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📅 March 9, 2026 👤 scraface

Mar 09, 2026

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Use the Chinese Remainder Theorem. Let $ L = 7k $. Substitute into the...
7k \equiv 1 \pmod{13}
We solve for $ k $: find the modular inverse of 7 modulo 13.
$ 7 \cdot 1 = 7
$ 7 \cdot 2 = 14 \equiv 1 \pmod{13} $
So $ 7^{-1} \equiv 2 \pmod{13} $
k \equiv 2 \pmod{13} \Rightarrow k = 13m + 2
Then $ L = 7k = 7(13m + 2) = 91m + 14 $
For $ m = 0 $, $ L = 14 $ (two-digit)
Check: Is 14 $ \equiv 1 \pmod{13} $?

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