2012 AIME I Problems/Problem 1
Find the number of positive integers with three not necessarily distinct digits, , with and such that both and are multiples of .
A positive integer is divisible by if and only if its last two digits are divisible by For any value of , there are two possible values for and , since we find that if is even, and must be either or , and if is odd, and must be either or . There are thus ways to choose and for each and ways to choose since can be any digit. The final answer is then .
A number is divisible by four if its last two digits are divisible by 4. Thus, we require that and are both divisible by . If is odd, then and must both be meaning that and are or . If is even, then and must be meaning that and are or . For each choice of there are choices for and for for a total of numbers.
For this number to fit the requirements and must be divisible by 4. So and so must for each two digits of . There are two possibilities for if is odd and three possibilities if is even. So there are possibilities but this overcounts when or . So when and the corresponding should be removed, so . But we are still overcounting when is even because then can be 0. So the answer is
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