Answer
The number $45^{8}\times88^{5}$ has exactly 8 final zeros.
We see that $45^{8}\times88^{5}=(3\times3\times5)^{8}(2\times2\times11)^{5}=(2^{10})(3^{18})(5^{8})(11^{5})$, where the last expression is in standard factored form. Since the only positive integral solution to $ab=10$ is $a=2$ and $b=5$ (or, equivalently, $a=5$ and $b=2$), and since an integer has exactly one final zero for every factor of $10$, it must be that $45^{8}\times88^{5}$ has 8 final zeros, because it has 8 factors of $10$. (We can pull $10$ from $(2^{10})(3^{18})(5^{8})(11^{5})$ eight times, because there are ten twos but only eight fives.)
Work Step by Step
Justifying the fact that the number of final zeros of an integer is equal to the number of factors of ten it contains can be done most easily by mathematical induction, a topic not covered until much later in the text. However, it is not an implausible assertion, so we take it here on intuition.
