TheoremLlama
TheoremLlama is a fine-tuned model for Lean4 proof writing. Using Open Bootstrapped Theorem dataset for fine-tuning, TheoremLlama can achieve considerably good performance in writing Lean4 proof based on Natural Language theorem statements and proof. This model is based on Llama-3-8B-Instruct.
Model Usage:
Usage(prompt in below):
from transformers import AutoTokenizer, AutoModelForCausalLM
import torch
model = AutoModelForCausalLM.from_pretrained("RickyDeSkywalker/TheoremLlama")
model = model.to(torch.device("cuda"))
tokenizer = AutoTokenizer.from_pretrained("RickyDeSkywalker/TheoremLlama")
terminators = [tokenizer.eos_token_id,
tokenizer.convert_tokens_to_ids("<|eot_id|>"),
tokenizer.convert_tokens_to_ids("<|reserved_special_token_26|>")]
results = model.generate(tokenized_prompt["input_ids"].to(torch.device("cuda")),
max_new_tokens=1024,
eos_token_id=terminators,
do_sample=True,
temperature=0.85,
top_p=0.9)
result_str = tokenizer.decode(results[0])
print(result_str[len(prompt):])
Prompt style
<|start_header_id|>system<|end_header_id|>You are a Lean4 expert who can write good Lean4 code based on natural language mathematical theorem and proof<|eot_id|><|start_header_id>user<|end_header_id|>
Example1
---
Example2
---
...
---
Question
\#\#\# Lean4 version of theorem and proof:\n
<|eot_id|><|start_header_id|>assistant<|end_header_id|>
example of prompt:
<|start_header_id|>system<|end_header_id|>You are a Lean4 expert who can write good Lean4 code based on natural language mathematical theorem and proof<|eot_id|><|start_header_id>user<|end_header_id|>
Natural language version of theorem and proof:
mathd_algebra_55
What fraction is the same as \[
\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?
\] Show that it is \frac{2}{3}.
We have \begin{align*}
&\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21} \\
& \qquad = \frac{2(1-2+3-4+5-6+7)}{3(1-2+3-4+5-6+7)} \\
& \qquad = \frac{2}{3}.
\end{align*}
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_55 (q p : ℝ) (h₀ : q = 2 - 4 + 6 - 8 + 10 - 12 + 14)
(h₁ : p = 3 - 6 + 9 - 12 + 15 - 18 + 21) : q / p = 2 / 3 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_55 (q p : ℝ) (h₀ : q = 2 - 4 + 6 - 8 + 10 - 12 + 14)
(h₁ : p = 3 - 6 + 9 - 12 + 15 - 18 + 21) : q / p = 2 / 3 := by
-- Proof:
-- We have q = 2(1-2+3-4+5-6+7) and p = 3(1-2+3-4+5-6+7)
-- Therefore, q / p = 2 / 3.
aesop_subst [h₀, h₁] -- substitute q and p
norm_num -- compute the result
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_132
What is the remainder of 2004 divided by 12? Show that it is 0.
An integer is divisible by 12 if and only if it is divisible by both 3 and 4. Because $2+0+0+4=6$ is divisible by 3, 2004 is divisible by 3. Also, the last two digits of 2004 form a multiple of 4, so 2004 is divisible by 4 as well. Therefore, 2004 is divisible by 12 and hence leaves a remainder of $0$ when divided by 12.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_132 : 2004 % 12 = 0 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_132 : 2004 % 12 = 0 := by
-- 2004 is divisible by 12, so the remainder is 0
apply Eq.refl
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_143
If $f(x) = x + 1$ and $g(x) = x^2 + 3$, what is the value of $f(g(2))$? Show that it is 8.
We are asked to apply the function $f$ to the number $g(2)$. First, we need to find $g(2)$. We substitute $x=2$ into the expression given for $g$ to find that $g(2)=2^2+3=7$. Then we substitute $x=7$ into the expression for $f$ to find $f(7)=7+1=8$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_143
(f g : ℝ → ℝ)
(h₀ : ∀ x, f x = x + 1)
(h₁ : ∀ x, g x = x^2 + 3) :
f (g 2) = 8 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_143
(f g : ℝ → ℝ)
(h₀ : ∀ x, f x = x + 1)
(h₁ : ∀ x, g x = x^2 + 3) :
f (g 2) = 8 := by
simp only [h₀, h₁, pow_two, one_mul]
ring
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_85
My father's age is $1222_{3}$, in base three to represent his three lower limbs -- two legs and a cane. How old is he in base ten? Show that it is 53.
