[content] evaluation/easy_problems.jsonl (jsonl)
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

Find the smallest prime number \\( p \\) such that \\( p \\) divides \\( n^2 + 7n + 23 \\)."}], "output": "<think>
Okay, so I need to find the smallest prime number p that divides n² + 7n + 23 for some integer n. Hmm, let me try to figure this out step by step.

First, the problem is asking for the smallest prime p such that there exists an integer n where p divides n² + 7n + 23. That means for that prime p, the equation n² + 7n + 23 ≡ 0 mod p has a solution. So essentially, I need to check the primes starting from the smallest ones (2, 3, 5, 7, 11, etc.) and see if there's an n that makes the quadratic expression divisible by each prime. The first prime for which this is possible will be the answer.

Let me start with the smallest prime, which is 2. Let's check if there exists an n such that n² + 7n + 23 ≡ 0 mod 2. To do this, I can evaluate the expression modulo 2. Let's consider n modulo 2. If n is even (n ≡ 0 mod 2), then n² ≡ 0, 7n ≡ 0, so 0 + 0 + 23 ≡ 1 mod 2. If n is odd (n ≡ 1 mod 2), then n² ≡ 1, 7n ≡ 1, so 1 + 1 + 23 ≡ 1 + 1 + 1 ≡ 3 ≡ 1 mod 2. So in both cases, the expression is congruent to 1 mod 2. Therefore, there's no n for which 2 divides the expression. So 2 is out.

Next prime is 3. Let's check modulo 3. Let's consider n ≡ 0, 1, 2 mod 3.

If n ≡ 0 mod 3:
n² + 7n + 23 ≡ 0 + 0 + 23 ≡ 23 mod 3. 23 ÷ 3 is 7 with remainder 2, so 23 ≡ 2 mod 3.

If n ≡ 1 mod 3:
1 + 7*1 + 23 ≡ 1 + 7 + 23 ≡ 1 + 1 + 2 ≡ 4 ≡ 1 mod 3.

If n ≡ 2 mod 3:
4 + 14 + 23 ≡ 4 + 2 + 2 ≡ 8 ≡ 2 mod 3.

So none of the residues mod 3 give 0. Therefore, 3 doesn't divide the expression for any n. So 3 is also out.

Next prime is 5. Let's check modulo 5. n can be 0,1,2,3,4 mod 5.

n ≡ 0:
0 + 0 + 23 ≡ 23 mod 5 → 23 - 4*5 = 23 -20 = 3 → 3 mod 5.

n ≡ 1:
1 + 7 + 23 = 31 mod 5. 31 ÷5 is 6*5=30, remainder 1. So 1 mod 5.

n ≡ 2:
4 + 14 +23 = 41 mod5. 41-8*5=41-40=1. So 1 mod5.

n ≡3:
9 +21 +23 =53. 53-10*5=53-50=3. 3 mod5.

n ≡4:
16 +28 +23=67. 67-13*5=67-65=2. So 2 mod5.

So none of the residues mod 5 give 0. So 5 is out.

Next prime is 7. Let's check modulo7. n from 0 to6.

n=0: 0 +0 +23=23 mod7. 23-3*7=2. 2 mod7.

n=1:1+7+23=31. 31-4*7=31-28=3. 3 mod7.

n=2:4+14+23=41. 41-5*7=41-35=6. 6 mod7.

n=3:9+21+23=53. 53-7*7=53-49=4. 4 mod7.

n=4:16+28+23=67. 67-9*7=67-63=4. 4 mod7.

n=5:25+35+23=83. 83-11*7=83-77=6. 6 mod7.

n=6:36+42+23=101. 101-14*7=101-98=3. 3 mod7.

So none of the residues give 0 mod7. So 7 is out.

Next prime is 11. Wait, but maybe I should check primes in order. The next prime after 7 is 11? Wait, no, primes are 2,3,5,7,11,13,17,19,23,29, etc. So next is 11.

Check modulo11. Let's see:

n can be 0 to 10.

Calculate n² +7n +23 mod11.

First, simplify coefficients modulo11.

7 mod11 is 7.

23 mod11=23-2*11=23-22=1. So expression becomes n² +7n +1 mod11.

