| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the remainder when $2^{87} + 3$ is divided by $7$, we can use the concept of modular arithmetic and look for a pattern in the powers of $2$ when divided by $7$. Let's calculate the remainders of the first few powers of $2$ when divided by $7$: - $2^1 \equiv 2 \mod 7$ - $2^2 \equiv 4 \mod 7$ - $2^3 \equiv 8 \equiv 1 \mod 7$ - $2^4 \equiv 16 \equiv 2 \mod 7$ - $2^5 \equiv 32 \equiv 4 \mod 7$ - $2^6 \equiv 64 \equiv 1 \mod 7$ We can see that the remainders cycle every $3$ powers: $2, 4, 1, 2, 4, 1, \ldots$ Since $87$ is a multiple of $3$ ($87 = 29 \times 3$), we know that $2^{87}$ will have the same remainder as $2^3$ when divided by $7$, which is $1$. Therefore, $2^{87} + 3 \equiv 1 + 3 \equiv 4 \mod 7$. The final answer is $\boxed{4}$. ANSWER: D |
D |
| deepseekr1 | To determine the remainder when \(2^{87} + 3\) is divided by 7, we first analyze the pattern of \(2^k \mod 7\). Observing the powers of 2 modulo 7: \[ \begin{align*} 2^1 &\equiv 2 \mod 7, \\ 2^2 &\equiv 4 \mod 7, \\ 2^3 &\equiv 8 \equiv 1 \mod 7. \end{align*} \] The cycle repeats every 3 exponents: \(2, 4, 1\). For \(2^{87}\), since \(87 = 3 \times 29\), it corresponds to the third term in the cycle. Thus, \(2^{87} \equiv 1 \mod 7\). Adding 3 to this result gives \(1 + 3 = 4\). Therefore, the remainder when \(2^{87} + 3\) is divided by 7 is **4**. **Answer:** D |
D |
| gpto1 | D | D |
| GPT-4o | To solve the problem, we need to find the remainder of \(2^{87} + 3\) when divided by 7. We can use modular arithmetic and properties of exponents. First, we analyze \(2^{87} \mod 7\). Notice that powers of 2 modulo 7 repeat in a cycle. Let's find the cycle: \[ 2^1 \equiv 2 \mod 7 \] \[ 2^2 \equiv 4 \mod 7 \] \[ 2^3 \equiv 8 \equiv 1 \mod 7 \] \[ 2^4 \equiv 2 \cdot 2^3 \equiv 2 \cdot 1 \equiv 2 \mod 7 \] \[ 2^5 \equiv 2 \cdot 2^4 \equiv 2 \cdot 2 \equiv 4 \mod 7 \] \[ 2^6 \equiv 2 \cdot 2^5 \equiv 2 \cdot 4 \equiv 8 \equiv 1 \mod 7 \] We observe that \(2^3 \equiv 1 \mod 7\). This means that every third power of 2 returns to 1 modulo 7. Since \(2^3 \equiv 1 \mod 7\), we can express \(2^{87}\) as \((2^3)^{29} \times 2^0 \mod 7\): \[ 2^{87} \equiv (2^3)^{29} \equiv 1^{29} \equiv 1 \mod 7 \] Now we need to find the remainder of \(2^{87} + 3 \mod 7\): \[ 2^{87} + 3 \equiv 1 + 3 \equiv 4 \mod 7 \] Thus, the remainder when \(2^{87} + 3\) is divided by 7 is 4. The answer is: D. 4 |
D |
| llama3.1-70B-Instruct | ## Step 1: Identify the problem and the key elements We need to find the remainder when $2^{87} + 3$ is divided by $7$. ## Step 2: Look for a pattern in the powers of 2 when divided by 7 Let's calculate the remainders of some powers of 2 when divided by 7 to see if there's a pattern. ## Step 3: Calculate the remainders of powers of 2 $2^1 \equiv 2 \mod 7$ $2^2 \equiv 4 \mod 7$ $2^3 \equiv 8 \equiv 1 \mod 7$ $2^4 \equiv 16 \equiv 2 \mod 7$ $2^5 \equiv 32 \equiv 4 \mod 7$ $2^6 \equiv 64 \equiv 1 \mod 7$ ## Step 4: Identify the pattern We see that the remainders of powers of 2 repeat every 3 powers: 2, 4, 1, 2, 4, 1, ... ## Step 5: Determine the remainder for $2^{87}$ Since $87 \equiv 0 \mod 3$, the remainder for $2^{87}$ is the same as the remainder for $2^0$, which is $1$. ## Step 6: Calculate the remainder for $2^{87} + 3$ Adding 3 to the remainder, we get $1 + 3 = 4$. The final answer is: $\boxed{4}$ |