$1222_{3} = 2\cdot3^{0}+2\cdot3^{1}+2\cdot3^{2}+1\cdot3^{3} = 2+6+18+27 = 53$.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_85 :
1 * 3^3 + 2 * 3^2 + 2*3 + 2 = 53 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_85 :
1 * 3^3 + 2 * 3^2 + 2*3 + 2 = 53 := by
-- Calculate the sum of digits in base 3 representation of the number 1222.
-- This represents my father's age with two legs and a cane in base 3.
-- Convert the sum to base 10.
-- The result is 53.
calc
1 * 3^3 + 2 * 3^2 + 2*3 + 2 =
(1 * 3^3 + 2 * 3^2) + 2*3 + 2 := by ring
_ = (1 * 27 + 2 * 9) + 2*3 + 2 := by ring
_ = 27 + 18 + 6 + 2 := by ring
_ = 53 := by ring
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
amc12a_2003_p1
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$ \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 $ Show that it is \mathrm{(D)}\ 2003.
The first $2003$ even counting numbers are $2,4,6,...,4006$.
The first $2003$ odd counting numbers are $1,3,5,...,4005$.
Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$.
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) $
$= 1+1+1+...+1 = \mathrm{(D)}\ 2003$
### Lean4 version of theorem statement:
```lean
theorem amc12a_2003_p1 (u v : ℕ → ℕ) (h₀ : ∀ n, u n = 2 * n + 2) (h₁ : ∀ n, v n = 2 * n + 1) :
((∑ k in Finset.range 2003, u k) - ∑ k in Finset.range 2003, v k) = 2003 :=
```
### Lean4 version of theorem and proof:
```lean
theorem amc12a_2003_p1 (u v : ℕ → ℕ) (h₀ : ∀ n, u n = 2 * n + 2) (h₁ : ∀ n, v n = 2 * n + 1) :
((∑ k in Finset.range 2003, u k) - ∑ k in Finset.range 2003, v k) = 2003 := by
-- u is the function that generates the sequence of even numbers starting from 2.
-- v is the function that generates the sequence of odd numbers starting from 1.
-- h₀ defines u(n) = 2n + 2, which represents the nth even number.
-- h₁ defines v(n) = 2n + 1, which represents the nth odd number.
-- This theorem aims to prove that the difference between the sum of the first 2003 even numbers and the sum of the first 2003 odd numbers is 2003.
-- aesop?
simp_all only [ge_iff_le] -- Simplify both sides of the equation.
rfl -- Both sides of the equation are equal by reflexivity.
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_119
Solve for $e$, given that $2d$ is $8$ less than $17e$, and $2e$ is $9$ less than $d$. Show that it is 2.
We begin with a system of two equations \begin{align*}
2d&=17e-8
\\2e&=d-9
\end{align*}Since the second equation can also be rewritten as $d=2e+9$, we can plug this expression for $d$ back into the first equation and solve for $e$ \begin{align*}
2d&=17e-8
\\\Rightarrow \qquad 2(2e+9)&=17e-8
\\\Rightarrow \qquad 4e+18&=17e-8
\\\Rightarrow \qquad -13e&=-26
\\\Rightarrow \qquad e&=2.
\end{align*}
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_119 (d e : ℝ) (h₀ : 2 * d = 17 * e - 8) (h₁ : 2 * e = d - 9) : e = 2 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_119 (d e : ℝ) (h₀ : 2 * d = 17 * e - 8) (h₁ : 2 * e = d - 9) : e = 2 := by
-- Proof:
-- We are given two equations:
-- h₀ : 2 * d = 17 * e - 8
-- h₁ : 2 * e = d - 9, which is equivalent to d = 2 * e + 9
-- We can substitute the expression for d from h₁ into h₀ and solve for e.
linarith
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_342
The sum of the first 5 terms of an arithmetic series is $70$. The sum of the first 10 terms of this arithmetic series is $210$. What is the first term of the series? Show that it is \frac{42}{5}.
Let the first term be $a$ and the common difference be $d$. The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms. The fifth term is $a + 4d$, so the sum of the first five terms is \[\frac{a + (a + 4d)}{2} \cdot 5 = 5a + 10d = 70,\]which implies that $a + 2d = 14$, so $2d = 14 - a$.
The tenth term is $a + 9d$, so the sum of the first ten terms is \[\frac{a + (a + 9d)}{2} \cdot 10 = 10a + 45d = 210,\]which implies that $2a + 9d = 42$, so $9d = 42 - 2a$.
From the equation $2d = 14 - a$, $18d = 126 - 9a$, and from the equation $9d = 42 - 2a$, $18d = 84 - 4a$, so \[126 - 9a = 84 - 4a.\]Then $5a = 42$, so $a = \frac{42}{5}$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_342
(a d: ℝ)
(h₀ : ∑ k in (Finset.range 5), (a + k * d) = 70)
(h₁ : ∑ k in (Finset.range 10), (a + k * d) = 210) :
a = 42/5 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_342
(a d: ℝ)
(h₀ : ∑ k in (Finset.range 5), (a + k * d) = 70)
(h₁ : ∑ k in (Finset.range 10), (a + k * d) = 210) :
a = 42/5 := by
simp_all [add_comm, ← sub_eq_iff_eq_add', add_assoc, Finset.sum_range_succ, add_comm,
Finset.sum_range_succ']; linarith
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_107
Find the radius of the circle with equation $x^2 + 8x + y^2 - 6y = 0$. Show that it is 5.
Completing the square gives us $(x +4)^2 + (y -3)^2 -25 = 0$. Rearranging terms, we have $(x +4)^2 + (y -3)^2 = 25$. It follows that the square of the radius is 25, so the radius must be $5$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_107
(x y : ℝ)
(h₀ : x^2 + 8 * x + y^2 - 6 * y = 0) :
(x + 4)^2 + (y-3)^2 = 5^2 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_107
(x y : ℝ)
(h₀ : x^2 + 8 * x + y^2 - 6 * y = 0) :
(x + 4)^2 + (y-3)^2 = 5^2 := by
-- Completing the square gives us (x +4)^2 + (y -3)^2 -25 = 0.
-- Rearranging terms, we have (x +4)^2 + (y -3)^2 = 25.
-- It follows that the square of the radius is 25, so the radius must be 5.
simp_all only [sub_self, add_comm, mul_right_comm, ← add_assoc, ← add_assoc, ← add_assoc]
linarith
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_251
Three plus the reciprocal of a number equals 7 divided by that number. What is the number? Show that it is 2.
Let $x$ be the number. Converting the words in the problem into an equation gives us $3+\dfrac{1}{x} = \dfrac{7}{x}$. Subtracting $\dfrac{1}{x}$ from both sides gives $3 = \dfrac{6}{x}$. Multiplying both sides of this equation by $x$ gives $3x =6$, and dividing both sides of this equation by 3 gives $x = 2$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_251 (x : ℝ) (h₀ : x ≠ 0) (h₁ : 3 + 1 / x = 7 / x) : x = 2 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_251 (x : ℝ) (h₀ : x ≠ 0) (h₁ : 3 + 1 / x = 7 / x) : x = 2 := by
-- Proof of the statement: Three plus the reciprocal of a number equals 7 divided by that number. What is the number? Show that it is 2.
-- Let $x$ be the number. Converting the words in the problem into an equation gives us $3+\dfrac{1}{x} = \dfrac{7}{x}$.
-- Subtracting $\dfrac{1}{x}$ from both sides gives $3 = \dfrac{6}{x}$.
-- Multiplying both sides of this equation by $x$ gives $3x =6$,
-- and dividing both sides of this equation by 3 gives $x = 2$.
-- aesop?
simp_all only [ne_eq, one_div]
field_simp [h₀] at h₁
linarith
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
amc12b_2020_p2
What is the value of the following expression?
$\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}$
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80} $ Show that it is \textbf{(A) } 1.
