| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to find the values of \( n \) for which the expression \( n^2 + 3n + 2 \) is divisible by 6. Given expression: \( n^2 + 3n + 2 \) We can factorize this expression as \( (n + 1)(n + 2) \). For \( (n + 1)(n + 2) \) to be divisible by 6, at least one of the factors must be divisible by 2 and at least one of the factors must be divisible by 3. Let's analyze the factors: 1. For \( n + 1 \) to be divisible by 2, \( n \) must be odd. There are 12 odd numbers between 1 and 25 (inclusive). 2. For \( n + 2 \) to be divisible by 2, \( n \) must be even. There are 12 even numbers between 1 and 25 (inclusive). 3. For \( n + 1 \) to be divisible by 3, \( n \) must be 2 or 5 modulo 3. There are 8 numbers that satisfy this condition between 1 and 25. 4. For \( n + 2 \) to be divisible by 3, \( n \) must be 1 or 4 modulo 3. There are 8 numbers that satisfy this condition between 1 and 25. Now, let's find the intersection of these conditions: 1. For \( n + 1 \) to be divisible by 6 (i.e., both 2 and 3), \( n \) must be 5 modulo 6. There are 4 numbers that satisfy this condition between 1 and 25. 2. For \( n + 2 \) to be divisible by 6 (i.e., both 2 and 3), \( n \) must be 4 modulo 6. There are 4 numbers that satisfy this condition between 1 and 25. To find the total number of integers \( n \) for which \( n^2 + 3n + 2 \) is divisible by 6, we need to consider the union of the sets of numbers that satisfy conditions 3 and 4. Since the sets of numbers that are 1 or 4 modulo 3 and 2 or 5 modulo 3 are disjoint, the total number of integers \( n \) for which \( n^2 + 3n + 2 \) is divisible by 6 is the sum of the numbers that satisfy conditions 3 and 4, minus the numbers that satisfy both conditions (to avoid double counting). Total number of integers \( n \) = 8 (condition 3) + 8 (condition 4) - 4 (both conditions) - 4 (both conditions) = 8. Therefore, the number of integers \( n \), 1 ≤ \( n \) ≤ 25, such that \( n^2 + 3n + 2 \) is divisible by 6 is 8. |
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| deepseekr1 | The expression \( n^2 + 3n + 2 \) factors into \((n + 1)(n + 2)\). Since \(n + 1\) and \(n + 2\) are consecutive integers, their product is always even, ensuring divisibility by 2. For divisibility by 6, we need the product to also be divisible by 3. This occurs when either \(n + 1\) or \(n + 2\) is divisible by 3: - \(n + 1 \equiv 0 \pmod{3}\) ⟹ \(n \equiv 2 \pmod{3}\) - \(n + 2 \equiv 0 \pmod{3}\) ⟹ \(n \equiv 1 \pmod{3}\) Thus, \(n\) must be congruent to **1** or **2** modulo 3. Between 1 and 25, numbers **not** divisible by 3 are those not congruent to 0 modulo 3. There are \( \left\lfloor \frac{25}{3} \right\rfloor = 8 \) multiples of 3 (3, 6, ..., 24). Therefore, the count of valid \(n\) is \(25 - 8 = 17\). **Answer:** \boxed{17} |
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| gpto1 | To solve this problem, we need to find the number of integers \( n \) in the range \( 1 \leq n \leq 25 \) such that \( n^2 + 3n + 2 \) is divisible by 6. First, note that for a number to be divisible by 6, it must be divisible by both 2 and 3. Therefore, we need to check the divisibility of \( n^2 + 3n + 2 \) by 2 and by 3 separately. **Divisibility by 2:** Let's consider \( n^2 + 3n + 2 \) modulo 2. - For any integer \( n \), \( n \mod 2 \) is either 0 or 1. **Case 1:** \( n \equiv 0 \mod 2 \) (n is even) - \( n^2 \equiv 0^2 \equiv 0 \mod 2 \) - \( 3n \equiv 3 \cdot 0 \equiv 0 \mod 2 \) - \( n^2 + 3n + 2 \equiv 0 + 0 + 2 \equiv 2 \mod 2 \) - Since \( 2 \equiv 0 \mod 2 \), the expression is divisible by 2. **Case 2:** \( n \equiv 1 \mod 2 \) (n is odd) - \( n^2 \equiv 1^2 \equiv 1 \mod 2 \) - \( 3n \equiv 3 \cdot 1 \equiv 1 \mod 2 \) - \( n^2 + 3n + 2 \equiv 1 + 1 + 2 \equiv 4 \mod 2 \) - Since \( 4 \equiv 0 \mod 2 \), the expression is also divisible by 2. **Conclusion:** \( n^2 + 3n + 2 \) is always even, regardless of \( n \). **Divisibility by 3:** Next, consider \( n^2 + 3n + 2 \) modulo 3. - For any integer \( n \), \( n \mod 3 \) is 0, 1, or 2. **Case 1:** \( n \equiv 0 \mod 3 \) - \( n^2 \equiv 0^2 \equiv 0 \mod 3 \) - \( 3n \equiv 3 \cdot 0 \equiv 0 \mod 3 \) - \( n^2 + 3n + 2 \equiv 0 + 0 + 2 \equiv 2 \mod 3 \) - The expression is not divisible by 3. **Case 2:** \( n \equiv 1 \mod 3 \) - \( n^2 \equiv 1^2 \equiv 1 \mod 3 \) - \( 3n \equiv 3 \cdot 1 \equiv 0 \mod 3 \) - \( n^2 + 3n + 2 \equiv 1 + 0 + 2 \equiv 3 \mod 3 \) - Since \( 3 \equiv 0 \mod 3 \), the expression is divisible by 3. **Case 3:** \( n \equiv 2 \mod 3 \) - \( n^2 \equiv 2^2 \equiv 4 \equiv 1 \mod 3 \) - \( 3n \equiv 3 \cdot 2 \equiv 0 \mod 3 \) - \( n^2 + 3n + 2 \equiv 1 + 0 + 2 \equiv 3 \mod 3 \) - The expression is divisible by 3. **Conclusion:** \( n^2 + 3n + 2 \) is divisible by 3 if \( n \not\equiv 0 \mod 3 \). **Combining both conditions:** - Since \( n^2 + 3n + 2 \) is always divisible by 2, we only need to consider when it's divisible by 3. - The expression is divisible by both 2 and 3 (hence by 6) when \( n \not\equiv 0 \mod 3 \). **Counting the numbers:** - Between 1 and 25, the numbers divisible by 3 are: \( 3, 6, 9, 12, 15, 18, 21, 24 \) (8 numbers). - Total numbers in the range: 25. - Numbers not divisible by 3: \( 25 - 8 = 17 \). **Therefore, there are 17 integers \( n \) between 1 and 25 such that \( n^2 + 3n + 2 \) is divisible by 6.** **Answer:** 17 |
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| GPT-4o | To solve the problem of finding the number of integers \( n \) in the range \( 1 \leq n \leq 25 \) such that \( n^2 + 3n + 2 \) is divisible by 6, we need to check the divisibility of the expression \( n^2 + 3n + 2 \) by 2 and 3, since 6 is the product of these two primes. ### Divisibility by 2 First, let's check whether \( n^2 + 3n + 2 \) is divisible by 2. An integer is divisible by 2 if it is even. - If \( n \) is even, \( n^2 \) is even, \( 3n \) is even, and \( n^2 + 3n + 2 \) is even (since the sum of even numbers is even). - If \( n \) is odd, \( n^2 \) is odd, \( 3n \) is odd, and \( n^2 + 3n + 2 \) is even (since the sum of two odd numbers is even, and adding 2 to an even number remains even). Therefore, \( n^2 + 3n + 2 \) is always even regardless of whether \( n \) is even or odd. ### Divisibility by 3 Next, let's check whether \( n^2 + 3n + 2 \) is divisible by 3. We will consider the possible remainders when \( n \) is divided by 3, i.e., \( n \equiv 0 \pmod{3} \), \( n \equiv 1 \pmod{3} \), or \( n \equiv 2 \pmod{3} \). - If \( n \equiv 0 \pmod{3} \), then \( n = 3k \) for some integer \( k \): \[ n^2 + 3n + 2 = (3k)^2 + 3(3k) + 2 = 9k^2 + 9k + 2 \equiv 2 \pmod{3} \] - If \( n \equiv 1 \pmod{3} \), then \( n = 3k + 1 \): \[ n^2 + 3n + 2 = (3k + 1)^2 + 3(3k + 1) + 2 = 9k^2 + 6k + 1 + 9k + 3 + 2 = 9k^2 + 15k + 6 \equiv 0 \pmod{3} \] - If \( n \equiv 2 \pmod{3} \), then \( n = 3k + 2 \): \[ n^2 + 3n + 2 = (3k + 2)^2 + 3(3k + 2) + 2 = 9k^2 + 12k + 4 + 9k + 6 + 2 = 9k^2 + 21k + 12 \equiv 0 \pmod{3} \] Thus, \( n^2 + 3n + 2 \) is divisible by 3 if \( n \equiv 1 \pmod{3} \) or \( n \equiv 2 \pmod{3} \). ### Conclusion We need to count the values of \( n \) from 1 to 25 that are congruent to 1 or 2 modulo 3: - \( n \equiv 1 \pmod{3} \): \( 1, 4, 7, 10, 13, 16, 19, 22, 25 \) (9 values) - \( n \equiv 2 \pmod{3} \): \( 2, 5, 8, 11, 14, 17, 20, 23 \) (8 values) Summing these, we get 9 + 8 = 17 values of \( n \) for which \( n^2 + 3n + 2 \) is divisible by 6. Therefore, there are \( \boxed{17} \) integers \( n \) in the range \( 1 \leq n \leq 25 \) such that \( n^2 + 3n + 2 \) is divisible by 6. |
