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.

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

C++

Java

Python3

C#

PHP

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