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Suppose we have two numbers N and D. We have to find N digit number, that is divisible by D. If N is 3, and D is 5, then the number can be 500. This can be solved easily. If D is 10 and N is 1, then it will be impossible. We can put D, and suppose the D has m number of digits, then attach N – m number of 0s to make it N digit number and divisible by D.

#include<iostream> using namespace std; string nDigitDivByD(int n, int d) { string ans = ""; if (d < 10) { ans += to_string(d); for (int i = 1; i < n; i++) ans += "0"; } else { if (n == 1) return "Cannot find any number"; else { string temp = to_string(d); ans += to_string(d); for (int i = 0; i < n-temp.length(); i++) ans += "0"; } } return ans; } int main() { int n = 5, d = 15; cout << nDigitDivByD(n, d); }

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