| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | A. By Fermat's theorem, we know that a^(p-1) ≡ 1 (mod p) for any integer a and prime number p. In this case, we have a = 3 and p = 23. So, 3^(23-1) ≡ 1 (mod 23) 3^22 ≡ 1 (mod 23) Now, we want to find the remainder of 3^47 when divided by 23. We can write 3^47 as: 3^47 = 3^22 × 3^22 × 3^3 Using the fact that 3^22 ≡ 1 (mod 23), we have: 3^47 ≡ 1 × 1 × 3^3 (mod 23) 3^47 ≡ 3^3 (mod 23) 3^47 ≡ 27 (mod 23) Since 27 ≡ 4 (mod 23), we have: 3^47 ≡ 4 (mod 23) Therefore, the remainder of 3^47 when divided by 23 is 4. |
D |
| deepseekr1 | To solve the problem using Fermat's Little Theorem, we first note that 23 is prime, so by the theorem, \( 3^{22} \equiv 1 \mod 23 \). The exponent \( 47 \) can be expressed as \( 47 = 22 \times 2 + 3 \), so: \[ 3^{47} = 3^{22 \times 2 + 3} = \left(3^{22}\right)^2 \times 3^3. \] Applying Fermat's theorem, \( \left(3^{22}\right)^2 \equiv 1^2 \equiv 1 \mod 23 \). Thus: \[ 3^{47} \equiv 1 \times 3^3 \mod 23. \] Calculating \( 3^3 = 27 \), we find: \[ 27 \mod 23 = 4. \] Therefore, the remainder of \( 3^{47} \) divided by 23 is **4**. The correct answer is **D**. **Answer:** D |
D |
| gpto1 | D | D |
| GPT-4o | To find the remainder of \(3^{47}\) when divided by 23 using Fermat's Little Theorem, we can proceed as follows: Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is an integer such that \( p \) does not divide \( a \), then: \[ a^{p-1} \equiv 1 \pmod{p} \] In this case, \( p = 23 \) and \( a = 3 \). According to Fermat's Little Theorem: \[ 3^{22} \equiv 1 \pmod{23} \] We need to find \( 3^{47} \mod 23 \). We can express 47 in terms of multiples of 22: \[ 47 = 2 \cdot 22 + 3 \] Thus: \[ 3^{47} = 3^{2 \cdot 22 + 3} = (3^{22})^2 \cdot 3^3 \] Using Fermat's Little Theorem: \[ (3^{22})^2 \equiv 1^2 \equiv 1 \pmod{23} \] So: \[ 3^{47} \equiv 1 \cdot 3^3 \equiv 3^3 \pmod{23} \] Now, calculate \( 3^3 \): \[ 3^3 = 27 \] Finally, find the remainder when 27 is divided by 23: \[ 27 \equiv 4 \pmod{23} \] Therefore, the remainder of \( 3^{47} \) when divided by 23 is 4. Answer: D |
D |
| llama3.1-70B-Instruct | A |
D |
| llama3.1-405B-Instruct | A. By Fermat's theorem, 3^22 = 1 (mod 23). Thus 3^47 = 3^22 * 3^22 * 3^3 = 1 * 1 * 27 = 4 (mod 23). |
D |
| DeepSeek-Coder-V2 | C | D |
| Qwen2-72B-Instruct | B. 2 | D |
| baichuan4 | Fermat's little theorem states that if p is a prime number and a is any integer not divisible by p, then a raised to the power of p - 1, when divided by p, has a remainder of 1. In this case, p = 23, so we want to calculate 3^(23-1) = 3^22. However, the question asks for 3^47. We can rewrite 47 as 22 * 2 + 3, so 3^47 = (3^22)^2 * 3^3. By Fermat's little theorem, (3^22)^2 gives a remainder of 1 when divided by 23. So we are left with 3^3 = 27, which gives a remainder of 4 when divided by 23. Answer: D. 4 |
D |
| doubao | A. 1 Fermat's theorem states that if p is a prime number and a is an integer not divisible by p, then a^(p-1) ≡ 1 (mod p). Here, 23 is a prime number and 3 is not divisible by 23. So, 3^22 ≡ 1 (mod 23). Now, 47 = 22×2 + 3. So, 3^47 = 3^(22×2 + 3) = (3^22)^2 × 3^3 ≡ 1^2 × 27 ≡ 27 ≡ 1 (mod 23). |
D |
| minimax | Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^{(p-1)} \equiv 1 \mod p \). Given \( p = 23 \) (which is prime) and \( a = 3 \) (which is not divisible by 23), we can apply Fermat's Little Theorem to find that: \( 3^{(23-1)} \equiv 3^{22} \equiv 1 \mod 23 \) Now, we want to find the remainder of \( 3^{47} \) when divided by 23. We can write \( 47 \) as \( 22 \times 2 + 3 \), which allows us to express \( 3^{47} \) as: \( 3^{47} = 3^{22 \times 2 + 3} = (3^{22})^2 \cdot 3^3 \) Using Fermat's Little Theorem, we know that \( (3^{22})^2 \equiv 1^2 \mod 23 \), which simplifies to \( 1 \). So we have: \( (3^{22})^2 \cdot 3^3 \equiv 1 \cdot 3^3 \mod 23 \) Now we just need to compute \( 3^3 \mod 23 \): \( 3^3 = 27 \) \( 27 \mod 23 = 4 \) Therefore, the remainder of \( 3^{47} \) when divided by 23 is \( 4 \). The correct answer is: D. 4 |
