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

• Difficulty Level : Basic
• Last Updated : 13 May, 2021

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 the 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 the same as the value of sum sumdigits(x). If both are the same then increase the counter and store the result in an array.
Below is the implementation of the above approach:

## C++

 `// C++ program to find the numbers of``// values that satisfy the equation``#include ``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;``}`

## Java

 `// 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 v = ``new` `Vector();` `    ``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`

## Python 3

 `# 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`

## C#

 `// 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`

## PHP

 ` 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``?>`

## Javascript

 ``

Output:

`1 31 337 967`

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

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