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I got this question, and I'm totally lost as to how I solve it! Any help is appreciated :)

When 100! is written out in full, it equals 100! = 9332621...000000. Without using a calculator, determine the number of 0 digits at the end of this number

EDIT: Just want to confirm this is okay --

I got 24 by splitting products into 2 cases 1) multiples of 10 and 2) multiples of 5 Case I (1*3*4*6*7*8*9*10)(100,000,000,000)--> 12 zeroes

Similarly got 12 zeroes for Case 2.

So 24 in total? Is that correct?

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marked as duplicate by David, hardmath, Joffan, Micah, N. F. Taussig Feb 12 '15 at 0:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Here's a hint: in what ways can you multiply to get $10$? $\endgroup$ – Cameron Williams Feb 12 '15 at 0:37
  • $\begingroup$ Possible duplicate of Derive a formula to find the number of trailing zeroes in n! [duplicate] math.stackexchange.com/q/111385/752 $\endgroup$ – Américo Tavares Feb 12 '15 at 0:42
  • $\begingroup$ Generic question and analysis at Highest power of a prime $p$ dividing $N!$ $\endgroup$ – Joffan Feb 12 '15 at 0:45
  • $\begingroup$ There are plenty of powers of $2$ in $100!$ - the limiting factor for trailing zeros is the powers of $5$. $\endgroup$ – Joffan Feb 12 '15 at 0:46