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 $\boxed{B}$ ~samrocksnature

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2020 AMC 8 Problems/Problem 19

Problem 19

Solution 1

See also