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by this_time_i_will » Fri Sep 24, 2010 7:29 pm
i have edited this post. though my approach was right, i made mistake in calculation. For all such questions, there is a quick method:
to find number of zero's at the end of x!, we do:
[x/5]+[x/5^2]+....+[x/5^n], where n = 1,2,3...etc. , 5^n <x<5^(n+1) and [y] means greates integer less than or equal to y.

so applying above formulae:
[60/5]+[60/5^2] = 12+2 = 14.
Last edited by this_time_i_will on Sat Sep 25, 2010 12:24 am, edited 1 time in total.
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by tlt2372 » Fri Sep 24, 2010 7:31 pm
There is a trick called "properties of the ones digit" that you can use to solve this problem. I cant remember how to do it though........
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by Rahul@gurome » Fri Sep 24, 2010 8:11 pm
Solution:
The number of zeroes at the end of 60! depends on the number of times 60! is a multiple of 10.
Now 10 = 2*5.
So what we need to do is to find out how many pairs of 2"s and 5's are contained in 60!.
Since every second integer in 60! = 1*2*3*4...*59*60 is a multiple of 2, we only need to calculate the number of 5's in 60!.
Each of the 5's will combine with a 2 to give 10.

Multiples of 5 in 60! are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60. They are 12 in number but 25 = 5^2, and so it has two 5's and 50 = (5^2)*2 , which again has two 5's.
So in total there are 12+2 = 14 fives.
Or 60! will have 14 zeroes in the end.
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