| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the digit at the 1001st place after the decimal point in the decimal expansion of 9/28, we must first understand the decimal expansion of 9/28. When dividing 9 by 28, the decimal expansion is 0.32142857142857..., where "142857" repeats indefinitely. This is a repeating cycle of 6 digits. To find the 1001st digit, we can consider the position of this digit within the repeating cycle. Since the cycle is 6 digits long, we can divide 1001 by 6 to find out how many complete cycles there are and where within the cycle the 1001st digit falls. 1001 divided by 6 is 166 with a remainder of 5. This means that the 1001st digit is the 5th digit in the repeating cycle of "142857". Thus, the 1001st digit after the decimal point in the decimal expansion of 9/28 is 5. |
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| deepseekr1 | To find the digit at the 1001th place after the decimal point in the decimal expansion of the fraction \( \frac{9}{28} \): 1. **Perform the long division of 9 by 28**: - The decimal expansion of \( \frac{9}{28} \) is found to be \( 0.3214285714285714285\ldots \). - The expansion starts with \( 0.32 \) followed by the repeating cycle \( 142857 \). 2. **Identify the repeating cycle**: - The repeating cycle is \( 142857 \), which has a length of 6 digits. - The non-repeating part of the decimal is \( 0.32 \), which occupies the first 2 decimal places. 3. **Determine the position within the repeating cycle**: - Subtract the non-repeating part (2 digits) from the target position: \( 1001 - 2 = 999 \). - Compute the position within the repeating cycle: \( 999 \mod 6 \). - \( 999 \div 6 = 166 \) remainder \( 3 \). 4. **Find the corresponding digit in the cycle**: - The repeating cycle \( 142857 \) at position 3 (0-based index) is \( 2 \). Thus, the digit at the 1001th place after the decimal point in the decimal expansion of \( \frac{9}{28} \) is \(\boxed{2}\). |
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| gpto1 | To find the 1001st digit after the decimal point in the decimal expansion of \(\frac{9}{28}\), let's first understand the repeating nature of this decimal expansion. **Step 1: Long Division of 9 by 28** We perform long division of 9 divided by 28 to find the decimal expansion: 1. **Divide 9 by 28**: Since 9 is less than 28, we add a decimal point and a zero to make it 90. 2. **Divide 28 into 90**: - \(28 \times 3 = 84\) - Remainder: \(90 - 84 = 6\) - Quotient so far: 0.3 3. **Bring down a zero to make the new dividend 60**. - \(28 \times 2 = 56\) - Remainder: \(60 - 56 = 4\) - Quotient so far: 0.32 4. **Bring down a zero to make the new dividend 40**. - \(28 \times 1 = 28\) - Remainder: \(40 - 28 = 12\) - Quotient so far: 0.321 5. **Bring down a zero to make the new dividend 120**. - \(28 \times 4 = 112\) - Remainder: \(120 - 112 = 8\) - Quotient so far: 0.3214 6. **Bring down a zero to make the new dividend 80**. - \(28 \times 2 = 56\) - Remainder: \(80 - 56 = 24\) - Quotient so far: 0.32142 7. **Bring down a zero to make the new dividend 240**. - \(28 \times 8 = 224\) - Remainder: \(240 - 224 = 16\) - Quotient so far: 0.321428 8. **Bring down a zero to make the new dividend 160**. - \(28 \times 5 = 140\) - Remainder: \(160 - 140 = 20\) - Quotient so far: 0.3214285 9. **Bring down a zero to make the new dividend 200**. - \(28 \times 7 = 196\) - Remainder: \(200 - 196 = 4\) - Quotient so far: 0.32142857 10. **Notice the remainder repeats**: At this point, the remainder is 4, which we had earlier. This indicates that the decimal expansion starts repeating from here. **Step 2: Identify the Repeating Sequence** From the division steps, we get the repeating sequence of digits after the decimal point: \[ \frac{9}{28} = 0.\overline{32142857} \] The sequence \(32142857\) repeats indefinitely, and the period of repetition is **8 digits**. **Step 3: Find the 1001st Digit** To find the 1001st digit after the decimal point, we use the fact that the decimal repeats every 8 digits. - **Calculate the position within the repeating sequence**: - Divide 1001 by 8: \(1001 \div 8 = 125\) with a remainder of \(1\). - **Interpret the remainder**: - The remainder of 1 indicates that the 1001st digit corresponds to the **first digit** in the repeating sequence. **Step 4: Determine the 1001st Digit** The repeating sequence is \(3 \, 2 \, 1 \, 4 \, 2 \, 8 \, 5 \, 7\). Since the 1001st digit corresponds to the first digit in this sequence, the digit is **3**. **Answer:** 3 |
