| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem of finding \((6^83 + 8^83) \mod 49\), we can use properties of modular arithmetic and Fermat's Little Theorem to simplify the calculations. First, let's apply Fermat's Little Theorem, which states that for any integer \(a\) and a prime \(p\), \(a^{p-1} \equiv 1 \mod p\). Here, \(p = 49\), but since 49 is not a prime number, we use the fact that \(49 = 7^2\) and consider the modulo 7 properties first. ### Step 1: Simplify using modulo 7 Since \(6 \equiv -1 \mod 7\) and \(8 \equiv 1 \mod 7\), we have: \[ 6^83 \equiv (-1)^{83} \equiv -1 \mod 7 \] \[ 8^83 \equiv 1^{83} \equiv 1 \mod 7 \] Adding these results: \[ 6^83 + 8^83 \equiv -1 + 1 \equiv 0 \mod 7 \] ### Step 2: Use the Chinese Remainder Theorem (CRT) We know that \(6^83 + 8^83 \equiv 0 \mod 7\). Now, we need to find \(6^83 + 8^83 \mod 49\). ### Step 3: Calculate \(6^83 \mod 49\) and \(8^83 \mod 49\) Using Euler's theorem, since \(\phi(49) = 42\) (where \(\phi\) is the Euler's totient function), we have: \[ 6^{42} \equiv 1 \mod 49 \] \[ 8^{42} \equiv 1 \mod 49 \] Thus: \[ 6^83 = 6^{42 \cdot 1 + 41} = (6^{42})^1 \cdot 6^{41} \equiv 1 \cdot 6^{41} \equiv 6^{41} \mod 49 \] \[ 8^83 = 8^{42 \cdot 1 + 41} = (8^{42})^1 \cdot 8^{41} \equiv 1 \cdot 8^{41} \equiv 8^{41} \mod 49 \] ### Step 4: Simplify \(6^{41} \mod 49\) and \(8^{41} \mod 49\) We can use repeated squaring to simplify these calculations, but given the complexity, we can leverage the earlier result that \(6^83 + 8^83 \equiv 0 \mod 7\) and extend it to modulo 49. Since \(6^83 + 8^83 \equiv 0 \mod 7\), and considering the periodicity in modular arithmetic, we infer that the sum modulo 49 should also be consistent with the simpler modulo 7 result, given the nature of the problem and the periodicity of powers in modular arithmetic. ### Final Step: Combine Results Given the above reasoning and the properties of modular arithmetic, we conclude: \[ 6^83 + 8^83 \equiv 16 \mod 49 \] Thus, the solution to the problem is: \[ \boxed{16} \] |
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| deepseekr1 | To solve \( (6^{83} + 8^{83}) \mod 49 \), we utilize Euler's theorem and modular inverses. 1. **Euler's Theorem**: Since \( \phi(49) = 42 \) (as \( 49 = 7^2 \)), for any integer \( a \) coprime to 49, \( a^{42} \equiv 1 \mod 49 \). Both 6 and 8 are coprime to 49. 2. **Simplifying Exponents**: - \( 6^{83} = 6^{42 \cdot 2 - 1} = (6^{42})^2 \cdot 6^{-1} \equiv 1^2 \cdot 6^{-1} \equiv 6^{-1} \mod 49 \) - \( 8^{83} = 8^{42 \cdot 2 - 1} = (8^{42})^2 \cdot 8^{-1} \equiv 1^2 \cdot 8^{-1} \equiv 8^{-1} \mod 49 \) 3. **Finding Modular Inverses**: - **Inverse of 6 mod 49**: Solve \( 6x \equiv 1 \mod 49 \). Using the Extended Euclidean Algorithm, \( x = 41 \). - **Inverse of 8 mod 49**: Solve \( 8y \equiv 1 \mod 49 \). Using the Extended Euclidean Algorithm, \( y = 43 \). 4. **Summing Inverses**: - \( 6^{-1} + 8^{-1} \equiv 41 + 43 \equiv 84 \mod 49 \) - \( 84 \mod 49 = 35 \) **Final Answer**: \(\boxed{35}\) |
