a1_solution::"nat \<times> nat \<times> nat \<times> nat" where "putnam_1985_a1_solution \<equiv> undefined"
(* 10, 10, 0, 0 *)
theorem putnam_1985_a1:
shows "let (a, b, c, d) = putnam_1985_a1_solution in
(2^a * 3^b * 5^c * 7^d = card {(A1::nat set, A2::nat set, A3::nat set). (A1\<union>A2\<union>A3 = {1..10::nat} \<and> A1\<inter>A2\<inter>A3 = {}) })"
sorry
end | null |
putnam_1985_a3 | abbrev putnam_1985_a3_solution : β β β := sorry
-- fun d β¦ exp d - 1
theorem putnam_1985_a3
(d : β)
(a : β β β β β)
(ha0 : β m : β, a m 0 = d / 2 ^ m)
(ha : β m : β, β j : β, a m (j + 1) = (a m j) ^ 2 + 2 * a m j)
: (Tendsto (fun n β¦ a n n) β€ (π (putnam_1985_a3_solution d))) :=
sorry | Let $d$ be a real number. For each integer $m \geq 0$, define a sequence $\{a_m(j)\}$, $j=0,1,2,\dots$ by the condition
\begin{align*}
a_m(0) &= d/2^m, \\
a_m(j+1) &= (a_m(j))^2 + 2a_m(j), \qquad j \geq 0.
\end{align*}
Evaluate $\lim_{n \to \infty} a_n(n)$. | Show that the limit equals $e^d - 1$. | ['analysis'] | Section putnam_1985_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1985_a3_solution (x: R) := exp x - 1.
Theorem putnam_1985_a3:
forall (x: R),
let A :=
fix a (i j: nat) :=
match (i,j) with
| (i, 0) => x/pow 2 i
| (i, S j') => pow (a i j') 2 + 2 * a i j'
end in
Lim_seq (fun n => A n n) = putnam_1985_a3_solution x.
Proof. Admitted.
End putnam_1985_a3. | theory putnam_1985_a3 imports Complex_Main
begin
definition putnam_1985_a3_solution::"real\<Rightarrow>real" where "putnam_1985_a3_solution \<equiv> undefined"
(* \<lambda>x. exp x - 1 *)
theorem putnam_1985_a3:
fixes d::real and a::"nat\<Rightarrow>nat\<Rightarrow>real"
assumes ha0 : "\<forall>m::nat. a m 0 = d / 2^m"
and ha : "\<forall>m::nat. \<forall>j::nat. a m (j+1) = (a m j)^2 + 2 * a m j"
shows "(\<lambda>n. a n n) \<longlonglongrightarrow> (putnam_1985_a3_solution d)"
sorry
end | null |
putnam_1985_a4 | abbrev putnam_1985_a4_solution : Set (Fin 100) := sorry
-- {87}
theorem putnam_1985_a4
(a : β β β)
(ha1 : a 1 = 3)
(ha : β i β₯ 1, a (i + 1) = 3 ^ a i)
: ({k : Fin 100 | β N : β, β i β₯ N, a i % 100 = k} = putnam_1985_a4_solution) :=
sorry | Define a sequence $\{a_i\}$ by $a_1=3$ and $a_{i+1}=3^{a_i}$ for $i \geq 1$. Which integers between $00$ and $99$ inclusive occur as the last two digits in the decimal expansion of infinitely many $a_i$? | Prove that the only number that occurs infinitely often is $87$. | ['number_theory'] | Section putnam_1985_a4.
Require Import Ensembles Nat Coquelicot.Coquelicot.
Definition putnam_1985_a4_solution := fun k => k = 87.
Theorem putnam_1985_a4
(a : nat -> nat := fix a (n: nat) :=
match n with
| O => 3
| S n' => 3 ^ (a n')
end)
: let E: Ensemble nat := fun k => k < 100 /\ forall (n: nat), n < 100 -> exists (i: nat), ge i n /\ a i mod 100 = k in
E = putnam_1985_a4_solution.
Proof. Admitted.
End putnam_1985_a4. | theory putnam_1985_a4 imports Complex_Main
begin
definition putnam_1985_a4_solution::"nat set" where "putnam_1985_a4_solution \<equiv> undefined"
(* {87} *)
theorem putnam_1985_a4:
fixes a::"nat\<Rightarrow>nat"
assumes ha1 : "a 1 = 3"
and ha : "\<forall>i \<ge> 1. a (i+1) = 3^ (a i)"
shows "putnam_1985_a4_solution = {k::nat. k < 100 \<and> k \<ge> 0 \<and> (\<forall>N::nat. \<exists>i::nat \<ge> N. (a i) mod 100 = k)}"
sorry
end | null |
putnam_1985_a5 | abbrev putnam_1985_a5_solution : Set β := sorry
-- {3, 4, 7, 8}
theorem putnam_1985_a5
(I : β β β := fun m β¦ β« x in (0)..(2 * Real.pi), β k in Finset.Icc 1 m, cos (k * x))
: ({m β Finset.Icc 1 10 | I m β 0} = putnam_1985_a5_solution) :=
sorry | Let $I_m = \int_0^{2\pi} \cos(x)\cos(2x)\cdots \cos(mx)\,dx$. For which integers $m$, $1 \leq m \leq 10$ is $I_m \neq 0$? | Prove that the integers $m$ with $1 \leq m \leq 10$ and $I_m \neq 0$ are $m = 3, 4, 7, 8$. | ['analysis'] | Section putnam_1985_a5.
Require Import Nat Reals List Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1985_a5_solution (n: nat) := Nat.eq n 3 \/ Nat.eq n 4 \/ Nat.eq n 7 \/ Nat.eq n 8.
Theorem putnam_1985_a5:
let F (n: nat) := fun x => let f (i: nat):= cos (INR i * x) in
let coeffs := map f (seq 1 n) in
fold_right Rmult 1 coeffs in
forall (n: nat), and (le 1 n) (le n 10) ->
RInt (F n) 0 2*PI <> 0 <-> putnam_1985_a5_solution n.
Proof. Admitted.
End putnam_1985_a5. | theory putnam_1985_a5 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1985_a5_solution::"nat set" where "putnam_1985_a5_solution \<equiv> undefined"
(* 3, 4, 7, 8 *)
theorem putnam_1985_a5:
fixes I::"nat\<Rightarrow>real"
defines "I \<equiv> \<lambda>m. interval_lebesgue_integral lebesgue 0 (2 * pi) (\<lambda>x. (\<Prod>k=1..m. cos (k * x)))"
shows "putnam_1985_a5_solution = {m. m \<in> {1..10} \<and> I m \<noteq> 0}"
sorry
end | null |
putnam_1985_a6 | abbrev putnam_1985_a6_solution : Polynomial β := sorry
-- 6 * X ^ 2 + 5 * X + 1
theorem putnam_1985_a6
(Ξ : Polynomial β β β := fun p β¦ β k in Finset.range (p.natDegree + 1), coeff p k ^ 2)
(f : Polynomial β := 3 * X ^ 2 + 7 * X + 2)
: (let g := putnam_1985_a6_solution; g.eval 0 = 1 β§ β n : β, n β₯ 1 β Ξ (f ^ n) = Ξ (g ^ n)) :=
sorry | If $p(x)= a_0 + a_1 x + \cdots + a_m x^m$ is a polynomial with real coefficients $a_i$, then set
\[
\Gamma(p(x)) = a_0^2 + a_1^2 + \cdots + a_m^2.
\]
Let $F(x) = 3x^2+7x+2$. Find, with proof, a polynomial $g(x)$ with real coefficients such that
\begin{enumerate}
\item[(i)] $g(0)=1$, and
\item[(ii)] $\Gamma(f(x)^n) = \Gamma(g(x)^n)$
\end{enumerate}
for every integer $n \geq 1$. | Show that $g(x) = 6x^2 + 5x + 1$ satisfies the conditions. | ['algebra'] | Section putnam_1985_a6.
From mathcomp Require Import ssralg ssrnum fintype seq poly.
Local Open Scope ring_scope.
Variable (R: numDomainType).
Definition putnam_1985_a6_solution : {poly R} := 6%:R *: 'X^2 + 5%:R *: 'X + 1%:R.
Theorem putnam_1985_a6:
let g : {poly R} := 3%:R *: 'X^2 + 7%:R *: 'X + 2%:R in
let Comp_poly_n :=
fix comp_poly_n (p : {poly R}) (n : nat) : {poly R} :=
match n with
| O => 1
| S n' => comp_poly (comp_poly_n p n') p
end in
forall (f: {poly R}), f`_0 = 0 ->
forall (n: nat),
let F : {poly R} := Comp_poly_n f n in
let G : {poly R} := Comp_poly_n g n in
(\sum_(i < size F) F`_i) = (\sum_(i < size G) G`_i)
<-> f = putnam_1985_a6_solution.
Proof. Admitted.
End putnam_1985_a6. | theory putnam_1985_a6 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1985_a6_solution::"real poly" where "putnam_1985_a6_solution \<equiv> undefined"
(* Poly [1, 5, 6] *)
theorem putnam_1985_a6:
fixes \<Gamma>::"(real poly) \<Rightarrow> real" and f::"real poly"
defines "\<Gamma> \<equiv> \<lambda>P. (\<Sum>i=0..(degree P). (coeff P i)^2)"
and "f \<equiv> Poly [2, 7, 3]"
shows "poly putnam_1985_a6_solution 0 = 1 \<and> (\<forall>n::nat \<ge> 1. \<Gamma> (f^n) = \<Gamma> (putnam_1985_a6_solution^n))"
sorry
end | null |
putnam_1985_b1 | abbrev putnam_1985_b1_solution : Fin 5 β β€ := sorry
-- fun i β¦ i - 2
theorem putnam_1985_b1
(p : (Fin 5 β β€) β (Polynomial β) := fun m β¦ β i : Fin 5, ((X : Polynomial β) - m i))
(numnzcoeff : Polynomial β β β := fun p β¦ {j β Finset.range (p.natDegree + 1) | coeff p j β 0}.ncard)
: (Injective putnam_1985_b1_solution β§ β m : Fin 5 β β€, Injective m β numnzcoeff (p putnam_1985_b1_solution) β€ numnzcoeff (p m)) :=
sorry | Let $k$ be the smallest positive integer for which there exist distinct integers $m_1, m_2, m_3, m_4, m_5$ such that the polynomial
\[
p(x) = (x-m_1)(x-m_2)(x-m_3)(x-m_4)(x-m_5)
\]
has exactly $k$ nonzero coefficients. Find, with proof, a set of integers $m_1, m_2, m_3, m_4, m_5$ for which this minimum $k$ is achieved. | Show that the minimum $k = 3$ is obtained for $\{m_1, m_2, m_3, m_4, m_5\} = \{-2, -1, 0, 1, 2\}$. | ['algebra'] | null | theory putnam_1985_b1 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1985_b1_solution::"int \<times> int \<times> int \<times> int \<times> int" where "putnam_1985_b1_solution \<equiv> undefined"
(* (-2, -1, 0, 1, 2) *)
theorem putnam_1985_b1:
fixes p::"int\<Rightarrow>int\<Rightarrow>int\<Rightarrow>int\<Rightarrow>int\<Rightarrow>(real poly)" and numnzcoeff::"(real poly) \<Rightarrow> nat"
defines "p \<equiv> \<lambda>m1. \<lambda>m2. \<lambda>m3. \<lambda>m4. \<lambda>m5. [: -m1, 1 :] * [: -m2, 1 :] * [: -m3, 1 :] * [: -m4, 1 :] * [: -m5, 1 :]"
and "numnzcoeff \<equiv> \<lambda>P. card {j::nat. j \<ge> 0 \<and> j \<le> degree P \<and> coeff P j \<noteq> 0}"
shows "let (m1, m2, m3, m4, m5) = putnam_1985_b1_solution in ((card {m1, m2, m3, m4, m5} = 5) \<and>
(\<forall>m. let (a, b, c, d, e) = m in (card {a, b, c, d, e} = 5 \<longrightarrow> numnzcoeff (p m1 m2 m3 m4 m5) \<le> numnzcoeff (p a b c d e))))"
sorry
end | null |
putnam_1985_b2 | abbrev putnam_1985_b2_solution : β β β := sorry
-- fun n β¦ ite (n = 101) 99 0
theorem putnam_1985_b2
(f : β β β β β)
(hf0x : β x : β, f 0 x = 1)
(hfn0 : β n β₯ 1, f n 0 = 0)
(hfderiv : β x : β, β n, deriv (f (n + 1)) x = (n + 1) * f n (x + 1))
: (β a : β, a = f 100 1 β§ Nat.factorization a = putnam_1985_b2_solution) :=
sorry | Define polynomials $f_n(x)$ for $n \geq 0$ by $f_0(x)=1$, $f_n(0)=0$ for $n \geq 1$, and
\[
\frac{d}{dx} f_{n+1}(x) = (n+1)f_n(x+1)
\]
for $n \geq 0$. Find, with proof, the explicit factorization of $f_{100}(1)$ into powers of distinct primes. | Show that $f_{100}(1) = 101^{99}$. | ['algebra'] | Section putnam_1985_b2.
Require Import Nat List ZArith Znumtheory.
Open Scope nat_scope.
Definition putnam_1985_b2_solution := repeat 101 99.
Theorem putnam_1985_b2:
let P :=
fix p (n x: nat) : nat :=
match (n,x) with
| (O, x) => 1
| (S n', x) => (n' + 1) * p n' (x + 1)
end in
let val := P 100 1 in
exists (l: list nat), forall (x: nat), (In x l -> prime (Z.of_nat x)) ->
fold_left mul l 1 = val
<-> l = putnam_1985_b2_solution.
Proof. Admitted.
End putnam_1985_b2. | theory putnam_1985_b2 imports
Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Computational_Algebra.Primes" "HOL-Analysis.Derivative"
begin
definition putnam_1985_b2_solution::"nat multiset" where "putnam_1985_b2_solution \<equiv> undefined"
(* (replicate_mset 99 101) *)
theorem putnam_1985_b2:
fixes f::"nat\<Rightarrow>real\<Rightarrow>real"
assumes hf0x : "\<forall>x::real. f 0 x = 1"
and hfn0 : "\<forall>n::nat \<ge> 1. f n 0 = 0"
and hfderiv : "\<forall>x::real. \<forall>n::nat. (deriv (f (n+1)) x) = (n+1) * f n (x+1)"
shows "\<exists>a::nat. a = f 100 1 \<and> prime_factorization a = putnam_1985_b2_solution"
sorry
end | null |
putnam_1985_b3 | theorem putnam_1985_b3
(a : β β β β β)
(apos : β m n : β, a m n > 0)
(ha : β k : β, k > 0 β {(m, n) : β Γ β | m > 0 β§ n > 0 β§ a m n = k}.encard = 8)
: (β m n, m > 0 β§ n > 0 β§ a m n > m * n) :=
sorry | Let
\[
\begin{array}{cccc} a_{1,1} & a_{1,2} & a_{1,3} & \dots \\
a_{2,1} & a_{2,2} & a_{2,3} & \dots \\
a_{3,1} & a_{3,2} & a_{3,3} & \dots \\
\vdots & \vdots & \vdots & \ddots
\end{array}
\]
be a doubly infinite array of positive integers, and suppose each positive integer appears exactly eight times in the array. Prove that $a_{m,n} > mn$ for some pair of positive integers $(m,n)$. | null | ['algebra'] | Section putnam_1985_b3.
Require Import Nat List.