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| GPT-4o | To determine the digit at the 1001st place after the decimal point in the decimal expansion of the fraction \( \frac{9}{28} \), we first need to find the repeating decimal expansion of \( \frac{9}{28} \). ### Step 1: Long Division to Determine the Decimal Expansion Perform long division of 9 by 28. 1. 9 divided by 28 goes 0 times, so we write 0 and a decimal point. 2. Bring down a zero to make it 90. 3. 90 divided by 28 goes 3 times (since \( 28 \times 3 = 84 \)). - Subtract 84 from 90 to get a remainder of 6. 4. Bring down a zero to make it 60. 5. 60 divided by 28 goes 2 times (since \( 28 \times 2 = 56 \)). - Subtract 56 from 60 to get a remainder of 4. 6. Bring down a zero to make it 40. 7. 40 divided by 28 goes 1 time (since \( 28 \times 1 = 28 \)). - Subtract 28 from 40 to get a remainder of 12. 8. Bring down a zero to make it 120. 9. 120 divided by 28 goes 4 times (since \( 28 \times 4 = 112 \)). - Subtract 112 from 120 to get a remainder of 8. 10. Bring down a zero to make it 80. 11. 80 divided by 28 goes 2 times (since \( 28 \times 2 = 56 \)). - Subtract 56 from 80 to get a remainder of 24. 12. Bring down a zero to make it 240. 13. 240 divided by 28 goes 8 times (since \( 28 \times 8 = 224 \)). - Subtract 224 from 240 to get a remainder of 16. 14. Bring down a zero to make it 160. 15. 160 divided by 28 goes 5 times (since \( 28 \times 5 = 140 \)). - Subtract 140 from 160 to get a remainder of 20. 16. Bring down a zero to make it 200. 17. 200 divided by 28 goes 7 times (since \( 28 \times 7 = 196 \)). - Subtract 196 from 200 to get a remainder of 4. Notice that we have a remainder of 4 again, which means the decimal repeats from this point. Therefore, the repeating sequence is: \[ 0.\overline{32142857} \] ### Step 2: Determine the Length of the Repeating Sequence The repeating decimal sequence "32142857" has a length of 8 digits. ### Step 3: Find the Digit at the 1001st Position To find the digit at the 1001st position, we need to determine the position within the repeating sequence. This can be done using the modulo operation: \[ 1001 \mod 8 = 1 \] So, the 1001st digit corresponds to the 1st digit in the repeating sequence "32142857". ### Conclusion The digit at the 1001st place after the decimal point in the decimal expansion of \( \frac{9}{28} \) is: \[ \boxed{3} \] |
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| llama3.1-70B-Instruct | To find the digit at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28, we can follow these steps: 1. Convert the fraction 9/28 to a decimal using long division. 2. Identify the repeating pattern in the decimal expansion. 3. Determine the length of the repeating pattern. 4. Calculate the position of the 1001th digit within the repeating pattern. Step 1: Convert 9/28 to a decimal using long division. 9 ÷ 28 = 0.321428571... Step 2: Identify the repeating pattern in the decimal expansion. The repeating pattern is 321428. Step 3: Determine the length of the repeating pattern. The length of the repeating pattern is 6 digits. Step 4: Calculate the position of the 1001th digit within the repeating pattern. To find the position of the 1001th digit, we can divide 1001 by the length of the repeating pattern (6) and take the remainder. 1001 ÷ 6 = 166 with a remainder of 5 This means that the 1001th digit is the 5th digit in the repeating pattern. The 5th digit in the repeating pattern is 2. Therefore, the digit at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28 is 2. |