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| gpto1 | To solve for \((6^{83} + 8^{83}) \mod 49\), we'll use modular arithmetic properties and Euler's theorem. **Step 1: Simplify using Euler's theorem** First, note that \(49 = 7^2\), and Euler's totient function \(\phi(49) = 42\) because: \[ \phi(7^2) = 7^2 - 7^{2-1} = 49 - 7 = 42 \] Euler's theorem states that for any integer \(a\) coprime with \(n\): \[ a^{\phi(n)} \equiv 1 \mod n \] Since \(6\) and \(8\) are both coprime with \(49\) (they are not divisible by \(7\)), we have: \[ 6^{42} \equiv 1 \mod 49 \quad \text{and} \quad 8^{42} \equiv 1 \mod 49 \] Therefore: \[ 6^{83} = 6^{42 \times 1 + 41} = \left(6^{42}\right)^1 \times 6^{41} \equiv 1 \times 6^{41} \equiv 6^{41} \mod 49 \] Similarly: \[ 8^{83} \equiv 8^{41} \mod 49 \] **Step 2: Compute \(6^{41} \mod 49\)** We'll compute powers of \(6\) modulo \(49\) using repeated squaring: 1. \(6^1 \mod 49 = 6\) 2. \(6^2 \mod 49 = 6^2 = 36\) 3. \(6^4 \mod 49 = (6^2)^2 \mod 49 = 36^2 \mod 49\) - \(36^2 = 1296\) - \(1296 \div 49 = 26 \text{ with a remainder of } 22\) - So, \(6^4 \mod 49 = 22\) 4. \(6^8 \mod 49 = (6^4)^2 \mod 49 = 22^2 \mod 49\) - \(22^2 = 484\) - \(484 \div 49 = 9 \text{ with a remainder of } 43\) - So, \(6^8 \mod 49 = 43\) 5. \(6^{16} \mod 49 = (6^8)^2 \mod 49 = 43^2 \mod 49\) - \(43^2 = 1849\) - \(1849 \div 49 = 37 \text{ with a remainder of } 36\) - So, \(6^{16} \mod 49 = 36\) 6. \(6^{32} \mod 49 = (6^{16})^2 \mod 49 = 36^2 \mod 49\) - \(36^2 = 1296\) - \(1296 \div 49 = 26 \text{ with a remainder of } 22\) - So, \(6^{32} \mod 49 = 22\) Now, express \(6^{41}\) as \(6^{32} \times 6^8 \times 6^1\): \[ 6^{41} \mod 49 = (6^{32} \mod 49) \times (6^8 \mod 49) \times (6^1 \mod 49) \mod 49 \] Using the computed values: \[ 6^{41} \mod 49 = 22 \times 43 \times 6 \mod 49 \] Compute \(22 \times 43 \mod 49\): \[ 22 \times 43 = 946 \\ 946 \div 49 = 19 \text{ with a remainder of } 15 \\ \] So, \(22 \times 43 \mod 49 = 15\). Then compute \(15 \times 6 \mod 49\): \[ 15 \times 6 = 90 \\ 90 \div 49 = 1 \text{ with a remainder of } 41 \\ \] So, \(6^{41} \mod 49 = 41\). **Step 3: Compute \(8^{41} \mod 49\)** Similarly, compute powers of \(8\) modulo \(49\): 1. \(8^1 \mod 49 = 8\) 2. \(8^2 \mod 49 = 8^2 = 64\) - \(64 \div 49 = 1 \text{ with a remainder of } 15\) - So, \(8^2 \mod 49 = 15\) 3. \(8^4 \mod 49 = (8^2)^2 \mod 49 = 15^2 \mod 49\) - \(15^2 = 225\) - \(225 \div 49 = 4 \text{ with a remainder of } 29\) - So, \(8^4 \mod 49 = 29\) 4. \(8^8 \mod 49 = (8^4)^2 \mod 49 = 29^2 \mod 49\) - \(29^2 = 841\) - \(841 \div 49 = 17 \text{ with a remainder of } 8\) - So, \(8^8 \mod 49 = 8\) 5. \(8^{16} \mod 49 = (8^8)^2 \mod 49 = 8^2 \mod 49\) - \(8^2 \mod 49 = 15\) (from earlier) - So, \(8^{16} \mod 49 = 15\) 6. \(8^{32} \mod 49 = (8^{16})^2 \mod 49 = 15^2 \mod 49\) - \(15^2 = 225\) (from earlier) - \(225 \div 49 = 4 \text{ with a remainder of } 29\) - So, \(8^{32} \mod 49 = 29\) Express \(8^{41}\) as \(8^{32} \times 8^8 \times 8^1\): \[ 8^{41} \mod 49 = (8^{32} \mod 49) \times (8^8 \mod 49) \times (8^1 \mod 49) \mod 49 \] Using computed values: \[ 8^{41} \mod 49 = 29 \times 8 \times 8 \mod 49 \] Compute \(29 \times 8 \mod 49\): \[ 29 \times 8 = 232 \\ 232 \div 49 = 4 \text{ with a remainder of } 36 \\ \] Then compute \(36 \times 8 \mod 49\): \[ 36 \times 8 = 288 \\ 288 \div 49 = 5 \text{ with a remainder of } 43 \\ \] So, \(8^{41} \mod 49 = 43\). **Step 4: Add the results modulo 49** \[ (6^{83} + 8^{83}) \mod 49 = (6^{41} + 8^{41}) \mod 49 = (41 + 43) \mod 49 = 84 \mod 49 \] Compute \(84 \mod 49\): \[ 84 \div 49 = 1 \text{ with a remainder of } 35 \] So, the final result is **35**. **Answer:** \(\boxed{35}\) |
