| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the sum of the primes between 100 and 200, inclusive, that are 1 or 2 more than a perfect square, we need to identify the perfect squares in that range and then check if the numbers 1 or 2 greater than those squares are prime. The perfect squares between 100 and 200 are 100, 121, 144, 169, and 196. 1. For 100, the numbers 1 and 2 more than 100 are 101 and 102. 101 is prime, but 102 is not. 2. For 121, the numbers 1 and 2 more than 121 are 122 and 123. Neither 122 nor 123 is prime. 3. For 144, the numbers 1 and 2 more than 144 are 145 and 146. Neither 145 nor 146 is prime. 4. For 169, the numbers 1 and 2 more than 169 are 170 and 171. Neither 170 nor 171 is prime. 5. For 196, the numbers 1 and 2 more than 196 are 197 and 198. 197 is prime, but 198 is not. Therefore, the primes between 100 and 200 that are 1 or 2 more than a perfect square are 101 and 197. The sum of these primes is 101 + 197 = 298. The correct answer is B. 298. |
B |
| deepseekr1 | To find the sum of primes between 100 and 200 that are 1 or 2 more than a perfect square, we analyze perfect squares \( n^2 \) for \( n \) such that \( n^2 + 1 \) or \( n^2 + 2 \) falls within the range. 1. **Determine valid \( n \):** - Lower bound: \( n^2 + 1 \geq 100 \Rightarrow n \geq 10 \) (since \( 10^2 = 100 \)). - Upper bound: \( n^2 + 2 \leq 200 \Rightarrow n \leq 14 \) (since \( 14^2 = 196 \)). 2. **Check primes for \( n = 10 \) to \( 14 \):** - **\( n = 10 \):** \( 10^2 = 100 \). \( 100 + 1 = 101 \) (prime). \( 100 + 2 = 102 \) (not prime). - **\( n = 11 \):** \( 11^2 = 121 \). \( 121 + 1 = 122 \) (even, not prime). \( 121 + 2 = 123 \) (divisible by 3, not prime). - **\( n = 12 \):** \( 12^2 = 144 \). \( 144 + 1 = 145 \) (divisible by 5, not prime). \( 144 + 2 = 146 \) (even, not prime). - **\( n = 13 \):** \( 13^2 = 169 \). \( 169 + 1 = 170 \) (even, not prime). \( 169 + 2 = 171 \) (divisible by 3, not prime). - **\( n = 14 \):** \( 14^2 = 196 \). \( 196 + 1 = 197 \) (prime). \( 196 + 2 = 198 \) (even, not prime). 3. **Sum the valid primes:** The primes are **101** and **197**. Sum: \( 101 + 197 = 298 \). **Answer:** B. 298 |
B |
| gpto1 | B | B |
| GPT-4o | To solve this problem, we need to identify the prime numbers between 100 and 200 that are either 1 or 2 more than a perfect square. First, let's generate the perfect squares within the range that when 1 or 2 is added, they fall between 100 and 200. The perfect squares between 100 and 200, inclusive, are: - \(10^2 = 100\) - \(11^2 = 121\) - \(12^2 = 144\) - \(13^2 = 169\) - \(14^2 = 196\) Now, we add 1 and 2 to each of these perfect squares: - \(100 + 1 = 101\) - \(100 + 2 = 102\) - \(121 + 1 = 122\) - \(121 + 2 = 123\) - \(144 + 1 = 145\) - \(144 + 2 = 146\) - \(169 + 1 = 170\) - \(169 + 2 = 171\) - \(196 + 1 = 197\) - \(196 + 2 = 198\) Next, we need to determine which of these numbers are prime: - 101 is prime. - 102 is not prime (divisible by 2). - 122 is not prime (divisible by 2). - 123 is not prime (divisible by 3). - 145 is not prime (divisible by 5). - 146 is not prime (divisible by 2). - 170 is not prime (divisible by 2). - 171 is not prime (divisible by 3). - 197 is prime. - 198 is not prime (divisible by 2). The prime numbers that are either 1 or 2 more than a perfect square between 100 and 200 are 101 and 197. Now, we calculate the sum of these prime numbers: \[ 101 + 197 = 298 \] Thus, the correct answer is: B. 298 |
B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | The perfect squares between 100 and 200 are $11^2=121$, $12^2=144$, $13^2=169$, and $14^2=196$. Of the numbers one more or two more than these, we see that $121+2=123$, $144+1=145$, $169+2=171$ are not prime, but $196+1=197$ is prime. Thus the sum of all primes between 100 and 200 that are one or two more than a perfect square is $197=\boxed{B}$. The best answer is B. |
B |
| DeepSeek-Coder-V2 | B. 298 | B |
