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Find M-th number whose repeated sum of digits of a number is N in C++
In this problem, we are given two positive numbers N and M. Our task is to find the M-th number whose repeated sum of digits of a number is N.
Problem description: Here, we need to find the Mth number whose sum of digits till the sum becomes single digit is equal to N.
Let’s take an example to understand the problem,
Input: N = 4 M = 6
Output: 49
Solution Approach
A simple solution of the problem, is by finding all numbers and count the number whose sum of digits is N, and return m-th number.
Another solution to the problem is using formula to find M-th number whose sum of digits is equal to N,
M-th number = (m-1)*9 + N
Program to illustrate the working of our solution,
Example
#include <bits/stdc++.h>
using namespace std;
int main() {
int n = 4, m = 6;
int mNumber = (m - 1) * 9 + n;
cout<<m<<"-th number whose repeated sum of digits of a number is "<<n<<" is "<<mNumber;
return 0;
}Output
6-th number whose repeated sum of digits of a number is 4 is 49
Updated on: 25-Jan-2021
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