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

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

**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** - Calculate
**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 <bits/stdc++.h>` `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` |

## Javascript

`<script>` `// Javascript Program to calculate` `// the sum of first K` `// numbers not divisible by N` `// Function to find the sum` `function` `findSum(n, k)` `{` ` ` ` ` `// Find the last multiple of N` ` ` `var` `val = parseInt(k / (n - 1)) * n;` ` ` `var` `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` ` ` `var` `sum = parseInt((val * (val + 1)) / 2);` ` ` `// Calculate the sum of` ` ` `// all multiples of N` ` ` `// between 1 to val` ` ` `var` `x = parseInt(k / (n - 1));` ` ` `var` `sum_of_multiples = parseInt(` ` ` `(x * (x + 1) * n) / 2);` ` ` ` ` `sum -= sum_of_multiples;` ` ` `return` `sum;` `}` `// Driver code` `var` `n = 7, k = 13;` `document.write(findSum(n, k))` `// This code is contributed by rrrtnx` `</script>` |

**Output:**

99

**Time Complexity:** *O(1)* **Auxiliary Space Complexity:** *O(1)*