Theorem putnam_1985_b3:
let exactly_equal (a: (nat*nat) -> nat) (ij: (nat*nat)) :=
(exists (l8: list (nat*nat)), length l8 = 8 /\ NoDup l8 /\ (forall (n: (nat*nat)), In n l8 -> a n = a ij)) /\
(~exists (l9: list (nat*nat)), length l9 = 9 /\ NoDup l9 /\ (forall (n: (nat*nat)), In n l9 -> a n = a ij)) in
forall (a: (nat*nat) -> nat),
(forall (ij: (nat*nat)), exactly_equal a ij) /\
(exists (ij: (nat*nat)), a ij > fst ij*snd ij).
Proof. Admitted.
End putnam_1985_b3. | theory putnam_1985_b3 imports Complex_Main
begin
theorem putnam_1985_b3:
fixes a::"nat\<Rightarrow>nat\<Rightarrow>nat"
assumes apos : "\<forall>m n::nat. a m n > 0"
and ha : "\<forall>k::nat. k > 0 \<longrightarrow> card {(m::nat, n::nat). m > 0 \<and> n > 0 \<and> a m n = k} = 8"
shows "\<exists>m n::nat. m > 0 \<and> n > 0 \<and> a m n > m * n"
sorry
end | null |
putnam_1985_b5 | abbrev putnam_1985_b5_solution : β := sorry
-- sqrt (Real.pi / 1985) * exp (-3970)
theorem putnam_1985_b5
(fact : β« x in univ, exp (- x ^ 2) = sqrt (Real.pi))
: (β« t in Set.Ioi 0, t ^ (- (1 : β) / 2) * exp (-1985 * (t + t ^ (-(1 : β)))) = putnam_1985_b5_solution) :=
sorry | Evaluate $\int_0^\infty t^{-1/2}e^{-1985(t+t^{-1})}\,dt$. You may assume that $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$. | Show that the integral evaluates to $\sqrt{\frac{\pi}{1985}}e^{-3970}$. | ['analysis'] | Section putnam_1985_b5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1985_b5_solution := sqrt (PI / 1985) * exp (-3970).
Theorem putnam_1985_b5:
Lim_seq (fun n => RInt (fun x => Rpower x (-1/2) * exp (-1985 * (x + 1/x))) 0 (INR n)) = putnam_1985_b5_solution.
Proof. Admitted.
End putnam_1985_b5. | theory putnam_1985_b5 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1985_b5_solution::real where "putnam_1985_b5_solution \<equiv> undefined"
(* sqrt (pi / 1985) * exp (-3970) *)
theorem putnam_1985_b5:
assumes "interval_lebesgue_integral lebesgue MInfty PInfty (\<lambda>x. exp (- (x^2))) = sqrt (pi)"
shows "interval_lebesgue_integral lebesgue 0 PInfty (\<lambda>t. t powr (-1/2) * exp (-1985 * (t + t powi (-1)))) = putnam_1985_b5_solution"
sorry
end | null |
putnam_1985_b6 | theorem putnam_1985_b6
(n : β)
(npos : n > 0)
(G : Finset (Matrix (Fin n) (Fin n) β))
(groupG : (β g β G, β h β G, g * h β G) β§ 1 β G β§ (β g β G, β h β G, g * h = 1))
(hG : β M in G, Matrix.trace M = 0)
: (β M in G, M = 0) :=
sorry | Let $G$ be a finite set of real $n\times n$ matrices $\{M_i\}$, $1 \leq i \leq r$, which form a group under matrix
multiplication. Suppose that $\sum_{i=1}^r \mathrm{tr}(M_i)=0$, where $\mathrm{tr}(A)$ denotes the trace of the matrix $A$. Prove that $\sum_{i=1}^r M_i$ is the $n \times n$ zero matrix. | null | ['abstract_algebra', 'linear_algebra'] | null | theory putnam_1985_b6 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Analysis.Determinants"
begin
theorem putnam_1985_b6:
fixes n::nat and G::"(real^'a^'a) set"
assumes npos : "n > 0"
and acard : "CARD('a) = n"
and Gfin : "finite G"
and groupG : "(\<forall>g \<in> G. \<forall>h \<in> G. g ** h \<in> G) \<and> (1 \<in> G) \<and> (\<forall>g \<in> G. \<exists>h \<in> G. g ** h = 1)"
and hG : "(\<Sum>M \<in> G. trace M) = 0"
shows "(\<Sum>M \<in> G. M) = 0"
sorry
end | null |
putnam_1994_a1 | theorem putnam_1994_a1
(a : β β β)
(ha : β n β₯ 1, 0 < a n β§ a n β€ a (2 * n) + a (2 * n + 1))
: Β¬(β s : β, Tendsto (fun N : β => β n : Set.Icc 1 N, a n) atTop (π s)) :=
sorry | Suppose that a sequence $a_1,a_2,a_3,\dots$ satisfies $0<a_n \leq a_{2n}+a_{2n+1}$ for all $n \geq 1$. Prove that the series $\sum_{n=1}^\infty a_n$ diverges. | null | ['analysis'] | Section putnam_1994_a1.
Require Import Nat Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1994_a1:
exists (a: nat -> R), forall (n: nat), gt n 0 -> 0 < a n <= a (mul 2 n) + a (add (mul 2 n) 1) ->
~ ex_lim_seq (fun n => sum_n (fun m => a m) n).
Proof. Admitted.
End putnam_1994_a1. | theory putnam_1994_a1 imports Complex_Main
begin
theorem putnam_1994_a1:
fixes a :: "nat \<Rightarrow> real"
assumes ha: "\<forall>n::nat\<ge>1. 0 < a n \<and> a n \<le> a (2*n) + a (2*n+1)"
shows "\<not>(\<exists>s::real. filterlim (\<lambda>N::nat. (\<Sum>n::nat=1..N. a n)) (nhds s) at_top)"
sorry
end
| null |
putnam_1994_a4 | theorem putnam_1994_a4
(A B : Matrix (Fin 2) (Fin 2) β€)
(ABinv : Nonempty (Invertible A) β§ Nonempty (Invertible (A + B)) β§ Nonempty (Invertible (A + 2 * B)) β§ Nonempty (Invertible (A + 3 * B)) β§ Nonempty (Invertible (A + 4 * B)))
: Invertible (A + 5 * B) :=
sorry | Let $A$ and $B$ be $2 \times 2$ matrices with integer entries such that $A$, $A+B$, $A+2B$, $A+3B$, and $A+4B$ are all invertible matrices whose inverses have integer entries. Show that $A+5B$ is invertible and that its inverse has integer entries. | null | ['linear_algebra'] | null | theory putnam_1994_a4 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1994_a4:
fixes A B :: "int^2^2"
assumes ABinv: "invertible A \<and> invertible (A + B) \<and> invertible (A + 2*B) \<and> invertible (A + 3*B) \<and> invertible (A + 4*B)"
shows "invertible (A + 5*B)"
sorry
end
| null |
putnam_1994_a5 | theorem putnam_1994_a5
(r : β β β)
(S : Set β)
(rpos : r > 0)
(rlim : Tendsto r atTop (π 0))
(hS : S = {x : β | β i : Fin 1994 β β, (β j k : Fin 1994, j < k β i j < i k) β§ (x = β j : Fin 1994, r (i j))})
: β a b : β, a < b β (β c d : β, a β€ c β§ c < d β§ d β€ b β§ (Set.Ioo c d) β© S = β
) :=
sorry | Let $(r_n)_{n \geq 0}$ be a sequence of positive real numbers such that $\lim_{n \to \infty} r_n=0$. Let $S$ be the set of numbers representable as a sum $r_{i_1}+r_{i_2}+\cdots+r_{i_{1994}}$, with $i_1<i_2<\cdots<i_{1994}$. Show that every nonempty interval $(a,b)$ contains a nonempty subinterval $(c,d)$ that does not intersect $S$. | null | ['analysis'] | null | theory putnam_1994_a5 imports Complex_Main
begin
(* uses (nat \<Rightarrow> nat) instead of (Fin 1994 \<Rightarrow> nat) *)
theorem putnam_1994_a5:
fixes r :: "nat \<Rightarrow> real"
and S :: "real set"
assumes rpos: "\<forall>i::nat. r i > 0"
and rlim: "filterlim r (nhds 0) at_top"
and hS: "S \<equiv> {x::real. (\<exists>i::nat\<Rightarrow>nat. (\<forall>j::nat\<in>{0..1993}. \<forall>k::nat\<in>{0..1993}. (j < k \<longrightarrow> i j < i k)) \<and> (x = (\<Sum>j::nat=0..1993. r (i j))))}"
shows "\<forall>a b::real. a < b \<longrightarrow> (\<exists>c d::real. a \<le> c \<and> c < d \<and> d \<le> b \<and> {c<..<d} \<inter> S = {})"
sorry
end
| null |
putnam_1994_a6 | theorem putnam_1994_a6
(f : Fin 10 β Equiv.Perm β€)
(mijcomp : β β (β β Fin 10) β β β (β€ β β€))
(F : Finset (β€ β β€))
(hf: β n : β€, β m : β, β i : β β Fin 10, m β₯ 1 β§ (mijcomp m i 0) 0 = n)
(hmijcomp : β m β₯ 1, β (i : β β Fin 10) (j : Fin m), mijcomp m i j = if (j = m - 1) then (f (i j) : β€ β β€) else (f (i j) β mijcomp m i (j + 1)))
(hF : F = {g : β€ β β€ | β e : Fin 10 β Fin 2, g = (f 0)^[e 0] β (f 1)^[e 1] β (f 2)^[e 2] β (f 3)^[e 3] β (f 4)^[e 4] β (f 5)^[e 5] β (f 6)^[e 6] β (f 7)^[e 7] β (f 8)^[e 8] β (f 9)^[e 9]})
: β A : Finset β€, A.Nonempty β {g β F | g '' A = A}.encard β€ 512 :=
sorry | Let $f_1,\dots,f_{10}$ be bijections of the set of integers such that for each integer $n$, there is some composition $f_{i_1} \circ f_{i_2} \circ \cdots \circ f_{i_m}$ of these functions (allowing repetitions) which maps 0 to $n$. Consider the set of $1024$ functions $\mathcal{F}=\{f_1^{e_1} \circ f_2^{e_2} \circ \cdots \circ f_{10}^{e_{10}}\}$, $e_i=0$ or $1$ for $1 \leq i \leq 10$. ($f_i^0$ is the identity function and $f_i^1=f_i$.) Show that if $A$ is any nonempty finite set of integers, then at most $512$ of the functions in $\mathcal{F}$ map $A$ to itself. | null | ['algebra'] | null | theory putnam_1994_a6 imports Complex_Main
"HOL-Combinatorics.Permutations"
begin
(* uses (nat \<Rightarrow> (int \<Rightarrow> int)) instead of (Fin 10 \<Rightarrow> (int \<Rightarrow> int)) and (nat \<Rightarrow> nat) instead of (Fin m \<Rightarrow> Fin 10) and (Fin 10 \<Rightarrow> Fin 2) *)
theorem putnam_1994_a6:
fixes f :: "nat \<Rightarrow> (int \<Rightarrow> int)"
and mijcomp :: "nat \<Rightarrow> (nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> (int \<Rightarrow> int)"
and F :: "(int \<Rightarrow> int) set"
assumes hfperm: "\<forall>i::nat\<in>{0..9}. (f i) permutes UNIV"
and hfcomp: "\<forall>n::int. \<exists>m::nat\<ge>1. \<exists>i::nat\<Rightarrow>nat. (\<forall>j::nat\<in>{0..(m-1)}. i j \<in> {0..9}) \<and> (mijcomp m i 0) 0 = n"
and hmijcomp: "\<forall>m::nat\<ge>1. \<forall>(i::nat\<Rightarrow>nat). \<forall>j::nat\<in>{0..(m-1)}. mijcomp m i j = (if (j = m-1) then (f (i j)) else (f (i j) \<circ> mijcomp m i (j+1)))"
and hF: "F \<equiv> {g::int\<Rightarrow>int. (\<exists>e::nat\<Rightarrow>nat. (\<forall>i::nat\<in>{0..9}. e i = 0 \<or> e i = 1) \<and> (g = (f 0)^^(e 0) \<circ> (f 1)^^(e 1) \<circ> (f 2)^^(e 2) \<circ> (f 3)^^(e 3) \<circ> (f 4)^^(e 4) \<circ> (f 5)^^(e 5) \<circ> (f 6)^^(e 6) \<circ> (f 7)^^(e 7) \<circ> (f 8)^^(e 8) \<circ> (f 9)^^(e 9)))}"
shows "\<forall>A::int set. (finite A \<and> A \<noteq> {}) \<longrightarrow> (card {g\<in>F. g ` A = A} \<le> 512)"
sorry
end
| null |
putnam_1994_b1 | abbrev putnam_1994_b1_solution : Set β€ := sorry
-- {n : β€ | (315 β€ n β§ n β€ 325) β¨ (332 β€ n β§ n β€ 350)}
theorem putnam_1994_b1
(n : β€)
(nwithin : Prop)
(hnwithin : nwithin = ({m : β | |n - m ^ 2| β€ 250}.encard = 15))
: (n > 0 β§ nwithin) β n β putnam_1994_b1_solution :=
sorry | Find all positive integers $n$ that are within $250$ of exactly $15$ perfect squares. | Show that an integer $n$ is within $250$ of exactly $15$ perfect squares if and only if either $315 \leq n \leq 325$ or $332 \leq n \leq 350$. | ['algebra'] | Section putnam_1994_b1.
Require Import Ensembles Finite_sets ZArith.
Open Scope Z.
Definition putnam_1994_b1_solution (n: Z) := 315 <= n <= 325 \/ 332 <= n <= 350.
Theorem putnam_1994_b1:
forall (n: Z), exists (E: Ensemble Z), cardinal Z E 15 ->
forall (m: Z), E m -> Z.abs (m * m - n) <= 250 ->
putnam_1994_b1_solution n.
Proof. Admitted.
End putnam_1994_b1. | theory putnam_1994_b1 imports Complex_Main
begin
definition putnam_1994_b1_solution :: "nat set" where "putnam_1994_b1_solution \<equiv> undefined"
(* {n::nat. (315 \<le> n \<and> n \<le> 325) \<or> (332 \<le> n \<and> n \<le> 350)} *)
theorem putnam_1994_b1:
fixes n :: nat
and nwithin :: bool
assumes hnwithin: "nwithin \<equiv> (card {m::nat. \<bar>n - m^2\<bar> \<le> 250} = 15)"
shows "(n > 0 \<and> nwithin) \<longleftrightarrow> n \<in> putnam_1994_b1_solution"
sorry
end
| null |
putnam_1994_b2 | abbrev putnam_1994_b2_solution : Set β := sorry
-- {c : β | c < 243 / 8}
theorem putnam_1994_b2
(c : β)
(inter : Prop)
(hinter : inter = β m b : β, {x : β | m * x + b = x ^ 4 + 9 * x ^ 3 + c * x ^ 2 + 9 * x + 4}.encard = 4)
: inter β c β putnam_1994_b2_solution :=
sorry | For which real numbers $c$ is there a straight line that intersects the curve $x^4+9x^3+cx^2+9x+4$ in four distinct points? | Show that there exists such a line if and only if $c<243/8$. | ['geometry', 'algebra'] | Section putnam_1994_b2.
Require Import List Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1994_b2_solution (c: R) := c < 243 / 8.
Theorem putnam_1994_b2:
forall (c: R), exists(m b: R),
let f (x: R) := pow x 4 + 9 * pow x 3 + c * pow x 2 + 9 * x + 4 in
let g (x: R) := m * x + b in
exists (l: list R), eq (length l) 4%nat /\ NoDup l /\ forall (r: R), In r l -> f r = g r
<-> putnam_1994_b2_solution c.
