# Sum of all numbers in the given range which are divisible by M

Given three numbers A, B and M such that A < B, the task is to find the sum of numbers divisible by M in the range [A, B].

Examples:

Input: A = 25, B = 100, M = 30
Output: 180
Explanation:
In the given range [25, 100] 30, 60 and 90 are the numbers which are divisible by M = 30
Therefore, sum of these numbers = 180.

Input: A = 6, B = 15, M = 3
Output: 42
Explanation:
In the given range [6, 15] 6, 9, 12 and 15 are the numbers which are divisible by M = 3.
Therefore, sum of these numbers = 42.

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

Naive Approach: Check for each number in the range [A, B] if they are divisible by M or not. And finally, add all the numbers that are divisible by M.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the sum of numbers ` `// divisible by M in the given range ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the sum of numbers ` `// divisible by M in the given range ` `int` `sumDivisibles(``int` `A, ``int` `B, ``int` `M) ` `{ ` `    ``// Variable to store the sum ` `    ``int` `sum = 0; ` ` `  `    ``// Running a loop from A to B and check ` `    ``// if a number is divisible by i. ` `    ``for` `(``int` `i = A; i <= B; i++) ` ` `  `        ``// If the number is divisible, ` `        ``// then add it to sum ` `        ``if` `(i % M == 0) ` `            ``sum += i; ` ` `  `    ``// Return the sum ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// A and B define the range ` `    ``// M is the dividend ` `    ``int` `A = 6, B = 15, M = 3; ` ` `  `    ``// Printing the result ` `    ``cout << sumDivisibles(A, B, M) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the sum of numbers ` `// divisible by M in the given range ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to find the sum of numbers ` `// divisible by M in the given range ` `static` `int` `sumDivisibles(``int` `A, ``int` `B, ``int` `M) ` `{ ` `    ``// Variable to store the sum ` `    ``int` `sum = ``0``; ` `  `  `    ``// Running a loop from A to B and check ` `    ``// if a number is divisible by i. ` `    ``for` `(``int` `i = A; i <= B; i++) ` `  `  `        ``// If the number is divisible, ` `        ``// then add it to sum ` `        ``if` `(i % M == ``0``) ` `            ``sum += i; ` `  `  `    ``// Return the sum ` `    ``return` `sum; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// A and B define the range ` `    ``// M is the dividend ` `    ``int` `A = ``6``, B = ``15``, M = ``3``; ` `  `  `    ``// Printing the result ` `    ``System.out.print(sumDivisibles(A, B, M) +``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python 3 program to find the sum of numbers ` `# divisible by M in the given range ` ` `  `# Function to find the sum of numbers ` `# divisible by M in the given range ` `def` `sumDivisibles(A, B, M): ` ` `  `    ``# Variable to store the sum ` `    ``sum` `=` `0` ` `  `    ``# Running a loop from A to B and check ` `    ``# if a number is divisible by i. ` `    ``for` `i ``in` `range``(A, B ``+` `1``): ` ` `  `        ``# If the number is divisible, ` `        ``# then add it to sum ` `        ``if` `(i ``%` `M ``=``=` `0``): ` `            ``sum` `+``=` `i ` ` `  `    ``# Return the sum ` `    ``return` `sum` ` `  `# Driver code ` `if` `__name__``=``=``"__main__"``: ` `     `  `    ``# A and B define the range ` `    ``# M is the dividend ` `    ``A ``=` `6` `    ``B ``=` `15` `    ``M ``=` `3` ` `  `    ``# Printing the result ` `    ``print``(sumDivisibles(A, B, M)) ` `     `  `# This code is contributed by chitranayal `

## C#

 `// C# program to find the sum of numbers ` `// divisible by M in the given range ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Function to find the sum of numbers ` `// divisible by M in the given range ` `static` `int` `sumDivisibles(``int` `A, ``int` `B, ``int` `M) ` `{ ` `    ``// Variable to store the sum ` `    ``int` `sum = 0; ` `   `  `    ``// Running a loop from A to B and check ` `    ``// if a number is divisible by i. ` `    ``for` `(``int` `i = A; i <= B; i++) ` `   `  `        ``// If the number is divisible, ` `        ``// then add it to sum ` `        ``if` `(i % M == 0) ` `            ``sum += i; ` `   `  `    ``// Return the sum ` `    ``return` `sum; ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// A and B define the range ` `    ``// M is the dividend ` `    ``int` `A = 6, B = 15, M = 3; ` `   `  `    ``// Printing the result ` `    ``Console.Write(sumDivisibles(A, B, M) +``"\n"``); ` `} ` `} ` `  `  `// This code is contributed by sapnasingh4991 `

Output:

```42
```

Time Complexity: O(N).

Efficient Approach: The idea is to use the concept of Arithmetic Progression and divisibility.

