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use formula

no=(a/5^1)+(a/5^2)+.......

where no=number of zero
a=no. for which no.of zeros is to be calculated

take only int value of no.

Each occurence of 5 with an even number gives the 0 at the end.
So, in 100!, first find the occurences of 5 . i.e. how many powers of 5 are contained in 100!

Powers of 5 in 100! = 100/5 + 100/(5^2) + 100/(5^3) + .....
= 20 + 4 + 0..
= 24

SO the ans is 24.
@avatarz: 100! = 9.33262154 × 10 ^ 157 is the way of representing the final ans but it does not provide the 0s at the end.

For ex: 1930000 can be written as 1.93 * 10^6 or 19.3 * 10^5 or 193 * 10^4.... it's not giving the no. of 0s at the end.

found one more.... when there is zeros at unit place, 2 or 4 or 6 with 5 at tens will contribute to one more zero.. Thus total zeros are 22....

the number of 0s in n! is the power of 5 in the prime factorization of n
n! = 2^k1 * 3^k2 * 5^k3 * 7^k4 * ...
to get k3, one will have to:

k3 = 0
for i=5:n
 j = i
 while j % 5 == 0
  k3++
  j /= 5
return k3

Time complexity:
O(n) = sum over i (log_5 i) <= n * log_5(n) = O(n * log_5 n)
log_5 i = log in base 5 of i

answer is 24 because 1 zero is added for every factorial of a multiples of 5 and 2 zeros are added for multiples of 25

50! alone has 50 zeros!!! Just use excel and try it out..

Each occurence of 5 with an even number gives the 0 at the end.
So, in 100!, first find the occurences of 5 . i.e. how many powers of 5 are contained in 100!

Powers of 5 in 100! = 100/5 + 100/(5^2) + 100/(5^3) + .....
= 20 + 4 + 0..
= 24

SO the ans is 24.
@avatarz: 100! = 9.33262154 × 10 ^ 157 is the way of representing the final ans but it does not provide the 0s at the end.

For ex: 1930000 can be written as 1.93 * 10^6 or 19.3 * 10^5 or 193 * 10^4.... it's not giving the no. of 0s at the end.

if you search in wolfram alpha, it gives an answer of 30 zeros, 24 is the last zeros, there are other zeros inside,I did not understand

A function E(p){n!}=[n/p]+[n/p^2]+[n/p^3]+.... where (p) is the prime factor. Eg: 10 = 2, 5. Prime factor. So, P =2,5.The soln is as follows: Prime factors of 10=2*5 and E(2){50!}=[50/2]+[50/4]+[50/8]+[50/16]+[… or E(2){50!}=47 and E(5){50!}=[50/5]+[50/25]+[50/125] or E(5){50!}=12. Then E(10){50!}=the number of zeros in 50!=min of (47,12). Hence number of zeros in 50! is 12.

A function E(p){n!}=[n/p]+[n/p^2]+[n/p^3]+.... where (p) is the prime factor. Eg: 10 = 2, 5. Prime factor. So, P =2,5.The soln is as follows: Prime factors of 10=2*5 and E(2){50!}=[50/2]+[50/4]+[50/8]+[50/16]+[… or E(2){50!}=47 and E(5){50!}=[50/5]+[50/25]+[50/125] or E(5){50!}=12. Then E(10){50!}=the number of zeros in 50!=min of (47,12). Hence number of zeros in 50! is 12.

100! = 1x2x3x..x98x99x100
all the zeros at unit places and 5*2 or (5*4) will contribute towards number of zeros.
11 zeros - for zeros at unit place
10 zeros - for (5*2) or (5*4)

Total 21 zeros.........

with 100 there are 20 number with facter 5, and 4 number with factor 25. So there are 24 zeros in 100!.

100 has two zeros!!! one 1 and two 0---100

Google it...and you get this answer :
100 ! = 9.33262154 × 10 ^ 157