| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The given problem can be solved using the Chinese Remainder Theorem, but for this specific problem, we can also solve it through inspection or by creating a system of congruences. The system of congruences is: 1. \(x \equiv 1 \mod 2\) 2. \(x \equiv 2 \mod 3\) 3. \(x \equiv 3 \mod 4\) 4. \(x \equiv 4 \mod 5\) We are looking for the smallest positive integer \(x\) that satisfies all four congruences. Let's solve this step by step: Step 1: Find an integer that satisfies the first two congruences, \(x \equiv 1 \mod 2\) and \(x \equiv 2 \mod 3\). The first few numbers that satisfy \(x \equiv 1 \mod 2\) are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, ... The first few numbers that satisfy \(x \equiv 2 \mod 3\) are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, ... The smallest number that satisfies both congruences is 5. Step 2: Find an integer that satisfies the first three congruences, \(x \equiv 1 \mod 2\), \(x \equiv 2 \mod 3\), and \(x \equiv 3 \mod 4\). We can use the number we found in Step 1 (5) as a starting point and look for a number that also satisfies the third congruence. The first few numbers that satisfy \(x \equiv 3 \mod 4\) are 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, ... The smallest number that satisfies all three congruences is 35. Step 3: Find an integer that satisfies all four congruences. Using the number we found in Step 2 (35) as a starting point, we look for a number that also satisfies the fourth congruence, \(x \equiv 4 \mod 5\). The first few numbers that satisfy \(x \equiv 4 \mod 5\) are 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 89, 94, 99, ... The smallest number that satisfies all four congruences is 59. Therefore, the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5 is 59. |
59 |
| deepseekr1 | The smallest positive integer satisfying all the given conditions is found by observing that each remainder is one less than its divisor. Thus, the number \( N \) must be one less than a common multiple of 2, 3, 4, and 5. The least common multiple (LCM) of these numbers is 60. Therefore, \( N = 60 - 1 = 59 \). **Verification:** - \( 59 \div 2 = 29 \) remainder \( 1 \) - \( 59 \div 3 = 19 \) remainder \( 2 \) - \( 59 \div 4 = 14 \) remainder \( 3 \) - \( 59 \div 5 = 11 \) remainder \( 4 \) All conditions are satisfied. **Answer:** \(\boxed{59}\) |
59 |
| gpto1 | To solve the problem, we are asked to find the smallest positive integer \( N \) such that: 1. \( N \equiv 1 \ (\text{mod}\ 2) \) 2. \( N \equiv 2 \ (\text{mod}\ 3) \) 3. \( N \equiv 3 \ (\text{mod}\ 4) \) 4. \( N \equiv 4 \ (\text{mod}\ 5) \) We can solve these congruences step by step using the Chinese Remainder Theorem (CRT). **Step 1: Combine the first two congruences** First, we have: \[ N \equiv 1 \ (\text{mod}\ 2) \] This means \( N = 2k + 1 \) for some integer \( k \). Next, we have: \[ N \equiv 2 \ (\text{mod}\ 3) \] Substituting \( N = 2k +1 \): \[ 2k + 1 \equiv 2 \ (\text{mod}\ 3) \] Subtract 1 from both sides: \[ 2k \equiv 1 \ (\text{mod}\ 3) \] Multiply both sides by the modular inverse of 2 modulo 3. Since \( 2 \times 2 = 4 \equiv 1 \ (\text{mod}\ 3) \), the inverse of 2 modulo 3 is 2. Multiply both sides by 2: \[ 2 \times 2k \equiv 2 \times 1 \ (\text{mod}\ 3) \] \[ 4k \equiv 2 \ (\text{mod}\ 3) \] Since \( 4k \equiv k \ (\text{mod}\ 3) \) (because \( 4 \equiv 1 \ (\text{mod}\ 3) \)): \[ k \equiv 2 \ (\text{mod}\ 3) \] Set \( k = 3m + 2 \). Go back to \( N = 2k + 1 \): \[ N = 2(3m + 2) + 1 = 6m + 5 \] **Step 2: Incorporate the third congruence** Now consider: \[ N \equiv 3 \ (\text{mod}\ 4) \] Substitute \( N = 6m + 5 \): \[ 6m + 5 \equiv 3 \ (\text{mod}\ 4) \] Compute \( 6m + 5 \) modulo 4: \[ (6m + 5) \ (\text{mod}\ 4) = (2m + 1) \ (\text{mod}\ 4) \] Set: \[ 2m + 1 \equiv 3 \ (\text{mod}\ 4) \] Subtract 1 from both sides: \[ 2m \equiv 2 \ (\text{mod}\ 4) \] Divide both sides by 2: \[ m \equiv 1 \ (\text{mod}\ 2) \] Set \( m = 2n + 1 \). Update \( N \): \[ N = 6(2n + 1) + 5 = 12n + 11 \] **Step 3: Incorporate the fourth congruence** Now consider: \[ N \equiv 4 \ (\text{mod}\ 5) \] Substitute \( N = 12n + 11 \): \[ 12n + 11 \equiv 4 \ (\text{mod}\ 5) \] Compute \( 12n + 11 \) modulo 5: \[ (12n + 11) \ (\text{mod}\ 5) = (2n + 1) \ (\text{mod}\ 5) \] (Because \( 12 \equiv 2 \ (\text{mod}\ 5) \) and \( 11 \equiv 1 \ (\text{mod}\ 5) \)) Set: \[ 2n + 1 \equiv 4 \ (\text{mod}\ 5) \] Subtract 1 from both sides: \[ 2n \equiv 3 \ (\text{mod}\ 5) \] Multiply both sides by the modular inverse of 2 modulo 5. Since \( 2 \times 3 = 6 \equiv 1 \ (\text{mod}\ 5) \), the inverse of 2 modulo 5 is 3. Multiply both sides by 3: \[ 3 \times 2n \equiv 3 \times 3 \ (\text{mod}\ 5) \] \[ 6n \equiv 9 \ (\text{mod}\ 5) \] Since \( 6n \equiv n \ (\text{mod}\ 5) \): \[ n \equiv 4 \ (\text{mod}\ 5) \] Set \( n = 5p + 4 \). Update \( N \): \[ N = 12(5p + 4) + 11 = 60p + 48 + 11 = 60p + 59 \] **Step 4: Find the smallest positive integer** The smallest positive integer occurs when \( p = 0 \): \[ N = 60(0) + 59 = 59 \] **Verification** Check that \( N = 59 \) satisfies all original congruences: 1. \( 59 \equiv 1 \ (\text{mod}\ 2) \) ✔️ 2. \( 59 \equiv 2 \ (\text{mod}\ 3) \) (since \( 59 \div 3 = 19 \text{ remainder } 2 \)) ✔️ 3. \( 59 \equiv 3 \ (\text{mod}\ 4) \) (since \( 59 \div 4 = 14 \text{ remainder } 3 \)) ✔️ 4. \( 59 \equiv 4 \ (\text{mod}\ 5) \) (since \( 59 \div 5 = 11 \text{ remainder } 4 \)) ✔️ **Answer:** 59 |
59 |
| GPT-4o | To solve the problem, we need to find the smallest positive integer \( x \) that satisfies the following set of congruences: \[ x \equiv 1 \pmod{2} \] \[ x \equiv 2 \pmod{3} \] \[ x \equiv 3 \pmod{4} \] \[ x \equiv 4 \pmod{5} \] Notice that each of these congruences can be rewritten to show a pattern: \[ x \equiv -1 \pmod{2} \] \[ x \equiv -1 \pmod{3} \] \[ x \equiv -1 \pmod{4} \] \[ x \equiv -1 \pmod{5} \] This implies: \[ x + 1 \equiv 0 \pmod{2} \] \[ x + 1 \equiv 0 \pmod{3} \] \[ x + 1 \equiv 0 \pmod{4} \] \[ x + 1 \equiv 0 \pmod{5} \] In other words, \( x + 1 \) must be a common multiple of 2, 3, 4, and 5. The smallest positive common multiple of these numbers is the least common multiple (LCM). First, we find the LCM of 2, 3, 4, and 5: - The prime factorization of the numbers is: - 2: \(2\) - 3: \(3\) - 4: \(2^2\) - 5: \(5\) - The LCM is the product of the highest powers of all prime factors present in the factorizations: \[ \text{LCM}(2, 3, 4, 5) = 2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60 \] Thus, \( x + 1 \) must be 60. Therefore, we have: \[ x + 1 = 60 \] \[ x = 60 - 1 \] \[ x = 59 \] The smallest positive integer \( x \) that meets the given conditions is \( \boxed{59} \). |
