Compute the sum of all two-digit positive integers $x$ such that for all three-digit (base 10) positive integers $\underline{a} \underline{b} \underline{c}$, if $\underline{a} \underline{b} \underline{c}$ is a multiple of $x$, then the three-digit (base 10) number $\underline{b} \underline{c} \underline{a}$ is also a multiple of $x$.
Solution: Note that $\overline{a b c 0}-\overline{b c a}=a\left(10^{4}-1\right)$ must also be a multiple of $x$. Choosing $a=1$ means that $x$ divides $10^{3}-1$, and this is clearly a necessary and sufficient condition. The only two-digit factors of $10^{3}-1$ are 27 and 37 , so our answer is $27+37=64$. $\fbox{64}$.