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

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2020 AMC 8 Problems/Problem 19

Problem 19

Solution 1