You can rewrite a number whose decimal form is $a_1a_2\cdots a_m$ as $a_1.a_2\cdots a_m\times 10^{m-1}$, so it is of the form $a\times 10^{m-1}$ with $1\leq a<10$. (This is "scientific notation.")
If $1\leq a<10$ and $1\leq b<10$, then
$(a\times 10^{m-1})\times (b\times 10^{n-1})=a\times b\times 10^{m+n-2}.$ Multiplying by the power of $10$ just shifts the digits, so what is needed is to determine how many digits $a\times b$ can have to the left of the decimal point, when $1\leq a<10$ and $1\leq b<10$. There are just two possibilities depending on whether or not $ab$ is less than $10$.