| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | A Final Answer: The final answer is A. I hope it is correct. |
B |
| deepseekr1 | To solve \((2^{20} + 2^{20} + 2^{20} + 2^{21}) \div 2^{17}\): 1. **Simplify the numerator**: - Combine the three \(2^{20}\) terms: \(3 \times 2^{20}\). - Express \(2^{21}\) as \(2 \times 2^{20}\). - Add all terms: \(3 \times 2^{20} + 2 \times 2^{20} = (3 + 2) \times 2^{20} = 5 \times 2^{20}\). 2. **Divide by \(2^{17}\)**: - Use the property \(2^a / 2^b = 2^{a-b}\): \(5 \times 2^{20} / 2^{17} = 5 \times 2^{20-17} = 5 \times 2^3 = 5 \times 8 = 40\). **Answer**: B. 40 |
B |
| gpto1 | B | B |
| GPT-4o | To solve the given expression \((2^{20} + 2^{20} + 2^{20} + 2^{21}) \div 2^{17}\), we can simplify it step by step. First, combine the terms inside the parentheses: \[ 2^{20} + 2^{20} + 2^{20} + 2^{21} = 3 \cdot 2^{20} + 2^{21} \] Next, factor out the common term \(2^{20}\): \[ 3 \cdot 2^{20} + 2^{21} = 3 \cdot 2^{20} + 2 \cdot 2^{20} = (3 + 2) \cdot 2^{20} = 5 \cdot 2^{20} \] Now, we need to divide this by \(2^{17}\): \[ \frac{5 \cdot 2^{20}}{2^{17}} \] Using the properties of exponents, we can simplify this as: \[ 5 \cdot 2^{20 - 17} = 5 \cdot 2^3 = 5 \cdot 8 = 40 \] Therefore, the answer is: \[ B. 40 \] |
B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | We have that \begin{align*} (2^{20} + 2^{20} + 2^{20} +2^{21}) \div 2^{17} &= (2^{20} + 2^{20} + 2^{20} +2^{21}) \cdot 2^{-17}\\ &= 2^{20} \cdot 2^{-17} + 2^{20} \cdot 2^{-17} + 2^{20} \cdot 2^{-17} + 2^{21} \cdot 2^{-17}\\ &= 2^{20-17} + 2^{20-17} + 2^{20-17} + 2^{21-17}\\ &= 2^3 + 2^3 + 2^3 + 2^4\\ &= 8 + 8 + 8 + 16\\ &= \boxed{40}. \end{align*} The best answer is B. |
B |
| DeepSeek-Coder-V2 | B. 40 | B |
| Qwen2-72B-Instruct | B. 40 | B |
| baichuan4 | B | B |
| doubao | \[ \begin{align*} &(2^{20}+2^{20}+2^{20}+2^{21})\div2^{17}\\ =&(3\times2^{20}+2\times2^{20})\div2^{17}\\ =&(3+2)\times2^{20}\div2^{17}\\ =&5\times2^{20}\div2^{17}\\ =&5\times2^{20-17}\\ =&5\times2^3\\ =&5\times8\\ =&40 \end{align*} \] So the answer is B. |
B |
| minimax | First, let's simplify the expression inside the parentheses: $(2^{20} + 2^{20} + 2^{20} + 2^{21})$ Since $2^{20}$ is repeated three times, we can factor it out: $2^{20} \times (1 + 1 + 1) + 2^{21}$ $2^{20} \times 3 + 2^{21}$ Now, we can rewrite $2^{21}$ as $2^{20} \times 2$ to combine the terms: $2^{20} \times 3 + 2^{20} \times 2$ $2^{20} \times (3 + 2)$ $2^{20} \times 5$ Now, we divide this by $2^{17}$: $(2^{20} \times 5) \div 2^{17}$ Using the properties of exponents, we can subtract the exponents when dividing with the same base: $2^{20-17} \times 5$ $2^3 \times 5$ $8 \times 5$ $40$ So the answer is: B. 40 |
B |
| kimi | First, let's simplify the expression: \[ (2^{20} + 2^{20} + 2^{20} + 2^{21}) \div 2^{17} \] Combine the like terms: \[ 3 \cdot 2^{20} + 2^{21} \div 2^{17} \] Now, divide each term by \(2^{17}\): \[ 3 \cdot \frac{2^{20}}{2^{17}} + \frac{2^{21}}{2^{17}} \] Simplify the exponents: \[ 3 \cdot 2^{20-17} + 2^{21-17} \] \[ 3 \cdot 2^3 + 2^4 \] Calculate the powers of 2: \[ 3 \cdot 8 + 16 \] \[ 24 + 16 \] \[ 40 \] So the correct answer is: B. 40 |
B |
| wenxinyiyan | 首先，将表达式中的相同项合并： $2^{20} + 2^{20} + 2^{20} + 2^{21} = 3 \times 2^{20} + 2^{21}$ 然后，将上述结果除以 $2^{17}$： $\frac{3 \times 2^{20} + 2^{21}}{2^{17}}$ 由于 $2^{20} = 2^{17} \times 2^{3}$ 和 $2^{21} = 2^{17} \times 2^{4}$，我们可以将分子中的每一项都拆分成 $2^{17}$ 的倍数： $\frac{3 \times 2^{17} \times 2^{3} + 2^{17} \times 2^{4}}{2^{17}}$ 然后，将 $2^{17}$ 提取出来作为公因子： $2^{17} \times \frac{3 \times 2^{3} + 2^{4}}{2^{17}}$ 由于分子和分母都含有 $2^{17}$，它们可以相互抵消： $3 \times 2^{3} + 2^{4}$ 计算上述表达式： $3 \times 8 + 16 = 24 + 16 = 40$ 所以，答案是 B. $40$。 |
B |
| zhipu | To solve the expression \((2^{20} + 2^{20} + 2^{20} + 2^{21}) \div 2^{17}\), we first simplify the numerator: 1. Combine the like terms in the numerator: \[ 2^{20} + 2^{20} + 2^{20} + 2^{21} \] This can be rewritten as: \[ 3 \cdot 2^{20} + 2^{21} \] 2. Notice that \(2^{21}\) can be written as \(2 \cdot 2^{20}\). Therefore, the expression becomes: \[ 3 \cdot 2^{20} + 2 \cdot 2^{20} = (3 + 2) \cdot 2^{20} = 5 \cdot 2^{20} \] 3. Now we need to divide this by \(2^{17}\): \[ \frac{5 \cdot 2^{20}}{2^{17}} \] 4. Use the property of exponents \(\frac{a^m}{a^n} = a^{m-n}\): \[ \frac{5 \cdot 2^{20}}{2^{17}} = 5 \cdot 2^{20-17} = 5 \cdot 2^3 = 5 \cdot 8 = 40 \] So, the answer is: B. 40 |
B |
| gpto1 | B | B |
| deepseekv3 | To solve the expression \((2^{20} + 2^{20} + 2^{20} + 2^{21}) \div 2^{17}\), follow these steps: 1. **Combine like terms**: \[ 2^{20} + 2^{20} + 2^{20} + 2^{21} = 3 \times 2^{20} + 2^{21} \] 2. **Factor out \(2^{20}\)**: \[ 3 \times 2^{20} + 2^{21} = 3 \times 2^{20} + 2 \times 2^{20} = (3 + 2) \times 2^{20} = 5 \times 2^{20} \] 3. **Divide by \(2^{17}\)**: \[ \frac{5 \times 2^{20}}{2^{17}} = 5 \times 2^{20-17} = 5 \times 2^{3} = 5 \times 8 = 40 \] Therefore, the answer is \(\boxed{B}\). |
B |