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| llama3.1-70B-Instruct | To find the number of integers n, 1 ≤ n ≤ 25 such that n^2 + 3n + 2 is divisible by 6, we can use the following approach: 1. Factorize the expression n^2 + 3n + 2: n^2 + 3n + 2 = (n + 1)(n + 2) 2. Since we want the expression to be divisible by 6, we need to find the values of n for which (n + 1)(n + 2) is divisible by 6. 3. Note that 6 = 2 × 3, so we need to find the values of n for which (n + 1)(n + 2) is divisible by both 2 and 3. 4. Since (n + 1)(n + 2) is the product of two consecutive integers, one of them must be even, so (n + 1)(n + 2) is always divisible by 2. 5. Now, we need to find the values of n for which (n + 1)(n + 2) is divisible by 3. This happens when either (n + 1) or (n + 2) is divisible by 3. 6. Let's count the number of values of n for which (n + 1) is divisible by 3: n + 1 = 3, 6, 9, 12, 15, 18, 21, 24 There are 8 values of n for which (n + 1) is divisible by 3. 7. Now, let's count the number of values of n for which (n + 2) is divisible by 3: n + 2 = 3, 6, 9, 12, 15, 18, 21, 24 There are 8 values of n for which (n + 2) is divisible by 3. 8. However, we have counted some values of n twice, so we need to subtract the number of values of n for which both (n + 1) and (n + 2) are divisible by 3. 9. The values of n for which both (n + 1) and (n + 2) are divisible by 3 are: n + 1 = 3, 6, 9, 12, 15, 18, 21, 24 n + 2 = 3, 6, 9, 12, 15, 18, 21, 24 There are 8 values of n for which both (n + 1) and (n + 2) are divisible by 3. 10. Therefore, the total number of values of n for which (n + 1)(n + 2) is divisible by 6 is: 8 + 8 - 8 = 8 So, there are 8 integers n, 1 ≤ n ≤ 25 such that n^2 + 3n + 2 is divisible by 6. |
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| llama3.1-405B-Instruct | To find the number of integers n, 1 ≤ n ≤ 25, such that n^2 + 3n + 2 is divisible by 6, we can use the following approach: 1. Factorize the expression n^2 + 3n + 2: n^2 + 3n + 2 = (n + 1)(n + 2) 2. Since we want the expression to be divisible by 6, we need to find the values of n for which (n + 1)(n + 2) is divisible by 6. 3. We can start by finding the values of n for which (n + 1) is divisible by 2 and (n + 2) is divisible by 3, or vice versa. 4. For (n + 1) to be divisible by 2, n must be odd. For (n + 2) to be divisible by 3, n must be of the form 3k - 2, where k is an integer. 5. Similarly, for (n + 1) to be divisible by 3, n must be of the form 3k - 1, where k is an integer. For (n + 2) to be divisible by 2, n must be even. 6. Now, we can list out the values of n that satisfy these conditions: For (n + 1) divisible by 2 and (n + 2) divisible by 3: * n = 1 (1 + 1 = 2, 1 + 2 = 3) * n = 7 (7 + 1 = 8, 7 + 2 = 9) * n = 13 (13 + 1 = 14, 13 + 2 = 15) * n = 19 (19 + 1 = 20, 19 + 2 = 21) * n = 25 (25 + 1 = 26, 25 + 2 = 27) For (n + 1) divisible by 3 and (n + 2) divisible by 2: * n = 2 (2 + 1 = 3, 2 + 2 = 4) * n = 8 (8 + 1 = 9, 8 + 2 = 10) * n = 14 (14 + 1 = 15, 14 + 2 = 16) * n = 20 (20 + 1 = 21, 20 + 2 = 22) 7. There are 9 values of n that satisfy the conditions. Therefore, the number of integers n, 1 ≤ n ≤ 25, such that n^2 + 3n + 2 is divisible by 6 is 9. |
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| DeepSeek-Coder-V2 | To solve the problem of finding the number of integers \( n \) such that \( 1 \leq n \leq 25 \) and \( n^2 + 3n + 2 \) is divisible by 6, we can follow these steps: 1. **Rewrite the expression**: \[ n^2 + 3n + 2 \] We can factor this expression: \[ n^2 + 3n + 2 = (n+1)(n+2) \] This shows that \( n^2 + 3n + 2 \) is the product of two consecutive integers, \( n+1 \) and \( n+2 \). 2. **Divisibility by 6**: For \( (n+1)(n+2) \) to be divisible by 6, it must be divisible by both 2 and 3. 3. **Divisibility by 2**: Since \( n+1 \) and \( n+2 \) are consecutive integers, one of them must be even. Therefore, \( (n+1)(n+2) \) is always divisible by 2. 4. **Divisibility by 3**: Among any three consecutive integers, one of them must be divisible by 3. Therefore, \( n \), \( n+1 \), and \( n+2 \) include one number that is divisible by 3. 