Proof. Admitted.
End putnam_1994_b2. | theory putnam_1994_b2 imports Complex_Main
begin
definition putnam_1994_b2_solution :: "real set" where "putnam_1994_b2_solution \<equiv> undefined"
(* {c::real. c < 243/8} *)
theorem putnam_1994_b2:
fixes c :: real
and inter :: bool
assumes hinter: "inter \<equiv> (\<exists>m b::real. card {x::real. m*x + b = x^4 + 9*x^3 + c*x^2 + 9*x + 4} = 4)"
shows "inter \<longleftrightarrow> c \<in> putnam_1994_b2_solution"
sorry
end
| null |
putnam_1994_b3 | abbrev putnam_1994_b3_solution : Set β := sorry
-- Set.Iio 1
theorem putnam_1994_b3
(k : β)
(allfexN : Prop)
(hallfexN : allfexN = β f : β β β, (f > 0 β§ Differentiable β f β§ β x : β, deriv f x > f x) β (β N : β, β x > N, f x > Real.exp (k * x)))
: allfexN β k β putnam_1994_b3_solution :=
sorry | Find the set of all real numbers $k$ with the following property: For any positive, differentiable function $f$ that satisfies $f'(x)>f(x)$ for all $x$, there is some number $N$ such that $f(x)>e^{kx}$ for all $x>N$. | Show that the desired set is $(-\infty,1)$. | ['analysis'] | Section putnam_1994_b3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1994_b3_solution (k: R) := k <= 1.
Theorem putnam_1994_b3:
forall (k: R) (f: R -> R) (x: R), f x > 0 /\ ex_derive f x /\
(Derive f) x > f x ->
exists (N: R), x > N ->
f x > exp (k * x)
<-> putnam_1994_b3_solution k.
Proof. Admitted.
End putnam_1994_b3. | theory putnam_1994_b3 imports Complex_Main
"HOL-Analysis.Derivative"
begin
definition putnam_1994_b3_solution :: "real set" where "putnam_1994_b3_solution \<equiv> undefined"
(* {..<1} *)
theorem putnam_1994_b3:
fixes k :: real
and allfexN :: bool
assumes hallfexN: "allfexN \<equiv> (\<forall>f::real\<Rightarrow>real. (((\<forall>x::real. f x > 0) \<and> f differentiable_on UNIV \<and> (\<forall>x::real. deriv f x > f x)) \<longrightarrow> (\<exists>N::real. \<forall>x::real>N. f x > exp (k*x))))"
shows "allfexN \<longleftrightarrow> k \<in> putnam_1994_b3_solution"
sorry
end
| null |
putnam_1994_b4 | theorem putnam_1994_b4
(matgcd : Matrix (Fin 2) (Fin 2) β€ β β€)
(A : Matrix (Fin 2) (Fin 2) β€)
(d : β β β€)
(hmatgcd : β M : Matrix (Fin 2) (Fin 2) β€, matgcd M = Int.gcd (Int.gcd (Int.gcd (M 0 0) (M 0 1)) (M 1 0)) (M 1 1))
(hA : A 0 0 = 3 β§ A 0 1 = 2 β§ A 1 0 = 4 β§ A 1 1 = 3)
(hd : β n β₯ 1, d n = matgcd (A ^ n - 1))
: Tendsto d atTop atTop :=
sorry | For $n \geq 1$, let $d_n$ be the greatest common divisor of the entries of $A^n-I$, where $A=\begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ and $I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Show that $\lim_{n \to \infty} d_n=\infty$. | null | ['linear_algebra', 'number_theory', 'analysis'] | Section putnam_1994_b4.
Require Import Nat List Reals Coquelicot.Coquelicot.
Import ListNotations.
Open Scope R.
Theorem putnam_1994_b4:
let fix gcd_n (args : list nat) : nat :=
match args with
| nil => 0%nat
| h :: args' => gcd h (gcd_n args')
end in
let fix Mmult_n {T : Ring} {n : nat} (A : matrix n n) (p : nat) :=
match p with
| O => A
| S p' => @Mmult T n n n A (Mmult_n A p')
end in
let A := mk_matrix 2 2 (fun i j =>
match i, j with
| 0, 0 => 3 | 0, 1 => 2
| 1, 0 => 4 | 1, 1 => 3
| _, _ => 0
end) in
let I := mk_matrix 2 2 (fun i j =>
match i, j with
| 0, 0 => 1 | 0, 1 => 0
| 1, 0 => 0 | 1, 1 => 1
| _, _ => 0
end) in
let dn_mat (n: nat) := Mplus (Mmult_n A n) (opp I) in
let dn (n: nat) := gcd_n [Z.to_nat (floor (coeff_mat 0 (dn_mat n) 0 0));
Z.to_nat (floor (coeff_mat 0 (dn_mat n) 0 1));
Z.to_nat (floor (coeff_mat 0 (dn_mat n) 1 0));
Z.to_nat (floor (coeff_mat 0 (dn_mat n) 1 1))] in
~ ex_lim_seq (fun n => INR (dn n)).
Proof. Admitted.
End putnam_1994_b4. | theory putnam_1994_b4 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1994_b4:
fixes matgcd :: "(int^2^2) \<Rightarrow> nat"
and A :: "int^2^2"
and matpow :: "int^2^2 \<Rightarrow> nat \<Rightarrow> int^2^2"
and d :: "nat \<Rightarrow> nat"
assumes hmatgcd: "\<forall>M::int^2^2. matgcd M = Gcd {M$1$1, M$1$2, M$2$1, M$2$2}"
and hA: "A$1$1 = 3 \<and> A$1$2 = 2 \<and> A$2$1 = 4 \<and> A$2$2 = 3"
and hmatpow: "\<forall>M::int^2^2. matpow M 0 = mat 1 \<and> (\<forall>k::nat>0. matpow M k = matpow M (k-1) ** M)"
and hd: "\<forall>n::nat\<ge>1. d n = matgcd (matpow A n - mat 1)"
shows "filterlim d at_top at_top"
sorry
end
| null |
putnam_1994_b5 | theorem putnam_1994_b5
(f : β β β β β€)
(n : β)
(hf : β (Ξ± : β) (x : β), f Ξ± x = Int.floor (Ξ± * x))
(npos : n > 0)
: β Ξ± : β, β k β Set.Icc 1 n, (((f Ξ±) ^ k) (n ^ 2) = n ^ 2 - k) β§ (f (Ξ± ^ k) (n ^ 2) = n ^ 2 - k) :=
sorry | For any real number $\alpha$, define the function $f_\alpha(x)=\lfloor \alpha x \rfloor$. Let $n$ be a positive integer. Show that there exists an $\alpha$ such that for $1 \leq k \leq n$, $f_\alpha^k(n^2)=n^2-k=f_{\alpha^k}(n^2)$. | null | ['algebra'] | Section putnam_1994_b5.
Require Import Basics ZArith Zpower Reals Coquelicot.Coquelicot.
Theorem putnam_1994_b5:
let fix compose_n {A: Type} (f : A -> A) (n : nat) :=
match n with
| O => fun x => x
| S n' => compose f (compose_n f n')
end in
let fa (a x: R) := IZR (floor (a * x)) in
forall (n: Z), Z.gt n 0 ->
exists (a: R), forall (k: Z), and (Z.ge 1 k) (Z.ge k n) ->
(compose_n (fa a) (Z.to_nat k)) (IZR (Z.pow n 2)) = IZR (Z.pow n 2 - k) /\ IZR (Z.pow n 2 - k) = fa (Rpower a (IZR k)) (IZR (Z.pow n 2)).
Proof. Admitted.
End putnam_1994_b5. | theory putnam_1994_b5 imports Complex_Main
begin
theorem putnam_1994_b5:
fixes f :: "real \<Rightarrow> nat \<Rightarrow> int"
and n :: nat
assumes hf: "\<forall>(\<alpha>::real)(x::nat). f \<alpha> x = \<lfloor>\<alpha>*x\<rfloor>"
and npos: "n > 0"
shows "\<exists>\<alpha>::real. \<forall>k::nat\<in>{1..n}. ((f \<alpha> (n^2))^k = n^2 - k) \<and> (f (\<alpha>^k) (n^2) = n^2 - k)"
sorry
end
| null |
putnam_1994_b6 | theorem putnam_1994_b6
(n : β β β€)
(hn : β a : β, n a = 101 * a - 100 * 2 ^ a)
: β a b c d : Set.Icc 0 99, (n a + n b β‘ n c + n d [ZMOD 10100]) β (({a, b} : Set (Set.Icc 0 99)) = {c, d}) :=
sorry | For any integer $a$, set $n_a=101a-100 \cdot 2^a$. Show that for $0 \leq a,b,c,d \leq 99$, $n_a+n_b \equiv n_c+n_d \pmod{10100}$ implies $\{a,b\}=\{c,d\}$. | null | ['number_theory'] | Section putnam_1994_b6.
Require Import Nat.
Theorem putnam_1994_b6:
let n (a: nat) := 101 * a - 100 * pow 2 a in
forall (a b c d: nat), 0 <= a <= 99 /\ 0 <= b <= 99 /\ 0 <= c <= 99 /\ 0 <= d <= 99 /\ n a + n b mod 10100 = n c + n d ->
(a,b) = (c,d).
Proof. Admitted.
End putnam_1994_b6. | theory putnam_1994_b6 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
theorem putnam_1994_b6:
fixes n :: "nat \<Rightarrow> int"
assumes hn: "\<forall>a::nat. n a = 101*a - 100 * 2^a"
shows "\<forall>a::nat\<in>{0..99}. \<forall>b::nat\<in>{0..99}. \<forall>c::nat\<in>{0..99}. \<forall>d::nat\<in>{0..99}. (([n a + n b = n c + n d] (mod 10100)) \<longrightarrow> {a, b} = {c, d})"
sorry
end
| null |
putnam_1979_a1 | abbrev putnam_1979_a1_solution : Multiset β := sorry
-- Multiset.replicate 659 3 + {2}
theorem putnam_1979_a1
(P : Multiset β β Prop := fun a => Multiset.card a > 0 β§ (β i β a, i > 0) β§ a.sum = 1979)
: P putnam_1979_a1_solution β§ β a : Multiset β, P a β putnam_1979_a1_solution.prod β₯ a.prod :=
sorry | For which positive integers $n$ and $a_1, a_2, \dots, a_n$ with $\sum_{i = 1}^{n} a_i = 1979$ does $\prod_{i = 1}^{n} a_i$ attain the greatest value? | $n$ equals $660$; all but one of the $a_i$ equal $3$ and the remaining $a_i$ equals $2$. | ['algebra'] | Section putnam_1979_a1.
Require Import Nat List.
Definition putnam_1979_a1_solution := 2 :: repeat 3 659.
Theorem putnam_1979_a1:
forall (l m: list nat),
fold_left add l 0 = 1979 /\
fold_left add m 0 = 1979 /\
fold_left mul l 1 >= fold_left mul m 1
<->
l = putnam_1979_a1_solution.
Proof. Admitted.
End putnam_1979_a1. | theory putnam_1979_a1 imports
Complex_Main
"HOL-Library.Multiset"
begin
definition putnam_1979_a1_solution :: "nat multiset" where "putnam_1979_a1_solution \<equiv> undefined"
(* add_mset 2 (replicate_mset 659 3) *)
theorem putnam_1979_a1:
fixes P :: "nat multiset \<Rightarrow> bool"
defines "P \<equiv> \<lambda> a :: nat multiset. size a > 0 \<and> (\<forall> i \<in># a. i > 0) \<and> sum_mset a = 1979"
shows "P putnam_1979_a1_solution \<and> (\<forall> a :: nat multiset. P a \<longrightarrow> prod_mset putnam_1979_a1_solution \<ge> prod_mset a)"
sorry
end | null |
putnam_1979_a2 | abbrev putnam_1979_a2_solution : β β Prop := sorry
-- fun k : β => k β₯ 0
theorem putnam_1979_a2
: β k : β, (β f : β β β, Continuous f β§ β x : β, f (f x) = k*x^9) β putnam_1979_a2_solution k :=
sorry | For which real numbers $k$ does there exist a continuous function $f : \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = kx^9$ for all real $x$? | Such a function exists if and only if $k \ge 0$. | ['analysis', 'algebra'] | Section putnam_1979_a2.
Require Import Basics Reals Coquelicot.Coquelicot.
Theorem putnam_1979_a2:
forall (k: R), exists (f: R -> R), continuity f ->
forall x, (compose f f) x = k * pow x 9.
Proof. Admitted.
End putnam_1979_a2. | theory putnam_1979_a2 imports
Complex_Main
begin
definition putnam_1979_a2_solution :: "real \<Rightarrow> bool" where "putnam_1979_a2_solution \<equiv> undefined"
(* \<lambda> k :: real. k \<ge> 0 *)
theorem putnam_1979_a2:
shows "\<forall> k :: real. (\<exists> f :: real \<Rightarrow> real. continuous_on UNIV f \<and> (\<forall> x :: real. f (f x) = k * x ^ 9)) \<longleftrightarrow> putnam_1979_a2_solution k"
sorry
end | null |
putnam_1979_a3 | abbrev putnam_1979_a3_solution : (β Γ β) β Prop := sorry
-- fun (a, b) => β m : β€, a = m β§ b = m
theorem putnam_1979_a3
(x : β β β)
(hx : β n : β, x n β 0 β§ (n β₯ 3 β x n = (x (n - 2))*(x (n - 1))/(2*(x (n - 2)) - (x (n - 1)))))
: (β m : β, β n : β, n > m β§ β a : β€, a = x n) β putnam_1979_a3_solution (x 1, x 2) :=
sorry | Let $x_1, x_2, x_3, \dots$ be a sequence of nonzero real numbers such that $$x_n = \frac{x_{n-2}x_{n-1}}{2x_{n-2}-x_{n-1}}$$ for all $n \ge 3$. For which real values of $x_1$ and $x_2$ does $x_n$ attain integer values for infinitely many $n$? | We must have $x_1 = x_2 = m$ for some integer $m$. | ['algebra'] | Section putnam_1979_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1979_a3_solution (x y: R) := x = IZR (floor x) /\ y = IZR (floor y).
Theorem putnam_1979_a3:
let A :=
fix a (x y: R) (n: nat) :=
match n with
| O => x
| S O => y
| S ((S n'') as n') => (a x y n'' * a x y n') / (2 * a x y n'' - a x y n')
end in
forall (n: nat), exists (x y: R), (A x y n+1 <> 2 * A x y n) /\
(~ exists (r: R), A x y n = IZR (floor (A x y n)) /\ INR n < r)
<->
putnam_1979_a3_solution x y.
Proof. Admitted.
