Skip to content
Related Articles

Related Articles

Find M-th number whose repeated sum of digits of a number is N

View Discussion
Improve Article
Save Article
  • Last Updated : 16 Jun, 2022
View Discussion
Improve Article
Save Article

Given two positive integers N and M, The task is to find the M-th number whose sum of digits of a number until the sum becomes a single digit is N.
Examples: 
 

Input: N = 1, M = 3
Output: 19 
The first two numbers being 1 and 9.

Input: N = 2, M = 5
Output:  38 
The first four numbers being 2, 11, 20 and 29.

 

A naive approach is to iterate for all numbers and keep a count of numbers whose sum returns N. 
An efficient approach is to find the summation of digits till it becomes single digits in O(1) which has been discussed here. Hence the formula to find the M-th number will be: 
 

Mth number: (M-1)*9 + N

Below is the implementation of the above approach: 
 

C++




// C++ program to Find m-th number whose
// sum of digits of a number until
// sum becomes single digit is N
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the M-th
// number whosesum till one digit is N
int findNumber(int n, int m)
{
    int num = (m - 1) * 9 + n;
    return num;
}
 
// Driver Code
int main()
{
 
    int n = 2, m = 5;
    cout << findNumber(n, m);
    return 0;
}

Java




// Java program to Find m-th number whose
// sum of digits of a number until
// sum becomes single digit is N
class GFG
{
 
// Function to find the M-th
// number whosesum till one digit is N
static int findNumber(int n, int m)
{
    int num = (m - 1) * 9 + n;
    return num;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 2, m = 5;
    System.out.print(findNumber(n, m));
}
}
 
// This code is contributed
// by Akanksha Rai

Python3




# Python3 program to Find m-th number
# whose sum of digits of a number
# until sum becomes single digit is N
 
# Function to find the M-th
# number whosesum till one digit is N
def findNumber(n, m) :
     
    num = (m - 1) * 9 + n;
    return num;
 
# Driver Code
if __name__ == "__main__" :
 
    n = 2 ;
    m = 5 ;
    print(findNumber(n, m))
     
# This code is contributed by Ryuga

C#




// C# program to Find m-th number whose
// sum of digits of a number until
// sum becomes single digit is N
using System;
 
class GFG
{
 
// Function to find the M-th
// number whosesum till one digit is N
static int findNumber(int n, int m)
{
    int num = (m - 1) * 9 + n;
    return num;
}
 
// Driver Code
public static void Main()
{
    int n = 2, m = 5;
    Console.Write(findNumber(n, m));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// PHP program to Find m-th number whose
// sum of digits of a number until
// sum becomes single digit is N
 
// number whosesum till one digit is N
function findNumber($n, $m)
{
    $num = ($m - 1) * 9 + $n;
    return $num;
}
 
// Driver Code
$n = 2; $m = 5;
echo findNumber($n, $m);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
 
// JavaScript program to Find m-th number whose
// sum of digits of a number until
// sum becomes single digit is N   
 
    // Function to find the M-th
    // number whosesum till one digit is N
    function findNumber(n , m) {
        var num = (m - 1) * 9 + n;
        return num;
    }
 
    // Driver Code
     
        var n = 2, m = 5;
        document.write(findNumber(n, m));
 
// This code contributed by Rajput-Ji
 
</script>

Output: 

38

 

Time Complexity: O(1), as we are doing constant time operations without using any loops or recursion.

Auxiliary Space: O(1), as we are not using any extra space.


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!