Find a number x such that sum of x and its digits is equal to given n.

Given a positive number n. We need to find a number x such that sum of digits of x to itself is equal to n.
If no such x is possible print -1.

Examples:

Input : n = 21
Output : x = 15
Explanation : x + its digit sum = 15 + 1 + 5 = 21 

Input : n = 5
Output : -1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We iterate from 1 to n and for each intermediate number x find its digit sum and then add that to x, if that is equal to n then x will be our required answer.

    // iterate from 1 to n. For every no.
    // check if its digit sum with it is
    // equal to n.
    for (int i = 0; i <= n; i++)
        if (i + digSum(i)  == n)
            return i;

    return -1;

C++

// CPP program to find x such that x + 
// digSum(x) is equal to n. 
#include <bits/stdc++.h> 
using namespace std; 
  
// utility function for digit sum 
int digSum(int n) 
{ 
    int sum = 0, rem = 0; 
    while (n) { 
        rem = n % 10; 
        sum += rem; 
        n /= 10; 
    } 
    return sum; 
} 
  
// function for finding x 
int findX(int n) 
{ 
    // iterate from 1 to n. For every no. 
    // check if its digit sum with it is 
    // equal to n. 
    for (int i = 0; i <= n; i++)  
        if (i + digSum(i) == n) 
            return i; 
      
    // if no such i found return -1 
    return -1; 
} 
  
// driver function 
int main() 
{ 
    int n = 43; 
    cout << "x = " << findX(n); 
    return 0; 
} 

Java

// Java program to find x such that x + 
// digSum(x) is equal to n. 
class GFG 
{ 
      
    // utility function for digit sum 
    static int digSum(int n) 
    { 
        int sum = 0, rem = 0; 
          
        while (n>0) { 
            rem = n % 10; 
            sum += rem; 
            n /= 10; 
        } 
          
        return sum; 
    } 
      
    // function for finding x 
    static int findX(int n) 
    { 
          
        // iterate from 1 to n. For every no. 
        // check if its digit sum with it is 
        // equal to n. 
        for (int i = 0; i <= n; i++)  
            if (i + digSum(i) == n) 
                return i; 
          
        // if no such i found return -1 
        return -1; 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    { 
        int n = 43; 
          
        System.out.println("x = "+findX(n)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python3 program to find  
# x such that dx + igSum(x)  
# is equal to n. 
  
# utility function  
# for digit sum 
def digSum(n): 
    sum = 0; 
    rem = 0; 
    while(n): 
        rem = n % 10; 
        sum = sum + rem; 
        n = int(n / 10); 
    return sum; 
  
# function for finding x 
def findX(n): 
      
    # iterate from 1 to n. 
    # For every no. 
    # check if its digit 
    # sum with it is# equal to n. 
    for i in range(n + 1): 
        if (i + digSum(i) == n): 
            return i; 
      
    # if no such i 
    # found return -1 
    return -1; 
  
# Driver Code 
n = 43; 
print("x = ", findX(n)); 
  
# This code is contributed by mits 

C#

// C# program to find x such that  
// x + digSum(x) is equal to n. 
using System; 
  
class GFG { 
      
    // utility function for digit sum 
    static int digSum(int n) 
    { 
        int sum = 0, rem = 0; 
          
        while (n > 0)  
        { 
            rem = n % 10; 
            sum += rem; 
            n /= 10; 
        } 
          
        return sum; 
    } 
      
    // function for finding x 
    static int findX(int n) 
    { 
          
        // iterate from 1 to n. For every no. 
        // check if its digit sum with it is 
        // equal to n. 
        for (int i = 0; i <= n; i++)  
            if (i + digSum(i) == n) 
                return i; 
          
        // if no such i found return -1 
        return -1; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int n = 43; 
          
        Console.Write("x = " + findX(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find x such that  
// dx + igSum(x) is equal to n. 
  
// utility function  
// for digit sum 
function digSum($n) 
{ 
    $sum = 0; $rem = 0; 
    while ($n)  
    { 
        $rem = $n % 10; 
        $sum += $rem; 
        $n /= 10; 
    } 
    return $sum; 
} 
  
// function for finding x 
function findX($n) 
{ 
      
    // iterate from 1 to n. 
    // For every no. 
    // check if its digit 
    // sum with it is 
    // equal to n. 
    for ($i = 0; $i <= $n; $i++)  
        if ($i + digSum($i) == $n) 
            return $i; 
      
    // if no such i  
    // found return -1 
    return -1; 
} 
  
    // Driver Code 
    $n = 43; 
    echo "x = " , findX($n); 
  
// This code is contributed by vt_m. 
?> 

Output:

x = 35

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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