Using difference of squares to factor the left term, we get
$\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.$
Cancelling all the terms, we get $\textbf{(A) } 1$ as the answer.
### Lean4 version of theorem statement:
```lean
theorem amc12b_2020_p2 :
((100 ^ 2 - 7 ^ 2):ℝ) / (70 ^ 2 - 11 ^ 2) * ((70 - 11) * (70 + 11) / ((100 - 7) * (100 + 7))) = 1 :=
```
### Lean4 version of theorem and proof:
```lean
theorem amc12b_2020_p2 :
((100 ^ 2 - 7 ^ 2):ℝ) / (70 ^ 2 - 11 ^ 2) * ((70 - 11) * (70 + 11) / ((100 - 7) * (100 + 7))) = 1 := by
ring_nf
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_15
If $a * b = a^b + b^a$, for all positive integer values of $a$ and $b$, then what is the value of $2 * 6$? Show that it is 100.
We can see that $2 * 6 = 2^6 + 6^2 = 64 + 36 = 100$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_15 (s : ℕ → ℕ → ℕ)
(h₀ : ∀ a b, 0 < a ∧ 0 < b → s a b = a ^ (b : ℕ) + b ^ (a : ℕ)) : s 2 6 = 100 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_15 (s : ℕ → ℕ → ℕ)
(h₀ : ∀ a b, 0 < a ∧ 0 < b → s a b = a ^ (b : ℕ) + b ^ (a : ℕ)) : s 2 6 = 100 := by
-- Given (a * b) is defined as a ^ b + b ^ a
-- calculate the value of 2 * 6
-- aesop?
simp_all only [and_imp, zero_lt_two, zero_lt_succ]
apply Eq.refl -- 2 * 6 = 2 ^ 6 + 6 ^ 2 = 64 + 36 = 100
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_400
Five plus $500\%$ of $10$ is the same as $110\%$ of what number? Show that it is 50.
We have $5+\frac{500}{100}\cdot10=5+5\cdot10=55$ equal to $110\%$ of the number $x$. $$\frac{110}{100}x=\frac{11}{10}x=55\qquad\Rightarrow x=55\cdot\frac{10}{11}=5\cdot10=50$$ The number is $50$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_400
(x : ℝ)
(h₀ : 5 + 500 / 100 * 10 = 110 / 100 * x) :
x = 50 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_400
(x : ℝ)
(h₀ : 5 + 500 / 100 * 10 = 110 / 100 * x) :
x = 50 := by
-- Show that Five plus $500\%$ of $10$ is the same as $110\%$ of $50$.
linarith
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_229
What is the remainder when $5^{30}$ is divided by 7? Show that it is 1.
Let us look at the powers of $5$: \begin{align*}
5^1 &\equiv 5 \pmod{7} \\
5^2 &\equiv 4 \pmod{7} \\
5^3 &\equiv 6 \pmod{7} \\
5^4 &\equiv 2 \pmod{7} \\
5^5 &\equiv 3 \pmod{7} \\
5^6 &\equiv 1 \pmod{7}.
\end{align*} Since $5^6 \equiv 1 \pmod{7},$ we see that $5^{30} \equiv (5^6)^5 \equiv 1 \pmod{7},$ hence our desired remainder is $1.$
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_229 :
(5^30) % 7 = 1 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_229 :
(5^30) % 7 = 1 := by
norm_num
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_155
How many positive three-digit integers are congruent to $7 \pmod{19}?$ Show that it is 48.
An integer congruent to $7 \pmod{19}$ can be written as $19n+7$. Therefore, we have the inequality $$100 \le 19n+7 \le 999.$$We solve for the inequality by subtracting each term by $7$ and then dividing by $19$ to get $$93\le 19n \le 992 \implies \frac{93}{19} \le n \le \frac{992}{19}.$$The smallest integer greater than $\frac{93}{19}$ is $5$ and the largest integer less than $\frac{992}{19}$ is $52$. There are $52-4=48$ integers from $5$ to $52$ inclusive.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_155 :
Finset.card (Finset.filter (fun x => x % 19 = 7) (Finset.Icc 100 999)) = 48 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_155 :
Finset.card (Finset.filter (fun x => x % 19 = 7) (Finset.Icc 100 999)) = 48 := by
-- An integer congruent to 7 modulo 19 can be written as 19n+7.
