Number of integral solutions for equation x = b*(sumofdigits(x)^a)+c

Given a, b and c which are part of the equation x = b * ( sumdigits(x) ^ a ) + c.

Where sumdigits(x) determines the sum of all digits of the number x. The task is to find out all integer solutions for x that satisfy the equation and print them in increasing order.

Given that, 1<=x<=109

Examples:

Input: a = 3, b = 2, c = 8
Output: 10 2008 13726
Values of x are: 10 2008 13726. For 10, s(x) is 1; Putting value of s(x) in equation b*(s(x)^a)+c we get 10, and as 10 lies in range 0<x<1e+9, therefore 10 is a possible answer, similar for 2008 and 13726. No other value of x satisfies the equation for the given value of a, b and c

Input: a = 2, b = 2, c = -1
Output: 1 31 337 967
Values of x that satisfy the equation are: 1 31 337 967


Approach:
sumdigits(x) can be in the range of 1<=s(X)<=81 for the given range of x i.e 0<x<1e+9. This is because value of x can be minimum 0 where sumdigits(x)=0 and maximum 999999999 where sumdigits(x) is 81. So first iterate through 1 to 81 to find x from the given equation, then cross check if the sum of digits of the number found, is same as the value of sum sumdigits(x). If both are same then increase the counter and store the result in an array.

Below is the implementation of the above approach:

C++

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// C++ program to find the numbers of
// values that satisfy the equation
#include <bits/stdc++.h>
using namespace std;
  
// This function returns the sum of
// the digits of a number
int getsum(int a)
{
    int r = 0, sum = 0;
    while (a > 0) {
        r = a % 10;
        sum = sum + r;
        a = a / 10;
    }
    return sum;
}
  
// This function creates
// the array of valid numbers
void value(int a, int b, int c)
{
    int co = 0, p = 0;
    int no, r = 0, x = 0, q = 0, w = 0;
    vector<int> v;
  
    for (int i = 1; i < 82; i++) {
  
        // this computes s(x)^a
        no = pow((double)i, double(a));
  
        // this gives the result of equation
        no = b * no + c;
  
        if (no > 0 && no < 1000000000) {
            x = getsum(no);
  
            // checking if the sum same as i
            if (x == i) {
  
                // counter to keep track of numbers
                q++;
  
                // resultant array
                v.push_back(no);
                w++;
            }
        }
    }
  
    // prints the number
    for (int i = 0; i < v.size(); i++) {
        cout << v[i] << " ";
    }
}
  
// Driver Code
int main()
{
    int a = 2, b = 2, c = -1;
  
    // calculate which value
    // of x are possible
    value(a, b, c);
  
    return 0;
}

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Java

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// Java program to find the numbers of
// values that satisfy the equation
import java.util.Vector;
  
class GFG
{
  
// This function returns the sum of
// the digits of a number
static int getsum(int a)
{
    int r = 0, sum = 0;
    while (a > 0
    {
        r = a % 10;
        sum = sum + r;
        a = a / 10;
    }
    return sum;
}
  
// This function creates
// the array of valid numbers
static void value(int a, int b, int c)
{
    int co = 0, p = 0;
    int no, r = 0, x = 0, q = 0, w = 0;
    Vector<Integer> v = new Vector<Integer>();
  
    for (int i = 1; i < 82; i++) 
    {
  
        // this computes s(x)^a
        no = (int) Math.pow(i, a);
  
        // this gives the result of equation
        no = b * no + c;
  
        if (no > 0 && no < 1000000000)
        {
            x = getsum(no);
  
            // checking if the sum same as i
            if (x == i)
            {
  
                // counter to keep track of numbers
                q++;
  
                // resultant array
                v.add(no);
                w++;
            }
        }
    }
  
    // prints the number
    for (int i = 0; i < v.size(); i++) 
    {
        System.out.print(v.get(i)+" ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int a = 2, b = 2, c = -1;
  
    // calculate which value
    // of x are possible
    value(a, b, c);
    }
  
// This code is contributed by 
// PrinciRaj1992

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Python 3

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# Python 3 program to find the numbers 
# of values that satisfy the equation
  