End putnam_1979_a3. | theory putnam_1979_a3 imports
Complex_Main
begin
definition putnam_1979_a3_solution :: "(real \<times> real) \<Rightarrow> bool" where "putnam_1979_a3_solution \<equiv> undefined"
(* \<lambda> (a :: real, b :: real). \<exists> m :: int. a = m \<and> b = m *)
theorem putnam_1979_a3:
fixes x :: "nat \<Rightarrow> real"
assumes hx: "\<forall> n :: nat. x n \<noteq> 0 \<and> (n \<ge> 3 \<longrightarrow> x n = x (n - 2) * x (n - 1) / (2 * x (n - 2) - x (n - 1)))"
shows "(\<forall> m :: nat. \<exists> n :: nat. n > m \<and> (\<exists> a :: int. a = x n)) \<longleftrightarrow> putnam_1979_a3_solution (x 1, x 2)"
sorry
end | null |
putnam_1979_a4 | abbrev putnam_1979_a4_solution : Prop := sorry
-- True
theorem putnam_1979_a4
(A : Finset (Fin 2 β β) Γ Finset (Fin 2 β β) β Prop := fun (R, B) => R.card = B.card β§ R β© B = β
β§
β u : Finset (Fin 2 β β), u β R βͺ B β§ u.card = 3 β Β¬Collinear β (u : Set (Fin 2 β β)))
(w : (Fin 2 β β) Γ (Fin 2 β β) β β β (Fin 2 β β) := fun (P, Q) => fun x : β => fun i : Fin 2 => x * P i + (1 - x) * Q i)
: (β R : Finset (Fin 2 β β), β B : Finset (Fin 2 β β), A (R, B) β β v : Finset ((Fin 2 β β) Γ (Fin 2 β β)),
(β L β v, β M β v, L β M β β x β Icc 0 1, β y β Icc 0 1,
Euclidean.dist (w (L.1, L.2) x) (w (M.1, M.2) y) β 0) β§
v.card = R.card β§ β L β v, L.1 β R β§ L.2 β B) β putnam_1979_a4_solution :=
sorry | Let $A$ be a set of $2n$ points in the plane, $n$ colored red and $n$ colored blue, such that no three points in $A$ are collinear. Must there exist $n$ closed straight line segments, each connecting one red and one blue point in $A$, such that no two of the $n$ line segments intersect? | Such line segments must exist. | ['geometry', 'combinatorics'] | null | theory putnam_1979_a4 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
definition putnam_1979_a4_solution :: bool where "putnam_1979_a4_solution \<equiv> undefined"
(* True *)
theorem putnam_1979_a4:
fixes A :: "(real^2) set \<times> (real^2) set \<Rightarrow> bool"
and w :: "(real^2) \<times> (real^2) \<Rightarrow> real \<Rightarrow> real^2"
defines "A \<equiv> \<lambda> (R, B). finite R \<and> finite B \<and> card R = card B \<and> R \<inter> B = {} \<and>
\<not>(\<exists> P \<in> R \<union> B. \<exists> Q \<in> R \<union> B. \<exists> N \<in> R \<union> B. Q \<noteq> P \<and> N \<noteq> P \<and> N \<noteq> Q \<and> (\<exists> c :: real. Q - P = c *s (N - P)))"
and "w \<equiv> \<lambda> (P, Q) x. x *s P + (1 - x) *s Q"
shows "(\<forall> R B :: (real^2) set. A (R, B) \<longrightarrow> (\<exists> v :: ((real^2) \<times> (real^2)) set. (card v = card R) \<and>
(\<forall> L \<in> v. fst L \<in> R \<and> snd L \<in> B \<and>
(\<forall> M \<in> v. L \<noteq> M \<longrightarrow> ((fst L \<noteq> fst M) \<and> (snd L \<noteq> snd M) \<and>
(\<forall> x \<in> {0..1}. \<forall> y \<in> {0..1}. w L x \<noteq> w M y)))))) \<longleftrightarrow> putnam_1979_a4_solution"
sorry
end
| null |
putnam_1979_a5 | theorem putnam_1979_a5
(S : β β β β β€ := fun x : β => fun n : β => Int.floor (n*x))
(P : β β Prop := fun x : β => x^3 - 10*x^2 + 29*x - 25 = 0)
: β Ξ± Ξ² : β, Ξ± β Ξ² β§ P Ξ± β§ P Ξ² β§ β n : β, β m : β€, m > n β§ β c d : β, S Ξ± c = m β§ S Ξ² d = m :=
sorry | Let $S(x)$ denote the sequence $\lfloor 0 \rfloor, \lfloor x \rfloor, \lfloor 2x \rfloor, \lfloor 3x \rfloor, \dots$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. Prove that there exist distinct real roots $\alpha$ and $\beta$ of $x^3 - 10x^2 + 29x - 25$ such that infinitely many positive integers appear in both $S(\alpha)$ and $S(\beta)$. | null | ['algebra'] | Section putnam_1979_a5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1979_a5:
let f (x: R) := Rpower x 3 - 10 * pow x 2 + 29 * x - 25 in
exists (r1 r2: R), r1 <> r2 /\ f r1 = 0 /\ f r2 = 0 /\
~ exists (r: R), forall (n: nat), exists (p q: Z), n = Z.to_nat (floor (IZR p * r1)) /\ n = Z.to_nat (floor (IZR q * r2))
/\ INR n < r.
Proof. Admitted.
End putnam_1979_a5. | theory putnam_1979_a5 imports
Complex_Main
begin
theorem putnam_1979_a5:
fixes S :: "real \<Rightarrow> nat \<Rightarrow> int"
and P :: "real \<Rightarrow> bool"
defines "S \<equiv> \<lambda> (x :: real) (n :: nat). floor (n * x)"
and "P \<equiv> \<lambda> x :: real. x ^ 3 - 10 * x ^ 2 + 29 * x - 25 = 0"
shows "\<exists> \<alpha> \<beta> :: real. \<alpha> \<noteq> \<beta> \<and> P \<alpha> \<and> P \<beta> \<and> (\<forall> n :: nat. \<exists> m :: int. m > n \<and> (\<exists> c d :: nat. S \<alpha> c = m \<and> S \<beta> d = m))"
sorry
end | null |
putnam_1979_a6 | theorem putnam_1979_a6
(n : β)
(p : β β β)
(hp : β i β Finset.range n, p i β Icc 0 1)
: β x β Icc 0 1, (β i β Finset.range n, x β p i) β§ β i in Finset.range n, 1/|x - p i| β€ 8*n*β i in Finset.range n, (1 : β)/(2*i + 1) :=
sorry | For all $i \in \{0, 1, \dots, n - 1\}$, let $p_i \in [0, 1]$. Prove that there exists some $x \in [0, 1]$ such that $$\sum_{i = 0}^{n - 1} \frac{1}{|x - p_i|} \le 8n\left(\sum_{i = 0}^{n-1} \frac{1}{2i + 1}\right).$$ | null | ['algebra'] | Section putnam_1979_a6.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1979_a6:
forall (a: nat -> R) (n: nat),
0 <= a n <= 1 ->
exists (b: R), 0 <= b <= 1 ->
sum_n (fun n => 1/(b - a n)) n <= 8 * INR n * sum_n (fun i => 1/(2*(INR i+1) - 1)) n.
Proof. Admitted.
End putnam_1979_a6. | theory putnam_1979_a6 imports
Complex_Main
begin
theorem putnam_1979_a6:
fixes n :: nat
and p :: "nat \<Rightarrow> real"
assumes hp: "\<forall> i \<in> {1..n}. p i \<in> {0..1}"
shows "\<exists> x \<in> {0..1}. (\<forall> i \<in> {1..n}. x \<noteq> p i) \<and> (\<Sum> i = 1..n. 1 / \<bar>x - p i\<bar>) \<le> 8 * n * (\<Sum> i = 1..n. 1 / (2 * i - 1))"
sorry
end | null |
putnam_1979_b2 | abbrev putnam_1979_b2_solution : β Γ β β β := sorry
-- fun (a, b) => (Real.exp (-1))*(b^b/a^a)^(1/(b-a))
theorem putnam_1979_b2
: β a b : β, 0 < a β§ a < b β Tendsto (fun t : β => (β« x in Icc 0 1, (b*x + a*(1 - x))^t)^(1/t)) (π[β ] 0) (π (putnam_1979_b2_solution (a, b))) :=
sorry | If $0 < a < b$, find $$\lim_{t \to 0} \left( \int_{0}^{1}(bx + a(1-x))^t dx \right)^{\frac{1}{t}}$$ in terms of $a$ and $b$. | The limit equals $$e^{-1}\left(\frac{b^b}{a^a}\right)^{\frac{1}{b-a}}.$$ | ['analysis'] | Section putnam_1979_b2.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1979_b2_solution (a b: R):= (Rpower b (b/(b-a))) / ((exp 1) * Rpower a (a/(b-a))).
Theorem putnam_1979_b2:
forall (a b: R), 0 < a < b ->
Lim_seq (fun lam => Rpower (RInt (fun x => Rpower (b*x + a*(1-x)) (INR lam)) 0 1) 1/INR lam)
= putnam_1979_b2_solution a b.
Proof. Admitted.
End putnam_1979_b2. | theory putnam_1979_b2 imports
Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1979_b2_solution :: "real \<Rightarrow> real \<Rightarrow> real" where "putnam_1979_b2_solution \<equiv> undefined"
(* \<lambda> a b. exp (-1) * (b powr b / a powr a) powr (1 / (b - a)) *)
theorem putnam_1979_b2:
shows "\<forall> a b :: real. 0 < a \<and> a < b \<longrightarrow> ((\<lambda> t :: real. (interval_lebesgue_integral lebesgue 0 1 (\<lambda> x. (b * x + a * (1 - x)) powr t)) powr (1 / t)) \<longlongrightarrow> putnam_1979_b2_solution a b) (at 0)"
sorry
end | null |
putnam_1979_b3 | abbrev putnam_1979_b3_solution : β β β€ := sorry
-- fun n β¦ (n - 1) / 2
theorem putnam_1979_b3
(F : Type*) [Field F] [Fintype F]
(n : β := Fintype.card F)
(nodd : Odd n)
(b c : F)
(p : Polynomial F := X ^ 2 + (C b) * X + (C c))
(hp : Irreducible p)
: ({d : F | Irreducible (p + (C d))}.ncard = putnam_1979_b3_solution n) :=
sorry | Let $F$ be a finite field with $n$ elements, and assume $n$ is odd. Suppose $x^2 + bx + c$ is an irreducible polynomial over $F$. For how many elements $d \in F$ is $x^2 + bx + c + d$ irreducible? | Show that there are $\frac{n - 1}{2}$ such elements $d$. | ['abstract_algebra'] | null | theory putnam_1979_b3 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1979_b3_solution :: "nat \<Rightarrow> nat" where
"putnam_1979_b3_solution \<equiv> undefined"
(* \<lambda> n :: nat. nat \<lfloor>(n - 1) / 2\<rfloor> *)
theorem putnam_1979_b3:
fixes F (structure)
and n :: "nat"
and b c :: "'a :: field"
and p :: "'a poly"
defines "p \<equiv> [:c, b, 1:]"
assumes hFfin : "finite (carrier F) \<and> card (carrier F) = n"
and Ffield : "field F"
and nodd : "odd n"
and hp : "irreducible p"
shows "card {d :: ('a :: field). irreducible (p + monom d 0)} = putnam_1979_b3_solution n"
sorry
end | null |
putnam_1979_b6 | theorem putnam_1979_b6
(n : β)
(z : Fin n β β)
: (|((β i : Fin n, (z i) ^ 2) ^ ((1 : β) / 2)).re| β€ β i : Fin n, |(z i).re|) :=
sorry | Let $z_i$ be complex numbers for $i = 1, 2, \dots, n$. Show that
\[
\left \lvert \mathrm{Re} \, [(z_1^2 + z_2^2 + \dots + z_n^2)^{1/2} ] \right \rvert \leq \lvert \mathrm{Re} \, z_1 \rvert + \lvert \mathrm{Re} \, z_2 \rvert + \dots + \lvert \mathrm{Re} \, z_n \rvert.
\] | null | ['analysis'] | Section putnam_1979_b6.
Require Import Reals List Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1979_b6:
forall (n: nat) (l: list C), length l = n ->
let sum1 := fold_left (fun acc x => Cplus acc (Cmult x x)) l 0 in
let sum2 := fold_left (fun acc x => Re x) l 0 in
sqrt (Re sum1) <= sum2.
Proof. Admitted.
End putnam_1979_b6. | theory putnam_1979_b6 imports Complex_Main
begin
(* Note: Problem dimension boosted from Fin n *)
theorem putnam_1979_b6:
fixes n :: "nat"
and z :: "nat \<Rightarrow> complex"
shows "\<bar>Re (csqrt (\<Sum> i \<in> {1::nat..n}. (z i)^2))\<bar> \<le> (\<Sum> i \<in> {1..n}. \<bar>Re (z i)\<bar>)"
sorry
end | null |
putnam_1987_a1 | theorem putnam_1987_a1
(A B C D : Set (β Γ β))
(hA : A = {(x, y) : β Γ β | x ^ 2 - y ^ 2 = x / (x ^ 2 + y ^ 2)})
(hB : B = {(x, y) : β Γ β | 2 * x * y + y / (x ^ 2 + y ^ 2) = 3})
(hC : C = {(x, y) : β Γ β | x ^ 3 - 3 * x * y ^ 2 + 3 * y = 1})
(hD : D = {(x, y) : β Γ β | 3 * x ^ 2 * y - 3 * x - y ^ 3 = 0})
: A β© B = C β© D := sorry | Curves $A$, $B$, $C$, and $D$ are defined in the plane as follows:
\begin{align*}
A&=\left\{ (x,y):x^2-y^2=\frac{x}{x^2+y^2} \right\}, \\
B&=\left\{ (x,y):2xy+\frac{y}{x^2+y^2}=3 \right\}, \\
C&=\left\{ (x,y):x^3-3xy^2+3y=1 \right\}, \\
D&=\left\{ (x,y):3x^2y-3x-y^3=0 \right\}.
