# factorial help

• Mar 21st 2013, 07:22 AM
hitesh091
factorial help
20! has 4 zeros at its end whereas 25! has 6 zeros..

how come..
somebody help.. (Thinking)
• Mar 21st 2013, 07:25 AM
Punch
Re: factorial help
• Mar 21st 2013, 07:31 AM
MAX09
Re: factorial help
Hi hitesh091,

i'm assuming that you need to reason out only the increase in the no. of zeroes in 25! from the 4 zeroes in 20!

in order to find the 25! you'll have to multiply 20! with 21, 22, 23, 24 and 25, right?

consider the product 24x25 = 600, the two zeroes in this product will add to the already existing 4 zeroes of 20!. This leads to the increase of the number of zeroes from 4 to 6 when you compute 25!

hope it helped....
• Mar 21st 2013, 07:32 AM
ibdutt
Re: factorial help
There is nothing peculiar or special ig you observe 20! you can find that there are only 4 numbers which can give zero like, 10, 20,, 15 *4, 12*6 and similarly for 25! you can work out.
• Mar 21st 2013, 07:34 AM
Plato
Re: factorial help
Quote:

Originally Posted by hitesh091
20! has 4 zeros at its end whereas 25! has 6 zeros..
how come..
somebody help.. (Thinking)

Every trailing zero in $n!$ results in the number of factors of fives in $n!$.

$26!$ is divisible by $5,~10,~15,~20,~\&~25=5^2$ that is six factors of five.

$50!$ has 12 trailing zeros. Fifty is has 10 factors of 5 and 2 factors of 25.
• Mar 22nd 2013, 02:34 AM
hitesh091
Re: factorial help
@ max09 thanks i got it.. (Evilgrin)