# This function returns the sum 
# of the digits of a number
def getsum(a):
  
    r = 0
    sum = 0
    while (a > 0) :
        r = a % 10
        sum = sum + r
        a = a // 10
      
    return sum
  
# This function creates
# the array of valid numbers
def value(a, b, c):
  
    x = 0
    q = 0
    w = 0
    v = []
  
    for i in range(1, 82) :
  
        # this computes s(x)^a
        no = pow(i, a)
  
        # this gives the result 
        # of equation
        no = b * no + c
  
        if (no > 0 and no < 1000000000) :
            x = getsum(no)
              
            # checking if the sum same as i
            if (x == i) :
  
                # counter to keep track 
                # of numbers
                q += 1
  
                # resultant array
                v.append(no)
                w += 1
                  
    # prints the number
    for i in range(len(v)) :
        print(v[i], end = " ")
  
# Driver Code
if __name__ == "__main__":
      
    a = 2
    b = 2
    c = -1
  
    # calculate which value
    # of x are possible
    value(a, b, c)
  
# This code is contributed 
# by ChitraNayal

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C#

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// C# program to find the numbers of 
// values that satisfy the equation 
using System;
using System.Collections.Generic;
  
class GFG 
  
    // This function returns the sum of 
    // the digits of a number 
    static int getsum(int a) 
    
        int r = 0, sum = 0; 
        while (a > 0) 
        
            r = a % 10; 
            sum = sum + r; 
            a = a / 10; 
        
        return sum; 
    
  
    // This function creates 
    // the array of valid numbers 
    static void value(int a, int b, int c) 
    
        int no, x = 0, q = 0, w = 0; 
        List<int> v = new List<int>(); 
  
        for (int i = 1; i < 82; i++) 
        
  
            // this computes s(x)^a 
            no = (int) Math.Pow(i, a); 
  
            // this gives the result of equation 
            no = b * no + c; 
  
            if (no > 0 && no < 1000000000) 
            
                x = getsum(no); 
  
                // checking if the sum same as i 
                if (x == i) 
                
  
                    // counter to keep track of numbers 
                    q++; 
  
                    // resultant array 
                    v.Add(no); 
                    w++; 
                
            
        
  
        // prints the number 
        for (int i = 0; i < v.Count; i++) 
        
            Console.Write(v[i]+" "); 
        
    
  
    // Driver Code 
    public static void Main(String[] args) 
    
        int a = 2, b = 2, c = -1; 
  
        // calculate which value 
        // of x are possible 
        value(a, b, c); 
    
}
  
// This code has been contributed by Rajput-Ji

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PHP

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<?php
// PHP program to find the numbers of
// values that satisfy the equation
  
// This function returns the sum of
// the digits of a number
function getsum($a)
{
    $r = 0;
    $sum = 0;
    while ($a > 0) 
    {
        $r = $a % 10;
        $sum = $sum + $r;
        $a = (int)($a / 10);
    }
    return $sum;
}
  
// This function creates
// the array of valid numbers
function value($a, $b, $c)
{
    $co = 0;
    $p = 0;
    $no;
    $r = 0;
    $x = 0;
    $q = 0;
    $w = 0;
    $v = array();
    $u = 0;
  
    for ($i = 1; $i < 82; $i++) 
    {
  
        // this computes s(x)^a
        $no = pow($i, $a);
  
        // this gives the result
        // of equation
        $no = $b * $no + $c;
  
        if ($no > 0 && $no < 1000000000) 
        {
            $x = getsum($no);
  
            // checking if the 
            // sum same as i
            if ($x == $i
            {
  
                // counter to keep
                // track of numbers
                $q++;
  
                // resultant array
                $v[$u++] = $no;
                $w++;
            }
        }
    }
  
    // prints the number
    for ($i = 0; $i < $u; $i++) 
    {
        echo $v[$i] . " ";
    }
}
  
// Driver Code
$a = 2;
$b = 2;
$c = -1;
  
// calculate which value
// of x are possible
value($a, $b, $c);
  
// This code is contributed 
// by mits
?>

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Output:

1 31 337 967

Time Complexity: O(N)
Auxiliary Space: O(N)



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