59 |
| llama3.1-70B-Instruct | The smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5 is 59. To find this solution, we can use the Chinese Remainder Theorem (CRT). The CRT states that if we have a system of congruences: x ≡ a1 (mod n1) x ≡ a2 (mod n2) ... x ≡ ak (mod nk) where ni are pairwise coprime (i.e., gcd(ni, nj) = 1 for i ≠ j), then there exists a unique solution modulo N = n1n2...nk. In this case, we have: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) We can rewrite these congruences as: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ -1 (mod 4) x ≡ -1 (mod 5) Using the CRT, we can combine these congruences to get: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ -1 (mod 20) Now, we can use the extended Euclidean algorithm to find the solution. We can write: 1 = 2(1) - 1(2) 2 = 3(1) - 1(3) -1 = 4(1) - 1(5) Using these equations, we can find the solution: x = 1(2) + 2(3) - 1(4) - 1(5) = 2 + 6 - 4 - 5 = 59 Therefore, the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5 is 59. |
59 |
| llama3.1-405B-Instruct | To find the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5, we can use the Chinese Remainder Theorem (CRT). First, we need to find the least common multiple (LCM) of 2, 3, 4, and 5, which is 60. Next, we need to find the solutions to the following congruences: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 4) x ≡ 4 (mod 5) Using the CRT, we can find the solution to the system of congruences as follows: x ≡ 1 (mod 2) => x = 2k + 1 for some integer k x ≡ 2 (mod 3) => x = 3m + 2 for some integer m x ≡ 3 (mod 4) => x = 4n + 3 for some integer n x ≡ 4 (mod 5) => x = 5p + 4 for some integer p Now, we need to find the smallest positive integer that satisfies all of these congruences. We can do this by finding the smallest positive integer that is a solution to all of the congruences. After some trial and error, we find that the smallest positive integer that satisfies all of the congruences is x = 59. Therefore, the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5 is 59. |
59 |
| DeepSeek-Coder-V2 | To solve this problem, we need to find the smallest positive integer \( x \) that satisfies the following system of congruences: \[ \begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{4} \\ x \equiv 4 \pmod{5} \end{cases} \] First, observe that the congruences can be rewritten in a more convenient form. Notice that: \[ \begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{4} \\ x \equiv 4 \pmod{5} \end{cases} \] can be transformed to: \[ \begin{cases} x \equiv -1 \pmod{2} \\ x \equiv -1 \pmod{3} \\ x \equiv -1 \pmod{4} \\ x \equiv -1 \pmod{5} \end{cases} \] This transformation shows that \( x + 1 \) must be divisible by 2, 3, 4, and 5. Therefore, \( x + 1 \) is a common multiple of these numbers. The least common multiple (LCM) of 2, 3, 4, and 5 is: \[ \text{LCM}(2, 3, 4, 5) = 60 \] Thus, \( x + 1 \) must be a multiple of 60. The smallest positive multiple of 60 is 60 itself. Therefore: \[ x + 1 = 60 \implies x = 60 - 1 = 59 \] So, the smallest positive integer \( x \) that satisfies all the given conditions is: \[ \boxed{59} \] |
59 |
| Qwen2-72B-Instruct | The given problem can be solved using the Chinese Remainder Theorem, but for this specific problem, we can also solve it through inspection or by creating a system of congruences. The system of congruences is: 1. \(x \equiv 1 \mod 2\) 2. \(x \equiv 2 \mod 3\) 3. \(x \equiv 3 \mod 4\) 4. \(x \equiv 4 \mod 5\) We are looking for the smallest positive integer \(x\) that satisfies all four congruences. Let's solve this step by step: Step 1: Find an integer that satisfies the first two congruences, \(x \equiv 1 \mod 2\) and \(x \equiv 2 \mod 3\). The first few numbers that satisfy \(x \equiv 1 \mod 2\) are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, ... The first few numbers that satisfy \(x \equiv 2 \mod 3\) are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, ... The smallest number that satisfies both congruences is 5. Step 2: Find an integer that satisfies the first three congruences, \(x \equiv 1 \mod 2\), \(x \equiv 2 \mod 3\), and \(x \equiv 3 \mod 4\). We can use the number we found in Step 1 (5) as a starting point and look for a number that also satisfies the third congruence. The first few numbers that satisfy \(x \equiv 3 \mod 4\) are 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, ... The smallest number that satisfies all three congruences is 35. Step 3: Find an integer that satisfies all four congruences. Using the number we found in Step 2 (35) as a starting point, we look for a number that also satisfies the fourth congruence, \(x \equiv 4 \mod 5\). The first few numbers that satisfy \(x \equiv 4 \mod 5\) are 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 89, 94, 99, ... The smallest number that satisfies all four congruences is 59. Therefore, the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5 is 59. |
59 |
| baichuan4 | To find the smallest positive integer that satisfies these conditions, we can use the Chinese Remainder Theorem (CRT). The theorem states that if one wants to solve a system of congruences with pairwise coprime moduli, then there exists a unique solution modulo the product of the moduli. Given the remainders and divisors, we have the following system of congruences: 1. x ≡ 1 (mod 2) 2. x ≡ 2 (mod 3) 3. x ≡ 3 (mod 4) 4. x ≡ 4 (mod 5) The moduli (2, 3, 4, 5) are pairwise coprime, which means they have no common factors other than 1. This allows us to apply the Chinese Remainder Theorem. We need to find the solution modulo the product of the moduli, which is 2 * 3 * 4 * 5 = 120. Now, we can solve each congruence separately and then combine them using the CRT. 1. For x ≡ 1 (mod 2), possible solutions are x = 1, 3, 5, ..., 119 (all odd numbers less than 120). 2. For x ≡ 2 (mod 3), possible solutions are x = 2, 5, 8, ..., 118 (numbers ending in 2 or 5 less than 120). 3. For x ≡ 3 (mod 4), possible solutions are x = 3, 7, 11, ..., 119 (numbers ending in 3 or 7 less than 120). 4. For x ≡ 4 (mod 5), possible solutions are x = 4, 9, 14, ..., 119 (numbers ending in 4 or 9 less than 120). The smallest positive integer that satisfies all these conditions is the first number that appears in all four lists. By inspection, we can see that the number 19 satisfies all the conditions: 1. 19 ≡ 1 (mod 2) 2. 19 ≡ 2 (mod 3) 3. 19 ≡ 3 (mod 4) 4. 19 ≡ 4 (mod 5) Therefore, the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5 is 19. |
59 |
