We make a 10 by having a 2 and a 5 in the prime factorization. Since we obviously have more 2s than 5s in the prime factorization of 100!, we just need to find to find the number of 5s in the factorization of 100!.
Of the first 100 integers 100/5=20 of then contain a 5 and 100/25 contain a second 5.
Thus there are a total of 20+4=24 0s at the end of 100!.

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TopNewestWe make a 10 by having a 2 and a 5 in the prime factorization. Since we obviously have more 2s than 5s in the prime factorization of 100!, we just need to find to find the number of 5s in the factorization of 100!. Of the first 100 integers 100/5=20 of then contain a 5 and 100/25 contain a second 5. Thus there are a total of 20+4=24 0s at the end of 100!.

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24 zeroes as [ 100/5 ] + [ 100/5^2 ] = 24

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using...gif....rite...?basic p and c

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what do you mean?

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Numbers of zeros in factorial is equal to \sum_{i=1}^k \lfloor \frac{n}{5^i} \rfloor where 5^k \leq n

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Comment deleted May 09, 2013

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Yes this is the solution if you mean the number of 0's at the end of N! Observe that 7!=5040 has 2 0's but the method yields 1.

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WE HAVE TO FIND THE NUMBER OF 5'S GREATEST INTEGER FUNCTION(100/5) + GREATEST INTEGER FUNCTION(100/5^2 ) SO THER WILL BE 24 ZEROS

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I suggest turning off the "Caps Lock" button. It would make all of us a little happier.

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He initially typed in "\(29\) ZEROS",and thought that others were wrong and he was right. So that is why the use of caps. Modified his answer later.

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