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| llama3.1-405B-Instruct | To find the digit at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28, we can use the following steps: 1. Convert the fraction 9/28 to a decimal by dividing 9 by 28: 9 ÷ 28 = 0.321428571... 2. Identify the repeating pattern in the decimal expansion. In this case, the pattern is 321428571. 3. Determine the length of the repeating pattern. In this case, the length is 6 digits (321428). 4. Divide 1001 by the length of the repeating pattern (6) to find the number of complete cycles: 1001 ÷ 6 = 166 with a remainder of 5 5. Since there are 166 complete cycles, we can ignore them and focus on the remaining 5 digits. 6. Count 5 digits from the beginning of the repeating pattern: 321428571... The 5th digit is 2. Therefore, the digit at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28 is 2. |
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| DeepSeek-Coder-V2 | To find which digit is at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28, we need to first determine the repeating decimal pattern of 9/28. Step 1: Perform long division of 9 by 28 to find the decimal expansion. 9 ÷ 28 = 0.3214285714285714... We observe that the decimal expansion of 9/28 is 0.3214285714285714... with a repeating block "3214285714285714". Step 2: Identify the length of the repeating block. The repeating block "3214285714285714" has a length of 16 digits. Step 3: Determine the position of the 1001th digit within the repeating block. To find the 1001th digit, we need to find the remainder when 1001 is divided by the length of the repeating block (16). 1001 ÷ 16 = 62 R 9 This means that the 1001th digit corresponds to the 9th digit in the repeating block. Step 4: Identify the 9th digit in the repeating block. The repeating block is "3214285714285714". The 9th digit in this block is '7'. Therefore, the digit at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28 is 7. |
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| Qwen2-72B-Instruct | To find the digit at the 1001st place after the decimal point in the decimal expansion of 9/28, we must first understand the decimal expansion of 9/28. When dividing 9 by 28, the decimal expansion is 0.32142857142857..., where "142857" repeats indefinitely. This is a repeating cycle of 6 digits. To find the 1001st digit, we can consider the position of this digit within the repeating cycle. Since the cycle is 6 digits long, we can divide 1001 by 6 to find out how many complete cycles there are and where within the cycle the 1001st digit falls. 1001 divided by 6 is 166 with a remainder of 5. This means that the 1001st digit is the 5th digit in the repeating cycle of "142857". Thus, the 1001st digit after the decimal point in the decimal expansion of 9/28 is 5. |
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| baichuan4 | To find the 1001th digit after the decimal point in the decimal expansion of the fraction 9/28, we need to determine the pattern of the repeating decimal. First, let's calculate 9/28: ``` 9 / 28 = 0.32142857142857... ``` We can see that the decimal expansion repeats the sequence "142857" indefinitely. The length of this repeating sequence is 6 digits. To find the 1001th digit, we need to determine how many times the repeating sequence fits into 1001 digits: ``` 1001 ÷ 6 = 166...5 ``` This means the 1001th digit is the 5th digit in the repeating sequence. Since the repeating sequence is "142857", the 5th digit is "5". Therefore, the 1001th digit after the decimal point in the decimal expansion of 9/28 is "5". |
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| doubao | 9/28 = 0.321428571428571428... The pattern 142857 repeats. The length of this pattern is 6. To find the digit at the 1001st place after the decimal point, we divide 1001 by 6: 1001 ÷ 6 = 166 remainder 5. This means that the pattern repeats 166 times and then we need the 5th digit in the pattern, which is 7. So the digit at the 1001st place after the decimal point is 7. |
