AMC 8 2020 Problem 19

AMC8 2020 年第 19 题

余数与同余★★★☆☆

A number is called flippy if its digits alternate between two distinct digits. For example, 2020 and 37373 are flippy, but 3883 and 123123 are not. How many five-digit flippy numbers are divisible by $15?$

若一个数的各个位上数字在 2 个不同的数字之间交替出现，那么这个数称为 flippy.例如， 2020 和 37373 都是 flippy 数，但是 3883 和 123123 不是。有多少个 5 位 flippy 数能被 15 整除？

正确答案 · Correct Answer

解析 · Solution

Solution 1

A number is divisible by 15 precisely if it is divisible by 3 and 5. The latter means the last digit must be either 5 or 0, and the former means the sum of the digits must be divisible by 3. If the last digit is 0, the first digit would be 0 (because the digits alternate), which is not possible. Hence the last digit must be 5, and the number is of the form

5\square 5\square 5

. If the unknown digit is

x

, we deduce

5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}

. We know

2^{-1}

exists modulo 3 because 2 is relatively prime to 3, so we conclude that

x

(i.e. the second and fourth digit of the number) must be a multiple of 3. It can be 0, 3, 6, or 9, so there are 4 options: 50505, 53535, 56565, and 59595. As in, we find that such numbers must start with 5 and alternate with 5 (i.e. must be of the form

5\square 5\square 5

), where the two digits between the 5 s need to be the same. Call that digit

x

. For the number to be divisible by 3, the sum of the digits must be divisible by 3; since the sum of the three 5 s is 15, which is already a multiple of 3, it must also be the case that x+x=2x is a multiple of 3. Thus, the problem reduces to finding the number of digits from 0 to 9 for which

2x

is a multiple of 3. This leads to x=0, 3, 6, or 9, so there are 4 possible numbers (namely 50505, 53535, 56565, and 59595 ).

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