Recall that each 5-and-2 pair (which makes 10 when multiplied) in the factorization of a number results in one trailing zero. Thus, we need to determine the number of 5-and-2 pairs within the prime factorization of that number.
Since we know there are fewer 5s than 2s in 200!, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.
To determine the number of 5s within 200!, we can use the following shortcut in which we divide 200 by 5, then divide the quotient of 200/5 by 5 and continue this process until we no longer get a nonzero quotient.
200/5 = 40
40/5 = 8
8/5 = 1 (we can ignore the remainder)
Since 1/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 200!.
Thus, there are 40 + 8 + 1 = 49 factors of 5 within 200!
Since there are 49 factors of 5 within 200!, there are 49 5-and-2 pairs and thus
49 trailing zeros.
