The number of [digits](https://artofproblemsolving.com/wiki/index.php/Digit) in $4^{16}5^{25}$ (when written in the usual [base](https://artofproblemsolving.com/wiki/index.php/Base_number) $10$ form) is
$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$
We can rewrite this as $2^{32}5^{25}$. We can also combine some of the factors to introduce factors of $10$, whose digit count is simple to evaluate because it simply adds $0$s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$. We can see that this final number is $2^7$ with $25$ $0$s annexed onto it. $2^7=128$, which has $3$ digits, so the entire number has $25+3=28$ digits, $\boxed{\text{D}}$.