-- We need to find n such that 100 <= 19n + 7 <= 999.
-- This is equivalent to 93 / 19 <= n <= 992 / 19
-- The smallest integer greater than 93 / 19 is 5 and the largest integer less than 992 / 19 is 52.
-- There are 52 - 5 + 1 = 48 integers from 5 to 52 inclusive.
apply Eq.refl
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
amc12a_2013_p4
What is the value of $\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?$
$ \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} $ Show that it is \textbf{(C)} \frac{5}{3}.
$\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}$
We can factor a ${2^{2012}}$ out of the numerator and denominator to obtain
$\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}$
The ${2^{2012}}$ cancels, so we get
$\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}$, which is $C$
### Lean4 version of theorem statement:
```lean
theorem amc12a_2013_p4 :
(2^2014 + 2^2012) / (2^2014 - 2^2012) = (5:ℝ) / 3 :=
```
### Lean4 version of theorem and proof:
```lean
theorem amc12a_2013_p4 :
(2^2014 + 2^2012) / (2^2014 - 2^2012) = (5:ℝ) / 3 := by
-- Factor out 2^2012 from the numerator and denominator
field_simp
-- Cancel the common factor
norm_num
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_algebra_35
If $p (x) = 2-x^2$ and $q(x) = \frac{6}{x}$, what is the value of $p (q(2))$? Show that it is -7.
Since $q(2) = \frac62 = 3$, we have $p(q(2)) = p(3) = 2-3^2 = -7$.
### Lean4 version of theorem statement:
```lean
theorem mathd_algebra_35 (p q : ℝ → ℝ) (h₀ : ∀ x, p x = 2 - x ^ 2)
(h₁ : ∀ x : ℝ, x ≠ 0 → q x = 6 / x) : p (q 2) = -7 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_algebra_35 (p q : ℝ → ℝ) (h₀ : ∀ x, p x = 2 - x ^ 2)
(h₁ : ∀ x : ℝ, x ≠ 0 → q x = 6 / x) : p (q 2) = -7 := by
-- We need to show that p (q 2) = -7
-- Since q(x) = 6/x, we know that q(2) = 3
-- Thus, we need to show that p(3) = -7
-- Since p(x) = 2 - x^2, we know that p(3) = -7
-- simp_all can be used to simplify the expression
simp_all only [rpow_two, ne_eq, OfNat.ofNat_ne_zero, not_false_eq_true, div_pow]
-- norm_num can be used to compute the final result
norm_num
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_269
When the expression $2005^2 + 2005^0 + 2005^0 + 2005^5$ is evaluated, what are the final two digits? Show that it is 52.
First, $2005^2 = 4020025$, so the last two digits of $2005^2$ are 25.
We need to look at $2005^5$, but since we only need the final two digits, we don't actually have to calculate this number entirely.
Consider $2005^3 = 2005^2 \times 2005 = 4020025 \times 2005$. When we carry out this multiplication, the last two digits of the product will only depend on the last two digits of the each of the two numbers being multiplied (try this by hand!), so the last two digits of $2005^3$ are the same as the last two digits of $25 \times 5 = 125$, ie. are 25.
Similarly, to calculate $2005^4$, we multiply $2005^3$ (which ends in 25) by $2005$, so by the same reasoning $2005^4$ ends in 25. Similarly, $2005^5$ ends in 25.
Therefore, $2005^2$ and $2005^5$ both end in 25.
Also, $2005^0 = 1$, so the expression overall is equal to $$\ldots 25 + 1 + 1 + \ldots 25 = \ldots 52.$$Therefore, the final two digits are $52$.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_269 : (2005 ^ 2 + 2005 ^ 0 + 2005 ^ 0 + 2005 ^ 5) % 100 = 52 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_269 : (2005 ^ 2 + 2005 ^ 0 + 2005 ^ 0 + 2005 ^ 5) % 100 = 52 := by
-- We only need the last two digits, so we can reduce the powers of 2005 modulo 100
-- aesop?
simp_all only [_root_.pow_zero] -- simplify the powers of 2005
apply Eq.refl -- the result is then by definition
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_728
Compute $29^{13} - 5^{13}$ modulo 7. Show that it is 3.
