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？

A. 3

B. 4

C. 5

D. 6

E. 8

1. 2020 AMC 8 官方解析

A five-digit fippy number looks like $\underline{A}\,\underline{B}\,\underline{A}\,\underline{B}\,\underline{A}$ for some distinct digits $A$ and $B$ with $A \neq 0$ (otherwise it would not be a five-digit number). In order for this number to be divisible by $15$, it must be divisible by $5$ and $3$. Any number divisible by $5$ will have a units digit of $0$ or $5$. Because $A \neq 0$, A must equal $5$. This means the flippy number must have the form $\underline{5}\,\underline{B}\,\underline{5}\,\underline{B}\,\underline{5}$. For a number to be divisible by $3$, the sum of its digits must be divisible by $3$. The sum of the digits of $\underline{5}\,\underline{B}\,\underline{5}\,\underline{B}\,\underline{5}$ is $15 + 2B$. This means $2B$ must be divisible by $3$, so $B$ is divisible by $3$. There are $4$ possible digits for $B : 0, 3, 6$, or $9$. Therefore there are $4$ five-digit flippy numbers divisible by $15 : 50505, 53535, 56565$, and $59595$.

一个五位交替数看起来是这样的形式$\underline{A}\,\underline{B}\,\underline{A}\,\underline{B}\,\underline{A}$，其中$A \neq 0$（否则将不是五位数）。为了使此数字可以被$15$整除，必须使其可以被$5$和$3$整除。可被$5$整除的任何数字的个位数必须为$0$或$5$。因为$A \neq 0$，所以$A$必须等于$5$。因此交替数可以表示为$\underline{5}\,\underline{B}\,\underline{5}\,\underline{B}\,\underline{5}$。要使数字可被$3$整除，其所有位数之和必须能被$3$整除。$\underline{5}\,\underline{B}\,\underline{5}\,\underline{B}\,\underline{5}$的数字总和为$15 + 2B$。这意味着$2B$必须可被$3$整除，因此$B$可被$3$整除。$B$的值可能为: $0, 3, 6$或$9$。因此，有$4$个五位交替数可以被$15$整除，分别为: $50505, 53535, 56565$和$59595$。