# Find the minimum number to be added to N to make it a power of K

Given two positive integers N and K, the task is to find the minimum number to be added to N to make it a power of K.

Examples:

Input: N = 9, K = 10
Output: 1
Explanation:
9 + 1 = 10 = 101

Input: N = 20, K = 5
Output: 5
Explanation:
20 + 5 = 25 = 52

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

Approach: The idea to solve this problem is to observe that the minimum power of K which can be formed from N is the next greater power of K. So, the idea is to find the next greater power of K and find the difference between N and this number. The next greater power of K can be found by the formula,

Kint(log(N)/log(K)) + 1

Therefore, the minimum number to be added can be computed by:

Minimum Number = Kint(log(N)/log(K)) + 1 – N

Below is the implementation of the above approach:

## C++

 `// C++ program to find the minimum number ` `// to be added to N to make it a power of K ` ` `  `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to return the minimum number ` `// to be added to N to make it a power of K. ` `int` `minNum(``int` `n, ``int` `k) ` `{ ` `    ``int` `x = (``int``)(``log``(n) / ``log``(k)) + 1; ` ` `  `    ``// Computing the difference between ` `    ``// then next greater power of K ` `    ``// and N ` `    ``int` `mn = ``pow``(k, x) - n; ` `    ``return` `mn; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 20, k = 5; ` `    ``cout << minNum(n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the minimum number ` `// to be added to N to make it a power of K ` ` `  `class` `GFG{ ` `  `  `// Function to return the minimum number ` `// to be added to N to make it a power of K. ` `static` `int` `minNum(``int` `n, ``int` `k) ` `{ ` `    ``int` `x = (``int``)(Math.log(n) / Math.log(k)) + ``1``; ` `  `  `    ``// Computing the difference between ` `    ``// then next greater power of K ` `    ``// and N ` `    ``int` `mn = (``int``) (Math.pow(k, x) - n); ` `    ``return` `mn; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``20``, k = ``5``; ` `    ``System.out.print(minNum(n, k)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## Python3

 `# Python3 program to find the minimum number  ` `# to be added to N to make it a power of K  ` `import` `math ` ` `  `# Function to return the minimum number  ` `# to be added to N to make it a power of K. ` `def` `minNum(n, k): ` `     `  `    ``x ``=` `int``((math.log(n) ``/``/` `math.log(k))) ``+` `1` `     `  `    ``# Computing the difference between  ` `    ``# then next greater power of K  ` `    ``# and N  ` `    ``mn ``=` `pow``(k, x) ``-` `n ` `    ``return` `mn ` `     `  `# Driver code  ` `if` `__name__``=``=``'__main__'``: ` `     `  `    ``n ``=` `20` `    ``k ``=` `5` `    ``print``(minNum(n, k)) ` ` `  `# This code is contributed by rutvik_56 `

Output:

```5
```

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Improved By : amit143katiyar, rutvik_56