# Sum of first K numbers which are not divisible by N

Given two numbers N and K, the task is to find the sum of first K numbers which are not divisible by N.

Examples:

Input: N = 5, K = 10
Output: 63
Explanation: Sum of { 1, 2, 3, 4, 6, 7, 8, 9, 11, 12 } is 63.

Input: N = 3, k = 13
Output: 127
Explanation: Sum of { 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19 } is 127.

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

Approach: In order to solve this problem, we ned to follow the following steps:

• Calculate the last multiple of N by (K / (N – 1)) * N
• Calculte K%(N – 1). If the remainder is 0, (K / (N – 1)) * N – 1 gives the last value which is not divisible by N. Otherwise, we need to add the remainder to (K / (N – 1)) * N.
• Calculate the sum up to the last value which is not divisible by N obtained in the above step.
• Calculate the sum of multiples of N and subtract from the sum calculated in the above step to get the desired result.

Below code is the implementation of the above approach:

## C++

 `// C++ Program to calculate ` `// the sum of first K ` `// numbers not divisible by N ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the sum ` `int` `findSum(``int` `n, ``int` `k) ` `{ ` `    ``// Find the last multiple of N ` `    ``int` `val = (k / (n - 1)) * n; ` ` `  `    ``int` `rem = k % (n - 1); ` ` `  `    ``// Find the K-th non-multiple of N ` `    ``if` `(rem == 0) { ` `        ``val = val - 1; ` `    ``} ` `    ``else` `{ ` `        ``val = val + rem; ` `    ``} ` ` `  `    ``// Calculate the  sum of ` `    ``// all elements from 1 to val ` `    ``int` `sum = (val * (val + 1)) / 2; ` ` `  `    ``// Calculate the sum of ` `    ``// all multiples of N ` `    ``// between 1 to val ` `    ``int` `x = k / (n - 1); ` `    ``int` `sum_of_multiples ` `        ``= (x ` `           ``* (x + 1) * n) ` `          ``/ 2; ` `    ``sum -= sum_of_multiples; ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 7, k = 13; ` `    ``cout << findSum(n, k) ` `         ``<< endl; ` `} `

## Java

 `// Java program to calculate  ` `// the sum of first K numbers ` `// not divisible by N ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `// Function to find the sum ` `static` `int` `findSum(``int` `n, ``int` `k) ` `{ ` `     `  `    ``// Find the last multiple of N ` `    ``int` `val = (k / (n - ``1``)) * n; ` ` `  `    ``int` `rem = k % (n - ``1``); ` ` `  `    ``// Find the K-th non-multiple of N ` `    ``if` `(rem == ``0``) ` `    ``{ ` `        ``val = val - ``1``; ` `    ``} ` `    ``else`  `    ``{ ` `        ``val = val + rem; ` `    ``} ` ` `  `    ``// Calculate the sum of ` `    ``// all elements from 1 to val ` `    ``int` `sum = (val * (val + ``1``)) / ``2``; ` ` `  `    ``// Calculate the sum of ` `    ``// all multiples of N ` `    ``// between 1 to val ` `    ``int` `x = k / (n - ``1``); ` `    ``int` `sum_of_multiples = (x * (x + ``1``) * n) / ``2``; ` `    ``sum -= sum_of_multiples; ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``7``, k = ``13``; ` ` `  `    ``System.out.println(findSum(n, k)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 Program to calculate ` `# the sum of first K ` `# numbers not divisible by N ` ` `  `# Function to find the sum ` `def` `findSum(n, k): ` ` `  `    ``# Find the last multiple of N ` `    ``val ``=` `(k ``/``/` `(n ``-` `1``)) ``*` `n; ` ` `  `    ``rem ``=` `k ``%` `(n ``-` `1``); ` ` `  `    ``# Find the K-th non-multiple of N ` `    ``if` `(rem ``=``=` `0``): ` `        ``val ``=` `val ``-` `1``; ` `     `  `    ``else``: ` `        ``val ``=` `val ``+` `rem; ` `     `  `    ``# Calculate the sum of ` `    ``# all elements from 1 to val ` `    ``sum` `=` `(val ``*` `(val ``+` `1``)) ``/``/` `2``; ` ` `  `    ``# Calculate the sum of ` `    ``# all multiples of N ` `    ``# between 1 to val ` `    ``x ``=` `k ``/``/` `(n ``-` `1``); ` `    ``sum_of_multiples ``=` `(x ``*` `(x ``+` `1``) ``*` `n) ``/``/` `2``; ` `    ``sum` `-``=` `sum_of_multiples; ` ` `  `    ``return` `sum``; ` ` `  `# Driver code ` `n ``=` `7``; k ``=` `13``; ` `print``(findSum(n, k)) ` ` `  `# This code is contributed by Code_Mech `

## C#

 `// C# program to calculate  ` `// the sum of first K numbers ` `// not divisible by N ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find the sum ` `static` `int` `findSum(``int` `n, ``int` `k) ` `{ ` `     `  `    ``// Find the last multiple of N ` `    ``int` `val = (k / (n - 1)) * n; ` ` `  `    ``int` `rem = k % (n - 1); ` ` `  `    ``// Find the K-th non-multiple of N ` `    ``if` `(rem == 0) ` `    ``{ ` `        ``val = val - 1; ` `    ``} ` `    ``else` `    ``{ ` `        ``val = val + rem; ` `    ``} ` ` `  `    ``// Calculate the sum of ` `    ``// all elements from 1 to val ` `    ``int` `sum = (val * (val + 1)) / 2; ` ` `  `    ``// Calculate the sum of ` `    ``// all multiples of N ` `    ``// between 1 to val ` `    ``int` `x = k / (n - 1); ` `    ``int` `sum_of_multiples = (x * (x + 1) * n) / 2; ` `    ``sum -= sum_of_multiples; ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 7, k = 13; ` ` `  `    ``Console.WriteLine(findSum(n, k)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```99
```

Time Complexity: O(1)
Auxilary Space Complexity: O(1)

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Improved By : offbeat, 29AjayKumar, Code_Mech

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