2015 AIME II 第 3 题

A number is congruent to its digit sum modulo $9,$ so $m = 17n$ requires $17n \equiv 17 \pmod 9$

That forces $n \equiv 1 \pmod 9;$ check the cases $n = 1,$ $n = 10,$ $n = 19,$ $n = 28,$ $\ldots$ in increasing order

Every number is congruent to its digit sum modulo $9,$ so $m = 17n$ must satisfy $17n \equiv 17 \pmod 9,$ that is $8n \equiv 8 \pmod 9,$ which gives $n \equiv 1 \pmod 9.$

Checking the candidates in increasing order: $n = 1,$ $n = 10,$ and $n = 19$ give $17,$ $170,$ and $323,$ with digit sums $8$ each, but $n = 28$ gives $17 \cdot 28 = 476$ with digit sum $4 + 7 + 6 = 17.$ So $m = 476.$