| Qwen2-72B-Instruct | B. 298 | B |
| baichuan4 | To solve this problem, we need to find the primes between 100 and 200 (inclusive) that are 1 or 2 more than a perfect square. First, let's list the perfect squares between 10^2 and 14^2 (since 14^2 = 196, which is the largest perfect square less than 200): 10^2 = 100 11^2 = 121 12^2 = 144 13^2 = 169 Now, we need to find the primes that are 1 or 2 more than these perfect squares: 100 + 1 = 101 (prime) 100 + 2 = 102 (not prime) 121 + 1 = 122 (not prime) 121 + 2 = 123 (not prime) 144 + 1 = 145 (not prime) 144 + 2 = 146 (not prime) 169 + 1 = 170 (not prime) 169 + 2 = 171 (not prime) The only prime number that is 1 or 2 more than a perfect square between 100 and 200 is 101. So, the sum of these primes is 101. Therefore, the correct answer is: A. 102 (which is incorrect because we found the sum to be 101, not 102). Since none of the options match our calculation, it seems there might be an error in the options provided. Based on our calculation, the correct sum is 101, which is not listed among the options. |
B |
| doubao | The primes between 100 and 200 are 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199. Among them, 101 = 100 + 1 (100 is a perfect square), 103 = 100 + 3 (not 1 or 2 more than a perfect square), 107 = 100 + 7 (not 1 or 2 more than a perfect square), 109 = 100 + 9 (not 1 or 2 more than a perfect square), 113 = 100 + 13 (not 1 or 2 more than a perfect square), 127 = 121 + 6 (121 is a perfect square), 131 = 121 + 10 (not 1 or 2 more than a perfect square), 137 = 144 - 7 (144 is a perfect square), 139 = 144 - 5 (not 1 or 2 more than a perfect square), 149 = 144 + 5 (not 1 or 2 more than a perfect square), 151 = 144 + 7 (not 1 or 2 more than a perfect square), 157 = 169 - 12 (169 is a perfect square), 163 = 169 - 6 (not 1 or 2 more than a perfect square), 167 = 169 - 2 (fits), 173 = 169 + 4 (not 1 or 2 more than a perfect square), 179 = 196 - 17 (196 is a perfect square), 181 = 196 - 15 (not 1 or 2 more than a perfect square), 191 = 196 - 5 (not 1 or 2 more than a perfect square), 193 = 196 - 3 (not 1 or 2 more than a perfect square), 197 = 225 - 28 (225 is a perfect square), 199 = 225 - 26 (not 1 or 2 more than a perfect square). So the primes that meet the condition are 101 and 167, and their sum is 268. None of the given options match. So the answer is none of the above. |
B |
| minimax | To solve this problem, we need to find the primes between 100 and 200 that are either 1 or 2 more than a perfect square. Let's first list the perfect squares between 100 and 200 and then add 1 or 2 to each to see if we get a prime number. The perfect squares between 100 and 200 are: - 11^2 = 121 - 12^2 = 144 - 13^2 = 169 - 14^2 = 196 Now, let's add 1 or 2 to each of these perfect squares and check for primality: - 121 + 1 = 122 (not prime) - 121 + 2 = 123 (not prime) - 144 + 1 = 145 (not prime) - 144 + 2 = 146 (not prime) - 169 + 1 = 170 (not prime) - 169 + 2 = 171 (not prime) - 196 + 1 = 197 (prime) - 196 + 2 = 198 (not prime) The only prime number we found that is 1 or 2 more than a perfect square between 100 and 200 is 197. Therefore, the sum of the primes between 100 and 200 that are 1 or 2 more than a perfect square is simply 197, which is not listed in the options provided. It seems there might be an error in the question or the options given. However, if we must choose from the provided options, none of them is correct. |
B |
| kimi | To solve this problem, we need to find the prime numbers between 100 and 200 that are either 1 or 2 more than a perfect square. First, let's identify the perfect squares between 100 and 200: - \(10^2 = 100\) - \(11^2 = 121\) - \(12^2 = 144\) - \(13^2 = 169\) - \(14^2 = 196\) Now, we'll check the numbers that are 1 or 2 more than these perfect squares: - \(100 + 1 = 101\) (prime) - \(100 + 2 = 102\) (not prime) - \(121 + 1 = 122\) (not prime) - \(121 + 2 = 123\) (not prime) - \(144 + 1 = 145\) (not prime) - \(144 + 2 = 146\) (not prime) - \(169 + 1 = 170\) (not prime) - \(169 + 2 = 171\) (not prime) - \(196 + 1 = 197\) (prime) So, the prime numbers between 100 and 200 that are 1 or 2 more than a perfect square are 101 and 197. Now, let's sum them up: \(101 + 197 = 298\) Answer: B. 298 |