\end{align*}
Prove that $A \cap B=C \cap D$. | null | ['algebra'] | null | theory putnam_1987_a1 imports
Complex_Main
begin
theorem putnam_1987_a1:
fixes A B C D :: "(real \<times> real) set"
defines "A \<equiv> {(x, y). x ^ 2 - y ^ 2 = x / (x ^ 2 + y ^ 2)}"
defines "B \<equiv> {(x, y). 2 * x * y + y / (x ^ 2 + y ^ 2) = 3}"
defines "C \<equiv> {(x, y). x ^ 3 - 3 * x * y ^ 2 + 3 * y = 1}"
defines "D \<equiv> {(x, y). 3 * x ^ 2 * y - 3 * x - y ^ 3 = 0}"
shows "A \<inter> B = C \<inter> D"
sorry
end | null |
putnam_1987_a2 | abbrev putnam_1987_a2_solution : β := sorry
-- 1984
theorem putnam_1987_a2
(seqind : β β β)
(seqsize : β β β)
(f : β β β)
(hseqind : seqind 1 = 1 β§ β i β₯ 2, seqind i = seqind (i - 1) + (Nat.digits 10 (i - 1)).length)
(hseqsize : β i β₯ 1, β j : Fin ((Nat.digits 10 i).length), seqsize (seqind i + j) = (Nat.digits 10 i).length)
(hf : β n : β, f n = seqsize (10 ^ n))
: f 1987 = putnam_1987_a2_solution :=
sorry | The sequence of digits $123456789101112131415161718192021 \dots$ is obtained by writing the positive integers in order. If the $10^n$-th digit in this sequence occurs in the part of the sequence in which the $m$-digit numbers are placed, define $f(n)$ to be $m$. For example, $f(2)=2$ because the $100$th digit enters the sequence in the placement of the two-digit integer $55$. Find, with proof, $f(1987)$. | Show that the value of $f(1987)$ is $1984$. | ['algebra'] | null | theory putnam_1987_a2 imports
Complex_Main
begin
definition putnam_1987_a2_solution :: nat where "putnam_1987_a2_solution \<equiv> undefined"
(* 1984 *)
theorem putnam_1987_a2:
fixes seqind :: "nat \<Rightarrow> nat"
and seqsize :: "nat \<Rightarrow> nat"
and f :: "nat \<Rightarrow> nat"
and numdigits :: "nat \<Rightarrow> nat"
assumes hseqind: "seqind 1 = 1 \<and> (\<forall> i \<ge> 2. seqind i = seqind (i - 1) + numdigits (i - 1))"
and hseqsize: "\<forall> i \<ge> 1. \<forall> j \<in> {0 .. numdigits i - 1}. seqsize (seqind i + j) = numdigits i"
and hf: "\<forall> n :: nat. f n = seqsize (10 ^ n)"
and hnumdigits: "\<forall> n \<ge> 1. numdigits n = (GREATEST k :: nat. 10 ^ k \<le> n) + 1"
shows "f 1987 = putnam_1987_a2_solution"
sorry
end | null |
putnam_1987_a4 | abbrev putnam_1987_a4_solution : β := sorry
-- (5 / 3) * sqrt 30
theorem putnam_1987_a4
(P : MvPolynomial (Fin 3) β)
(hPreal : β i : Fin 3 ββ β, (coeff i P).im = 0)
(F : β β β β β)
(vars : β β β β β β (Fin 3 β β) := fun a b c β¦ fun i β¦ ite (i = 0) a (ite (i = 1) b c))
(h : β x y z u : β, eval (vars (u * x) (u * y) (u * z)) P = u ^ 2 * F (y - x) (z - x))
(hPval : eval (vars 1 0 0) P = 4 β§ eval (vars 0 1 0) P = 5 β§ eval (vars 0 0 1) P = 6)
(A B C : β)
(hPABC : eval (vars A B C) P = 0)
(habs : βB - Aβ = 10)
: (βC - Aβ = putnam_1987_a4_solution) :=
sorry | Let $P$ be a polynomial, with real coefficients, in three variables and $F$ be a function of two variables such that
\[
P(ux, uy, uz) = u^2 F(y-x,z-x) \quad \mbox{for all real $x,y,z,u$},
\]
and such that $P(1,0,0)=4$, $P(0,1,0)=5$, and $P(0,0,1)=6$. Also let $A,B,C$ be complex numbers with $P(A,B,C)=0$ and $|B-A|=10$. Find $|C-A|$. | Prove that $|C - A| = \frac{5}{3}\sqrt{30}$. | ['algebra'] | null | theory putnam_1987_a4 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1987_a4_solution::real where "putnam_1987_a4_solution \<equiv> undefined"
(* (5/3) * sqrt 30 *)
theorem putnam_1987_a4:
fixes P::"complex poly poly poly" and F::"real\<Rightarrow>real\<Rightarrow>real" and a b c::complex
assumes hPreal : "\<forall>i j k::nat. Im (coeff (coeff (coeff P k) j) i) = 0"
and h : "\<forall>x y z u::real. (poly (poly (poly P [:[: u * z :]:]) [: u * y :]) (u * x)) = u^2 * F (y-x) (z-x)"
and hPval1 : "(poly (poly (poly P [:[: 0 :] :]) [: 0 :]) 1) = 4"
and hPval2 : "(poly (poly (poly P [:[: 0 :] :]) [: 1 :]) 0) = 5"
and hPval3 : "(poly (poly (poly P [:[: 1 :] :]) [: 0 :]) 0) = 6"
and hPabc : "(poly (poly (poly P [:[: c :] :]) [: b :]) a) = 0"
and habs : "norm (b - a) = 10"
shows "norm (c - a) = putnam_1987_a4_solution"
sorry
end | null |
putnam_1987_a6 | abbrev putnam_1987_a6_solution : Set β := sorry
-- {x : β | x > 0 β§ x < 25}
theorem putnam_1987_a6
(a : β β β := fun n β¦ {i | (digits 3 n).get i = 0}.ncard)
: ({x : β | x > 0 β§ Summable (fun n β¦ x ^ (a n) / (n ^ 3))} = putnam_1987_a6_solution) :=
sorry | For each positive integer $n$, let $a(n)$ be the number of zeroes in the base $3$ representation of $n$. For which positive real numbers $x$ does the series
\[
\sum_{n=1}^\infty \frac{x^{a(n)}}{n^3}
\]
converge? | Show that for positive $x$, the series converges if and only if $x < 25$. | ['algebra', 'analysis'] | null | theory putnam_1987_a6 imports Complex_Main
begin
definition putnam_1987_a6_solution::"real set" where "putnam_1987_a6_solution \<equiv> undefined"
(* { x. x > 0 \<and> x < 25 } *)
fun digits_b3::"nat \<Rightarrow> nat list" where
"digits_b3 n = (if n < 3 then [n] else ([n mod 3::nat] @ digits_b3 (n div 3::nat)))"
theorem putnam_1987_a6:
fixes a::"nat\<Rightarrow>nat"
defines "a \<equiv> \<lambda>n. card {i::nat. i < length (digits_b3 n) \<and> (digits_b3 n)!i = 0}"
shows "putnam_1987_a6_solution = {x::real. x > 0 \<and> summable (\<lambda>n. x^(a n) / n^3)}"
sorry
end | null |
putnam_1987_b1 | abbrev putnam_1987_b1_solution : β := sorry
-- 1
theorem putnam_1987_b1
: (β« x in (2)..4, sqrt (log (9 - x)) / (sqrt (log (9 - x)) + sqrt (log (x + 3))) = putnam_1987_b1_solution) :=
sorry | Evaluate
\[
\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}.
\] | Prove that the integral evaluates to $1$. | ['analysis'] | Section putnam_1987_b1.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1987_b1_solution := 1.
Theorem putnam_1987_b1:
RInt (fun x => ln (9 - x) ^ (1/2) / ( ln (9 - x) ^ (1/2) + ln (x + 3) ^ (1/2))) 2 4 = putnam_1987_b1_solution.
Proof. Admitted.
End putnam_1987_b1. | theory putnam_1987_b1 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1987_b1_solution::real where "putnam_1987_b1_solution \<equiv> undefined"
(* 1 *)
theorem putnam_1987_b1:
shows "putnam_1987_b1_solution =
interval_lebesgue_integral lebesgue 2 4 (\<lambda>x. sqrt (ln (9-x)) / (sqrt (ln (9-x)) + sqrt (ln (x+3))))"
sorry
end | null |
putnam_1987_b2 | theorem putnam_1987_b2
(r s t : β)
(hsum : r + s β€ t)
: (β i : Finset.range (s + 1), (choose s i : β) / (choose t (r + i)) = ((t + 1) : β) / ((t + 1 - s) * choose (t - s) r)) :=
sorry | Let $r, s$ and $t$ be integers with $0 \leq r$, $0 \leq s$ and $r+s \leq t$. Prove that
\[
\frac{\binom s0}{\binom tr}
+ \frac{\binom s1}{\binom{t}{r+1}} + \cdots
+ \frac{\binom ss}{\binom{t}{r+s}}
= \frac{t+1}{(t+1-s)\binom{t-s}{r}}.
\] | null | ['algebra'] | Section putnam_1987_b2.
Require Import Binomial Reals Coquelicot.Coquelicot.
Theorem putnam_1987_b2:
forall (n r s: nat), ge n (r + s) ->
sum_n (fun i => Binomial.C s i / Binomial.C n (r + i)) s = (INR n + 1)/((INR n + 1 - INR s) * Binomial.C (n - s) r).
Proof. Admitted.
End putnam_1987_b2. | theory putnam_1987_b2 imports Complex_Main
begin
theorem putnam_1987_b2:
fixes r s t::nat
assumes hsum : "r + s \<le> t"
shows "(\<Sum>i=0..s. (s choose i) / (t choose (r+i))) = (t+1) / ((t+1-s) * ((t-s) choose r))"
sorry
end | null |
putnam_1987_b4 | abbrev putnam_1987_b4_solution : Prop Γ β Γ Prop Γ β := sorry
-- (True, -1, True, 0)
theorem putnam_1987_b4
(x y : β β β)
(hxy1 : (x 1, y 1) = (0.8, 0.6))
(hx : β n β₯ 1, x (n + 1) = (x n) * cos (y n) - (y n) * sin (y n))
(hy : β n β₯ 1, y (n + 1) = (x n) * sin (y n) + (y n) * cos (y n))
: let (existsx, limx, existsy, limy) := putnam_1987_b4_solution
(((β c : β, Tendsto x β€ (π c)) β existsx) β§ (existsx β Tendsto x β€ (π limx)) β§ ((β c : β, Tendsto y β€ (π c)) β existsy) β§ (existsy β Tendsto y β€ (π limy))) :=
sorry | Let $(x_1,y_1) = (0.8, 0.6)$ and let $x_{n+1} = x_n \cos y_n - y_n \sin y_n$ and $y_{n+1}= x_n \sin y_n + y_n \cos y_n$ for $n=1,2,3,\dots$. For each of $\lim_{n\to \infty} x_n$ and $\lim_{n \to \infty} y_n$, prove that the limit exists and find it or prove that the limit does not exist. | Show that $\lim_{n \to \infty} x_n = -1$ and $\lim_{n \to \infty} y_n = 0$. | ['analysis'] | Section putnam_1987_b4.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1987_b4_solution1 := -1.
Definition putnam_1987_b4_solution2 := PI.
Theorem putnam_1987_b4:
let A :=
fix a (i j: nat) : (R * R):=
match (i, j) with
| (O, O) => (0.8, 0.6)
| (S i', S j') =>
let xn := fst (a i' j') in
let yn := snd (a i' j') in
(xn * cos yn - yn * sin yn,xn * sin yn + yn * cos yn)
| (_, _) => (0, 0)
end in
Lim_seq (fun n => fst (A n 0%nat)) = putnam_1987_b4_solution1 /\
Lim_seq (fun n => snd (A 0%nat n)) = putnam_1987_b4_solution2.
Proof. Admitted.
End putnam_1987_b4. | theory putnam_1987_b4 imports Complex_Main
begin
definition putnam_1987_b4_solution::"(bool \<times> real \<times> bool \<times> real)" where "putnam_1987_b4_solution \<equiv> undefined"
(* (True, -1, True, 0) *)
theorem putnam_1987_b4:
fixes x y::"nat\<Rightarrow>real"
assumes hxy1 : "(x 1, y 1) = (0.8, 0.6)"
and hx : "\<forall>n \<ge> 1. x (n+1) = (x n) * cos (y n) - (y n) * sin (y n)"
and hy : "\<forall>n \<ge> 1. y (n+1) = (x n) * sin (y n) + (y n) * cos (y n)"
shows "let (existsx, limx, existsy, limy) = putnam_1987_b4_solution in
(((\<exists>c::real. (x \<longlonglongrightarrow> c)) \<longrightarrow> existsx) \<and> (existsx \<longrightarrow> (x \<longlonglongrightarrow> limx)) \<and>
((\<exists>c::real. (y \<longlonglongrightarrow> c)) \<longrightarrow> existsy) \<and> (existsy \<longrightarrow> (y \<longlonglongrightarrow> limy)))"
sorry
end | null |
putnam_1987_b5 | theorem putnam_1987_b5
(n : β)
(npos : n > 0)
(O : Matrix (Fin 1) (Fin n) β := 0)
(M : Matrix (Fin (2 * n)) (Fin n) β)
(hM : β z : Matrix (Fin 1) (Fin (2 * n)) β, z * M = O β (Β¬β i : Fin (2 * n), z 0 i = 0) β β i : Fin (2 * n), (z 0 i).im β 0)
: (β r : Matrix (Fin (2 * n)) (Fin 1) β, β w : Matrix (Fin n) (Fin 1) β, β i : (Fin (2 * n)), ((M * w) i 0).re = r i 0) :=
sorry | Let $O_n$ be the $n$-dimensional vector $(0,0,\cdots, 0)$. Let $M$ be a $2n \times n$ matrix of complex numbers such that whenever $(z_1, z_2, \dots, z_{2n})M = O_n$, with complex $z_i$, not all zero, then at least one of the $z_i$ is not real. Prove that for arbitrary real numbers $r_1, r_2, \dots, r_{2n}$, there are complex numbers $w_1, w_2, \dots, w_n$ such that
\[
\mathrm{re}\left[ M \left( \begin{array}{c} w_1 \\ \vdots \\ w_n \end{array} \right) \right] = \left( \begin{array}{c} r_1 \\ \vdots \\ r_{2n} \end{array} \right).
\]
(Note: if $C$ is a matrix of complex numbers, $\mathrm{re}(C)$ is the matrix whose entries are the real parts of the entries of $C$.) | null | ['linear_algebra'] | null | theory putnam_1987_b5 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1987_b5:
fixes n::nat and M::"complex^'a^'b"
assumes matsize : "CARD('a) = n \<and> CARD('b) = 2 * n"
and npos : "n > 0"
and hM : "\<forall>z::(complex^'b^1). (let prod = z ** M in (\<forall>i. prod$i = 0))
\<longrightarrow> (\<not>(\<forall>i. z$1$i = 0)) \<longrightarrow> (\<exists>i. Im (z$1$i) \<noteq> 0)"
shows "\<forall>r::(real^1^'b). \<exists>w::(complex^1^'a). \<forall>i. Re ((M**w)$i$1) = r$i$1"
sorry
end | null |
putnam_1987_b6 | theorem putnam_1987_b6
(p : β)
(podd : Odd p β§ Nat.Prime p)
(F : Type*) [Field F] [Fintype F]
(Fcard : Fintype.card F = p ^ 2)
(S : Set F)
(Snz : β x β S, x β 0)
(Scard : S.ncard = ((p : β€) ^ 2 - 1) / 2)
(hS : β a : F, a β 0 β Xor' (a β S) (-a β S))
(N : β := (S β© {x | β a β S, x = 2 * a}).ncard)
: (Even N) :=
sorry | Let $F$ be the field of $p^2$ elements, where $p$ is an odd prime. Suppose $S$ is a set of $(p^2-1)/2$ distinct nonzero elements of $F$ with the property that for each $a\neq 0$ in $F$, exactly one of $a$ and $-a$ is in $S$. Let $N$ be the number of elements in the intersection $S \cap \{2a: a \in S\}$. Prove that $N$ is even. | null | ['abstract_algebra'] | null | theory putnam_1987_b6 imports Complex_Main "HOL-Computational_Algebra.Primes" "HOL-Algebra.Ring"
begin
theorem putnam_1987_b6:
fixes p::nat and F (structure) and S::"'a set"
assumes podd : "odd p \<and> prime p"
and Ffield : "field F"
and Fcard : "finite (carrier F) \<and> card (carrier F) = p^2"
and Snz : "\<forall>x \<in> S. x \<noteq> \<zero>\<^bsub>F\<^esub>"
and Scard : "real_of_nat (card S) = (p^2 - 1::real) / 2"
and hS : "\<forall>a::'a. a \<in> carrier F \<longrightarrow> a \<noteq> \<zero>\<^bsub>F\<^esub> \<longrightarrow> \<not>((a \<in> S) \<longleftrightarrow> ((\<ominus>\<^bsub>F\<^esub> a) \<in> S))"
shows "even (card (S \<inter> {x. (\<exists>a \<in> S. x = a \<oplus>\<^bsub>F\<^esub> a)}))"
sorry
end | null |
putnam_2004_a1 | abbrev putnam_2004_a1_solution : Prop := sorry
-- True
theorem putnam_2004_a1
(S : (β β Fin 2) β β β β)
(hS : β attempts : β β Fin 2, β N β₯ 1, S attempts N = (β i : Fin N, (attempts i).1) / N)
: (β (attempts : β β Fin 2) (a b : β), (1 β€ a β§ a < b β§ S attempts a < 0.8 β§ S attempts b > 0.8) β (β c : β, a < c β§ c < b β§ S attempts c = 0.8)) β putnam_2004_a1_solution :=
sorry | Basketball star Shanille O'Keal's team statistician keeps track of the number, $S(N)$, of successful free throws she has made in her first $N$ attempts of the season. Early in the season, $S(N)$ was less than $80\%$ of $N$, but by the end of the season, $S(N)$ was more than $80\%$ of $N$. Was there necessarily a moment in between when $S(N)$ was exactly $80\%$ of $N$? | Show that the answer is yes. | ['probability'] | null | theory putnam_2004_a1 imports Complex_Main
begin
definition putnam_2004_a1_solution :: bool where "putnam_2004_a1_solution \<equiv> undefined"
(* True *)
theorem putnam_2004_a1:
fixes S :: "(nat \<Rightarrow> bool) \<Rightarrow> nat \<Rightarrow> real"
assumes hS: "\<forall>attempts::nat\<Rightarrow>bool. \<forall>N::nat\<ge>1. S attempts N = (\<Sum>i::nat=0..(N-1). of_bool (attempts i)) / N"
shows "(\<forall>(attempts::nat\<Rightarrow>bool)(a::nat)(b::nat). (1 \<le> a \<and> a < b \<and> S attempts a < 0.8 \<and> S attempts b > 0.8) \<longrightarrow> (\<exists>c::nat. a < c \<and> c < b \<and> S attempts c = 0.8)) \<longleftrightarrow> putnam_2004_a1_solution"
sorry
end
| null |
putnam_2004_a3 | theorem putnam_2004_a3
(u : β β β)
(hubase : u 0 = 1 β§ u 1 = 1 β§ u 2 = 1)
(hudet : β n : β, Matrix.det (fun i j : Finset.range 2 => u (n + i * 2 + j)) = (n)!)