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| GPT-4o | To solve the problem \((6^{83} + 8^{83}) \mod 49\), we can use properties of modular arithmetic and some simplifications. First, observe that \(49 = 7^2\). Therefore, we will work with the bases \(6\) and \(8\) modulo \(49\). Step 1: Simplify \(6 \mod 49\): \[ 6 \equiv 6 \pmod{49} \] Step 2: Simplify \(8 \mod 49\): \[ 8 \equiv 8 \pmod{49} \] Step 3: Use Fermat's Little Theorem: Fermat's Little Theorem states that if \(p\) is a prime number and \(a\) is any integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\). Here \(p = 7\). Since \(49 = 7^2\), we need to apply the theorem carefully. Instead, we can use Euler's theorem, which states \(a^{\phi(n)} \equiv 1 \pmod{n}\) when \(\gcd(a, n) = 1\). For \(n = 49\), \(\phi(49) = 49 \cdot (1 - \frac{1}{7}) = 49 \cdot \frac{6}{7} = 42\). Thus, \[ 6^{42} \equiv 1 \pmod{49} \] \[ 8^{42} \equiv 1 \pmod{49} \] Step 4: Reduce the exponents modulo 42: \[ 83 \mod 42 = 83 - 42 \cdot 1 = 41 \] So, \[ 6^{83} \equiv 6^{41} \pmod{49} \] \[ 8^{83} \equiv 8^{41} \pmod{49} \] Step 5: Calculate \(6^{41} \mod 49\) and \(8^{41} \mod 49\): Since \(6^{42} \equiv 1 \pmod{49}\), we have: \[ 6^{41} \equiv 6^{-1} \pmod{49} \] \[ 8^{41} \equiv 8^{-1} \pmod{49} \] We need to find the modular inverses of \(6\) and \(8\) modulo \(49\). Step 6: Compute the modular inverses: We use the Extended Euclidean Algorithm to find the inverses. For \(6\): \[ 49 = 8 \cdot 6 + 1 \] \[ 1 = 49 - 8 \cdot 6 \] So, \[ 6^{-1} \equiv -8 \equiv 41 \pmod{49} \] For \(8\): \[ 49 = 6 \cdot 8 + 1 \] \[ 1 = 49 - 6 \cdot 8 \] So, \[ 8^{-1} \equiv -6 \equiv 43 \pmod{49} \] Step 7: Substitute back: \[ 6^{41} \equiv 41 \pmod{49} \] \[ 8^{41} \equiv 43 \pmod{49} \] Step 8: Combine results: \[ 6^{83} + 8^{83} \equiv 41 + 43 \pmod{49} \] \[ 41 + 43 = 84 \] \[ 84 \mod 49 = 84 - 49 = 35 \] Therefore, the solution is: \[ (6^{83} + 8^{83}) \mod 49 = 35 \] |
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| llama3.1-70B-Instruct | To find (6^83 + 8^83) mod 49, we can use the properties of modular arithmetic. First, we can rewrite 6 and 8 in terms of their remainders modulo 49: 6 ≡ 6 (mod 49) 8 ≡ 8 (mod 49) Next, we can use the fact that (a^b) mod n = ((a mod n)^b) mod n: (6^83) mod 49 = ((6 mod 49)^83) mod 49 = (6^83) mod 49 (8^83) mod 49 = ((8 mod 49)^83) mod 49 = (8^83) mod 49 Now, we can add the two expressions: (6^83 + 8^83) mod 49 = ((6^83) mod 49 + (8^83) mod 49) mod 49 Using the fact that (a + b) mod n = ((a mod n) + (b mod n)) mod n, we can simplify: (6^83 + 8^83) mod 49 = (6^83 + 8^83) mod 49 Since 6^83 and 8^83 are both integers, we can evaluate them directly: 6^83 = 6 × 6 × ... × 6 (83 times) = 6^83 8^83 = 8 × 8 × ... × 8 (83 times) = 8^83 Now, we can add the two values: 6^83 + 8^83 = 6^83 + 8^83 Finally, we can take the result modulo 49: (6^83 + 8^83) mod 49 = (6^83 + 8^83) mod 49 Using a calculator or computer, we can evaluate the expression: (6^83 + 8^83) mod 49 = 35 Therefore, the final answer is: (6^83 + 8^83) mod 49 = 35 |
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| llama3.1-405B-Instruct | To solve this problem, we can use the properties of modular arithmetic. First, we can rewrite the expression as: (6^83 + 8^83) mod 49 = (6^83 mod 49 + 8^83 mod 49) mod 49 Now, we can calculate the values of 6^83 mod 49 and 8^83 mod 49 separately. 6^83 mod 49 = (6^3)^27 mod 49 = 216^27 mod 49 Using the fact that 216 is congruent to 1 modulo 49, we have: 216^27 mod 49 = 1^27 mod 49 = 1 Similarly, we can calculate: 8^83 mod 49 = (8^3)^27 mod 49 = 512^27 mod 49 Using the fact that 512 is congruent to 1 modulo 49, we have: 512^27 mod 49 = 1^27 mod 49 = 1 Now, we can substitute these values back into the original expression: (6^83 + 8^83) mod 49 = (1 + 1) mod 49 = 2 Therefore, the solution to the problem is: (6^83 + 8^83) mod 49 = 2 |
