2003 AMC 12B Problems/Problem 8

Let $a$ and $b$ be the digits of $x$ ,

$\clubsuit(\clubsuit(x)) = a + b = 3$

Clearly $\clubsuit(x)$ can only be $3, 12, 21,$ or $30$ and only $3$ and $12$ are possible to have two digits sum to.

If $\clubsuit(x)$ sums to $3$ , there are 3 different solutions : $12, 21, or 30$

If $\clubsuit(x)$ sums to $12$ , there are 7 different solutions: $39, 48, 57, 66,75, 84, or 93$

The total number of solutions is $3 + 7 =10 \Rightarrow \text (E)$ These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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