60!, no of 0's at tthe end

[rahulg83]

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

6
12
14
42
56

My approach: [spoiler]between 1 and 10, we can have 2 0's, one from 2*5 and another from 10.
similarly we can calculate 0's for numbers between 11 and 20 (2), 21 and 30(2), 31 and 40(2), 41 and 50(2) and finally 51 and 60(2).

So total numbers of 0's at the end = 2+2+2+2+2+2 = 12[/spoiler]
But OA is 14

Please tell me what i am missing here.

[PussInBoots]

1) In a chaos of prime numbers only a product of 2 and 5 can make 0.
2) Perform prime factorization of 60!. We will have a lot of prime numbers, when we multiply them all, we'll get 60!.
3) In that pool we have tons of 2s, but very few 5s. Count all 5s. That's how many 0s we'll get at the end:

5 10 15 ... 60 gives us 12 5s
25 and 50 give additional 2 5s {you didn't count those}

So in that pool of prime numbers we have 14 5s, tons of 2s, and a bunch of other prime numbers. Hence the answer is 14

[pops]

(first it is important to understand the basic as explained above. If thats clear then one can use this shortcut)

In such type of problem where one has to find number of zeroes at the end of N! keep dividing N by 5 and take the integer and add...i.e.
N/5 + (N/5)/5 + ((N/5)/5)/5...... so on till 1
(this makes it look horrible so let me state an example)
for 150!
150/5=30
30/5=6
6/5=1 (taking the integer part only)
so there are 30+6+1 = 37 zeroes at the end of 150!

hence at the end of 60! there should be
60/5=12
12/5=2
i.e. 12+2=14 zeroes