How many times does the digit 0 appear in the integer equal to $20^{10}$ ?
By factoring and using exponent rules, we have $20^{10}=(2 \times 10)^{10}=2^{10} \times 10^{10}$.

Therefore, $20^{10}=1024 \times 10^{10}$, which is the integer 1024 followed by ten zeros.

Thus, $20^{10}$ has eleven digits that are 0 . That is, 10 zeros at the end and one coming from the 1024 at the beginning.

ANSWER: 11