Let $ N = 100a + 10b + 3 $. We require $ N \equiv 0 \pmod9 $. Since $ 100 \equiv 1 \pmod9 $, $ 10 \equiv 1 \pmod9 $, so: - AMAZONAWS

📅 March 11, 2026 👤 scraface

Mar 11, 2026

$Let $ N = 100a + 10b + 3 $. We require $ N \equiv 0 \pmod{9} $. Since $ 100 \equiv 1 \pmod{9} $, $ 10 \equiv 1 \pmod{9} $, so:$