2015 AIME II Problem 3

Attempt Problem 3 of the 2015 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AIME II solutions, or check the answer key.

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3.

Let mm be the least positive integer divisible by 1717 whose digits sum to 17.17. Find m.m.

Answer: 476
Concepts:digitsmodular arithmeticsystematic listing
Difficulty rating: 2070
Solution:

Every number is congruent to its digit sum modulo 9,9, so m=17nm = 17n must satisfy 17n17(mod9),17n \equiv 17 \pmod 9, that is 8n8(mod9),8n \equiv 8 \pmod 9, which gives n1(mod9).n \equiv 1 \pmod 9.

Checking the candidates in increasing order: n=1,10,19n = 1, 10, 19 give 17,17, 170,170, 323,323, with digit sums 88 each, but n=28n = 28 gives 1728=47617 \cdot 28 = 476 with digit sum 4+7+6=17.4 + 7 + 6 = 17. So m=476.m = 476.

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