20! has 4 zeros at its end whereas 25! has 6 zeros..

how come..

somebody help.. (Thinking)

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- Mar 21st 2013, 06:22 AMhitesh091factorial help
20! has 4 zeros at its end whereas 25! has 6 zeros..

how come..

somebody help.. (Thinking) - Mar 21st 2013, 06:25 AMPunchRe: factorial help
i dont get your question

- Mar 21st 2013, 06:31 AMMAX09Re: 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, 06:32 AMibduttRe: 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, 06:34 AMPlatoRe: factorial help
Every trailing zero in $\displaystyle n!$ results in the number of factors of fives in $\displaystyle n!$.

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

$\displaystyle 50!$ has 12 trailing zeros. Fifty is has 10 factors of 5 and 2 factors of 25. - Mar 22nd 2013, 01:34 AMhitesh091Re: factorial help
@ max09 thanks i got it.. (Evilgrin)