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| minimax | To find the digit at the 1001st place after the decimal point in the decimal expansion of the fraction 9/28, we need to perform the division of 9 by 28. When we divide 9 by 28, we get a repeating decimal because 28 has factors other than 2 and 5, which means it cannot be expressed as a terminating decimal. Let's perform the long division to see the pattern of the repeating decimal: ``` 0.***... --------------- 28|9.*** *.4 ---- 60 56 ---- 40 28 ---- 120 112 ---- 80 56 ---- 240 224 ---- 160 140 ---- 200 196 ---- 40 28 ---- 120 ... ``` As we can see from the long division, the decimal expansion of 9/28 is 0.***..., and the repeating block is "***". This block has a length of 6 digits. To find the 1001st digit, we need to determine which digit in the repeating block corresponds to the 1001st position. Since the block repeats every 6 digits, we can divide 1001 by 6 to see how many full cycles of the repeating block we have and what the remainder is: 1001 ÷ 6 = 166 remainder 5 This means that after 166 full cycles of the repeating block, we will have reached the 996th digit (since 166 * 6 = 996), and we need to find the 5th digit in the block to get to the 1001st digit overall. The repeating block is "***", so the 5th digit in this block is "2". Therefore, the digit at the 1001st place after the decimal point in the decimal expansion of the fraction 9/28 is 2. |
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| kimi | To find the digit at the 1001st place after the decimal point in the decimal expansion of the fraction 9/28, we first need to determine the repeating decimal pattern of the fraction. Let's perform the division of 9 by 28: ``` 0.32142857142857... ----------- 28 | 9.000000000000... - 28 -------- 120 - 112 -------- 80 - 70 -------- 100 - 98 -------- 20 - 28 -------- 40 - 28 -------- 120 - 112 -------- 80 - 70 -------- 100 - 98 -------- 20 ``` As you can see, the decimal expansion of 9/28 is `0.32142857142857...`, which has a repeating cycle of 6 digits: `321428`. Now, to find the digit at the 1001st place, we can use the following steps: 1. Determine the length of the repeating cycle, which is 6. 2. Divide 1001 by the length of the cycle to find out how many complete cycles there are and what the remainder is. 3. The remainder will tell us the position within the cycle that corresponds to the 1001st digit. Let's do the calculation: 1001 ÷ 6 = 166 remainder 5 This means that after 166 complete cycles of `321428`, we are 5 digits into the next cycle. Now, we look at the repeating cycle `321428` and find the 5th digit: 1st digit: 3 2nd digit: 2 3rd digit: 1 4th digit: 4 5th digit: 2 So, the digit at the 1001st place after the decimal point in the decimal expansion of 9/28 is 2. |
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| wenxinyiyan | 要找到分数9/28的小数点后第1001位的数字，我们可以利用长除法来找到小数展开，但直接计算到第1001位是不切实际的。不过，我们可以观察这个分数的小数展开的周期性。 首先，我们计算9除以28： 9/28 = 0.3214285714285714... 这里我们可以看到小数部分有一个重复的模式：“142857”。这个模式有6位数字，所以我们可以利用这个周期性来找到第1001位的数字。 首先，我们找到1001位是在多少个完整的周期之后： 1001 ÷ 6 = 166 余 5 这意味着前1000位包含了166个完整的周期（每个周期6位数字），然后是下一个周期的前5位。 因此，第1001位数字是周期“142857”中的第5位，即5。 所以，分数9/28的小数点后第1001位的数字是5。 |
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| zhipu | To find the digit at the 1001th place after the decimal point in the decimal expansion of the fraction 9/28, we first need to understand the repeating decimal pattern of this fraction. Let's convert 9/28 to a decimal: Step 1: Perform the division of 9 by 28. When we divide 9 by 28, we get a decimal that starts with 0.321428571428571428571428571428... and continues repeating. The repeating block for 9/28 is "321428571", which has a length of 9 digits. Step 2: Determine the position within the repeating block. To find the 1001th digit after the decimal point, we need to determine which digit within the repeating block corresponds to this position. Since the block repeats every 9 digits, we can find the position within the block by taking the remainder of 1001 divided by 9. 1001 ÷ 9 = 111 remainder 2 This tells us that the 1001th digit corresponds to the 2nd digit in the repeating block. Step 3: Identify the 2nd digit in the repeating block. The repeating block is "321428571". The 2nd digit in this block is "2". Therefore, the digit at the 1001th place after the decimal point in the decimal expansion of 9/28 is **2**. |