| doubao | One way to approach this problem is to use the Chinese Remainder Theorem. First, we find the least common multiple of 2, 3, 4, and 5, which is 60. For the remainder of 1 when divided by 2, we can represent this as 2k + 1. For the remainder of 2 when divided by 3, we can represent this as 3m + 2. For the remainder of 3 when divided by 4, we can represent this as 4n + 3. For the remainder of 4 when divided by 5, we can represent this as 5p + 4. Now we need to find a number that satisfies all these conditions. We start by looking for a number that is 1 less than a multiple of 2, 2 less than a multiple of 3, 3 less than a multiple of 4, and 4 less than a multiple of 5. Trying some numbers, we find that 59 satisfies these conditions. So the smallest positive integer that leaves the required remainders is 59. |
59 |
| minimax | This problem can be solved using the Chinese Remainder Theorem (CRT), which provides a way to solve systems of simultaneous congruences with pairwise coprime moduli. However, since the problem is simple enough, we can also find a solution through a direct search. We are looking for a number \( x \) that satisfies the following system of congruences: 1. \( x \equiv 1 \mod 2 \) 2. \( x \equiv 2 \mod 3 \) 3. \( x \equiv 3 \mod 4 \) 4. \( x \equiv 4 \mod 5 \) We can start by listing out the numbers that satisfy each congruence and then look for the smallest number that satisfies all of them. For the first congruence, \( x \equiv 1 \mod 2 \), the numbers that satisfy this are all odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... For the second congruence, \( x \equiv 2 \mod 3 \), the numbers that satisfy this are: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, ... For the third congruence, \( x \equiv 3 \mod 4 \), the numbers that satisfy this are: 3, 7, 11, 15, 19, 23, 27, 31, ... For the fourth congruence, \( x \equiv 4 \mod 5 \), the numbers that satisfy this are: 4, 9, 14, 19, 24, 29, 34, ... Now, we look for the smallest number that appears in all four lists. Scanning through the lists, we find that the number 11 appears in the lists for the first three congruences, but not the fourth. The number 19 appears in the lists for the first and fourth congruences, but not the second and third. Continuing this process, we find that the number 43 appears in the lists for the first three congruences and is also one more than a multiple of 5, satisfying the fourth congruence. Therefore, the smallest positive integer that satisfies all the given conditions is 43. |
59 |
| kimi | This problem can be solved using the Chinese Remainder Theorem, which is a method for solving systems of simultaneous congruences with different moduli. Let's denote the number we are looking for as \( x \). The problem gives us the following system of congruences: \[ \begin{align*} x &\equiv 1 \pmod{2} \\ x &\equiv 2 \pmod{3} \\ x &\equiv 3 \pmod{4} \\ x &\equiv 4 \pmod{5} \end{align*} \] We can observe that each congruence is of the form \( x \equiv (modulus - 1) \pmod{modulus} \). This suggests that \( x + 1 \) is divisible by each of the moduli (2, 3, 4, 5). Therefore, \( x + 1 \) must be a multiple of the least common multiple (LCM) of these moduli. The LCM of 2, 3, 4, and 5 is calculated as follows: - The prime factors of 2 are \( 2 \) - The prime factors of 3 are \( 3 \) - The prime factors of 4 are \( 2^2 \) - The prime factors of 5 are \( 5 \) The LCM is the product of the highest powers of all prime factors present in the numbers, which is \( 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60 \). So, \( x + 1 = 60k \) for some integer \( k \). The smallest positive \( x \) occurs when \( k = 1 \), so: \[ x + 1 = 60 \times 1 \\ x = 60 - 1 \\ x = 59 \] Therefore, the smallest positive integer that satisfies all the given conditions is \( x = 59 \). |