5. **Check the range**: We need to check which values of \( n \) in the range \( 1 \leq n \leq 25 \) make \( n^2 + 3n + 2 \) divisible by 6. Since \( (n+1)(n+2) \) is always divisible by 2 and among any three consecutive integers, one is divisible by 3, we only need to ensure that \( n \) itself is not causing a conflict. 6. **Identify valid \( n \)**: We need to exclude \( n \) such that \( n \equiv 2 \pmod{3} \) because in this case, \( n+1 \equiv 0 \pmod{3} \) and \( n+2 \equiv 1 \pmod{3} \), making \( (n+1)(n+2) \) not divisible by 6. The values of \( n \) in the range \( 1 \leq n \leq 25 \) that are \( 2 \pmod{3} \) are: \[ 2, 5, 8, 11, 14, 17, 20, 23, 26 \] Since \( 26 \) is out of our range, we have: \[ 2, 5, 8, 11, 14, 17, 20, 23 \] There are 8 such values. 7. **Count valid \( n \)**: The total number of integers from 1 to 25 is 25. Subtracting the 8 invalid values, we get: \[ 25 - 8 = 17 \] Therefore, the number of integers \( n \) such that \( 1 \leq n \leq 25 \) and \( n^2 + 3n + 2 \) is divisible by 6 is: \[ \boxed{17} \] |
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| Qwen2-72B-Instruct | To solve this problem, we need to find the values of n for which the expression n^2 + 3n + 2 is divisible by 6. Given expression: n^2 + 3n + 2 We can factorize this expression as (n + 1)(n + 2). For (n + 1)(n + 2) to be divisible by 6, at least one of the factors must be divisible by 2 and at least one of the factors must be divisible by 3. Let's analyze the factors: 1. For n + 1 to be divisible by 2, n must be odd. There are 12 odd numbers between 1 and 25 (inclusive). 2. For n + 2 to be divisible by 2, n must be even. There are 12 even numbers between 1 and 25 (inclusive). 3. For n + 1 to be divisible by 3, n must be 2 or 5 modulo 3. There are 8 numbers that satisfy this condition between 1 and 25. 4. For n + 2 to be divisible by 3, n must be 1 or 4 modulo 3. There are 8 numbers that satisfy this condition between 1 and 25. Now, let's find the intersection of these conditions: 1. For n + 1 to be divisible by 6 (i.e., both 2 and 3), n must be 5 modulo 6. There are 4 numbers that satisfy this condition between 1 and 25. 2. For n + 2 to be divisible by 6 (i.e., both 2 and 3), n must be 4 modulo 6. There are 4 numbers that satisfy this condition between 1 and 25. To find the total number of integers n for which n^2 + 3n + 2 is divisible by 6, we need to consider the union of the sets of numbers that satisfy conditions 3 and 4. Since the sets of numbers that are 1 or 4 modulo 3 and 2 or 5 modulo 3 are disjoint, the total number of integers n for which n^2 + 3n + 2 is divisible by 6 is the sum of the numbers that satisfy conditions 3 and 4, minus the numbers that satisfy both conditions (to avoid double counting). Total number of integers n = 8 (condition 3) + 8 (condition 4) - 4 (both conditions) - 4 (both conditions) = 8. Therefore, the number of integers n, 1 ≤ n ≤ 25, such that n^2 + 3n + 2 is divisible by 6 is 8. Note: The final answer should be 8, but there seems to be a mistake in the calculation. The correct calculation should be: Total number of integers n = 8 (condition 3) + 8 (condition 4) - 4 (n = 5 modulo 6) - 4 (n = 4 modulo 6) = 8 + 8 - 4 - 4 = 8. So, the correct answer is 8. |
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| baichuan4 | To find the number of integers \(n\), where \(1 \leq n \leq 25\), such that \(n^2 + 3n + 2\) is divisible by 6, we can break down the problem into two parts: 1. Find the values of \(n\) for which \(n^2 + 3n + 2\) is divisible by 3. 