: β n : β, u n = round (u n) :=
sorry | Define a sequence $\{u_n\}_{n=0}^\infty$ by $u_0=u_1=u_2=1$, and thereafter by the condition that $\det \begin{pmatrix}
u_n & u_{n+1} \\
u_{n+2} & u_{n+3}
\end{pmatrix} = n!$ for all $n \geq 0$. Show that $u_n$ is an integer for all $n$. (By convention, $0!=1$.) | null | ['linear_algebra'] | Section putnam_2004_a3.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2004_a3:
let fix u (n: nat) : R:=
match n with
| O => 1
| S O => 1
| S (S O) => 1
| S ((S (S n''') as n'') as n') => (INR (fact n) + u n'' * u n') / u n'''
end in
forall (n: nat), u n = IZR (floor (u n)).
Proof. Admitted.
End putnam_2004_a3. | theory putnam_2004_a3 imports Complex_Main
"HOL-Analysis.Determinants"
begin
theorem putnam_2004_a3:
fixes u::"nat\<Rightarrow>real"
assumes hubase : "u 0 = 1 \<and> u 1 = 1 \<and> u 2 = 1"
and hudet : "\<forall>n::nat. det (vector[vector[u n, u (n+1)], vector[u (n+2), u (n+3)]]) = fact n"
shows "\<forall>n::nat. u n \<in> \<int>"
sorry
end
| null |
putnam_2004_a4 | theorem putnam_2004_a4
(n : β)
(x : Fin n β β)
(avals : β β (β β Fin n β β) β Prop)
(npos : n > 0)
(havals : β (N : β) (a : (β β Fin n β β)), avals N a = β (i : Fin N) (j : Fin n), (a i j = -1 β¨ a i j = 0 β¨ a i j = 1))
: β (N : β) (c : Fin N β β) (a : β β Fin n β β), avals N a β§ ((β i : Fin n, x i) = β i : Fin N, c i * (β j : Fin n, a i j * x j) ^ n) :=
sorry | Show that for any positive integer $n$ there is an integer $N$ such that the product $x_1x_2 \cdots x_n$ can be expressed identically in the form $x_1x_2 \cdots x_n=\sum_{i=1}^Nc_i(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^n$ where the $c_i$ are rational numbers and each $a_{ij}$ is one of the numbers $-1,0,1$. | null | ['algebra'] | Section putnam_2004_a4.
Require Import List Reals Coquelicot.Coquelicot.
Theorem putnam_2004_a4:
forall (n: nat), exists (N: nat) (a: nat -> nat -> R) (c: R), (exists (p q: Z), c = IZR (p / q)) /\ (forall (i j: nat), a i j = -1 \/ a i j = 0 \/ a i j = 1) ->
forall (x: list R), length x = n ->
fold_left Rmult x 1 = sum_n (fun i => c * (sum_n (fun j => a i j * nth j x 0) n) ^ (1 / n)) N.
Proof. Admitted.
End putnam_2004_a4. | theory putnam_2004_a4 imports Complex_Main
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_2004_a4:
fixes n::nat and x::"nat\<Rightarrow>real" and avals::"nat \<Rightarrow> (nat \<Rightarrow> nat \<Rightarrow> real) \<Rightarrow> bool"
defines "avals \<equiv> \<lambda>N. \<lambda>a. \<forall>i \<in> {0..<N}. \<forall>j \<in> {0..<n}. (a i j = -1 \<or> a i j = 0 \<or> a i j = 1)"
assumes npos : "n > 0"
shows "\<exists>N::nat. \<exists>c::nat\<Rightarrow>rat. \<exists>a. avals N a \<and> ((\<Prod>i=0..<n. x i) = (\<Sum>i=0..<N. (c i * (\<Sum>j=0..<n. a i j * x j)^n)))"
sorry
end
| null |
putnam_2004_a6 | theorem putnam_2004_a6
(f : Set.Icc (0 : β) 1 β Set.Icc (0 : β) 1 β β)
(fcont : Continuous f)
: (β« y : Set.Icc (0 : β) 1, (β« x : Set.Icc (0 : β) 1, f x y) ^ 2) + (β« x : Set.Icc (0 : β) 1, (β« y : Set.Icc (0 : β) 1, f x y) ^ 2) β€ (β« y : Set.Icc (0 : β) 1, (β« x : Set.Icc (0 : β) 1, f x y)) ^ 2 + (β« y : Set.Icc (0 : β) 1, (β« x : Set.Icc (0 : β) 1, (f x y) ^ 2)) :=
sorry | Suppose that $f(x,y)$ is a continuous real-valued function on the unit square $0 \leq x \leq 1,0 \leq y \leq 1$. Show that $\int_0^1 \left(\int_0^1 f(x,y)dx\right)^2dy+\int_0^1 \left(\int_0^1 f(x,y)dy\right)^2dx \leq \left(\int_0^1 \int_0^1 f(x,y)dx\,dy\right)^2+\int_0^1 \int_0^1 [f(x,y)]^2dx\,dy$. | null | ['analysis'] | Section putnam_2004_a6.
Require Import Basics Reals Coquelicot.Coquelicot.
Theorem putnam_2004_a6:
forall (f: R -> R -> R) (x y: R), 0 <= x <= 1 /\ 0 <= y <= 1 /\ continuity_2d_pt f x y ->
RInt (compose (fun y => RInt (fun x => f x y) 0 1) (fun x => RInt (fun y => f x y) 0 1)) 0 1 +
RInt (compose (fun x => RInt (fun y => f x y) 0 1) (fun x => RInt (fun y => f x y) 0 1)) 0 1 <=
(RInt (fun x => RInt (fun y => f x y) 0 1) 0 1) ^ 2 +
RInt (fun x => RInt (fun y => (f x y) ^ 2) 0 1) 0 1.
Proof. Admitted.
End putnam_2004_a6. | theory putnam_2004_a6 imports Complex_Main
"HOL-Analysis.Interval_Integral"
begin
theorem putnam_2004_a6:
fixes f::"(real\<times>real)\<Rightarrow>real" and usquare::"(real\<times>real) set" and Fx Fy::"real\<Rightarrow>real\<Rightarrow>real"
defines "usquare \<equiv> {(x, y). x \<ge> 0 \<and> y \<ge> 0 \<and> x \<le> 1 \<and> y \<le> 1}"
assumes fcont : "continuous_on usquare f"
and fxderiv : "\<forall>y::real \<in> {0<..<1}. \<forall>x::real \<in> {0<..<1}. ((Fx y) has_derivative (\<lambda>x. f (x, y))) (nhds x)"
and fyderiv : "\<forall>x::real \<in> {0<..<1}. \<forall>y::real \<in> {0<..<1}. ((Fy x) has_derivative (\<lambda>y. f (x, y))) (nhds y)"
shows "interval_lebesgue_integral lebesgue 0 1 (\<lambda>y. (Fx y 1 - Fx y 0)^2) +
interval_lebesgue_integral lebesgue 0 1 (\<lambda>x. (Fy x 1 - Fy x 0)^2) \<le>
(set_lebesgue_integral lebesgue usquare f)^2 + set_lebesgue_integral lebesgue usquare (\<lambda>(x, y). f (x, y)^2)"
sorry
end
| null |
putnam_2004_b1 | theorem putnam_2004_b1
(n : β)
(P : Polynomial β)
(isint : β β Prop)
(r : β)
(Pdeg : P.degree = n)
(hisint : β x : β, isint x = (x = round x))
(Pcoeff : β i : Fin (n + 1), isint (P.coeff i))
(Preq0 : P.eval (r : β) = 0)
: β i : Fin n, isint (β j : Fin (i + 1), (P.coeff (n - j) * r ^ ((i.1 + 1) - j))) :=
sorry | Let $P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_0$ be a polynomial with integer coefficients. Suppose that $r$ is a rational number such that $P(r)=0$. Show that the $n$ numbers $c_nr,\,c_nr^2+c_{n-1}r,\,c_nr^3+c_{n-1}r^2+c_{n-2}r,\dots,\,c_nr^n+c_{n-1}r^{n-1}+\cdots+c_1r$ are integers. | null | ['algebra'] | Section putnam_2004_b1.
Require Import Nat Reals Coquelicot.Coquelicot.
Theorem putnam_2004_b1:
forall (c: nat -> Z) (n: nat),
let P (x: R) := sum_n (fun i => IZR (c i) * x ^ i) (n + 1) in
forall (p q: Z), P (IZR (p / q)) = 0 -> let r := IZR (p / q) in
forall (i: nat), and (le 1 i) (le i n) -> sum_n (fun j => IZR (c (sub n j)) * r ^ (i - j)) i = IZR (floor (sum_n (fun j => IZR (c (sub n j)) * r ^ (i - j)) i)).
Proof. Admitted.
End putnam_2004_b1. | theory putnam_2004_b1 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_2004_b1:
fixes n::nat and P::"real poly" and r::rat
assumes Pdeg : "degree p = n"
and Pcoeff : "\<forall>i \<in> {0..n}. (coeff p i) \<in> \<int>"
and Preq0 : "poly p r = 0"
shows "\<forall>i \<in> {0..<n}. (\<Sum>j=0..i. (coeff p (n-j) * r^(i+1-j))) \<in> \<int>"
sorry
end
| null |
putnam_2004_b2 | theorem putnam_2004_b2
(m n : β)
(mnpos : m > 0 β§ n > 0)
: ((m + n)! / ((m + n) ^ (m + n) : β)) < (((m)! / (m ^ m : β)) * ((n)! / (n ^ n : β))) :=
sorry | Let $m$ and $n$ be positive integers. Show that $\frac{(m+n)!}{(m+n)^{m+n}}<\frac{m!}{m^m}\frac{n!}{n^n}$. | null | ['algebra'] | Section putnam_2004_b2.
Require Import Factorial Reals Coquelicot.Coquelicot.
Theorem putnam_2004_b2:
forall (m n: nat), ge m 0 /\ ge n 0 ->
INR (fact (m + n)) / INR (m + n) ^ (m + n) < INR (fact m) / INR m ^ m * INR (fact n) / INR n ^ n.
Proof. Admitted.
End putnam_2004_b2. | theory putnam_2004_b2 imports Complex_Main
begin
theorem putnam_2004_b2:
fixes m n::nat
assumes mnpos : "m > 0 \<and> n > 0"
shows "fact (m+n) / (m+n)^(m+n) < (fact m / m^m) * (fact n / n^n)"
sorry
end
| null |
putnam_2004_b5 | abbrev putnam_2004_b5_solution : β := sorry
-- 2 / Real.exp 1
theorem putnam_2004_b5
(xprod : β β β)
(hxprod : β x β₯ 0, Tendsto (fun N : β => β n : Fin N, ((1 + x ^ (n.1 + 1)) / (1 + x ^ n.1)) ^ (x ^ n.1)) atTop (π (xprod x)))
: Tendsto xprod (π[<] 1) (π putnam_2004_b5_solution) :=
sorry | Evaluate $\lim_{x \to 1^-} \prod_{n=0}^\infty \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$. | Show that the desired limit is $2/e$. | ['analysis'] | Section putnam_2004_b5.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2004_b5_solution := 2 / exp 1.
Theorem putnam_2004_b5:
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
filterlim (fun x => (Lim_seq (fun nInc => prod_n (fun n => Rpower ((1 + x ^ (n + 1)) / (1 + x ^ n)) (x ^ n) ) nInc))) (at_left 1) (locally 0).
Proof. Admitted.
End putnam_2004_b5. | theory putnam_2004_b5 imports Complex_Main
begin
definition putnam_2004_b5_solution where "putnam_2004_b5_solution \<equiv> undefined"
(* 2 / exp 1 *)
theorem putnam_2004_b5:
fixes xprod::"real\<Rightarrow>real"
defines "xprod \<equiv> \<lambda>x. lim (\<lambda>N::nat. (\<Prod>n=0..N. ((1+x^(n+1)) / (1 + x^n)) powr (x^n)))"
shows "(xprod \<longlongrightarrow> putnam_2004_b5_solution) (at_left 1)"
sorry
end
| null |
putnam_2004_b6 | theorem putnam_2004_b6
(A : Set β)
(N : β β β)
(B : Set β)
(b : β β β)
(Anempty : A.Nonempty)
(Apos : β a β A, a > 0)
(hN : β x : β, N x = Set.encard {a : A | a β€ x})
(hB : B = {b' > 0 | β a β A, β a' β A, b' = a - a'})
(hbB : Set.range b = B β§ β i : β, b i < b (i + 1))
: (β r : β, β i : β, (b (i + 1) - b i) β₯ r) β Tendsto (fun x => N x / x) atTop (π 0) :=
sorry | Let $\mathcal{A}$ be a non-empty set of positive integers, and let $N(x)$ denote the number of elements of $\mathcal{A}$ not exceeding $x$. Let $\mathcal{B}$ denote the set of positive integers $b$ that can be written in the form $b=a-a'$ with $a \in \mathcal{A}$ and $a' \in \mathcal{A}$. Let $b_1<b_2<\cdots$ be the members of $\mathcal{B}$, listed in increasing order. Show that if the sequence $b_{i+1}-b_i$ is unbounded, then $\lim_{x \to\infty} N(x)/x=0$. | null | ['analysis'] | null | theory putnam_2004_b6 imports Complex_Main
begin
theorem putnam_2004_b6:
fixes A B::"nat set" and N::"real\<Rightarrow>nat" and b::"nat\<Rightarrow>nat"
defines "N \<equiv> \<lambda>x::real. card {a\<in>A. a \<le> x}"
and "B \<equiv> {b'. b' > 0 \<and> (\<exists>a \<in> A. \<exists>a' \<in> A. b' = a - a')}"
assumes Anempty : "card A > 0"
and Apos : "\<forall>a \<in> A. a > 0"
and hbB : "B = image b {1..(card B)}"
and hbasc : "\<forall>i::nat. b i < b (i+1)"
shows "(\<forall>r::nat. \<exists>i::nat. (b (i+1) - b i) \<ge> r) \<longrightarrow> ((\<lambda>x::real. N x / x) \<longlongrightarrow> 0) at_top"
sorry
end
| null |
putnam_2018_a1 | abbrev putnam_2018_a1_solution : Set (β€ Γ β€) := sorry
-- {β¨673, 1358114β©, β¨674, 340033β©, β¨1009, 2018β©, β¨2018, 1009β©, β¨340033, 674β©, β¨1358114, 673β©}
theorem putnam_2018_a1 : β a b : β€, (a > 0 β§ b > 0 β§ ((1: β) / a + (1: β) / b = (3: β) / 2018)) β (β¨a, bβ© β putnam_2018_a1_solution) :=
sorry | Find all ordered pairs $(a,b)$ of positive integers for which $\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}$. | Show that all solutions are in the set of ${(673,1358114), (674,340033), (1009,2018), (2018,1009), (340033,674), (1358114,673)}$. | ['number_theory'] | Section putnam_2018_a1.