B |
| wenxinyiyan | B Explanation: To solve this problem, we need to find all prime numbers between 100 and 200 (inclusive) that are either 1 or 2 more than a perfect square. Let's go through the steps: 1. **Identify Perfect Squares**: * Perfect squares between 100 and 200 are: $100 (10^2)$, $121 (11^2)$, $144 (12^2)$, $169 (13^2)$, and $196 (14^2)$. 2. **Find Numbers 1 or 2 More than Perfect Squares**: * Numbers 1 more than these squares: $101, 122, 145, 170, 197$ * Numbers 2 more than these squares: $102, 123, 146, 171, 198$ 3. **Check for Primality**: * From the list above, the prime numbers are: $101, 145$ (Note: $145$ is actually not prime, it's a mistake in the original answer. The correct primes are $101$ and $197$.) * The sum of these primes is $101 + 197 = 298$. Therefore, the correct answer is B. $298$ Note: The original answer included $145$ as a prime, which is incorrect. $145$ is divisible by $5$ and thus not prime. The correct primes are $101$ and $197$, and their sum is $298$. |
B |
| zhipu | To solve this problem, we need to identify the prime numbers between 100 and 200 that are either 1 or 2 more than a perfect square. First, let's list the perfect squares between 100 and 200: \[ 10^2 = 100, \quad 11^2 = 121, \quad 12^2 = 144, \quad 13^2 = 169, \quad 14^2 = 196 \] Next, we'll find the numbers that are 1 or 2 more than these perfect squares: \[ 100 + 1 = 101, \quad 100 + 2 = 102, \quad 121 + 1 = 122, \quad 121 + 2 = 123, \quad 144 + 1 = 145, \quad 144 + 2 = 146, \quad 169 + 1 = 170, \quad 169 + 2 = 171, \quad 196 + 1 = 197, \quad 196 + 2 = 198 \] Now, we need to check which of these numbers are prime: - 101: Prime - 102: Not prime (divisible by 2) - 122: Not prime (divisible by 2) - 123: Not prime (divisible by 3) - 145: Not prime (divisible by 5) - 146: Not prime (divisible by 2) - 170: Not prime (divisible by 2) - 171: Not prime (divisible by 3) - 197: Prime - 198: Not prime (divisible by 2) The prime numbers we found are 101 and 197. Now, sum these prime numbers: \[ 101 + 197 = 298 \] Therefore, the correct answer is: B. 298 |
B |
| gpto1 | B | B |
| deepseekv3 | To solve this question, we need to find all prime numbers between 100 and 200 that are 1 or 2 more than a perfect square, and then sum them up. Here's the step-by-step solution: ### Step 1: Identify perfect squares in the given range First, determine the range of perfect squares such that their values plus 1 or 2 fall between 100 and 200. Let \( n^2 \) be a perfect square. Then: - \( n^2 + 1 \geq 100 \), so \( n^2 \geq 99 \) and \( n \geq 10 \). - \( n^2 + 2 \leq 200 \), so \( n^2 \leq 198 \) and \( n \leq 14 \). Thus, \( n \) ranges from 10 to 14. ### Step 2: Calculate \( n^2 + 1 \) and \( n^2 + 2 \) for \( n = 10 \) to \( 14 \) For each \( n \), compute \( n^2 + 1 \) and \( n^2 + 2 \), and check if they are prime numbers. 1. \( n = 10 \): - \( 10^2 + 1 = 101 \) (prime). - \( 10^2 + 2 = 102 \) (not prime). 2. \( n = 11 \): - \( 11^2 + 1 = 122 \) (not prime). - \( 11^2 + 2 = 123 \) (not prime). 3. \( n = 12 \): - \( 12^2 + 1 = 145 \) (not prime). - \( 12^2 + 2 = 146 \) (not prime). 4. \( n = 13 \): - \( 13^2 + 1 = 170 \) (not prime). - \( 13^2 + 2 = 171 \) (not prime). 5. \( n = 14 \): - \( 14^2 + 1 = 197 \) (prime). - \( 14^2 + 2 = 198 \) (not prime). ### Step 3: Identify the primes From the above calculations, the primes that satisfy the condition are: - 101 (from \( n = 10 \)). - 197 (from \( n = 14 \)). ### Step 4: Sum the primes The sum of these primes is: \[ 101 + 197 = 298. \] ### Final Answer: The correct answer is **B. 298**. |
B |