# Find the number of solutions to the given equation

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: 3
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: 1

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

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

 `

Output:

```3
```

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