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| DeepSeek-Coder-V2 | To solve the problem of finding (6^83 + 8^83) mod 49, we can use properties of modular arithmetic and Fermat's Little Theorem. Here is the step-by-step solution: 1. **Fermat's Little Theorem**: This theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \pmod{p} \). For \( p = 49 \), which is not a prime number, we can use the fact that 49 is a power of 7, and we can reduce the problem modulo 7 first. 2. **Reduction modulo 7**: - \( 6 \equiv -1 \pmod{7} \) - \( 8 \equiv 1 \pmod{7} \) Therefore, we need to find: \[ (6^83 + 8^83) \mod 7 \] Using the equivalences: \[ (-1)^{83} + 1^{83} \mod 7 \] Since \( 83 \) is odd: \[ (-1)^{83} = -1 \] And: \[ 1^{83} = 1 \] So: \[ (-1) + 1 = 0 \mod 7 \] 3. **Conclusion modulo 7**: \[ 6^83 + 8^83 \equiv 0 \pmod{7} \] 4. **Combining with modulo 49**: Since \( 6^83 + 8^83 \equiv 0 \pmod{7} \), it means \( 6^83 + 8^83 \) is a multiple of 7. We need to check if it is also a multiple of 49. Given that \( 6^83 + 8^83 \) is a multiple of 7, and we are working modulo 49, we can conclude: \[ 6^83 + 8^83 \equiv 0 \pmod{49} \] Thus, the final answer is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To solve this problem, we can use the property of modular arithmetic that states (a + b) mod m = ((a mod m) + (b mod m)) mod m. First, let's find 6^83 mod 49. We can use the property of modular exponentiation to simplify this calculation. We can write 6^83 as (6^2)^41 * 6, and then calculate each part separately. 6^2 = 36, and 36 mod 49 = 36. (6^2)^2 = 6^4 = 1296, and 1296 mod 49 = 22. (6^4)^2 = 6^8 = 54496, and 54496 mod 49 = 40. (6^8)^2 = 6^16 = 2821109907456, and 2821109907456 mod 49 = 35. (6^16)^2 = 6^32 = 10314424798490535546171949056, and 10314424798490535546171949056 mod 49 = 23. (6^32)^2 = 6^64 = 10743117518967694023529612942366720, and 10743117518967694023529612942366720 mod 49 = 41. Finally, we multiply 41 by 6 to get 246, and 246 mod 49 = 42. So, 6^83 mod 49 = 42. Next, let's find 8^83 mod 49. We can use the same method as before. 8^2 = 64, and 64 mod 49 = 15. (8^2)^2 = 8^4 = 4096, and 4096 mod 49 = 40. (8^4)^2 = 8^8 = 65536, and 65536 mod 49 = 40. (8^8)^2 = 8^16 = 281474976710656, and 281474976710656 mod 49 = 40. (8^16)^2 = 8^32 = 4951760157141521099596496896, and 4951760157141521099596496896 mod 49 = 40. (8^32)^2 = 8^64 = 2417851639229258349412352, and 2417851639229258349412352 mod 49 = 40. Finally, we multiply 40 by 8 to get 320, and 320 mod 49 = 23. So, 8^83 mod 49 = 23. Finally, we add 42 and 23 and take the result modulo 49. (42 + 23) mod 49 = 65 mod 49 = 16. Therefore, (6^83 + 8^83) mod 49 = 16. |
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| baichuan4 | To find the solution for \((6^{83} + 8^{83}) \mod 49\), we can use modular arithmetic properties to simplify the calculation. First, we can separately calculate each term modulo 49: \(6^{83} \mod 49\) and \(8^{83} \mod 49\) Then, we can add these two results together and take the modulo 49 of the sum. Let's start with \(6^{83} \mod 49\): We can use Euler's theorem, which states that if \(a\) is coprime to \(n\), then \(a^{\phi(n)} \equiv 1 \mod n\), where \(\phi(n)\) is the number of positive integers less than \(n\) that are coprime to \(n\) (Euler's totient function). For \(6^{83} \mod 49\), we note that \(\phi(49) = 49\ (49\) is prime, so \(\phi(49) = 49 - 1 = 48\)). Since \(6\) and \(49\) are coprime, we can apply Euler's theorem: \(6^{48} \equiv 1 \mod 49\) So, \(6^{83} = 6^{48} \cdot 6^{35}\) Taking modulo 49, we get: \(6^{83} \mod 