Given three positive integers** N**, **K**, and **M**, the task is to find the number to be added to **N** to obtain the sum of first **M** multiples of **K**.

**Examples:**

Input:N = 17, K = 3, M = 4Output:13Explanation:

Sum of first 4 multiples of 3 = (3 + 6 + 9 + 12) = 30.

Therefore, the value to be added to 17 is (30 – 17) = 13.

Therefore, the required output is 13.

Input:N = 5, K = 2, M = 1Output:-3Explanation:

Sum of first 1 multiples of 2 is 2.

The value to be added to 5 to get 2 is (2 – 5) = -3

**Approach: **Follow the steps below to solve the problem:

- Calculate the
**sum**of first**M**multiples of**K**, which will be equal to**K * (1 + 2 + 3 + … M)**=**K * M * (M + 1) / 2**. - Initialize a variable, say
**res**, to store the number required to be added to**N**to obtain**sum**. - Therefore,
**res**will be equal to**sum – N**. Pritn the value of**res**.

Below is the implementation of the above approach:

## C++

`// Python3 program for the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to print the value` `// to be added to N to obtain` `// sum of first M multiples of K` `static` `int` `printNumber(` `int` `N, ` `int` `K, ` `int` `M)` `{` ` ` ` ` `// Store the sum of the` ` ` `// first M multiples of K` ` ` `int` `sum = K * (M * (M + 1) / 2);` ` ` ` ` `// Store the value to be` ` ` `// added to obtain N` ` ` `return` `sum - N;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Input` ` ` `int` `N = 17;` ` ` `int` `K = 3;` ` ` `int` `M = 4;` ` ` `cout << printNumber(N, K, M);` ` ` `return` `0;` `}` `// This code is contributed by shubhamsingh10` |

## Java

`// Java code of above approach` `import` `java.util.*;` `class` `GFG` `{` ` ` `// Function to print the value` ` ` `// to be added to N to obtain` ` ` `// sum of first M multiples of K` ` ` `static` `int` `printNumber(` `int` `N, ` `int` `K, ` `int` `M)` ` ` `{` ` ` `// Store the sum of the` ` ` `// first M multiples of K` ` ` `int` `sum = K * (M * (M + ` `1` `) / ` `2` `);` ` ` `// Store the value to be` ` ` `// added to obtain N` ` ` `return` `sum - N;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Input` ` ` `int` `N = ` `17` `;` ` ` `int` `K = ` `3` `;` ` ` `int` `M = ` `4` `;` ` ` `System.out.print(printNumber(N, K, M));` ` ` `}` `}` `// This code is contributed by offbeat` |

## Python3

`# Python3 program for the above approach` `# Function to print the value` `# to be added to N to obtain` `# sum of first M multiples of K` `def` `printNumber(N, K, M):` ` ` `# Store the sum of the` ` ` `# first M multiples of K` ` ` `sum` `=` `K ` `*` `(M ` `*` `(M ` `+` `1` `) ` `/` `2` `)` ` ` `# Store the value to be` ` ` `# added to obtain N` ` ` `return` `sum` `-` `N` `# Driver Code` `# Input` `N ` `=` `17` `K ` `=` `3` `M ` `=` `4` `print` `(` `int` `(printNumber(N, K, M)))` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to print the value` ` ` `// to be added to N to obtain` ` ` `// sum of first M multiples of K` ` ` `static` `int` `printNumber(` `int` `N, ` `int` `K, ` `int` `M)` ` ` `{` ` ` `// Store the sum of the` ` ` `// first M multiples of K` ` ` `int` `sum = K * (M * (M + 1) / 2);` ` ` `// Store the value to be` ` ` `// added to obtain N` ` ` `return` `sum - N;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `// Input` ` ` `int` `N = 17;` ` ` `int` `K = 3;` ` ` `int` `M = 4;` ` ` `Console.Write(printNumber(N, K, M));` ` ` `}` `}` `// This code is contributed by shubhamsingh10` |

**Output**

13

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

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