2. Find the values of \(n\) for which \(n^2 + 3n + 2\) is divisible by 2. If a number is divisible by both 3 and 2, then it is divisible by 6 (since 3 and 2 are coprime). Let's start by finding the values of \(n\) for which \(n^2 + 3n + 2\) is divisible by 3: \[n^2 + 3n + 2 \equiv 0 \pmod{3}\] Expanding and simplifying modulo 3: \[n^2 + (3 \cdot n) + 2 \equiv n^2 + 0 \cdot n + 2 \pmod{3}\] \[n^2 + 2 \equiv 0 \pmod{3}\] \[n^2 \equiv 1 \pmod{3}\] The solutions to this congruence are \(n \equiv \pm1 \pmod{3}\), which means \(n = 1, 2\) (mod 3) within our range. Now, let's find the values of \(n\) for which \(n^2 + 3n + 2\) is divisible by 2: \[n^2 + 3n + 2 \equiv 0 \pmod{2}\] Expanding and simplifying modulo 2: \[n^2 + (3 \cdot n) + 2 \equiv n^2 + n + 0 \pmod{2}\] \[n^2 + n \equiv 0 \pmod{2}\] The solutions to this congruence are \(n \equiv 0, 1 \pmod{2}\), which means \(n = 0, 1\) (mod 2) within our range. Now, we need to find the integers \(n\) between 1 and 25 that satisfy both conditions. We can do this by finding the integers that are congruent to either 1 or 2 modulo 3 and are also even (since they need to be divisible by 2). The integers in the range [1, 25] that are congruent to 1 or 2 modulo 3 and are even are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24. There are 12 such integers. Therefore, there are 12 integers \(n\), where \(1 \leq n \leq 25\), such that \(n^2 + 3n + 2\) is divisible by 6. |
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| doubao | Factorize n^2 + 3n + 2 as (n + 1)(n + 2). For a number to be divisible by 6, it must be divisible by both 2 and 3. An integer is divisible by 2 if it is even. So, either n + 1 is even or n + 2 is even. This is true for all integers n. An integer is divisible by 3 if the sum of its digits is divisible by 3. Consider the cases: - If n = 1, (n + 1)(n + 2) = 2 × 3 = 6 which is divisible by 3. - If n = 2, (n + 1)(n + 2) = 3 × 4 = 12 which is divisible by 3. - If n = 4, (n + 1)(n + 2) = 5 × 6 = 30 which is divisible by 3. - If n = 5, (n + 1)(n + 2) = 6 × 7 = 42 which is divisible by 3. - If n = 7, (n + 1)(n + 2) = 8 × 9 = 72 which is divisible by 3. - If n = 8, (n + 1)(n + 2) = 9 × 10 = 90 which is divisible by 3. - If n = 10, (n + 1)(n + 2) = 11 × 12 = 132 which is divisible by 3. - If n = 11, (n + 1)(n + 2) = 12 × 13 = 156 which is divisible by 3. - If n = 13, (n + 1)(n + 2) = 14 × 15 = 210 which is divisible by 3. - If n = 14, (n + 1)(n + 2) = 15 × 16 = 240 which is divisible by 3. - If n = 16, (n + 1)(n + 2) = 17 × 18 = 306 which is divisible by 3. - If n = 19, (n + 1)(n + 2) = 20 × 21 = 420 which is divisible by 3. - If n = 20, (n + 1)(n + 2) = 21 × 22 = 462 which is divisible by 3. - If n = 22, (n + 1)(n + 2) = 23 × 24 = 552 which is divisible by 3. - If n = 23, (n + 1)(n + 2) = 24 × 25 = 600 which is divisible by 3. So, there are 16 integers n such that n^2 + 3n + 2 is divisible by 6. |
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| minimax | To find the number of integers \( n \) such that \( n^2 + 3n + 2 \) is divisible by 6, we need to analyze the expression modulo 6. First, let's factor the quadratic expression: \( n^2 + 3n + 2 = (n + 1)(n + 2) \) Now, let's consider the possible remainders when \( n \) is divided by 6, since we are looking for \( n \) in the range \( 1 \leq n \leq 25 \). The possible remainders are 0, 1, 2, 3, 4, or 5. We can now check each case modulo 6: 1. If \( n \equiv 0 \mod 6 \), then \( n = 6k \) for some integer \( k \). In this case, \( n^2 + 3n + 2 = (6k)^2 + 3(6k) + 2 = 36k^2 + 18k + 2 \). Since \( 36k^2 \) and \( 18k \) are both divisible by 6, the expression will be divisible by 6. 2. If \( n \equiv 1 \mod 6 \), then \( n = 6k + 1 \). In this case, \( n^2 + 3n + 2 = (6k + 1)^2 + 3(6k + 1) + 2 = 36k^2 + 12k + 1 + 18k + 3 + 2 = 36k^2 + 30k + 6 \). This is also divisible by 6. 3. If \( n \equiv 2 \mod 6 \), then \( n = 6k + 2 \). In this case, \( n^2 + 3n + 2 = (6k + 2)^2 + 3(6k + 2) + 2 = 36k^2 + 24k + 4 + 18k + 6 + 2 = 36k^2 + 42k + 12 \). This is divisible by 6. 4. If \( n \equiv 3 \mod 6 \), then \( n = 6k + 3 \). In this case, \( n^2 + 3n + 2 = (6k + 3)^2 + 3(6k + 3) + 2 = 36k^2 + 36k + 9 + 18k + 9 + 2 = 36k^2 + 54k + 20 \). This is not divisible by 6 because the last term, 20, is not divisible by 6. 