000000000000000...............
How many trailing 0's does 100! have? Generalize it for any N! after you have computed for 100! N! = Factorial N = N * N-1 * N-2 * ......2 * 1
Level: Medium to Hard
posted by Karthik Nagaraj at
8:48 AM
Saturday, May 13, 2006
How many trailing 0's does 100! have? Generalize it for any N! after you have computed for 100! N! = Factorial N = N * N-1 * N-2 * ......2 * 1
Level: Medium to Hard
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2 Comments:
Lets consider the first occurrence of a trailing 0 in n! which happens at 5! (5!=120) and from then on until 9! every n! has one trailing zero. Later on a 10! produces an additional 0, so does 15! So every multiple of 5 produces an additional 0. The trouble is that 25!=25*24*23! has two zeros more than 23! or 24!.
So, we need to count the number of times each number is completely divided by 5 and keep incrementing that.
So 100! has 20 zeros (20 multiples of 5) + 4 zeros (4 multiples of 25) i.e. a total of 24 zeros.
For any n: it can be generalized as ceil(n/5)+ceil(n/5^2)+.... until n>=5^k
Sanjay,
Interesting solution. I think you cracked it right.
Excellent! Think about a scenario if the number of 0's in the result is asked instead of the trailing 0's. Hmm, that would be toasty! But for now This puzzle is SOLVED!
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