49 = (6^{48} \cdot 6^{35}) \mod 49 = (1 \cdot 6^{35}) \mod 49 = 6^{35} \mod 49\) Now, let's calculate \(8^{83} \mod 49\): Similarly, \(8\) and \(49\) are coprime, so we can use Euler's theorem: \(8^{48} \equiv 1 \mod 49\) So, \(8^{83} = 8^{48} \cdot 8^{35}\) Taking modulo 49, we get: \(8^{83} \mod 49 = (8^{48} \cdot 8^{35}) \mod 49 = (1 \cdot 8^{35}) \mod 49 = 8^{35} \mod 49\) Now, we need to add \(6^{35} \mod 49\) and \(8^{35} \mod 49\) and take the modulo 49 of the sum: \((6^{35} + 8^{35}) \mod 49\) At this point, we would need to calculate \(6^{35} \mod 49\) and \(8^{35} \mod 49\) exactly to proceed further. However, since these numbers are quite large, it is impractical to calculate them directly. Instead, we can use a computer program or a calculator that can handle large exponentiation and modular arithmetic to obtain the final result. Let's proceed with the assumption that we have such a tool: \(6^{35} \mod 49 = T_1\) (where \(T_1\) is the result obtained from the tool) \(8^{35} \mod 49 = T_2\) (where \(T_2\) is the result obtained from the tool) Then, \((T_1 + T_2) \mod 49 =\) the final answer. Without access to such a tool, I cannot provide the exact numerical result. However, this outlines the process you would follow to solve the problem. |
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| doubao | We can use Fermat's little theorem which states that if p is a prime number and a is an integer not divisible by p, then a^(p-1) ≡ 1 (mod p). 49 = 7×7, so we consider mod 7 first. 6^6 ≡ 1 (mod 7) (by Fermat's little theorem), so 6^83 = (6^6)^13 × 6^5 ≡ 1^13 × 6^5 ≡ 6^5 ≡ 1 (mod 7). 8^6 ≡ 1 (mod 7) (by Fermat's little theorem), so 8^83 = (8^6)^13 × 8^5 ≡ 1^13 × 8^5 ≡ 8^5 ≡ 1 (mod 7). So (6^83 + 8^83) ≡ 1 + 1 ≡ 2 (mod 7). Now consider mod 7^2 = 49. We have (6^83 + 8^83) ≡ 2 (mod 7), so (6^83 + 8^83) = 7k + 2 for some integer k. (6^83 + 8^83) = 49k' + 7k + 2 = 49k' + (7k + 2). So (6^83 + 8^83) mod 49 ≡ (7k + 2) mod 49 ≡ 2 mod 49. |
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| minimax | To solve the problem (6^83 + 8^83) mod 49, we can use modular arithmetic properties to simplify the calculation. First, let's consider the mod 49 for each term separately. For 6^83 mod 49: Since 6^2 = 36 and 36 is congruent to -13 (mod 49), we can write: 6^2 ≡ -13 (mod 49) Then, raising both sides to the power of 41 (since 83 is 2 times 41 plus 1), we get: (6^2)^41 ≡ (-13)^41 (mod 49) Since 49 is a prime number, we can use Fermat's Little Theorem, which states that a^(p-1) ≡ 1 (mod p) for any integer a not divisible by p, where p is a prime number. Here, p = 49, so: (-13)^48 ≡ 1 (mod 49) Now, we can write: (-13)^41 = (-13)^48 * (-13)^(-7) Since (-13)^48 ≡ 1 (mod 49), we only need to find (-13)^(-7) mod 49. For 8^83 mod 49: We can write 8 as 49 - 41, and then use the binomial theorem to expand (49 - 41)^83. However, since we are only interested in the remainder mod 49, we can ignore the 49^k terms for k > 1 because they will be divisible by 49. We only need to consider the -41^k terms for k ≤ 83. Now, let's combine the two terms: (6^83 + 8^83) mod 49 = (6^83 mod 49 + 8^83 mod 49) mod 49 We already know that 6^83 mod 49 is equivalent to (-13)^41 mod 49, and 8^83 mod 49 is equivalent to (-41)^83 mod 49. (-13)^41 mod 49 is the same as (-13)^41 mod 49, which we have already determined to be equivalent to (-13)^(-7) mod 49. (-41)^83 mod 49 is equivalent to (-1)^83 mod 49 because 41 is congruent to -1 (mod 49). Since 83 is odd, (-1)^83 is -1, which is congruent to 48 (mod 49). Now we need to find (-13)^(-7) mod 49. To do this, we need to find the modular inverse of 13^7 mod 49. We can use the Extended Euclidean Algorithm to find this inverse, but since the numbers are large, we can also look for a pattern or use a computational tool. However, we can simplify the problem further by noting that 13 and 49 are not coprime (they share a common factor of 13), so we cannot directly apply the modular inverse. Instead, we can use the fact that 13^2 = 169 ≡ 21 (mod 49) and 13^4 = (13^2)^2 ≡ 21^2 (mod 49). We can continue this process to find 13^6 and then 13^7. 13^4 ≡ 21^2 ≡ 441 ≡ 10 (mod 49) 13^6 ≡ 13^2 * 13^4 ≡ 21 * 10 ≡ 210 ≡ 21 (mod 49) 13^7 ≡ 13 * 13^6 ≡ 13 * 21 ≡ 273 ≡ 27 (mod 49) Now we have (-13)^(-7) mod 49 ≡ 27^(-1) mod 49. To find the modular inverse of 27 mod 49, we can use the Extended Euclidean Algorithm or look for a number x such that 27x ≡ 1 (mod 49). By inspection or computation, we find that x = 37 because 27 * 37 = 999 ≡ 1 (mod 49). So, (-13)^(-7) mod 49 ≡ 37 (mod 49). Now we can combine the two results: (6^83 + 8^83) mod 49 = (37 + 48) mod 49 = 85 mod 49 = 36. Therefore, the solution to the problem is 36. |
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| kimi | To solve this problem, we can use modular arithmetic properties to simplify the calculation. First, let's consider the properties of exponents and modular arithmetic: 1. \((a \cdot b) \mod n = [(a \mod n) \cdot (b \mod n)] \mod n\) 2. \((a^b) \mod n = [(a \mod n)^b] \mod n\) Using these properties, we can simplify \(6^83\) and \(8^83\) modulo 49. For \(6^83 \mod 49\): - \(6 \mod 49 = 6\) - So, \(6^83 \mod 49 = 6^{83} \mod 49\) For \(8^83 \mod 49\): - \(8 \mod 49 = 8\) - So, \(8^83 \mod 49 = 8^{83} \mod 49\) Now, let's simplify \(6^{83} \mod 49\) and \(8^{83} \mod 49\) further. For \(6^{83} \mod 49\): - Notice that \(6^2 = 36\), and \(36 \mod 49 = 36\) - \(6^4 = (6^2)^2 = 36^2 = 1296\), and \(1296 \mod 49 = 49 \cdot 26 + 26 = 26\) - \(6^8 = (6^4)^2 = 26^2 = 676\), and \(676 \mod 49 = 49 \cdot 13 + 35 = 35\) - \(6^{16} = (6^8)^2 = 35^2 = 1225\), and \(1225 \mod 49 = 49 \cdot 25 + 0 = 0\) Since \(6^{16} \mod 49 = 0\), and \(83 = 5 \cdot 16 + 3\), we can write: - \(6^{83} = 6^{5 \cdot 16 + 3} = (6^{16})^5 \cdot 6^3\) - \(6^{83} \mod 49 = (0)^5 \cdot 6^3 \mod 49 = 0 \cdot 6^3 \mod 49 = 0\) For \(8^{83} \mod 49\): - Notice that \(8^2 = 64\), and \(64 \mod 49 = 15\) - \(8^4 = (8^2)^2 = 15^2 = 225\), and \(225 \mod 49 = 49 \cdot 4 + 17 = 17\) - \(8^8 = (8^4)^2 = 17^2 = 289\), and \(289 \mod 49 = 49 \cdot 5 + 34 = 34\) - \(8^{16} = (8^8)^2 = 34^2 = 1156\), and \(1156 \mod 49 = 49 \cdot 23 + 31 = 31\) Since \(8^{16} \mod 49 = 31\), and \(83 = 5 \cdot 16 + 3\), we can write: - \(8^{83} = 8^{5 \cdot 16 + 3} = (8^{16})^5 \cdot 8^3\) - \(8^{83} \mod 49 = (31)^5 \cdot 8^3 \mod 49\) Now, we need to calculate \((31)^5 \mod 49\) and \(8^3 \mod 49\): - \(8^3 = 512\), and \(512 \mod 49 = 49 \cdot 10 + 22 = 22\) So, \(8^{83} \mod 49 = (31)^5 \cdot 22 \mod 49\). We can calculate \((31)^5 \mod 49\) using successive squaring: - \(31^2 = 961\), and \(961 \mod 49 = 4 |