5. If \( n \equiv 4 \mod 6 \), then \( n = 6k + 4 \). In this case, \( n^2 + 3n + 2 = (6k + 4)^2 + 3(6k + 4) + 2 = 36k^2 + 48k + 16 + 18k + 12 + 2 = 36k^2 + 66k + 30 \). This is divisible by 6. 6. If \( n \equiv 5 \mod 6 \), then \( n = 6k + 5 \). In this case, \( n^2 + 3n + 2 = (6k + 5)^2 + 3(6k + 5) + 2 = 36k^2 + 60k + 25 + 18k + 15 + 2 = 36k^2 + 78k + 42 \). This is divisible by 6. Now, we need to count the number of integers \( n \) in the range \( 1 \leq n \leq 25 \) that fall into the categories where \( n^2 + 3n + 2 \) is divisible by 6. - For \( n \equiv 0 \mod 6 \), the possible values are 6, 12, 18, and 24. - For \( n \equiv 1 \mod 6 \), the possible values are 7, 13, 19, and 25. - For \( n \equiv 2 \mod 6 \), the possible values are 8, 14, and 20. - For \( n \equiv 3 \mod 6 \), there are no possible values in the range \( 1 \leq n \leq 25 \). - For \( n \equiv 4 \mod 6 \), the possible values are 10, 16, and 22. - For \( n \equiv 5 \mod 6 \), the possible values are 11, 17, and 23. Counting these up, we have: - 4 values for \( n \equiv 0 \mod 6 \) - 4 values for \( n \equiv 1 \mod 6 \) - 3 values for \( n \equiv 2 \mod 6 \) - 0 values for \( n \equiv 3 \mod 6 \) - 3 values for \( n \equiv 4 \mod 6 \) - 3 values for \( n \equiv 5 \mod 6 \) Adding these up gives us a total of \( 4 + 4 + 3 + 0 + 3 + 3 = 17 \) integers \( n \) such that \( n^2 + 3n + 2 \) is divisible by 6. |
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| kimi | To find the number of integers \( n \) such that \( n^2 + 3n + 2 \) is divisible by 6, we can first simplify the expression and then check for divisibility by 6. Let's simplify the expression: \[ n^2 + 3n + 2 \] We can factor the quadratic expression: \[ n^2 + 3n + 2 = (n + 1)(n + 2) \] Now, we want to find \( n \) such that \( (n + 1)(n + 2) \) is divisible by 6. Since 6 is the product of 2 and 3, the product \( (n + 1)(n + 2) \) must be divisible by both 2 and 3. Let's consider the divisibility by 2 first: - If \( n \) is even, then \( n + 1 \) is odd and \( n + 2 \) is even, so the product is divisible by 2. - If \( n \) is odd, then \( n + 1 \) is even and \( n + 2 \) is odd, so the product is also divisible by 2. Now, let's consider the divisibility by 3: - If \( n \) is a multiple of 3, then \( n + 1 \) is not a multiple of 3, but \( n + 2 \) is a multiple of 3, so the product is divisible by 3. - If \( n \) is one less than a multiple of 3, then \( n + 1 \) is a multiple of 3, and \( n + 2 \) is not a multiple of 3, so the product is divisible by 3. - If \( n \) is two less than a multiple of 3, then neither \( n + 1 \) nor \( n + 2 \) is a multiple of 3, so the product is not divisible by 3. Now, we can check each integer \( n \) from 1 to 25 to see if it satisfies the conditions: - \( n = 1 \): \( (1 + 1)(1 + 2) = 2 \cdot 3 = 6 \) (divisible by 6) - \( n = 2 \): \( (2 + 1)(2 + 2) = 3 \cdot 4 = 12 \) (divisible by 6) - \( n = 3 \): \( (3 + 1)(3 + 2) = 4 \cdot 5 = 20 \) (not divisible by 6) - \( n = 4 \): \( (4 + 1)(4 + 2) = 5 \cdot 6 = 30 \) (divisible by 6) - \( n = 5 \): \( (5 + 1)(5 + 2) = 6 \cdot 7 = 42 \) (divisible by 6) - \( n = 6 \): \( (6 + 1)(6 + 2) = 7 \cdot 8 = 56 \) (not divisible by 6) - ... - We continue this process until \( n = 25 \). To make this process more efficient, we can use the fact that \( n \) and \( n + 2 \) are always consecutive integers, so one of them will always be even, ensuring divisibility by 2. We only need to check for divisibility by 3. We can list the multiples of 3 between 1 and 25: - \( 3, 6, 9, 12, 15, 18, 21, 24 \) And we can list the numbers that are one less than a multiple of 3: - \( 2, 5, 8, 11, 14, 17, 20, 23 \) Now, we combine the lists, excluding duplicates: - \( 1, 2, 4, 5, 8, 11, 14, 17, 20, 23, 24 \) These are the values of \( n \) for which \( n^2 + 3n + 2 \) is divisible by 6. There are 11 such integers. |
