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We have to take a number Y, we will find smallest number X, such that X! contains at least Y number of training zeros. For example, if Y = 2, then the value of X = 10. As X! = 3228800. It has Y number of zeros.

We can solve this using binary search. The number of trailing zeros in N! is given by the count of the factors 5 in N!. X can be found using binary search in range [0, 5*Y]

#include<iostream> using namespace std; int factorCount(int n, int X) { if (X < n) return 0; return (X / n + factorCount(n, X / n)); } int findX(int Y) { int left = 0, right = 5 * Y; int N = 0; while (left <= right) { int mid = (right + left) / 2; if (factorCount(5, mid) < Y) { left = mid + 1; }else { N = mid; right = mid - 1; } } return N; } int main() { int Y = 4; cout << "Smallest value of X: " << findX(Y); }

Smallest value of X: 20

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