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📅 May 17, 2026 👤 scraface

May 17, 2026

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📌 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 $36 \cdot 7 = 252$.

📌 Let’s find $n$ such that:

📌 Let $n = 36k$. Then:

📌 36k \equiv 3 \pmod{7} \Rightarrow k \equiv 3 \cdot 36^{-1} \pmod{7}

📌 Since $36 \equiv 1 \pmod{7}$, its inverse modulo 7 is 1. So:

📌 k \equiv 3 \pmod{7} \Rightarrow k = 7m + 3

📌 Then $n = 36k = 36(7m + 3) = 252m + 108$.

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