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Sum of first K numbers which are not divisible by N
• Last Updated : 26 May, 2021

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.

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 ``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

 ``
Output:
`99`

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

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