Now check each n from 0 to10.

n=0: 0 +0 +1=1 mod11.

n=1:1 +7 +1=9 mod11.

n=2:4 +14 +1=19 mod11=19-11=8.

n=3:9 +21 +1=31 mod11=31-2*11=9.

n=4:16 +28 +1=45 mod11. 45-4*11=45-44=1.

n=5:25 +35 +1=61 mod11. 61-5*11=61-55=6.

n=6:36 +42 +1=79 mod11. 79-7*11=79-77=2.

n=7:49 +49 +1=99 mod11=0. Wait, 49 mod11: 44 is 4*11, so 49=44+5, so 5 mod11. So n=7:

n=7: 7²=49≡5, 7*7=49≡5, so 5 (from n²) + 5 (from 7n) +1 (constant term) =5+5+1=11≡0 mod11. Oh! So when n=7 mod11, the expression is 0 mod11. Therefore, 11 divides the expression when n=7. So 11 is a candidate. But we need to check if there is a smaller prime that works. Since we checked 2,3,5,7 and none worked, the next prime is 11. So the answer should be 11? But wait, maybe I made a mistake here. Let me double-check.

Wait, when n=7, we have n² +7n +23 = 49 + 49 +23 = 121. 121 is 11², so 11 divides 121. So yes, that works. So 11 is the prime here. But since the problem says \"the smallest prime number p such that p divides n² +7n +23\", and 11 is the first prime we found that works, after checking 2,3,5,7. Therefore, 11 is the answer. But wait, just to be thorough, maybe I missed a prime between 7 and 11? No, because 7 is followed by 11, as 8,9,10 aren't primes. So the answer is 11. Hmm. But wait, let me check n=7: 7² +7*7 +23 = 49 +49 +23=121, which is 11 squared. So yes, 11 divides 121. Therefore, the smallest prime is 11. Therefore, the answer should be 11. But let me check one more prime after 11 just to make sure I didn't miss anything. For example, maybe 13 is smaller? Wait, 11 is smaller than 13, so even if 13 works, 11 is the answer. So I think 11 is the answer. But let me check primes up to 11 again. 2,3,5,7 don't work. 11 works. Therefore, the answer is 11. Hmm. Wait, but I just thought of another approach. Maybe using the quadratic formula or discriminant. Let me think.

Alternatively, the equation n² +7n +23 ≡0 mod p. For this equation to have a solution, the discriminant must be a quadratic residue modulo p. The discriminant D = 7² -4*1*23 = 49 -92 = -43. So discriminant is -43. So for the equation to have a solution, -43 must be a quadratic residue modulo p. So (-43|p) =1, where (•|p) is the Legendre symbol. So primes p where -43 is a quadratic residue. Then, we need to find the smallest prime p where (-43|p) =1. Then, the primes in order would be checked. But maybe this is more complicated, but perhaps it can help. But I think the way I checked each prime up to 11 is sufficient. Since 11 works, and smaller primes don't. Therefore, the answer is 11. Wait, but let me verify again. For p=11, does -43 ≡ a square mod 11? Let's compute -43 mod11. 43 divided by 11 is 3*11=33, remainder 10. So 43 ≡10 mod11. So -43 ≡-10 ≡1 mod11. So -43 ≡1 mod11. 1 is a quadratic residue (since 1^2=1). Therefore, the discriminant is 1 mod11, which is a square. Therefore, the equation has solutions. Therefore, 11 is the smallest prime. Thus, the answer is 11. So \\boxed{11} is the answer. I think this is correct.
</think>

The smallest prime number \\( p \\) that divides \\( n^2 + 7n + 23 \\) for some integer \\( n \\) is found by checking primes in ascending order. 

1. **Prime 2**: The expression \\( n^2 + 7n + 23 \\equiv 1 \\mod 2 \\) for all \\( n \\), so 2 is excluded.
2. **Prime 3**: Testing \\( n \\equiv 0, 1, 2 \\mod 3 \\) yields residues 2, 1, and 2 respectively, so 3 is excluded.
3. **Prime 5**: Testing \\( n \\equiv 0, 1, 2, 3, 4 \\mod 5 \\) yields residues 3, 1, 1, 3, and 2 respectively, so 5 is excluded.
4. **Prime 7**: Testing \\( n \\equiv 0, 1, 2, 3, 4, 5, 6 \\mod 7 \\) yields residues 2, 3, 6, 4, 4, 6, and 3 respectively, so 7 is excluded.
5. **Prime 11**: Simplifying the expression modulo 11 gives \\( n^2 + 7n + 1 \\mod 11 \\). Testing \\( n = 7 \\) gives \\( 7^2 + 7 \\cdot 7 + 1 = 121 \\equiv 0 \\mod 11 \\). Thus, 11 divides the expression when \\( n = 7 \\).

Therefore, the smallest prime \\( p \\) is \\(\\boxed{11}\\).", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}