59 |
| wenxinyiyan | 要找到这样一个最小的正整数，我们可以使用中国剩余定理的一个特例，即找到满足以下条件的数： 当该数除以2时余1，即 x ≡ 1 (mod 2) 当该数除以3时余2，即 x ≡ 2 (mod 3) 当该数除以4时余3，即 x ≡ 3 (mod 4) 当该数除以5时余4，即 x ≡ 4 (mod 5) 我们可以从最小的公倍数开始，逐一检查每个数，直到找到满足上述所有条件的最小数。 首先，2、3、4和5的最小公倍数是2*3*4*5=120。 然后，我们从120开始逐一检查每个数，直到找到满足上述所有条件的最小数。 经过计算，我们发现179是满足这些条件的最小数： 当179除以2时余1，即 179 ≡ 1 (mod 2) 当179除以3时余2，即 179 ≡ 2 (mod 3) 当179除以4时余3，即 179 ≡ 3 (mod 4) 当179除以5时余4，即 179 ≡ 4 (mod 5) 因此，答案是179。 |
59 |
| zhipu | To solve this problem, we are looking for the smallest positive integer \( x \) that satisfies the following system of congruences: \[ \begin{align*} x &\equiv 1 \pmod{2}, \\ x &\equiv 2 \pmod{3}, \\ x &\equiv 3 \pmod{4}, \\ x &\equiv 4 \pmod{5}. \end{align*} \] This is a classic problem that can be solved using the Chinese Remainder Theorem (CRT), but given the specific nature of the remainders (each remainder is 1 less than its modulus), we can use a simpler approach. Notice that each remainder is 1 less than its respective divisor. This means that if we add 1 to our desired number \( x \), it should be divisible by 2, 3, 4, and 5. In other words, \( x + 1 \) must be a multiple of the least common multiple (LCM) of these numbers. Let's find the LCM of 2, 3, 4, and 5: - The prime factorization of 2 is \( 2 \). - The prime factorization of 3 is \( 3 \). - The prime factorization of 4 is \( 2^2 \). - The prime factorization of 5 is \( 5 \). The LCM is the product of the highest powers of all prime factors that appear in these factorizations: \[ LCM(2, 3, 4, 5) = 2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60. \] So, \( x + 1 \) must be a multiple of 60. The smallest such multiple is 60 itself, which means the smallest possible value for \( x \) is: \[ x = 60 - 1 = 59. \] Let's verify that 59 satisfies all the original congruences: \[ \begin{align*} 59 \div 2 &= 29 \text{ remainder } 1, \\ 59 \div 3 &= 19 \text{ remainder } 2, \\ 59 \div 4 &= 14 \text{ remainder } 3, \\ 59 \div 5 &= 11 \text{ remainder } 4. \end{align*} \] Indeed, 59 satisfies all the given conditions. Therefore, the smallest positive integer that meets the criteria is: \[ \boxed{59}. \] |
59 |
| gpto1 | To solve the problem, we are asked to find the smallest positive integer \( N \) such that: 1. \( N \equiv 1 \ (\text{mod}\ 2) \) 2. \( N \equiv 2 \ (\text{mod}\ 3) \) 3. \( N \equiv 3 \ (\text{mod}\ 4) \) 4. \( N \equiv 4 \ (\text{mod}\ 5) \) We can solve these congruences step by step using the Chinese Remainder Theorem (CRT). **Step 1: Combine the first two congruences** First, we have: \[ N \equiv 1 \ (\text{mod}\ 2) \] This means \( N = 2k + 1 \) for some integer \( k \). Next, we have: \[ N \equiv 2 \ (\text{mod}\ 3) \] Substituting \( N = 2k +1 \): \[ 2k + 1 \equiv 2 \ (\text{mod}\ 3) \] Subtract 1 from both sides: \[ 2k \equiv 1 \ (\text{mod}\ 3) \] Multiply both sides by the modular inverse of 2 modulo 3. Since \( 2 \times 2 = 4 \equiv 1 \ (\text{mod}\ 3) \), the inverse of 2 modulo 3 is 2. Multiply both sides by 2: \[ 2 \times 2k \equiv 2 \times 1 \ (\text{mod}\ 3) \] \[ 4k \equiv 2 \ (\text{mod}\ 3) \] Since \( 4k \equiv k \ (\text{mod}\ 3) \) (because \( 4 \equiv 1 \ (\text{mod}\ 3) \)): \[ k \equiv 2 \ (\text{mod}\ 3) \] Set \( k = 3m + 2 \). Go back to \( N = 2k + 1 \): \[ N = 2(3m + 2) + 1 = 6m + 5 \] **Step 2: Incorporate the third congruence** Now consider: \[ N \equiv 3 \ (\text{mod}\ 4) \] Substitute \( N = 6m + 5 \): \[ 6m + 5 \equiv 3 \ (\text{mod}\ 4) \] Compute \( 6m + 5 \) modulo 4: \[ (6m + 5) \ (\text{mod}\ 4) = (2m + 1) \ (\text{mod}\ 4) \] Set: \[ 2m + 1 \equiv 3 \ (\text{mod}\ 4) \] Subtract 1 from both sides: \[ 2m \equiv 2 \ (\text{mod}\ 4) \] Divide both sides by 2: \[ m \equiv 1 \ (\text{mod}\ 2) \] Set \( m = 2n + 1 \). Update \( N \): \[ N = 6(2n + 1) + 5 = 12n + 11 \] **Step 3: Incorporate the fourth congruence** Now consider: \[ N \equiv 4 \ (\text{mod}\ 5) \] Substitute \( N = 12n + 11 \): \[ 12n + 11 \equiv 4 \ (\text{mod}\ 5) \] Compute \( 12n + 11 \) modulo 5: \[ (12n + 11) \ (\text{mod}\ 5) = (2n + 1) \ (\text{mod}\ 5) \] (Because \( 12 \equiv 2 \ (\text{mod}\ 5) \) and \( 11 \equiv 1 \ (\text{mod}\ 5) \)) Set: \[ 2n + 1 \equiv 4 \ (\text{mod}\ 5) \] Subtract 1 from both sides: \[ 2n \equiv 3 \ (\text{mod}\ 5) \] Multiply both sides by the modular inverse of 2 modulo 5. Since \( 2 \times 3 = 6 \equiv 1 \ (\text{mod}\ 5) \), the inverse of 2 modulo 5 is 3. Multiply both sides by 3: \[ 3 \times 2n \equiv 3 \times 3 \ (\text{mod}\ 5) \] \[ 6n \equiv 9 \ (\text{mod}\ 5) \] Since \( 6n \equiv n \ (\text{mod}\ 5) \): \[ n \equiv 4 \ (\text{mod}\ 5) \] Set \( n = 5p + 4 \). Update \( N \): \[ N = 12(5p + 4) + 11 = 60p + 48 + 11 = 60p + 59 \] **Step 4: Find the smallest positive integer** The smallest positive integer occurs when \( p = 0 \): \[ N = 60(0) + 59 = 59 \] **Verification** Check that \( N = 59 \) satisfies all original congruences: 1. \( 59 \equiv 1 \ (\text{mod}\ 2) \) ✔️ 2. \( 59 \equiv 2 \ (\text{mod}\ 3) \) (since \( 59 \div 3 = 19 \text{ remainder } 2 \)) ✔️ 3. \( 59 \equiv 3 \ (\text{mod}\ 4) \) (since \( 59 \div 4 = 14 \text{ remainder } 3 \)) ✔️ 4. \( 59 \equiv 4 \ (\text{mod}\ 5) \) (since \( 59 \div 5 = 11 \text{ remainder } 4 \)) ✔️ **Answer:** 59 |
59 |
| deepseekv3 | To solve this problem, we need to find the smallest positive integer \( x \) that satisfies the following system of congruences: \[ \begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{4} \\ x \equiv 4 \pmod{5} \\ \end{cases} \] ### Step 1: Observe the Pattern Notice that in each congruence, the remainder is always one less than the modulus. This suggests that \( x + 1 \) is divisible by 2, 3, 4, and 5. ### Step 2: Find the Least Common Multiple (LCM) Let \( x + 1 = \text{LCM}(2, 3, 4, 5) \). The LCM of 2, 3, 4, and 5 is: \[ \text{LCM}(2, 3, 4, 5) = 60 \] ### Step 3: Solve for \( x \) Now, substitute \( x + 1 = 60 \) and solve for \( x \): \[ x = 60 - 1 = 59 \] ### Step 4: Verify the Solution Check that \( x = 59 \) satisfies all the given congruences: \[ \begin{cases} 59 \div 2 = 29 \text{ with a remainder of } 1 \\ 59 \div 3 = 19 \text{ with a remainder of } 2 \\ 59 \div 4 = 14 \text{ with a remainder of } 3 \\ 59 \div 5 = 11 \text{ with a remainder of } 4 \\ \end{cases} \] All conditions are satisfied, so \( x = 59 \) is the smallest positive integer that meets the requirements. ### Final Answer: \[ \boxed{59} \] |
59 |