Require Import Nat Factorial QArith.
Open Scope Q_scope.
Definition putnam_2018_a1_solution (a b: nat): Prop := (a = 673%nat /\ b = 1358114%nat) \/ (a = 674%nat /\ b = 340033%nat) \/ (a = 1009%nat /\ b = 2018%nat) \/ (a =2018%nat /\ b = 1009%nat) \/ (a = 340033%nat /\ b = 674%nat) \/ (a = 1358114%nat /\ b = 673%nat).
Theorem putnam_2018_a1:
forall (a b: nat), gt a 0 /\ gt b 0 ->
1/inject_Z (Z.of_nat a) + 1/inject_Z (Z.of_nat b) = 3/2018
<-> putnam_2018_a1_solution a b.
Proof. Admitted.
End putnam_2018_a1. | theory putnam_2018_a1 imports Complex_Main
begin
definition putnam_2018_a1_solution::"(nat \<times> nat) set" where "putnam_2018_a1_solution \<equiv> undefined"
(* {(673, 1358114), (674, 340033), (1009, 2018), (2018, 1009), (340033, 674), (1358114, 673)} *)
theorem putnam_2018_a1:
shows "\<forall>a b::nat. (a > 0 \<and> b > 0 \<and> (1 / a + 1 / b = 3 / 2018)) \<longleftrightarrow> (a, b) \<in> putnam_2018_a1_solution"
sorry
end
| null |
putnam_2018_a2 | abbrev putnam_2018_a2_solution : β β β := sorry
-- (fun n : β => if n = 1 then 1 else -1)
theorem putnam_2018_a2
(n : β)
(S : Fin (2 ^ n - 1) β Set β)
(M : Matrix (Fin (2 ^ n - 1)) (Fin (2 ^ n - 1)) β)
(npos : n β₯ 1)
(hS : Set.range S = (Set.Icc 1 n).powerset \ {β
})
(hM : β i j : Fin (2 ^ n - 1), M i j = if (S i β© S j = β
) = True then 0 else 1)
: M.det = putnam_2018_a2_solution n :=
sorry | Let \( S_1, S_2, \ldots, S_{2^n-1} \) be the nonempty subsets of \( \{1, 2, \ldots, n\} \) in some order, and let \( M \) be the \( (2^n - 1) \times (2^n - 1) \) matrix whose \((i, j)\) entry is $m_{ij} = \begin{cases} 0 & \text{if } S_i \\cap S_j = \\emptyset; \\ 1 & \text{otherwise}. \\end{cases} $ Calculate the determinant of \( M \). | Show that the solution is 1 if n = 1, and otherwise -1. | ['linear_algebra', 'number_theory'] | null | theory putnam_2018_a2 imports Complex_Main
"HOL-Analysis.Determinants"
begin
definition putnam_2018_a2_solution :: "nat \<Rightarrow> real" where "putnam_2018_a2_solution \<equiv> undefined"
(* (\<lambda>n::nat. if n = 1 then 1 else -1) *)
theorem putnam_2018_a2:
fixes n :: nat
and S :: "'n::finite \<Rightarrow> (nat set)"
and M :: "real^'n^'n"
assumes npos: "n \<ge> 1"
and pncard: "CARD('n) = 2^n - 1"
and hS: "range S = (Pow {1..n}) - {{}}"
and hM: "\<forall>i j::'n. M$i$j = (if (S i \<inter> S j = {}) then 0 else 1)"
shows "det M = putnam_2018_a2_solution n"
sorry
end
| null |
putnam_2018_a3 | abbrev putnam_2018_a3_solution : β := sorry
-- 480/49
theorem putnam_2018_a3
(P : Set (Fin 10 β β))
(f : (Fin 10 β β) β β β β := fun x => fun k => β i : Fin 10, Real.cos (k * (x i)))
(hP : β x β P, f x 1 = 0)
: β y β P, f y 3 β€ putnam_2018_a3_solution β§ β x β P, f x 3 = putnam_2018_a3_solution :=
sorry | Determine the greatest possible value of $\sum_{i=1}^{10} \cos(3x_i)$ for real numbers $x_1, x_2, \ldots, x_{10}$ satisfying $\sum_{i=1}^{10} \cos(x_i) = 0$. | Show that the solution is $\frac{480}{49}$ | ['number_theory'] | Section putnam_2018_a3.
Require Import Reals List Rtrigo_def Coquelicot.Derive.
Open Scope R.
Definition putnam_2018_a3_solution : R := 480/49.
Theorem putnam_2018_a3 :
forall (X: list R), length X = 10%nat ->
let f x := cos (INR 3 * x) in
let coeffs := map f X in
let val := fold_right Rmult 1 coeffs in
putnam_2018_a3_solution >= val
/\
exists (X: list R), length X = 10%nat ->
let f x := cos (INR 3 * x) in
let coeffs := map f X in
let val := fold_right Rmult 1 coeffs in
putnam_2018_a3_solution = val.
Proof. Admitted.
End putnam_2018_a3. | theory putnam_2018_a3 imports Complex_Main
begin
(* Note: Boosted domain to infinite set *)
definition putnam_2018_a3_solution::real where "putnam_2018_a3_solution \<equiv> undefined"
(* 480/49 *)
theorem putnam_2018_a3:
fixes f :: "(nat\<Rightarrow>real) \<Rightarrow> nat \<Rightarrow> real"
defines "f \<equiv> \<lambda>x. \<lambda>k. \<Sum>i=0..<10. cos (k * (x i))"
shows "putnam_2018_a3_solution = (GREATEST r. \<exists>x. f x 1 = 0 \<and> r = f x 3)"
sorry
end
| null |
putnam_2018_a4 | theorem putnam_2018_a4
(m n : β)
(a : β β β€)
(G : Type*) [Group G]
(g h : G)
(mnpos : m > 0 β§ n > 0)
(mngcd : Nat.gcd m n = 1)
(ha : β k : Set.Icc 1 n, a k = Int.floor (m * k / (n : β)) - Int.floor (m * ((k : β€) - 1) / (n : β)))
(ghprod : ((List.Ico 1 (n + 1)).map (fun k : β => g * h ^ (a k))).prod = 1)
: g * h = h * g :=
sorry | Let $m$ and $n$ be positive integers with $\gcd(m,n)=1$, and let $a_k=\left\lfloor \frac{mk}{n} \right\rfloor - \left\lfloor \frac{m(k-1)}{n} \right\rfloor$ for $k=1,2,\dots,n$. Suppose that $g$ and $h$ are elements in a group $G$ and that $gh^{a_1}gh^{a_2} \cdots gh^{a_n}=e$, where $e$ is the identity element. Show that $gh=hg$. (As usual, $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.) | null | ['abstract_algebra', 'number_theory'] | Section putnam_2018_a4.
Theorem putnam_2018_a4: True.
Proof. Admitted.
End putnam_2018_a4. | theory putnam_2018_a4 imports Complex_Main
begin
(* uses (nat \<Rightarrow> nat) instead of ({1..n} \<Rightarrow> nat) *)
theorem putnam_2018_a4:
fixes m n :: nat
and a :: "nat \<Rightarrow> nat"
and Gmul :: "'G \<Rightarrow> 'G \<Rightarrow> 'G" (infixl "\<^bold>*" 70)
and Gunit :: 'G ("\<^bold>1")
and Ginv :: "'G \<Rightarrow> 'G"
and g h :: 'G
and Gpow :: "'G \<Rightarrow> nat \<Rightarrow> 'G"
and ghprodrec :: "nat \<Rightarrow> 'G"
assumes mnpos: "m > 0 \<and> n > 0"
and mngcd: "gcd m n = 1"
and ha: "\<forall>k::nat\<in>{1..n}. (int (a k)) = \<lfloor>(real (m*k)) / n\<rfloor> - \<lfloor>(real (m*(k-1))) / n\<rfloor>"
and Ggroup: "group Gmul Gunit Ginv"
and hGpow: "\<forall>x::'G. Gpow x 0 = \<^bold>1 \<and> (\<forall>k::nat>0. Gpow x k = (Gpow x (k-1) \<^bold>* x))"
and hghprodrec: "ghprodrec 0 = \<^bold>1 \<and> (\<forall>k::nat\<in>{1..n}. ghprodrec k = ghprodrec (k-1) \<^bold>* g \<^bold>* Gpow h (a k))"
and ghprod: "ghprodrec n = \<^bold>1"
shows "g \<^bold>* h = h \<^bold>* g"
sorry
end
| null |
putnam_2018_a5 | theorem putnam_2018_a5
(f : β β β)
(h0 : f 0 = 0)
(h1 : f 1 = 1)
(hpos : β x : β, f x β₯ 0)
(hf : ContDiff β β€ f)
: β n > 0, β x : β, iteratedDeriv n f x < 0 :=
sorry | Let $f: \mathbb{R} \to \mathbb{R}$ be an infinitely differentiable function satisfying $f(0) = 0$, $f(1)= 1$, and $f(x) \geq 0$ for all $x \in \mathbb{R}$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x) < 0$. | null | ['analysis'] | null | theory putnam_2018_a5 imports Complex_Main
"HOL-Analysis.Derivative"
begin
theorem putnam_2018_a5:
fixes f :: "real\<Rightarrow>real"
assumes f_diff : "\<forall>n. (deriv^^n) f differentiable_on UNIV"
and f0 : "f 0 = 0"
and f1 : "f 1 = 1"
and fpos : "\<forall>x. f x \<ge> 0"
shows "\<exists>n > 0. \<exists>x::real. (deriv^^n) f x < 0"
sorry
end
| null |
putnam_2018_b1 | abbrev putnam_2018_b1_solution : Set (Vector β€ 2) := sorry
-- {v : Vector β€ 2 | β b : β€, 0 β€ b β§ b β€ 100 β§ Even b β§ v.toList = [1, b]}
theorem putnam_2018_b1
(P : Finset (Vector β€ 2))
(v : Vector β€ 2)
(vinP : Prop)
(Pvdiff : Finset (Vector β€ 2))
(Pvpart : Prop)
(hP : P = {v' : Vector β€ 2 | 0 β€ v'[0] β§ v'[0] β€ 2 β§ 0 β€ v'[1] β§ v'[1] β€ 100})
(hvinP : vinP = (v β P))
(hPvdiff : Pvdiff = P \ ({v} : Finset (Vector β€ 2)))
(hPvpart : Pvpart = (β Q R : Finset (Vector β€ 2), (Q βͺ R = Pvdiff) β§ (Q β© R = β
) β§ (Q.card = R.card) β§ (β q in Q, q[0] = β r in R, r[0]) β§ (β q in Q, q[1] = β r in R, r[1])))
: (vinP β§ Pvpart) β v β putnam_2018_b1_solution :=
sorry | Let $\mathcal{P}$ be the set of vectors defined by $\mathcal{P}=\left\{\left.\begin{pmatrix} a \\ b \end{pmatrix}\right| 0 \leq a \leq 2, 0 \leq b \leq 100,\text{ and }a,b \in \mathbb{Z}\right\}$. Find all $\mathbf{v} \in \mathcal{P}$ such that the set $\mathcal{P} \setminus \{\mathbf{v}\}$ obtained by omitting vector $\mathbf{v}$ from $\mathcal{P}$ can be partitioned into two sets of equal size and equal sum. | Show that the answer is the collection of vectors $\begin{pmatrix} 1 \\ b \end{pmatrix}$ where $0 \leq b \leq 100$ and $b$ is even. | ['algebra'] | Section putnam_2018_b1.
Require Import Logic Ensembles Finite_sets Nat List.
Open Scope nat_scope.
Definition putnam_2018_b1_solution : Ensemble (nat * nat) := fun v : nat * nat => exists (b : nat), 0 <= b <= 100 /\ even b = true /\ fst v = 1 /\ snd v = b.
Definition is_in_ensemble_fst (E : Ensemble (nat * nat)) (x : nat) : bool :=
match E (x, _) with
| True => true
end.
Definition is_in_ensemble_snd (E : Ensemble (nat * nat)) (y : nat) : bool :=
match E (_, y) with
| True => true
end.
Theorem putnam_2018_b1
(P : Ensemble (nat * nat))
(v : nat * nat)
(vinP : Prop)
(Pvdiff : Ensemble (nat * nat))
(Pvpart : Prop)
(hP : P = fun v': nat * nat => 0 <= fst v' <= 2 /\ 0 <= snd v' <= 100)
(hvinP : vinP = P v)
(hPvdiff : Pvdiff = fun v' => P v' /\ v' <> v)
(hPvpart : Pvpart =
(exists Q R : Ensemble (nat * nat),
(Union (nat * nat) Q R = Pvdiff) /\
(Intersection (nat * nat) Q R = Empty_set (nat * nat)) /\
(exists (n: nat), cardinal (nat * nat) Q n = cardinal (nat * nat) R n /\
(fold_right plus 0%nat (filter (fun x: nat => is_in_ensemble_fst Q x) (seq 0 3)) = fold_right plus 0%nat (filter (fun x: nat => is_in_ensemble_fst R x) (seq 0 3))) /\
(fold_right plus 0%nat (filter (fun y: nat => is_in_ensemble_snd Q y) (seq 0 101)) = fold_right plus 0%nat (filter (fun y: nat => is_in_ensemble_snd R y) (seq 0 101))))))
: (vinP /\ Pvpart) <-> putnam_2018_b1_solution v.
Proof. Admitted.