D |
| kimi | To solve this problem using Fermat's Little Theorem, we first need to understand the theorem itself. Fermat's Little Theorem states that if p is a prime number, then for any integer a, the number a^p - a is divisible by p. In other words, a^p ≡ a (mod p). In this case, we are asked to find the remainder of 3^47 when divided by 23. Since 23 is a prime number, we can apply Fermat's Little Theorem. According to Fermat's Little Theorem: 3^23 ≡ 3 (mod 23) Now, we need to express 3^47 in a form that we can use the theorem. We can write 47 as 23 * 2 + 1, so: 3^47 = 3^(23 * 2 + 1) = (3^23)^2 * 3^1 Using Fermat's Little Theorem, we know that (3^23) is congruent to 3 mod 23, so: (3^23)^2 ≡ 3^2 (mod 23) Now, we multiply this by 3^1: 3^2 * 3 ≡ 9 * 3 (mod 23) Calculating 9 * 3 gives us 27, and we need to find the remainder when 27 is divided by 23: 27 mod 23 = 4 So, the remainder of 3^47 when divided by 23 is 4. Answer: D. 4 |
D |
| wenxinyiyan | A Explanation: Fermat's Little Theorem states that if p is a prime number and a is an integer not divisible by p, then a^(p-1) ≡ 1 (mod p). In this case, p = 23 (which is prime), and a = 3 (which is not divisible by 23). So, by Fermat's Little Theorem, 3^22 ≡ 1 (mod 23). To find 3^47 (mod 23), we can write 3^47 as (3^22)^2 * 3^3. Then, using the result from Fermat's Little Theorem, we have: 3^47 ≡ (3^22)^2 * 3^3 ≡ 1^2 * 27 ≡ 27 ≡ 4 (mod 23). However, this is not one of the given options, so let's check our work. Ah, I made a mistake in the last step. 27 modulo 23 is actually 4, but we need to further reduce that modulo 23 since 4 is less than 23. So, the correct answer is 3^47 ≡ 4 (mod 23). But wait, the options are 1, 2, 3, and 4, and we just got 4 as the remainder. Did I make a mistake in my calculation? Let's recheck. No, the calculation is correct. The remainder of 3^47 when divided by 23 is indeed 4. Sorry for the confusion, the answer is D. 4. |
D |
| zhipu | To solve this problem using Fermat's Little Theorem, we recall that the theorem states: If \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \mod p \). Given the prime number \( p = 23 \) and \( a = 3 \), Fermat's Little Theorem tells us that: \[ 3^{22} \equiv 1 \mod 23 \] We need to find the remainder of \( 3^{47} \) when divided by 23. We can express 47 in terms of 22 (since \( 3^{22} \equiv 1 \mod 23 \)): \[ 47 = 2 \times 22 + 3 \] Therefore: \[ 3^{47} = 3^{2 \times 22 + 3} = (3^{22})^2 \times 3^3 \] Using Fermat's theorem: \[ (3^{22})^2 \equiv 1^2 \equiv 1 \mod 23 \] So: \[ 3^{47} \equiv 1 \times 3^3 \equiv 3^3 \mod 23 \] Now we compute \( 3^3 \): \[ 3^3 = 27 \] Next, we find the remainder of 27 when divided by 23: \[ 27 \div 23 = 1 \text{ remainder } 4 \] Thus: \[ 27 \equiv 4 \mod 23 \] So the remainder of \( 3^{47} \) when divided by 23 is: \[ \boxed{4} \] The correct answer is: D. 4 |
D |
| gpto1 | D | D |
| deepseekv3 | To solve the problem using Fermat's Little Theorem, follow these steps: 1. **Fermat's Little Theorem**: If \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then: \[ a^{p-1} \equiv 1 \pmod{p} \] Here, \( p = 23 \) and \( a = 3 \), so: \[ 3^{22} \equiv 1 \pmod{23} \] 2. **Simplify the exponent**: We need to find \( 3^{47} \mod 23 \). First, express 47 in terms of 22: \[ 47 = 2 \times 22 + 3 \] So: \[ 3^{47} = 3^{2 \times 22 + 3} = (3^{22})^2 \times 3^3 \] 3. **Apply Fermat's Little Theorem**: \[ (3^{22})^2 \times 3^3 \equiv 1^2 \times 3^3 \equiv 27 \pmod{23} \] 4. **Simplify 27 modulo 23**: \[ 27 \div 23 = 1 \text{ with a remainder of } 4 \] So: \[ 27 \equiv 4 \pmod{23} \] Thus, the remainder when \( 3^{47} \) is divided by 23 is **4**. **Answer:** D |
D |