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| gpto1 | To find the 1001st digit after the decimal point in the decimal expansion of \(\frac{9}{28}\), let's first understand the repeating nature of this decimal expansion. **Step 1: Long Division of 9 by 28** We perform long division of 9 divided by 28 to find the decimal expansion: 1. **Divide 9 by 28**: Since 9 is less than 28, we add a decimal point and a zero to make it 90. 2. **Divide 28 into 90**: - \(28 \times 3 = 84\) - Remainder: \(90 - 84 = 6\) - Quotient so far: 0.3 3. **Bring down a zero to make the new dividend 60**. - \(28 \times 2 = 56\) - Remainder: \(60 - 56 = 4\) - Quotient so far: 0.32 4. **Bring down a zero to make the new dividend 40**. - \(28 \times 1 = 28\) - Remainder: \(40 - 28 = 12\) - Quotient so far: 0.321 5. **Bring down a zero to make the new dividend 120**. - \(28 \times 4 = 112\) - Remainder: \(120 - 112 = 8\) - Quotient so far: 0.3214 6. **Bring down a zero to make the new dividend 80**. - \(28 \times 2 = 56\) - Remainder: \(80 - 56 = 24\) - Quotient so far: 0.32142 7. **Bring down a zero to make the new dividend 240**. - \(28 \times 8 = 224\) - Remainder: \(240 - 224 = 16\) - Quotient so far: 0.321428 8. **Bring down a zero to make the new dividend 160**. - \(28 \times 5 = 140\) - Remainder: \(160 - 140 = 20\) - Quotient so far: 0.3214285 9. **Bring down a zero to make the new dividend 200**. - \(28 \times 7 = 196\) - Remainder: \(200 - 196 = 4\) - Quotient so far: 0.32142857 10. **Notice the remainder repeats**: At this point, the remainder is 4, which we had earlier. This indicates that the decimal expansion starts repeating from here. **Step 2: Identify the Repeating Sequence** From the division steps, we get the repeating sequence of digits after the decimal point: \[ \frac{9}{28} = 0.\overline{32142857} \] The sequence \(32142857\) repeats indefinitely, and the period of repetition is **8 digits**. **Step 3: Find the 1001st Digit** To find the 1001st digit after the decimal point, we use the fact that the decimal repeats every 8 digits. - **Calculate the position within the repeating sequence**: - Divide 1001 by 8: \(1001 \div 8 = 125\) with a remainder of \(1\). - **Interpret the remainder**: - The remainder of 1 indicates that the 1001st digit corresponds to the **first digit** in the repeating sequence. **Step 4: Determine the 1001st Digit** The repeating sequence is \(3 \, 2 \, 1 \, 4 \, 2 \, 8 \, 5 \, 7\). Since the 1001st digit corresponds to the first digit in this sequence, the digit is **3**. **Answer:** 3 |
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| deepseekv3 | To find the digit at the 1001th place after the decimal point in the decimal expansion of the fraction \( \frac{9}{28} \), follow these steps: ### Step 1: Find the Decimal Expansion of \( \frac{9}{28} \) Divide 9 by 28 to find the decimal expansion: \[ \frac{9}{28} = 0.32142857142857142857 \ldots \] ### Step 2: Identify the Repeating Pattern The decimal expansion of \( \frac{9}{28} \) is a repeating decimal. The repeating sequence is **"142857"**, which has a length of **6 digits**. ### Step 3: Determine the Position in the Repeating Sequence The decimal starts with **"32"** (non-repeating part), and then the repeating sequence **"142857"** begins. To find the digit at the 1001th place: 1. Subtract the non-repeating part: \( 1001 - 2 = 999 \) 2. Find the position within the repeating sequence: \( 999 \mod 6 = 3 \) (since the repeating sequence has 6 digits). ### Step 4: Find the Digit The repeating sequence is **"142857"**. The 3rd digit in this sequence is **2**. ### Final Answer: The digit at the 1001th place after the decimal point in the decimal expansion of \( \frac{9}{28} \) is **2**. |
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