Firstly, note that $29 \equiv 1$ modulo 7, so $29^{13} \equiv 1$ modulo 7. Also, $5 \equiv (-2)$, so $1 - 5^{13} \equiv 1 + 2^{13}$ modulo 7. Finally, $2^3 \equiv 1$ modulo 7, so $2^{13} \equiv 2(2^3)^4 \equiv 2 \cdot 1 \equiv 2$. Thus $29^{13} - 5^{13} \equiv 1+2 \equiv 3$ modulo 7.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_728 :
(29^13 - 5^13) % 7 = 3 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_728 :
(29^13 - 5^13) % 7 = 3 := by
apply Eq.refl
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_321
Notice that \[35\cdot40=1400.\]Find some integer $n$ with $0\leq n<1399$ such that $n$ is the multiplicative inverse to 160 modulo 1399. Show that it is 1058.
Taking the given equation modulo 1399 gives \[35\cdot40\equiv1\pmod{1399},\]so we know that 35 is the multiplicative inverse to 40. We want to use this to find the multiplicative inverse to $4\cdot40=160$, so we want to try to "divide" 35 by 4.
The difficulty in dividing by 4 is that 35 is odd. We do know, though, that \[35\equiv35+1399\equiv1434\pmod{1399}\]and this number is even! Let's go even further, though, to find a multiple of 4: \[35\equiv35+3\cdot1399\equiv4232\pmod{1399}.\]Factoring 4 we get \[35\equiv4\cdot1058\pmod{1399}.\]Finally we multiply by 40: \[1\equiv 40\cdot35\equiv40\cdot4\cdot1058\equiv160\cdot1058\pmod{1399}.\]This argument is inelegant. Let's write it in a more clear order: \begin{align*}
1058\cdot160&\equiv1058\cdot(4\cdot40)\\
&\equiv(1058\cdot4)\cdot40\\
&\equiv35\cdot40\\
&\equiv1\pmod{1399}.
\end{align*}The multiplicative inverse to 160 modulo 1399 is $1058$.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_321
(n : ZMod 1399)
(h₁ : n = 160⁻¹) :
n = 1058 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_321
(n : ZMod 1399)
(h₁ : n = 160⁻¹) :
n = 1058 := by
simpa using h₁
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_66
Determine the remainder of 194 (mod 11). Show that it is 7.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_66 :
194 % 11 = 7 :=
```
### Lean4 version of theorem and proof:
```lean
theorem mathd_numbertheory_66 :
194 % 11 = 7 := by
ring
```<|reserved_special_token_26|>
---
Natural language version of theorem and proof:
mathd_numbertheory_233
Find $24^{-1} \pmod{11^2}$. That is, find the residue $b$ for which $24b \equiv 1\pmod{11^2}$.
Express your answer as an integer from $0$ to $11^2-1$, inclusive. Show that it is 116.
Since $5 \times 24 = 120 = 121 - 1$, it follows that $-5 \times 24 \equiv 1 \pmod{121}$. Adding 121 to $-5$ to make it positive, we find $(-5 + 121) \times 24 \equiv 116 \times 24 \equiv 1 \pmod{121}$, so it follows that the modular inverse of $24$ is $116$ when taken modulo $121$.
### Lean4 version of theorem statement:
```lean
theorem mathd_numbertheory_233
(b : ZMod (11^2))
(h₀ : b = 24⁻¹) :
b = 116 :=
```
### Lean4 version of theorem and proof:
<|eot_id|><|start_header_id|>assistant<|end_header_id|>
Bibtex citation
@misc{wang2024theoremllamatransforminggeneralpurposellms,
title={TheoremLlama: Transforming General-Purpose LLMs into Lean4 Experts},
author={Ruida Wang and Jipeng Zhang and Yizhen Jia and Rui Pan and Shizhe Diao and Renjie Pi and Tong Zhang},
year={2024},
eprint={2407.03203},
archivePrefix={arXiv},
primaryClass={cs.FL},
url={https://arxiv.org/abs/2407.03203},
}
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