D |
| llama3.1-405B-Instruct | To find the remainder when $2^{87} + 3$ is divided by $7$, we can use the concept of modular arithmetic and look for a pattern in the powers of $2$ when divided by $7$. Let's calculate the remainders of the first few powers of $2$ when divided by $7$: - $2^1 \equiv 2 \mod 7$ - $2^2 \equiv 4 \mod 7$ - $2^3 \equiv 8 \equiv 1 \mod 7$ - $2^4 \equiv 16 \equiv 2 \mod 7$ - $2^5 \equiv 32 \equiv 4 \mod 7$ - $2^6 \equiv 64 \equiv 1 \mod 7$ We can see that the remainders cycle every $3$ powers: $2, 4, 1, 2, 4, 1, \ldots$ Since $87$ is a multiple of $3$ ($87 = 29 \times 3$), we know that $2^{87}$ will have the same remainder as $2^3$ when divided by $7$, which is $1$. Therefore, $2^{87} + 3 \equiv 1 + 3 \equiv 4 \mod 7$. The final answer is $\boxed{4}$. The best answer is D. |
D |
| DeepSeek-Coder-V2 | To solve the problem of finding the remainder when \(2^{87} + 3\) is divided by 7, we can use modular arithmetic. First, let's find the pattern in the powers of 2 modulo 7: \[ 2^1 \equiv 2 \pmod{7} \] \[ 2^2 \equiv 4 \pmod{7} \] \[ 2^3 \equiv 8 \equiv 1 \pmod{7} \] We observe that \(2^3 \equiv 1 \pmod{7}\). This means that every third power of 2 will be congruent to 1 modulo 7. Therefore, we can express \(2^{87}\) in terms of this pattern: \[ 2^{87} = 2^{3 \cdot 29} = (2^3)^{29} \equiv 1^{29} \equiv 1 \pmod{7} \] Now, we need to find the remainder when \(2^{87} + 3\) is divided by 7: \[ 2^{87} + 3 \equiv 1 + 3 \equiv 4 \pmod{7} \] Thus, the remainder when \(2^{87} + 3\) is divided by 7 is 4. Therefore, the correct answer is: \[ \boxed{D} \] |
D |
| Qwen2-72B-Instruct | D. 4 | D |
| baichuan4 | B | D |
| doubao | We can use the concept of modular arithmetic. $2^3=8\equiv1\pmod{7}$. So $2^{87}=(2^3)^{29}\equiv1^{29}\pmod{7}\equiv1\pmod{7}$. Then $2^{87}+3\equiv1+3\pmod{7}\equiv4\pmod{7}$. So the answer is D. |
D |
| minimax | To find the remainder when $2^{87} + 3$ is divided by $7$, we can use modular arithmetic. First, let's find the remainder when $2^{87}$ is divided by $7$. We can look for a pattern in the powers of $2$ modulo $7$: $2^1 \equiv 2 \pmod{7}$ $2^2 \equiv 4 \pmod{7}$ $2^3 \equiv 8 \equiv 1 \pmod{7}$ (since $8 - 7 = 1$) Notice that $2^3 \equiv 1 \pmod{7}$. This means that every power of $2$ that is a multiple of $3$ will also be congruent to $1$ modulo $7$. Since $87$ is a multiple of $3$, we can write: $2^{87} \equiv (2^3)^{29} \equiv 1^{29} \equiv 1 \pmod{7}$ Now, we can add $3$ to both sides of the congruence: $2^{87} + 3 \equiv 1 + 3 \equiv 4 \pmod{7}$ Therefore, the remainder when $2^{87} + 3$ is divided by $7$ is $4$. The answer is D. 4. |
D |
