smc

Let $y=♣(x)$. Since $x\le 99$, we have $y\le 18$. Thus if $♣(y)=3$, then $y=3$ or $y=12$. The 3 values of $x$ for which $♣(x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $♣(x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\boxed{E}$ = $\boxed{10}$ values in all.