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

28 Apr, 2021
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
 
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++
Java
Python3
C#
PHP
Javascript
// 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 sumint digSum(int n){    int sum = 0, rem = 0;    while (n) {        rem = n % 10;        sum += rem;        n /= 10;    }    return sum;} // function for finding xint 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 functionint main(){    int n = 43;    cout << "x = " << findX(n);    return 0;}
// 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 program to find# x such that dx + igSum(x)# is equal to n. # utility function# for digit sumdef digSum(n):    sum = 0;    rem = 0;    while(n):        rem = n % 10;        sum = sum + rem;        n = int(n / 10);    return sum; # function for finding xdef 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 Coden = 43;print("x = ", findX(n)); # This code is contributed by mits
// 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 program to find x such that// dx + igSum(x) is equal to n. // utility function// for digit sumfunction digSum($n){    $sum = 0; $rem = 0;    while ($n)    {        $rem = $n % 10;        $sum += $rem;        $n /= 10;    }    return $sum;} // function for finding xfunction findX($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.?>
<script> // Javascript program to find x such that x +// digSum(x) is equal to n.      // utility function for digit sum    function digSum(n)    {        let sum = 0, rem = 0;                   while (n>0) {            rem = n % 10;            sum += rem;            n = Math.floor(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 (let i = 0; i <= n; i++)            if (i + digSum(i) == n)                return i;                   // if no such i found return -1        return -1;    } // driver program     let n = 43;               document.write("x = "+findX(n));     </script>
Output:  
x = 35
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