2012 AIME I Problems/Problem 1

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Problem 1

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

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 $6$. Thus, there are $2 * 2 = 4$ ways to choose $a$ and $c$, and $10$ ways to choose $b$ (since $b$ can be any digit). Therefore, the final answer is $4 * 10 = \boxed{040}$.

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions