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📅 May 4, 2026 👤 scraface

May 04, 2026

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Substitute into the second congruence:
9k - 1 \equiv 1 \pmod{7} \Rightarrow 9k \equiv 2 \pmod{7}
Since $ 9 \equiv 2 \pmod{7} $, this becomes:
2k \equiv 2 \pmod{7} \Rightarrow k \equiv 1 \pmod{7}
So $ k = 7m + 1 $, and:
x = 9(7m + 1) - 1 = 63m + 8
We want $ x $ to be a two-digit number:
\leq 63m + 8 \leq 99 \Rightarrow 2 \leq 63m \leq 91 \Rightarrow m = 1...
Try $ m = 1 $: $ x = 63(1) + 8 = 71 $
Try $ m = 2 $: $ x = 63(2) + 8 = 134 $ (too large)

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