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📅 March 11, 2026 👤 scraface

Mar 11, 2026

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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 $n = 36k = 36(7m + 3) = 252m + 108$.
We seek the smallest four-digit value of $n$:
252m + 108 \geq 1000 \Rightarrow 252m \geq 892 \Rightarrow m \geq \lef...
For $m = 4$:
n = 252 \cdot 4 + 108 = 1008 + 108 = 1116
$1116 \div 36 = 31$, so divisible by 36
$1116 \div 7 = 159$ remainder $3$, so satisfies the second condition

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