5. [5] Rachelle picks a positive integer $a$ and writes it next to itself to obtain a new positive integer $b$. For instance, if $a=17$, then $b=1717$. To her surprise, she finds that $b$ is a multiple of $a^{2}$. Find the product of all the possible values of $\frac{b}{a^{2}}$.
Answer: $\square$
Suppose $a$ has $k$ digits. Then $b=a\left(10^{k}+1\right)$. Thus $a$ divides $10^{k}+1$. Since $a \geq 10^{k-1}$, we have $\frac{10^{k}+1}{a} \leq 11$. But since none of 2,3 , or 5 divide $10^{k}+1$, the only possibilities are 7 and 11 . These values are obtained when $a=143$ and $a=1$, respectively.