I came across this problem when reading a math book. How many zeroes are at the end of 100!. One way I guess is to take out factors of 10 and see how factors of 10 arise. Any thoughts on a this?

Question

I came across this problem when reading a math book. How many zeroes are at the end of 100!. One way I guess is to take out factors of 10 and see how factors of 10 arise.  Any thoughts on a this?

cwrw238 · Answer

100! is boviously = y * 10^x where y does not end in a zero.

cwrw238 · Answer

* obviously

anonymous · Answer

Well we know 5x2 = 10, since we have more twos than fives we can just pay attention to the fives.

100/5 = 20, then 100, 75, 50, 25, all have two fives, which we already have included.

= 24

rational · Answer

factors of 10 is one way

rational · Answer

the only way i think*

cwrw238 · Answer

I couldn't get my head around iambatman's answer at first  but I see it now.  
24 is correct.

anonymous · Answer

Maybe I should've clarified by saying each set of 5 and 2 give an output of 0.

anonymous · Answer

There are 24 0's in the end of 100

anonymous · Answer

the highest power of 5 which divides 100! is given
by [100/5]+[100/25] +[100/125] + [100/625]+...
=20+4+0+0...
=24

thus the highest power of 5 which divides 100! is 5^24..
now since there are more 2s in 100! than 5s in 
so 100! conatins 24 0s at the end..