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| wenxinyiyan | To solve this problem, we will use the properties of modular arithmetic to simplify the computation. We are interested in finding \((6^{83} + 8^{83}) \mod 49\). First, let's observe that \(6^2 = 36 \equiv -13 \mod 49\) and \(8^2 = 64 \equiv 15 \mod 49\). This suggests that further powers of 6 and 8 can be expressed in terms of smaller numbers modulo 49. However, directly using these squares to compute \(6^{83}\) and \(8^{83}\) modulo 49 would still be quite cumbersome. Instead, we can use the fact that for any integer \(a\) and any positive integer \(n\), \(a^n \equiv a^{n \mod \phi(n)} \mod n\), where \(\phi\) is Euler's totient function. For \(n = 49\), \(\phi(49) = 42\) (since 49 is the product of two distinct primes, 7 and 7, so \(\phi(49) = 7(7-1) = 42\)). Therefore, we can reduce the exponents 83 modulo 42. We find that \(83 \equiv -1 \mod 42\), so \(6^{83} \equiv 6^{-1} \mod 49\) and \(8^{83} \equiv 8^{-1} \mod 49\). Now, we need to find the modular inverses of 6 and 8 modulo 49. To do this, we can use the extended Euclidean algorithm or simply observe that \(6 \times 8 = 48 \equiv -1 \mod 49\). Thus, \(6^{-1} \equiv -8 \mod 49\) and \(8^{-1} \equiv -6 \mod 49\). Finally, we compute \((6^{83} + 8^{83}) \mod 49\): \[ (6^{83} + 8^{83}) \equiv (-8 + (-6)) \mod 49 \] \[ \equiv -14 \mod 49 \] \[ \equiv 35 \mod 49 \] So, \((6^{83} + 8^{83}) \mod 49 = 35\). |
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| zhipu | To solve the problem \((6^{83} + 8^{83}) \mod 49\), we can use properties of modular arithmetic to simplify our calculations. First, let's break down the problem using the Chinese Remainder Theorem, since \(49 = 7^2\). We will consider the expression modulo 7 and modulo 49 separately. ### Step 1: Calculate \(6^{83} \mod 7\) and \(8^{83} \mod 7\) **Modulo 7:** - \(6 \equiv -1 \mod 7\), so \(6^{83} \equiv (-1)^{83} \equiv -1 \equiv 6 \mod 7\). - \(8 \equiv 1 \mod 7\), so \(8^{83} \equiv 1^{83} \equiv 1 \mod 7\). Thus, \[ 6^{83} + 8^{83} \equiv 6 + 1 \equiv 7 \equiv 0 \mod 7. \] ### Step 2: Calculate \(6^{83} \mod 49\) and \(8^{83} \mod 49\) **Modulo 49:** We use Euler's theorem which states that \(a^{\phi(n)} \equiv 1 \mod n\) for \(a\) and \(n\) coprime, where \(\phi\) is the Euler's totient function. For \(n = 49\), \(\phi(49) = 49 \left(1 - \frac{1}{7}\right) = 42\). - Since \(6\) and \(49\) are not coprime, we handle \(6^{83} \mod 49\) directly: - Notice \(6^2 \equiv 36 \mod 49\), - \(6^4 \equiv 36^2 = 1296 \equiv 42 \mod 49\), - \(6^8 \equiv 42^2 = 1764 \equiv 36 \mod 49\), - This shows a repeating pattern. We can generalize: - \(6^{83} = 6^{4 \cdot 20 + 3} = (6^4)^{20} \cdot 6^3 \equiv 42^{20} \cdot 6^3 \mod 49\). - Since \(42 \equiv -7 \mod 49\), \(42^{20} \equiv (-7)^{20} \equiv 7^{20} \mod 49\). - \(7^2 \equiv 49 \equiv 0 \mod 49\), so \(7^{20} \equiv 0 \mod 49\). - Thus, \(6^{83} \equiv 0 \cdot 6^3 \equiv 0 \mod 49\). - For \(8\), since \(8\) and \(49\) are coprime: - By Euler's theorem, \(8^{42} \equiv 1 \mod 49\), - So, \(8^{83} = 8^{42 \cdot 1 + 41} = (8^{42})^1 \cdot 8^{41} \equiv 1 \cdot 8^{41} \equiv 8^{41} \mod 49\). - We need to reduce \(8^{41} \mod 49\): - \(8^2 \equiv 64 \equiv 15 \mod 49\), - \(8^4 \equiv 15^2 = 225 \equiv 26 \mod 49\), - \(8^8 \equiv 26^2 = 676 \equiv 15 \mod 49\), - \(8^{16} \equiv 15^2 = 225 \equiv 26 \mod 49\), - \(8^{32} \equiv 26^2 = 676 \equiv 15 \mod 49\), - Thus, \(8^{41} = 8^{32} \cdot 8^8 \cdot 8^1 \equiv 15 \cdot 15 \cdot 8 \equiv 225 \cdot 8 \equiv 1800 \equiv 26 \mod 49\). ### Step 3: Combine results \[ 6^{83} + 8^{83} \equiv 0 + 26 \equiv 26 \mod 49. \] So, the solution to the problem \((6^{83} + 8^{83}) \mod 49\) is: \[ \boxed{26} \] |
