no of zeros in 27! = 27/5 +27/25 = 5+1=6

to figure out how many zeros there will be at the end of n!
1.Divide n by 10 and determine a quotient which we'll call q and a remainder which we'll call r.
The number of zeros at the end will be:
@ 2q if r < 5
@ 2q + 1 if r>-5

For 27!
27/10 = 2 Remainder 7
so the number of zeros at the end of 27! will be 2*2 + 1 or 5

23
from 5 factorial we get 0 s.

How about 100!

If we divide 100 by 5, the answer is 20. Now, what is the reason why 100 was divided by 25 in the solution above?

In 100!, we have factors that are multiples of 25. If we divide them by 5, (25, 50, 75, 100)/5 we get (5, 10, 15, 20), still multiples of 5. This means that we still have four 5's that needs an even pair, so we need to divide again by 5. Now, since we have already counted the 5’s in the first division, we need to count the second set of 5’s. Instead of dividing again by 5, we divide them by 25. That is the reason why we also divide 100 by 25.

100 factorial

So, the solution will be

displaystyle frac{100}{5} + frac{100}{25 }= 20 + 4 = 24 zeroes!

Large Numbers

In larger numbers, we divide the factors by 5 three times (5^3), four times (5^4), and so on, or simply powers of 5. For example, in 130!, 125 is a factor, so we divide it by 5, 5^2 and 5^3. That is, we get the sum of the integer quotients of

displaystyle frac{130}{5} + frac {130}{25} +frac {130}{125} = 32.