Given an integer Y, find the smallest number X such that X! contains at least Y trailing zeros.
Prerequisites – Count trailing zeroes in factorial of a number
Input : Y = 2
Output : 10
10! = 3628800, which has 2 trailing zeros. 9! = 362880, which has 1 trailing zero. Hence, 10 is the correct answer.
Input : Y = 6
Output : 25
25! = 15511210043330985984000000, which has 6 trailing zeros. 24! = 620448401733239439360000, which has 4 trailing zeros. Hence, 25 is the correct answer.
Approach: The problem can be easily solved by using Binary Search. The number of trailing zeros in N! is given by the count of the factors 5 in N!.Read this article for prerequisites. The countFactor(5, N) function returns the count of factor 5 in N! which is equal to count of trailing zeros in N!. The smallest number X such that X! contains at least Y trailing zeros can be computed quickly by using binary search on a range [0, 5 * Y] using this function.
Below is the implementation of above approach.
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- Count number of trailing zeros in Binary representation of a number using Bitset
- Smallest number with at least n trailing zeroes in factorial
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- Find Nth smallest number that is divisible by 100 exactly K times
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