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

📅 March 6, 2026 👤 scraface

Mar 06, 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:$