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# Mock AIME I 2015 Problems/Problem 9

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Since $a$ is a multiple of $b$, let $a=kb$.$\newline$ We can rewrite the first and second conditions as:$\newline$ (a) $(bk)bc$ is a perfect square, or $ck$ is a perfect square.$\newline$ (b) $b(k+7)c$ is a power of $2$, so it follows that $b$, $c$, and $k+7$ are all powers of $2$.$\newline$ Now we use casework on $k$. Since $k+7$ is a power of $2$, $k$ is $1, 9, 25, 57, 121,$ or $249$ or $k>500$. If $k>500$, then no value of b makes $1<=a, b<=500$.