# 2012 AIME I Problems/Problem 1

## Problem

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

## Solution 1

A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$, there are two possible values for $a$ and $c$, since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$, and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$. There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{040}$.

## Solution 2

A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$. If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$. If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$. For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{040}$ numbers.

## Solution 3

For this number to fit the requirements $bc$ and $ba$ must be divisible by 4. So $bc = 00, 04, 08, 12, 16, ... , 92, 96$ and so must $ba$ for each two digits of $bc$. There are two possibilities for $ba$ if $b$ is odd and three possibilities if $b$ is even. So there are $2^{2} \cdot 5+3^{2} \cdot 4 = 65$ possibilities but this overcounts when $a$ or $c = 0$. So when $bc = 00, 20, 40, 60, 80$ and the corresponding $ba$ should be removed, so $65 - 5 \cdot 3 = 50$. But we are still overcounting when $b$ is even because then $a$ can be 0. So the answer is $50 - 10 \cdot 1 = \boxed{040}$

~LuisFonseca123

~ pi_is_3.14