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!.

## Comments

Sort by:

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!.

Log in to reply

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

Log in to reply

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

Log in to reply

what do you mean?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

Comment deleted May 09, 2013

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply