99!, which, according to another contributor to WA, is
93326215443944152681699238856266700490715968264381621468592963895217
59999322991560894146397615651828625369792082722375825118521091686400 00000000000000000000
There are 24 trailing zeros in 100 factorial.
1.0309
9.33262222e+157
100*99*...*2*1= 93326215443944152681699238856266700490715968264381621468592963895217 59999322991560894146397615651828625369792082722375825118521091686400 0000000000000000000000
50
100!This was the original answer, which I left because it's funny. While the answer is 100, 100! represents one hundred factorial, a far, far greater number.
To calculate the number of zeros in a factorial number, we need to determine the number of factors of 5 in the factorial. In this case, we are looking at 10 to the power of 10 factorial. The number of factors of 5 in 10! is 2 (from 5 and 10). Therefore, the number of zeros in 10 to the power of 10 factorial would be 2.
18 factorial is equal to 6402373705728000 - with three consecutive zeroes at the end.
270
45
2625
The question, as asked, is difficult to answer, The number of zeros in factorial 100 is not the same as the number of 0s at the end of factorial 100 since there will be some before the end.The answer to the second question is easy:The number of zeros is determined by the number of 10s in the factors.Since 2s are common, this, in turn, depends on the number of 5s.In 100!, there are 20 multiples of 5 each of which will contribute a 5 to the factors of 100!.In addition there are 4 multiples of 52 = 25 each of which will contribute another 5 to the factors of 100!All in all, therefore, there are 24 5s giving 24 0s at the end of 100!