| kimi | To solve this problem, we can use the property of modular arithmetic, specifically Fermat's Little Theorem, which states that if p is a prime number, then for any integer a, the number a^p - a is an integer multiple of p. In other words, a^p ≡ a (mod p). Since 7 is a prime number, we can apply Fermat's Little Theorem to 2^87. According to the theorem, 2^7 ≡ 1 (mod 7). However, we need to find 2^87 (mod 7), not 2^7. We can break down 87 into a multiple of 7 plus a remainder: 87 = 7 * 12 + 3. So, we can write 2^87 as (2^7)^12 * 2^3. Using Fermat's Little Theorem, we know that (2^7)^12 ≡ 1^12 ≡ 1 (mod 7). Now we just need to find 2^3 (mod 7). 2^3 = 8, and when we divide 8 by 7, we get a quotient of 1 and a remainder of 1. So, 2^3 ≡ 1 (mod 7). Now we combine the results: (2^7)^12 * 2^3 ≡ 1 * 1 ≡ 1 (mod 7). So, 2^87 ≡ 1 (mod 7). Now we add 3 to both sides: 2^87 + 3 ≡ 1 + 3 ≡ 4 (mod 7). The remainder when 2^87 + 3 is divided by 7 is 4. Answer: D. 4 |
D |
| wenxinyiyan | D Explanation: We need to find the remainder when $2^{87} + 3$ is divided by 7. First, let's consider the powers of 2 modulo 7: * $2^1 \equiv 2 \pmod{7}$ * $2^2 \equiv 4 \pmod{7}$ * $2^3 \equiv 1 \pmod{7}$ Note that $2^3 \equiv 1 \pmod{7}$. This means that any power of 2 that is a multiple of 3 will also be equivalent to 1 modulo 7. Now, we can write $2^{87}$ as $(2^3)^{29}$. Since $2^3 \equiv 1 \pmod{7}$, we have: $(2^3)^{29} \equiv 1^{29} \equiv 1 \pmod{7}$ Therefore, $2^{87} \equiv 1 \pmod{7}$. Finally, we add 3 to both sides: $2^{87} + 3 \equiv 1 + 3 \equiv 4 \pmod{7}$ So, the remainder when $2^{87} + 3$ is divided by 7 is 4. |
D |
| zhipu | To determine the remainder when \(2^{87} + 3\) is divided by \(7\), we can use modular arithmetic. First, we need to find the remainder of \(2^{87}\) when divided by \(7\). To do this, we look for a pattern in the powers of \(2\) modulo \(7\): \[ \begin{align*} 2^1 &\equiv 2 \pmod{7}, \\ 2^2 &\equiv 4 \pmod{7}, \\ 2^3 &\equiv 8 \equiv 1 \pmod{7}. \end{align*} \] Notice that \(2^3 \equiv 1 \pmod{7}\). This implies that every power of \(2\) that is a multiple of \(3\) will be congruent to \(1\) modulo \(7\). Since \(87\) is a multiple of \(3\) (as \(87 = 3 \times 29\)), we have: \[ 2^{87} \equiv (2^3)^{29} \equiv 1^{29} \equiv 1 \pmod{7}. \] Now, we add \(3\) to \(2^{87}\) and find the remainder modulo \(7\): \[ 2^{87} + 3 \equiv 1 + 3 \equiv 4 \pmod{7}. \] Therefore, the remainder when \(2^{87} + 3\) is divided by \(7\) is \(4\). The correct answer is \( \boxed{4} \). So, the choice is: D. 4 |
D |
| gpto1 | D | D |
| deepseekv3 | To find the remainder when \( 2^{87} + 3 \) is divided by \( 7 \), we can use modular arithmetic. **Step 1: Find the remainder of \( 2^{87} \) when divided by 7.** Notice that the powers of \( 2 \) cycle every 3 when taken modulo \( 7 \): \[ \begin{align*} 2^1 &\equiv 2 \pmod{7}, \\ 2^2 &\equiv 4 \pmod{7}, \\ 2^3 &\equiv 1 \pmod{7}, \\ 2^4 &\equiv 2 \pmod{7}, \\ 2^5 &\equiv 4 \pmod{7}, \\ 2^6 &\equiv 1 \pmod{7}, \\ \end{align*} \] and so on. This pattern repeats every 3 powers. To find \( 2^{87} \pmod{7} \), determine the remainder when \( 87 \) is divided by \( 3 \): \[ 87 \div 3 = 29 \quad \text{with a remainder of } 0. \] Since the remainder is \( 0 \), \( 2^{87} \) corresponds to \( 2^3 \), which is: \[ 2^{87} \equiv 1 \pmod{7}. \] **Step 2: Add 3 and find the remainder modulo 7.** \[ 2^{87} + 3 \equiv 1 + 3 \equiv 4 \pmod{7}. \] **Final Answer:** \[ \boxed{D} \] |
D |