Latest TCS Aptitude Question SOLUTION: how many digit of xeros at the end of product of the number from 1 to 100?

Direct formulla of no of Zeroes is
1*2*3********************100
100!
No of Zeroes depend upon no of 5
[100/5]+[100/5^2]+[100/5^3] ([] indicate no just lower than fraction)
20+4+0
24
ans=24

It occurred to me that the only way you can get a product of numbers to end in 0 is if one of them is a multiple of a power of 10; for example 20, 100, 350, or some such number, or if one is a number that ends in 5 and is multiplied by an even number.

So, if I want to calculate the number of zeros at the end of 100! I count the numbers that end in zero, namely 10,20,30,...100 (that will put 11 zeros at the end of our answer). Then the numbers that end in 5 are 5,15,25,35,...95.

(There are 10 of them, but don't forget that 25 and 75 are multiples of 5^2, so we have to count them twice).

That will give us 12 more zeroes at the end of our number.

And finally, I almost forget the number 50 has to be counted twice because 50 is 5 x 10. For example, if you multiply 50 by an even number like 50 x 4, you get 200, which has 2 zeroes at the end.

To summarize:

There are 10 numbers that end in zero from 1 to 100, but 50 and 100 are also multiples of powers of 5, so that gives us 12 zeroes.

There are 10 numbers that end in 5 from 1 to 100, but 25 and 75 are also multiples of powers of 5, so that gives us 12 more zeroes.

Therefore 100! must end in 24 zeroes.

ans:24
means end of 100! = calculate no. of 5s in 100 factorial factors which comes out to be 24.

hai you may see thiss... link found useful

http://puzzles.nigelcoldwell.co.uk/nineteen.htm

24 zeros at the end

24 zeros....since in factorial of 100! there comes 24 zeros...
1) 100/5 = 20;
2) again 20/5 =4;
adding both results we get 24...hence 24 is d answer......

Direct formulla of no of Zeroes is
1*2*3********************100
100!
No of Zeroes depend upon no of 5
[100/5]+[100/5^2]+[100/5^3] ([] indicate no just lower than fraction)
20+4+0
24
ans=24