✓ Lean version verified to be leanprover/lean4:v4.15.0
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Instance Name: mathd_algebra_478_test

Informal Statement+Proof (Input): /-- The volume of a cone is given by the formula $V = \frac{1}{3}Bh$, where $B$ is the area of the base and $h$ is the height. The area of the base of a cone is 30 square units, and its height is 6.5 units. What is the number of cubic units in its volume? Show that it is 65.-/

We are given that $B = 30$ and $h = 6.5$ and asked to find $\frac{1}{3}Bh$.  We find that \[\frac{1}{3}Bh = \frac{1}{3}(30)(6.5) = (10)(6.5) = 65.\]

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem mathd_algebra_478 (b h v : ℝ) (h₀ : 0 < b ∧ 0 < h ∧ 0 < v) (h₁ : v = 1 / 3 * (b * h))
    (h₂ : b = 30) (h₃ : h = 13 / 2) : v = 65 := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: numbertheory_4x3m7y3neq2003_test

Informal Statement+Proof (Input): /-- Show that there are no integers $x$ and $y$ such that $4x^3 - 7y^3 = 2003$.-/

We have that $2003 = 1 + 7 \times 286$. So $4x^3 - 7y^3 = 2003 \iff 4x^3-1=7(y^3+286)$.
We also observe that for every integer $x$, the rest of the division of $x^3$ by $7$ is $0$, $1$, or $6$. So the rest of the division of $4x^3-1$ by $7$ is $6$, $3$, or $2$.
As a result, $7$ does not divide $4x^3-1$, but since $7(y^3+286)$ is divisible by $7$, the equation cannot have integers solutions.

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem numbertheory_4x3m7y3neq2003 (x y : ℤ) : 4 * x ^ 3 - 7 * y ^ 3 ≠ 2003 := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: aime_1983_p1_test

Informal Statement+Proof (Input): /-- Let $x$, $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$, $\log_y w = 40$ and $\log_{xyz} w = 12$. Find $\log_z w$. Show that it is 060.-/

The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.

$x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. If we now convert everything to a power of $120$, it will be easy to isolate $z$ and $w$.

$x^{120}=w^5$, $y^{120}=w^3$, and $(xyz)^{120}=w^{10}$.

With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\log_zw=060$.

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem aime_1983_p1 (x y z w : ℕ) (ht : 1 < x ∧ 1 < y ∧ 1 < z) (hw : 0 ≤ w)
    (h0 : Real.log w / Real.log x = 24) (h1 : Real.log w / Real.log y = 40)
    (h2 : Real.log w / Real.log (x * y * z) = 12) : Real.log w / Real.log z = 60 := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: amc12_2001_p5_test

Informal Statement+Proof (Input): /-- What is the product of all positive odd integers less than $10000$?

$\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad
\text{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad
\text{(E)}\ \dfrac{5000!}{2^{5000}}$ Show that it is \text{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}.-/

$1 \cdot 3 \cdot 5 \cdots 9999 = \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdots 10000}{2 \cdot 4 \cdot 6 \cdots 10000}= \dfrac{10000!}{2^{5000} \cdot 1 \cdot 2 \cdot 3 \cdots 5000}= \dfrac{10000!}{2^{5000}\cdot5000!}$

Therefore the answer is $\text{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}$.

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem amc12_2001_p5 :
    Finset.prod (Finset.filter (fun x => ¬Even x) (Finset.range 10000)) (id : ℕ → ℕ) =
      10000! / (2 ^ 5000 * 5000!) := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: mathd_algebra_141_test

Informal Statement+Proof (Input): /-- A rectangular patio has an area of $180$ square feet and a perimeter of $54$ feet. What is the length of the diagonal (in feet) squared? Show that it is 369.-/

Set one side of the patio equal to $a$ and the other equal to $b$, producing two equations: \begin{align*}
ab&=180,\text{ and}\\
2a+2b&=54.
\end{align*}The second equation can be rewritten as $b=27-a$. Substituting, we have \begin{align*}
180&=a\left(27-a\right) \quad \Rightarrow \\
180&=27a-a^2 \quad \Rightarrow \\
-180&=a^2-27a \quad \Rightarrow \\
0&=a^2-27a+180 \quad \Rightarrow \\
0&=\left(a-12\right)\left(a-15\right).
\end{align*}So $12$ feet and $15$ feet are the lengths of the two sides of the patio. Therefore, the diagonal is $\sqrt{12^2+15^2}$, or $\sqrt{369}$. Therefore, the length of the diagonal squared is $369$.

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem mathd_algebra_141 (a b : ℝ) (h₁ : a * b = 180) (h₂ : 2 * (a + b) = 54) :
    a ^ 2 + b ^ 2 = 369 := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: mathd_numbertheory_3_test

Informal Statement+Proof (Input): /-- What is the units digit of the sum of the squares of the first nine positive integers? Show that it is 5.-/

We can reduce the amount of work we have to do in this problem by realizing that the units digit of the sum of the squares is the units digit of the sum of the units digits of the squares. In other words, the units digit of $1^2+2^2+\ldots+9^2$ is the units digit of $1+4+9+6+5+6+9+4+1=45$, which is $5$.

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem mathd_numbertheory_3 : (∑ x in Finset.range 10, (x + 1) ^ 2) % 10 = 5 := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: imo_1969_p2_test

Informal Statement+Proof (Input): /-- Let $a_1, a_2,\cdots, a_n$ be real constants, $x$ a real variable, and $f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).$ Given that $f(x_1)=f(x_2)=0,$ prove that $x_2-x_1=m\pi$ for some integer $m.$-/

Because the period of $\cos(x)$ is $2\pi$, the period of $f(x)$ is also $2\pi$.
$f(x_1)=f(x_2)=f(x_1+x_2-x_1)$
We can get $x_2-x_1 = 2k\pi$ for $k\in N^*$. Thus, $x_2-x_1=m\pi$ for some integer $m.$

LEAN4 Statement+Proof (Output): import Mathlib
import Aesop

set_option maxHeartbeats 0

open BigOperators Real Nat Topology Rat


theorem imo_1969_p2 (m n : ℝ) (k : ℕ) (a : ℕ → ℝ) (y : ℝ → ℝ) (h₀ : 0 < k)
    (h₁ : ∀ x, y x = ∑ i in Finset.range k, Real.cos (a i + x) / 2 ^ i) (h₂ : y m = 0)
    (h₃ : y n = 0) : ∃ t : ℤ, m - n = t * Real.pi := by

sorry

LEAN4 check outcome: True
LEAN4 check error: None
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Instance Name: mathd_algebra_209_test

Informal Statement+Proof (Input): 