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| wenxinyiyan | 首先，我们将给定的表达式$n^2 + 3n + 2$进行因式分解，得到$(n+1)(n+2)$。这意味着$n^2 + 3n + 2 = (n+1)(n+2)$。 接下来，我们考虑$(n+1)(n+2)$被6整除的条件。由于$n+1$和$n+2$是连续的两个整数，它们之间必然有一个是偶数。而6是2和3的乘积，所以我们需要考虑以下两种情况： 1. 当$n+1$或$n+2$中有一个是3的倍数时，$(n+1)(n+2)$就能被6整除。在1到25的范围内，3的倍数有8个，分别是3, 6, 9, 12, 15, 18, 21, 24。但我们需要排除3和25，因为当$n=3$时，$n+2=5$不是3的倍数；当$n=25$时，$n+1=26$不是3的倍数。所以，这种情况下有6个整数满足条件。 2. 当$n+1$和$n+2$中有一个是2的倍数且另一个是3的倍数时，$(n+1)(n+2)$也能被6整除。在1到25的范围内，这样的组合有(4,5), (5,6), (10,11), (11,12), (16,17), (17,18), (22,23), (23,24)。这些组合都满足条件，所以有8个整数满足条件。 综上所述，满足条件的整数共有$6+8=14$个。 |
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| zhipu | To find the number of integers \( n \) such that \( 1 \leq n \leq 25 \) and \( n^2 + 3n + 2 \) is divisible by 6, we need to check when the expression \( n^2 + 3n + 2 \) yields a remainder of 0 when divided by 6. First, let's factor the quadratic expression: \[ n^2 + 3n + 2 = (n + 1)(n + 2) \] Now we need to determine when \( (n + 1)(n + 2) \) is divisible by 6. Since 6 is the product of 2 and 3, \( (n + 1)(n + 2) \) must be divisible by both 2 and 3. ### Divisibility by 2: - For \( (n + 1)(n + 2) \) to be divisible by 2, at least one of \( n + 1 \) or \( n + 2 \) must be even. - Since \( n \) and \( n + 2 \) have different parities (one is even, the other is odd), \( (n + 1)(n + 2) \) is always even, so it is always divisible by 2. ### Divisibility by 3: - For \( (n + 1)(n + 2) \) to be divisible by 3, at least one of \( n + 1 \) or \( n + 2 \) must be divisible by 3. - This means \( n \) must not be divisible by 3 (because if \( n \) were divisible by 3, neither \( n + 1 \) nor \( n + 2 \) would be). Let's count the integers \( n \) from 1 to 25 that are not divisible by 3: - The numbers divisible by 3 in this range are: 3, 6, 9, 12, 15, 18, 21, 24. - There are 8 such numbers. Since there are 25 numbers in total from 1 to 25, the number of integers \( n \) that are not divisible by 3 is: \[ 25 - 8 = 17 \] Thus, there are 17 integers \( n \) such that \( 1 \leq n \leq 25 \) and \( n^2 + 3n + 2 \) is divisible by 6. ### Verification: We can verify this by checking specific values: - For \( n = 1 \): \( (1 + 1)(1 + 2) = 2 \cdot 3 = 6 \) (divisible by 6) - For \( n = 2 \): \( (2 + 1)(2 + 2) = 3 \cdot 4 = 12 \) (divisible by 6) - For \( n = 3 \): \( (3 + 1)(3 + 2) = 4 \cdot 5 = 20 \) (not divisible by 6) - Continue this pattern, skipping \( n \) values that are multiples of 3. This confirms our approach is correct. ### Final Answer: The number of integers \( n \) such that \( 1 \leq n \leq 25 \) and \( n^2 + 3n + 2 \) is divisible by 6 is \( \boxed{17} \). |
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| gpto1 | To solve this problem, we need to find the number of integers \( n \) in the range \( 1 \leq n \leq 25 \) such that \( n^2 + 3n + 2 \) is divisible by 6. First, note that for a number to be divisible by 6, it must be divisible by both 2 and 3. Therefore, we need to check the divisibility of \( n^2 + 3n + 2 \) by 2 and by 3 separately. **Divisibility by 2:** Let's consider \( n^2 + 3n + 2 \) modulo 2. - For any integer \( n \), \( n \mod 2 \) is either 0 or 1. **Case 1:** \( n \equiv 0 \mod 2 \) (n is even) - \( n^2 \equiv 0^2 \equiv 0 \mod 2 \) - \( 3n \equiv 3 \cdot 0 \equiv 0 \mod 2 \) - \( n^2 + 3n + 2 \equiv 0 + 0 + 2 \equiv 2 \mod 2 \) - Since \( 2 \equiv 0 \mod 2 \), the expression is divisible by 2. **Case 2:** \( n \equiv 1 \mod 2 \) (n is odd) - \( n^2 \equiv 1^2 \equiv 1 \mod 2 \) - \( 3n \equiv 3 \cdot 1 \equiv 1 \mod 2 \) - \( n^2 + 3n + 2 \equiv 1 + 1 + 2 \equiv 4 \mod 2 \) - Since \( 4 \equiv 0 \mod 2 \), the expression is also divisible by 2. **Conclusion:** \( n^2 + 3n + 2 \) is always even, regardless of \( n \). **Divisibility by 3:** Next, consider \( n^2 + 3n + 2 \) modulo 3. - For any integer \( n \), \( n \mod 3 \) is 0, 1, or 2. **Case 1:** \( n \equiv 0 \mod 3 \) - \( n^2 \equiv 0^2 \equiv 0 \mod 3 \) - \( 