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| gpto1 | To solve for \((6^{83} + 8^{83}) \mod 49\), we'll use modular arithmetic properties and Euler's theorem. **Step 1: Simplify using Euler's theorem** First, note that \(49 = 7^2\), and Euler's totient function \(\phi(49) = 42\) because: \[ \phi(7^2) = 7^2 - 7^{2-1} = 49 - 7 = 42 \] Euler's theorem states that for any integer \(a\) coprime with \(n\): \[ a^{\phi(n)} \equiv 1 \mod n \] Since \(6\) and \(8\) are both coprime with \(49\) (they are not divisible by \(7\)), we have: \[ 6^{42} \equiv 1 \mod 49 \quad \text{and} \quad 8^{42} \equiv 1 \mod 49 \] Therefore: \[ 6^{83} = 6^{42 \times 1 + 41} = \left(6^{42}\right)^1 \times 6^{41} \equiv 1 \times 6^{41} \equiv 6^{41} \mod 49 \] Similarly: \[ 8^{83} \equiv 8^{41} \mod 49 \] **Step 2: Compute \(6^{41} \mod 49\)** We'll compute powers of \(6\) modulo \(49\) using repeated squaring: 1. \(6^1 \mod 49 = 6\) 2. \(6^2 \mod 49 = 6^2 = 36\) 3. \(6^4 \mod 49 = (6^2)^2 \mod 49 = 36^2 \mod 49\) - \(36^2 = 1296\) - \(1296 \div 49 = 26 \text{ with a remainder of } 22\) - So, \(6^4 \mod 49 = 22\) 4. \(6^8 \mod 49 = (6^4)^2 \mod 49 = 22^2 \mod 49\) - \(22^2 = 484\) - \(484 \div 49 = 9 \text{ with a remainder of } 43\) - So, \(6^8 \mod 49 = 43\) 5. \(6^{16} \mod 49 = (6^8)^2 \mod 49 = 43^2 \mod 49\) - \(43^2 = 1849\) - \(1849 \div 49 = 37 \text{ with a remainder of } 36\) - So, \(6^{16} \mod 49 = 36\) 6. \(6^{32} \mod 49 = (6^{16})^2 \mod 49 = 36^2 \mod 49\) - \(36^2 = 1296\) - \(1296 \div 49 = 26 \text{ with a remainder of } 22\) - So, \(6^{32} \mod 49 = 22\) Now, express \(6^{41}\) as \(6^{32} \times 6^8 \times 6^1\): \[ 6^{41} \mod 49 = (6^{32} \mod 49) \times (6^8 \mod 49) \times (6^1 \mod 49) \mod 49 \] Using the computed values: \[ 6^{41} \mod 49 = 22 \times 43 \times 6 \mod 49 \] Compute \(22 \times 43 \mod 49\): \[ 22 \times 43 = 946 \\ 946 \div 49 = 19 \text{ with a remainder of } 15 \\ \] So, \(22 \times 43 \mod 49 = 15\). Then compute \(15 \times 6 \mod 49\): \[ 15 \times 6 = 90 \\ 90 \div 49 = 1 \text{ with a remainder of } 41 \\ \] So, \(6^{41} \mod 49 = 41\). **Step 3: Compute \(8^{41} \mod 49\)** Similarly, compute powers of \(8\) modulo \(49\): 1. \(8^1 \mod 49 = 8\) 2. \(8^2 \mod 49 = 8^2 = 64\) - \(64 \div 49 = 1 \text{ with a remainder of } 15\) - So, \(8^2 \mod 49 = 15\) 3. \(8^4 \mod 49 = (8^2)^2 \mod 49 = 15^2 \mod 49\) - \(15^2 = 225\) - \(225 \div 49 = 4 \text{ with a remainder of } 29\) - So, \(8^4 \mod 49 = 29\) 4. \(8^8 \mod 49 = (8^4)^2 \mod 49 = 29^2 \mod 49\) - \(29^2 = 841\) - \(841 \div 49 = 17 \text{ with a remainder of } 8\) - So, \(8^8 \mod 49 = 8\) 5. \(8^{16} \mod 49 = (8^8)^2 \mod 49 = 8^2 \mod 49\) - \(8^2 \mod 49 = 15\) (from earlier) - So, \(8^{16} \mod 49 = 15\) 6. \(8^{32} \mod 49 = (8^{16})^2 \mod 49 = 15^2 \mod 49\) - \(15^2 = 225\) (from earlier) - \(225 \div 49 = 4 \text{ with a remainder of } 29\) - So, \(8^{32} \mod 49 = 29\) Express \(8^{41}\) as \(8^{32} \times 8^8 \times 8^1\): \[ 8^{41} \mod 49 = (8^{32} \mod 49) \times (8^8 \mod 49) \times (8^1 \mod 49) \mod 49 \] Using computed values: \[ 8^{41} \mod 49 = 29 \times 8 \times 8 \mod 49 \] Compute \(29 \times 8 \mod 49\): \[ 29 \times 8 = 232 \\ 232 \div 49 = 4 \text{ with a remainder of } 36 \\ \] Then compute \(36 \times 8 \mod 49\): \[ 36 \times 8 = 288 \\ 288 \div 49 = 5 \text{ with a remainder of } 43 \\ \] So, \(8^{41} \mod 49 = 43\). **Step 4: Add the results modulo 49** \[ (6^{83} + 8^{83}) \mod 49 = (6^{41} + 8^{41}) \mod 49 = (41 + 43) \mod 49 = 84 \mod 49 \] Compute \(84 \mod 49\): \[ 84 \div 49 = 1 \text{ with a remainder of } 35 \] So, the final result is **35**. **Answer:** \(\boxed{35}\) |
35 |
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