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Home » \( t = \frac4 \pm \sqrt16 + 402 \) - jntua results

\( t = \frac4 \pm \sqrt16 + 402 \) - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Solution: Compute $ 2^{10} = 1024 $. Divide 1024 by 7:

📌 Use modular arithmetic: $ 2^3 = 8 \equiv 1 \mod 7 $, so $ 2^3 \equiv 1 \Rightarrow 2^{10} = (2^3)^3 \cdot 2 = 1^3 \cdot 2 = 2 \mod 7 $.

📌 Thus, the remainder is $ \boxed{2} $.Question: What is the smallest four-digit number that is divisible by the least common multiple of 12 and 18, and also leaves a remainder of 3 when divided by 7?

📌 To find the smallest four-digit number satisfying the given conditions, we proceed in steps.

📌 \text{lcm}(12, 18) = \frac{12 \cdot 18}{\gcd(12, 18)} = \frac{216}{6} = 36

📌 So, the number must be divisible by 36 and leave a remainder of 3 when divided by 7. That is:

📌 n \equiv 0 \pmod{36}, \quad n \equiv 3 \pmod{7}

📌 We solve this system using the Chinese Remainder Theorem. Since 36 and 7 are coprime, there is a unique solution modulo $36 \cdot 7 = 252$.

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