3n \equiv 3 \cdot 0 \equiv 0 \mod 3 \) - \( n^2 + 3n + 2 \equiv 0 + 0 + 2 \equiv 2 \mod 3 \) - The expression is not divisible by 3. **Case 2:** \( n \equiv 1 \mod 3 \) - \( n^2 \equiv 1^2 \equiv 1 \mod 3 \) - \( 3n \equiv 3 \cdot 1 \equiv 0 \mod 3 \) - \( n^2 + 3n + 2 \equiv 1 + 0 + 2 \equiv 3 \mod 3 \) - Since \( 3 \equiv 0 \mod 3 \), the expression is divisible by 3. **Case 3:** \( n \equiv 2 \mod 3 \) - \( n^2 \equiv 2^2 \equiv 4 \equiv 1 \mod 3 \) - \( 3n \equiv 3 \cdot 2 \equiv 0 \mod 3 \) - \( n^2 + 3n + 2 \equiv 1 + 0 + 2 \equiv 3 \mod 3 \) - The expression is divisible by 3. **Conclusion:** \( n^2 + 3n + 2 \) is divisible by 3 if \( n \not\equiv 0 \mod 3 \). **Combining both conditions:** - Since \( n^2 + 3n + 2 \) is always divisible by 2, we only need to consider when it's divisible by 3. - The expression is divisible by both 2 and 3 (hence by 6) when \( n \not\equiv 0 \mod 3 \). **Counting the numbers:** - Between 1 and 25, the numbers divisible by 3 are: \( 3, 6, 9, 12, 15, 18, 21, 24 \) (8 numbers). - Total numbers in the range: 25. - Numbers not divisible by 3: \( 25 - 8 = 17 \). **Therefore, there are 17 integers \( n \) between 1 and 25 such that \( n^2 + 3n + 2 \) is divisible by 6.** **Answer:** 17 |
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| deepseekv3 | To solve the problem, we need to find the number of integers \( n \) within the range \( 1 \leq n \leq 25 \) such that the expression \( n^2 + 3n + 2 \) is divisible by 6. ### Step 1: Factor the Quadratic Expression First, let's factor the quadratic expression: \[ n^2 + 3n + 2 = (n + 1)(n + 2) \] So, \( n^2 + 3n + 2 \) is the product of two consecutive integers, \( n + 1 \) and \( n + 2 \). ### Step 2: Analyze Divisibility by 6 For the product \( (n + 1)(n + 2) \) to be divisible by 6, it must be divisible by both 2 and 3. 1. **Divisibility by 2**: Since \( n + 1 \) and \( n + 2 \) are two consecutive integers, one of them must be even. Therefore, the product \( (n + 1)(n + 2) \) is always divisible by 2. 2. **Divisibility by 3**: Among any three consecutive integers, one must be divisible by 3. So, in the sequence \( n, n + 1, n + 2 \), one of these three numbers is divisible by 3. Since \( n \) is not in the product \( (n + 1)(n + 2) \), we need to ensure that either \( n + 1 \) or \( n + 2 \) is divisible by 3. ### Step 3: Determine When the Product is Divisible by 3 The product \( (n + 1)(n + 2) \) is divisible by 3 unless \( n \) leaves a remainder of 1 when divided by 3. Specifically: - If \( n \equiv 0 \pmod{3} \), then \( n + 1 \equiv 1 \pmod{3} \) and \( n + 2 \equiv 2 \pmod{3} \). Neither \( n + 1 \) nor \( n + 2 \) is divisible by 3, so the product is **not** divisible by 3. - If \( n \equiv 1 \pmod{3} \), then \( n + 1 \equiv 2 \pmod{3} \) and \( n + 2 \equiv 0 \pmod{3} \). Here, \( n + 2 \) is divisible by 3, so the product is divisible by 3. - If \( n \equiv 2 \pmod{3} \), then \( n + 1 \equiv 0 \pmod{3} \) and \( n + 2 \equiv 1 \pmod{3} \). Here, \( n + 1 \) is divisible by 3, so the product is divisible by 3. Thus, the product \( (n + 1)(n + 2) \) is divisible by 3 unless \( n \equiv 0 \pmod{3} \). ### Step 4: Count the Valid \( n \) in the Range \( 1 \leq n \leq 25 \) We need to count the integers \( n \) in the range \( 1 \leq n \leq 25 \) such that \( n \not\equiv 0 \pmod{3} \). There are 25 integers in total. The number of integers divisible by 3 in this range is \( \left\lfloor \frac{25}{3} \right\rfloor = 8 \). Therefore, the number of integers not divisible by 3 is \( 25 - 8 = 17 \). ### Step 5: Conclusion Since the product \( (n + 1)(n + 2) \) is always divisible by 2 and is divisible by 3 unless \( n \equiv 0 \pmod{3} \), the number of integers \( n \) in the range \( 1 \leq n \leq 25 \) such that \( n^2 + 3n + 2 \) is divisible by 6 is **17**. ### Final Answer: \[ \boxed{17} \] |
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