End putnam_2018_b1. | theory putnam_2018_b1 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Library.Disjoint_Sets"
begin
definition putnam_2018_b1_solution :: "(int^2) set" where "putnam_2018_b1_solution \<equiv> undefined"
(* {v::int^2. (\<exists>b::int. b \<in> {0..100} \<and> even b \<and> v = vector [1, b])} *)
theorem putnam_2018_b1:
fixes P :: "(int^2) set"
and v :: "int^2"
and vinP :: bool
and Pvpart :: bool
assumes hP: "P \<equiv> {v'::int^2. v'$1 \<in> {0..2} \<and> v'$2 \<in> {0..100}}"
and hvinP: "vinP \<equiv> (v \<in> P)"
and hPvpart: "Pvpart = (\<exists>Q R::(int^2) set. (partition_on (P - {v}) {Q, R}) \<and> card Q = card R \<and> (\<Sum>q\<in>Q. q) = (\<Sum>r\<in>R. r))"
shows "(vinP \<and> Pvpart) \<longleftrightarrow> v \<in> putnam_2018_b1_solution"
sorry
end
| null |
putnam_2018_b2 | theorem putnam_2018_b2
(n : β)
(hn : n > 0)
(f : β β β β β)
(hf : β z : β, f n z = β i in Finset.range n, (n - i) * z^i)
: β z : β, βzβ β€ 1 β f n z β 0 :=
sorry | Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \cdots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \leq 1\}$. | null | ['analysis'] | null | theory putnam_2018_b2 imports Complex_Main
begin
theorem putnam_2018_b2:
fixes n :: nat and f_n :: "complex \<Rightarrow> complex" and z :: complex
defines "f_n \<equiv> \<lambda>x::complex. \<Sum>i=0..(n-1). (n - i) * x^i"
assumes npos: "n > 0" and zunit: "norm(z) \<le> 1"
shows "f_n(z) \<noteq> 0"
sorry
end
| null |
putnam_2018_b3 | abbrev putnam_2018_b3_solution : Set β := sorry
-- {2^2, 2^4, 2^8, 2^16}
theorem putnam_2018_b3
(n : β)
(hn : n > 0)
: ((n < 10^100 β§ ((n : β€) β£ (2^n : β€) β§ (n - 1 : β€) β£ (2^n - 1 : β€) β§ (n - 2 : β€) β£ (2^n - 2 : β€))) β n β putnam_2018_b3_solution) :=
sorry | Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$, and $n-2$ divides $2^n - 2$. | Show that the solution is the set $\{2^2, 2^4, 2^8, 2^16\}$. | ['number_theory'] | null | theory putnam_2018_b3 imports Complex_Main
begin
definition putnam_2018_b3_solution::"nat set" where "putnam_2018_b3_solution \<equiv> undefined"
(* {2^2, 2^4, 2^8, 2^16} *)
theorem putnam_2018_b3:
shows "putnam_2018_b3_solution \<equiv> {n::nat. n > 3 \<and> n < 10^100 \<and>
n dvd 2^n \<and> (n-1) dvd (2^n - 1) \<and> (n-2) dvd (2^n - 2)}"
sorry
end
| null |
putnam_2018_b4 | theorem putnam_2018_b4
(a : β)
(x : β β β)
(hx0 : x 0 = 1)
(hx12 : x 1 = a β§ x 2 = a)
(hxn : β n β₯ 2, x (n + 1) = 2 * (x n) * (x (n - 1)) - x (n - 2))
: (β n, x n = 0) β (β c, c > 0 β§ Function.Periodic x c) :=
sorry | Given a real number $a$, we define a sequence by $x_0 = 1$, $x_1 = x_2 = a$, and $x_{n+1} = 2x_n x_{n-1} - x_{n-2}$ for $n \geq 2$. Prove that if $x_n = 0$ for some $n$, then the sequence is periodic. | null | ['algebra'] | null | theory putnam_2018_b4 imports Complex_Main
begin
theorem putnam_2018_b4:
fixes a::real and f::"nat \<Rightarrow> real"
assumes f0 : "f 0 = 1"
and f1 : "f 1 = a"
and f2 : "f 2 = a"
and fn : "\<forall>n \<ge> 2. f (n+1) = 2 * f n * f (n-1) - f (n-2)"
and fex0 : "\<exists>n. f n = 0"
shows "\<exists>T>0. \<forall>n. f (n + T) = f n"
sorry
end
| null |
putnam_2018_b6 | theorem putnam_2018_b6
(S : Finset (Fin 2018 β β€))
(hS : S = {s : Fin 2018 β β€ | (β i : Fin 2018, s i β ({1, 2, 3, 4, 5, 6, 10} : Set β€)) β§ (β i : Fin 2018, s i) = 3860})
: S.card β€ 2 ^ 3860 * ((2018 : β) / 2048) ^ 2018 :=
sorry | Let $S$ be the set of sequences of length $2018$ whose terms are in the set $\{1,2,3,4,5,6,10\}$ and sum to $3860$. Prove that the cardinality of $S$ is at most $2^{3860} \cdot \left(\frac{2018}{2048}\right)^{2018}$. | null | ['algebra'] | Section putnam_2018_b6.
Require Import Nat List Ensembles Finite_sets Reals.
Theorem putnam_2018_b6:
forall (E: Ensemble (list nat)),
forall (l: list nat), (
E l ->
length l = 2018 /\
forall (n: nat), (List.In n l) -> (n = 1 \/ n = 2 \/ n = 3 \/ n = 4 \/ n = 5 \/ n = 6 \/ n = 10) /\
fold_left Nat.add l 0 = 3860
) ->
exists (n: nat), cardinal (list nat) E n /\ Rle (INR n) (Rmult (Rpower 2 3860) (Rpower (Rdiv (INR 2018) (INR 2048)) 2018)).
Proof. Admitted.
End putnam_2018_b6. | theory putnam_2018_b6 imports Complex_Main
begin
(* uses (nat \<Rightarrow> nat) instead of (Fin 2018 \<Rightarrow> nat) *)
theorem putnam_2018_b6:
fixes S :: "(nat \<Rightarrow> nat) set"
assumes hS: "S \<equiv> {s::nat\<Rightarrow>nat. (\<forall>i::nat\<in>{0..2017}. s i \<in> {1,2,3,4,5,6,10}) \<and> (\<Sum>i::nat=0..2017. s i) = 3860}"
shows "card S \<le> 2^3860 * (2018/2048)^2018"
sorry
end
| null |
putnam_1993_a1 | abbrev putnam_1993_a1_solution : β := sorry
-- 4 / 9
theorem putnam_1993_a1
: 0 < putnam_1993_a1_solution β§ putnam_1993_a1_solution < (4 * Real.sqrt 2) / 9 β§ (β« x in Set.Ioo 0 ((Real.sqrt 2) / 3), max (putnam_1993_a1_solution - (2 * x - 3 * x ^ 3)) 0) = (β« x in Set.Ioo 0 ((Real.sqrt 6) / 3), max ((2 * x - 3 * x ^ 3) - putnam_1993_a1_solution) 0) :=
sorry | The horizontal line $y=c$ intersects the curve $y=2x-3x^3$ in the first quadrant as in the figure. Find $c$ so that the areas of the two shaded regions are equal. [Figure not included. The first region is bounded by the $y$-axis, the line $y=c$ and the curve; the other lies under the curve and above the line $y=c$ between their two points of intersection.] | Show that the area of the two regions are equal when $c=4/9$. | ['analysis', 'algebra'] | null | theory putnam_1993_a1 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1993_a1_solution::real where "putnam_1993_a1_solution \<equiv> undefined"
(* 4/9 *)
theorem putnam_1993_a1:
shows "putnam_1993_a1_solution > 0 \<and> putnam_1993_a1_solution < (4 * (sqrt 2) / 9) \<and>
interval_lebesgue_integral lebesgue 0 ((sqrt 2) / 3) (\<lambda>x. max 0 (putnam_1993_a1_solution - (2 * x - 3 * x^3)))
= interval_lebesgue_integral lebesgue 0 ((sqrt 6) / 3) (\<lambda>x. max 0 ((2 * x - 3 * x^3) - putnam_1993_a1_solution))"
sorry
end | null |
putnam_1993_a2 | theorem putnam_1993_a2
(x : β β β)
(xnonzero : β n : β, x n β 0)
(hx : β n β₯ 1, (x n) ^ 2 - x (n - 1) * x (n + 1) = 1)
: β a : β, β n β₯ 1, x (n + 1) = a * x n - x (n - 1) :=
sorry | Let $(x_n)_{n \geq 0}$ be a sequence of nonzero real numbers such that $x_n^2-x_{n-1}x_{n+1}=1$ for $n=1,2,3,\dots$. Prove there exists a real number $a$ such that $x_{n+1}=ax_n-x_{n-1}$ for all $n \geq 1$. | null | ['algebra'] | Section putnam_1993_a2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1993_a2:
forall (x: nat -> R) (n: nat), gt n 0 ->
pow (x n) 2 - x (pred n) * x (S n) = 1 ->
exists (a: R), ge n 1 -> x (S n) = a * x n - x (pred n).
Proof. Admitted.
End putnam_1993_a2. | theory putnam_1993_a2 imports Complex_Main
begin
theorem putnam_1993_a2:
fixes x::"nat\<Rightarrow>real"
assumes xn0 : "\<forall>n::nat. x n \<noteq> 0"
and hx : "\<forall>n \<ge> 1. (x n)^2 - x (n-1) * x (n+1) = 1"
shows "\<exists>a::real. \<forall>n \<ge> 1. x (n+1) = a * x n - x (n-1)"
sorry
end | null |
putnam_1993_a3 | theorem putnam_1993_a3
(c : β β β β β)
(hc : β n β₯ 1, β m β₯ 1, c n m = {f : Finset (Fin n) β Fin m | β A B : Finset (Fin n), f (A β© B) = min (f A) (f B)}.encard)
: β n β₯ 1, β m β₯ 1, c n m = β' j : Set.Icc 1 m, (j : β€) ^ n :=
sorry | Let $\mathcal{P}_n$ be the set of subsets of $\{1,2,\dots,n\}$. Let $c(n,m)$ be the number of functions $f:\mathcal{P}_n \to \{1,2,\dots,m\}$ such that $f(A \cap B)=\min\{f(A),f(B)\}$. Prove that $c(n,m)=\sum_{j=1}^m j^n$. | null | ['algebra'] | null | theory putnam_1993_a3 imports Complex_Main
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_1993_a3:
fixes pown::"nat \<Rightarrow> ((nat set) set)" and c::"nat\<Rightarrow>nat\<Rightarrow>nat"
defines "pown \<equiv> \<lambda>n. Pow {1..n}"
and "c \<equiv> \<lambda>n. \<lambda>m. card {f::(nat set) \<Rightarrow> nat. (\<forall>s \<in> pown n. f s \<in> {1..m}) \<and> (\<forall>s \<in> - pown n. f s = 0) \<and>
(\<forall>A B. (A \<in> pown n \<and> B \<in> pown n) \<longrightarrow> f (A \<inter> B) = min (f A) (f B))}"
shows "\<forall>n \<ge> 1. \<forall>m \<ge> 1. c n m = (\<Sum>j=1..m. j^n)"
sorry
end | null |
putnam_1993_a4 | theorem putnam_1993_a4
(x : Fin 19 β β€)
(y : Fin 93 β β€)
(hx : β i : Fin 19, x i > 0 β§ x i β€ 93)
(hy : β j : Fin 93, y j > 0 β§ y j β€ 19)
: β (is : Finset (Fin 19)) (js : Finset (Fin 93)), is β β
β§ (β i : is, x i) = (β j : js, y j) :=
sorry | Let $x_1,x_2,\dots,x_{19}$ be positive integers each of which is less than or equal to $93$. Let $y_1,y_2,\dots,y_{93}$ be positive integers each of which is less than or equal to $19$. Prove that there exists a (nonempty) sum of some $x_i$'s equal to a sum of some $y_j$'s. | null | ['algebra'] | Section putnam_1993_a4.
Require Import List Bool Reals Peano_dec Coquelicot.Coquelicot.
Open Scope nat_scope.
Theorem putnam_1993_a4:
forall (x y: list nat), length x = 19 /\ length y = 93 /\ forall (n: nat), In n x -> 1 < n <= 93 /\ In n y -> 1 < n <= 19 ->
exists (presentx presenty : nat -> bool),
sum_n (fun n =>
if ((existsb (fun i => if eq_nat_dec n i then true else false) x) && presentx n)
then (INR n) else R0) 94 =
sum_n (fun n =>
if ((existsb (fun i => if eq_nat_dec n i then true else false) y) && presenty n)
then (INR n) else R0) 20.
Proof. Admitted.
End putnam_1993_a4. | theory putnam_1993_a4 imports Complex_Main
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_1993_a4:
fixes x y::"nat\<Rightarrow>nat"
assumes hx : "\<forall>i \<in> {0..<19}. x i > 0 \<and> x i \<le> 93"
and hy : "\<forall>j \<in> {0..<93}. y j > 0 \<and> y j \<le> 19"
shows "\<exists>is js. is \<in> (Pow {0..<19}) \<and> js \<in> (Pow {0..<93}) \<and> is \<noteq> {} \<and> (\<Sum>i \<in> is. x i) = (\<Sum>j \<in> js. y j)"
sorry
end | null |
putnam_1993_a5 | theorem putnam_1993_a5
: Β¬Irrational ((β« x in Set.Ioo (-100) (-10), (((x ^ 2 - x) / (x ^ 3 - 3 * x + 1)) ^ 2)) + (β« x in Set.Ioo (1 / 101) (1 / 11), (((x ^ 2 - x) / (x ^ 3 - 3 * x + 1)) ^ 2)) + (β« x in Set.Ioo (101 / 100) (11 / 10), (((x ^ 2 - x) / (x ^ 3 - 3 * x + 1)) ^ 2))) :=
sorry | Show that $\int_{-100}^{-10} (\frac{x^2-x}{x^3-3x+1})^2\,dx+\int_{\frac{1}{101}}^{\frac{1}{11}} (\frac{x^2-x}{x^3-3x+1})^2\,dx+\int_{\frac{101}{100}}^{\frac{11}{10}} (\frac{x^2-x}{x^3-3x+1})^2\,dx$ is a rational number. | null | ['analysis'] | Section putnam_1993_a5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1993_a5:
let f (x: R) := (pow x 2 - x) / (pow x 3 - 3 * x + 1) in
exists (p q: Z),
RInt (fun x => pow (f x) 2) (-100) (-10) +
RInt (fun x => pow (f x) 2) (1/101) (1/11) +
RInt (fun x => pow (f x) 2) (101/100) (11/10) =
IZR p /IZR q.
Proof. Admitted.
End putnam_1993_a5. | theory putnam_1993_a5 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
theorem putnam_1993_a5:
fixes f::"real\<Rightarrow>real"
defines "f \<equiv> \<lambda>x. ((x^2 - x) / (x^3 - 3 * x + 1))^2"
shows "(interval_lebesgue_integral lebesgue (-100) (-10) f +
interval_lebesgue_integral lebesgue (1/101) (1/11) f +
interval_lebesgue_integral lebesgue (101/100) (11/10) f ) \<in> \<rat>"
sorry
end | null |
putnam_1993_a6 | theorem putnam_1993_a6
(seq : β β β€)
(hseq23 : β n : β, seq n = 2 β¨ seq n = 3)
(hseq2inds : β n : β, seq n = 2 β (β N : β, n = β i : Fin N, (seq i + 1)))
: β r : β, β n : β, seq n = 2 β (β m : β€, n + 1 = 1 + Int.floor (r * m)) :=
sorry | The infinite sequence of $2$'s and $3$'s $2,3,3,2,3,3,3,2,3,3,3,2,3,3,2,3,3,3,2,3,3,3,2,3,3,3,2,3,3,2,3,3,3,2,\dots$ has the property that, if one forms a second sequence that records the number of $3$'s between successive $2$'s, the result is identical to the given sequence. Show that there exists a real number $r$ such that, for any $n$, the $n$th term of the sequence is $2$ if and only if $n=1+\lfloor rm \rfloor$ for some nonnegative integer $m$. (Note: $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$.) | null | ['algebra'] | null | theory putnam_1993_a6 imports Complex_Main
begin
theorem putnam_1993_a6:
fixes seq::"nat\<Rightarrow>nat"
assumes hseq23 : "\<forall>n::nat. seq n = 2 \<or> seq n = 3"
and hseq2inds : "\<forall>n::nat. (seq n = 2 \<longleftrightarrow> (\<exists>N::nat. n = (\<Sum>i=0..<N. (seq i + 1))))"
shows "\<exists>r::real. \<forall>n::nat. (seq n = 2 \<longleftrightarrow> (\<exists>m::nat. n+1 = 1+\<lfloor>r * m\<rfloor>))"
sorry
end | null |