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Find the number of solutions to the given equation
• Difficulty Level : Expert
• Last Updated : 03 May, 2021

Given three integers A, B and C, the task is to find the count of values of X such that the following condition is satisfied,
X = B * Sm(X)A + C where Sm(X) denotes the sum of digits of X and 1 < X < 109.
Examples:

Input: A = 3, B = 2, C = 8
Output:
For X = 10, 2 * (1)3 + 8 = 10
For X = 2008, 2 * (10)3 + 8 = 2008
For X = 13726, 2 * (19)3 + 8 = 13726
Input: A = 2, B = 3, C = 10
Output:

Approach: An important observation can be made here that the sum of digits can be atmost 81 (i.e. X = 999999999) and corresponding to each sum of digits, we get a single value of X. So we can iterate through each sum of digit and check if the resulting value of X is valid or not.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of``// valid values of X``int` `getCount(``int` `a, ``int` `b, ``int` `c)``{``    ``int` `count = 0;` `    ``// Iterate through all possible``    ``// sum of digits of X``    ``for` `(``int` `i = 1; i <= 81; i++) {` `        ``// Get current value of X for sum of digits i``        ``int` `cr = b * ``pow``(i, a) + c;` `        ``int` `tmp = cr;``        ``int` `sm = 0;` `        ``// Find sum of digits of cr``        ``while` `(tmp) {``            ``sm += tmp % 10;``            ``tmp /= 10;``        ``}` `        ``// If cr is a valid choice for X``        ``if` `(sm == i && cr < 1e9)``            ``count++;``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `a = 3, b = 2, c = 8;``    ``cout << getCount(a, b, c);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GfG``{` `// Function to return the count of``// valid values of X``static` `int` `getCount(``int` `a, ``int` `b, ``int` `c)``{``    ``int` `count = ``0``;` `    ``// Iterate through all possible``    ``// sum of digits of X``    ``for` `(``int` `i = ``1``; i <= ``81``; i++)``    ``{` `        ``// Get current value of X for sum of digits i``        ``int` `cr = b * (``int``)Math.pow(i, a) + c;` `        ``int` `tmp = cr;``        ``int` `sm = ``0``;` `        ``// Find sum of digits of cr``        ``while` `(tmp != ``0``)``        ``{``            ``sm += tmp % ``10``;``            ``tmp /= ``10``;``        ``}` `        ``// If cr is a valid choice for X``        ``if` `(sm == i && cr < 1e9)``            ``count++;``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a = ``3``, b = ``2``, c = ``8``;``    ``System.out.println(getCount(a, b, c));``}``}` `// This code is contributed by Prerna Saini.`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# valid values of X``def` `getCount(a, b, c):` `    ``count ``=` `0` `    ``# Iterate through all possible``    ``# sum of digits of X``    ``for` `i ``in` `range``(``1``, ``82``):` `        ``# Get current value of X for``        ``# sum of digits i``        ``cr ``=` `b ``*` `pow``(i, a) ``+` `c` `        ``tmp ``=` `cr``        ``sm ``=` `0` `        ``# Find sum of digits of cr``        ``while` `(tmp):``            ``sm ``+``=` `tmp ``%` `10``            ``tmp ``/``/``=` `10``        ` `        ``# If cr is a valid choice for X``        ``if` `(sm ``=``=` `i ``and` `cr < ``10``*``*``9``):``            ``count ``+``=` `1``    ` `    ``# Return the count``    ``return` `count` `# Driver code``a, b, c ``=` `3``, ``2``, ``8``print``(getCount(a, b, c))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the count of``// valid values of X``static` `int` `getCount(``int` `a, ``int` `b, ``int` `c)``{``    ``int` `count = 0;` `    ``// Iterate through all possible``    ``// sum of digits of X``    ``for` `(``int` `i = 1; i <= 81; i++)``    ``{` `        ``// Get current value of X for sum``        ``// of digits i``        ``int` `cr = b * (``int``)Math.Pow(i, a) + c;` `        ``int` `tmp = cr;``        ``int` `sm = 0;` `        ``// Find sum of digits of cr``        ``while` `(tmp != 0)``        ``{``            ``sm += tmp % 10;``            ``tmp /= 10;``        ``}` `        ``// If cr is a valid choice for X``        ``if` `(sm == i && cr < 1e9)``            ``count++;``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `a = 3, b = 2, c = 8;``    ``Console.Write(getCount(a, b, c));``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

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## Javascript

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Output:
`3`

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