HMMT 十一月 2020 · 冲刺赛 · 第 3 题

HMMT November 2020 — Guts Round — Problem 3

[5] How many six-digit multiples of 27 have only 3, 6, or 9 as their digits? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMO 2020, November 14–21, 2020 — GUTS ROUND Organization Team Team ID#

解析

[5] How many six-digit multiples of 27 have only 3, 6, or 9 as their digits? Proposed by: Daniel Zhu Answer: 51 Solution: Divide by 3. We now want to count the number of six-digit multiples of 9 that only have 1, 2, or 3 as their digits. Due to the divisibility rule for 9, we only need to consider when the digit sum is a multiple of 9. Note that 3 · 6 = 18 is the maximum digit sum. If the sum is 18, the only case is 333333. Otherwise, the digit sum is 9. The possibilities here, up to ordering of the digits, are 111222 and ( ) 6
The first has = 20 cases, while the second has 6 · 5 = 30. Thus the final answer is 3 1 + 20 + 30 = 51.