Count the number of 0 digits without stop in $10000!$

The answer is 2499.

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Each pair of a $5$ and $2$ that gets factored in is equivalent to factoring a $10,$ or adding a trailing zero. But there are always more than enough $2$ s, so the number of $5$ s is all we need to consider. So divide $10000$ by $5.$ It is still not enough, because that doesn't count cases like $25,$ $50,$ and $125$ as many times as their power of $5.$ The formula is $Z_T(n!) = \sum_{k=1}^{\lfloor \log_5 n\rfloor} \lfloor\frac{n}{5^k}\rfloor$ In this case $n=10000$ so the sum is $\boxed{2499}.$ This formula only counts trailing zeros.