# Difference between revisions of "2020 AMC 8 Problems/Problem 19"

## Problem 19

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 $15?$

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

## Solution 1

To be divisible by $15$, a number must first be divisible by $3$ and $5$. By divisibility rules, the last digit must be either $5$ or $0$, and 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). So, the last digit must be $5$, and we have $$5+x+5+x+5 \equiv 0 \pmod{3}$$ $$2x \equiv 0 \pmod{3}$$ We know the inverse exists because 2 is relatively prime to 3, and thus we can conclude that $x$ (or the second and fourth digits) is always a multiple of $3$. We have 4 options: $0, 3, 6, 9$, and our answer is $4$ and $\boxed{\textbf{(B) } 4}$ ~samrocksnature

## Solution 2

A number that is divisible by $15$ must be divisible by $3$ and $5$. To be divisible by $3$, the sum of its digits must be divisible by $3$ and to be divisible by $5$, it must end in a $5$ or a $0$. Observe that a five-digit flippy number must start and end with the same digit. Since a five-digit number cannot start with $0$, it also cannot end in $0$. This means that the numbers that we are looking for must end in $5$. This also means that they must start with $5$ and alternate with $5$ (i.e. the number must be of the form $5\square5\square5$). The two digits between the $5s$ must be the same. Let's call that digit $x$. We know that the sum of the digits must be a multiple of $3$. Since the sum of the three $5s$ is $15$ which is already a multiple of $3$, for the entire five-digit number to be a multiple of $3$, it must also be the case that $x+x=2x$ is also 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 $x=0,3,6,$ and $9$ and we have four valid answers (i.e. $50505,53535,56565,$ and $59595$) $\implies\boxed{\textbf{(B) } 4}$.
~ junaidmansuri