| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B |
B |
| deepseekr1 | To determine the exponent of \(109\) in the prime factorization of \(x^{11}\), start with the given factorization of \(x = 107 \cdot 109^5\). Raising \(x\) to the 11th power: \[ x^{11} = \left(107 \cdot 109^5\right)^{11} = 107^{11} \cdot \left(109^5\right)^{11}. \] Applying the exponentiation rule \((a^m)^n = a^{m \cdot n}\), the exponent of \(109\) becomes: \[ \left(109^5\right)^{11} = 109^{5 \times 11} = 109^{55}. \] Thus, the exponent of \(109\) in \(x^{11}\) is \(55\). **Answer: B** |
B |
| gpto1 | B | B |
| GPT-4o | To determine the exponent of \(109\) in the prime factorization of \(x^{11}\), we first consider the given information about \(x\). The number \(x\) factors as: \[ x = 107 \cdot 109^5 \] When raising \(x\) to the 11th power, we have: \[ x^{11} = (107 \cdot 109^5)^{11} \] According to the properties of exponents, we can distribute the exponent across the factors: \[ x^{11} = 107^{11} \cdot (109^5)^{11} \] Simplifying \( (109^5)^{11} \): \[ (109^5)^{11} = 109^{5 \cdot 11} = 109^{55} \] So, the exponent of \(109\) in the prime factorization of \(x^{11}\) is \(55\). Therefore, the answer is: B. 55 |
B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | B |
B |
| DeepSeek-Coder-V2 | B | B |
| Qwen2-72B-Instruct | B. 55 | B |
| baichuan4 | B. 55 | B |
| doubao | When we raise $x$ to the power of 11, we multiply the exponents of each prime factor by 11. So the exponent of 109 in $x^{11}$ is $5\times11=55$. So the answer is B. |
B |
| minimax | The exponent of $109$ in the prime factorization of $x$ is $5$. When we raise $x$ to the $11$th power, we raise the exponent of $109$ to the $11$th power as well. Therefore, the exponent of $109$ in the prime factorization of $x^{11}$ is $5 \times 11 = 55$. Answer: B. 55 |
B |
| kimi | To solve this problem, we first need to understand the prime factorization of the given number \( x \). The number \( x \) is given as \( 107 \cdot 109^5 \). This means that \( x \) has a prime factorization of \( 107^1 \cdot 109^5 \). Now, we want to find the exponent of \( 109 \) in the prime factorization of \( x^{11} \). To do this, we raise each prime factor in the factorization of \( x \) to the power of 11: \( (107^1 \cdot 109^5)^{11} = 107^{11} \cdot 109^{55} \) So, when \( x \) is raised to the power of 11, the exponent of \( 109 \) in the prime factorization of \( x^{11} \) is \( 55 \). Answer: B. 55 |
B |
| wenxinyiyan | B Explanation: Given that $x = 107 \cdot 109^{5}$, we can find $x^{11}$ by raising both factors to the power of 11: $x^{11} = (107 \cdot 109^{5})^{11}$ $= 107^{11} \cdot (109^{5})^{11}$ $= 107^{11} \cdot 109^{55}$ From this prime factorization of $x^{11}$, we can see that the exponent of $109$ is 55. |
B |
| zhipu | To find the exponent of \(109\) in the prime factorization of \(x^{11}\), we start by examining the given factorization of \(x\): \[ x = 107 \cdot 109^5 \] We need to raise \(x\) to the power of \(11\): \[ x^{11} = (107 \cdot 109^5)^{11} \] Using the properties of exponents, we can distribute the exponent \(11\) to each factor: \[ x^{11} = 107^{11} \cdot (109^5)^{11} \] Now, simplify the expression for \((109^5)^{11}\): \[ (109^5)^{11} = 109^{5 \cdot 11} = 109^{55} \] Therefore, in the prime factorization of \(x^{11}\), the exponent of \(109\) is \(55\). So, the correct answer is: B. 55 |
B |
| gpto1 | B | B |
| deepseekv3 | To determine the exponent of \(109\) in the prime factorization of \(x^{11}\), let's follow these steps: 1. **Express \(x\) in its prime factorization:** \[ x = 107 \cdot 109^5 \] 2. **Raise \(x\) to the power of 11:** \[ x^{11} = (107 \cdot 109^5)^{11} = 107^{11} \cdot (109^5)^{11} \] 3. **Simplify the exponents:** \[ x^{11} = 107^{11} \cdot 109^{5 \times 11} = 107^{11} \cdot 109^{55} \] 4. **Identify the exponent of \(109\):** The exponent of \(109\) in \(x^{11}\) is \(55\). Thus, the correct answer is: \boxed{B} |
B |