• Upon visualization, the multiples of M can be seen to form a series
```M, 2M, 3M, ...
```
• If we can find the value of K which is the first term in the range [A, B] which is divisible by M, then directly, the series would be:
```K, (K + M), (K + 2M), ------  (K + (N - 1)*M )
where N is the number of elements in the series.
```
• Therefore, the first term ‘K’ in the series is nothing but the largest number smaller than or equal to A that is divisible by M.
• Similarly, the last term is the smallest number greater than or equal B that is divisible by M.
• However, if any of the above numbers exceed out of the range, then we can directly subtract M from it to bring it into the range.
• And, the number of terms divisible by M can be found out by the formula:
```N = B / M - (A - 1)/ M
```
• Therefore, the sum of the elements can be found out by:
```sum = N * ( (first term + last term) / 2)
```

Below is the implementation of the above approach:

## C++

 `// C++ program to find the sum of numbers ` `// divisible by M in the given range ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the largest number ` `// smaller than or equal to N ` `// that is divisible by K ` `int` `findSmallNum(``int` `N, ``int` `K) ` `{ ` `    ``// Finding the remainder when N is ` `    ``// divided by K ` `    ``int` `rem = N % K; ` ` `  `    ``// If the remainder is 0, then the ` `    ``// number itself is divisible by K ` `    ``if` `(rem == 0) ` `        ``return` `N; ` `    ``else` ` `  `        ``// Else, then the difference between ` `        ``// N and remainder is the largest number ` `        ``// which is divisible by K ` `        ``return` `N - rem; ` `} ` ` `  `// Function to find the smallest number ` `// greater than or equal to N ` `// that is divisible by K ` `int` `findLargeNum(``int` `N, ``int` `K) ` `{ ` `    ``// Finding the remainder when N is ` `    ``// divided by K ` `    ``int` `rem = (N + K) % K; ` ` `  `    ``// If the remainder is 0, then the ` `    ``// number itself is divisible by K ` `    ``if` `(rem == 0) ` `        ``return` `N; ` `    ``else` ` `  `        ``// Else, then the difference between ` `        ``// N and remainder is the largest number ` `        ``// which is divisible by K ` `        ``return` `N + K - rem; ` `} ` ` `  `// Function to find the sum of numbers ` `// divisible by M in the given range ` `int` `sumDivisibles(``int` `A, ``int` `B, ``int` `M) ` `{ ` `    ``// Variable to store the sum ` `    ``int` `sum = 0; ` `    ``int` `first = findSmallNum(A, M); ` `    ``int` `last = findLargeNum(B, M); ` ` `  `    ``// To bring the smallest and largest ` `    ``// numbers in the range [A, B] ` `    ``if` `(first < A) ` `        ``first += M; ` ` `  `    ``if` `(last > B) ` `        ``first -= M; ` ` `  `    ``// To count the number of terms in the AP ` `    ``int` `n = (B / M) - (A - 1) / M; ` ` `  `    ``// Sum of n terms of an AP ` `    ``return` `n * (first + last) / 2; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// A and B define the range, ` `    ``// M is the dividend ` `    ``int` `A = 6, B = 15, M = 3; ` ` `  `    ``// Printing the result ` `    ``cout << sumDivisibles(A, B, M); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the sum of numbers ` `// divisible by M in the given range ` ` `  ` `  `class` `GFG{ ` `  `  `// Function to find the largest number ` `// smaller than or equal to N ` `// that is divisible by K ` `static` `int` `findSmallNum(``int` `N, ``int` `K) ` `{ ` `    ``// Finding the remainder when N is ` `    ``// divided by K ` `    ``int` `rem = N % K; ` `  `  `    ``// If the remainder is 0, then the ` `    ``// number itself is divisible by K ` `    ``if` `(rem == ``0``) ` `        ``return` `N; ` `    ``else` `  `  `        ``// Else, then the difference between ` `        ``// N and remainder is the largest number ` `        ``// which is divisible by K ` `        ``return` `N - rem; ` `} ` `  `  `// Function to find the smallest number ` `// greater than or equal to N ` `// that is divisible by K ` `static` `int` `findLargeNum(``int` `N, ``int` `K) ` `{ ` `    ``// Finding the remainder when N is ` `    ``// divided by K ` `    ``int` `rem = (N + K) % K; ` `  `  `    ``// If the remainder is 0, then the ` `    ``// number itself is divisible by K ` `    ``if` `(rem == ``0``) ` `        ``return` `N; ` `    ``else` `  `  `        ``// Else, then the difference between ` `        ``// N and remainder is the largest number ` `        ``// which is divisible by K ` `        ``return` `N + K - rem; ` `} ` `  `  `// Function to find the sum of numbers ` `// divisible by M in the given range ` `static` `int` `sumDivisibles(``int` `A, ``int` `B, ``int` `M) ` `{ ` `    ``// Variable to store the sum ` `    ``int` `first = findSmallNum(A, M); ` `    ``int` `last = findLargeNum(B, M); ` `  `  `    ``// To bring the smallest and largest ` `    ``// numbers in the range [A, B] ` `    ``if` `(first < A) ` `        ``first += M; ` `  `  `    ``if` `(last > B) ` `        ``first -= M; ` `  `  `    ``// To count the number of terms in the AP ` `    ``int` `n = (B / M) - (A - ``1``) / M; ` `  `  `    ``// Sum of n terms of an AP ` `    ``return` `n * (first + last) / ``2``; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// A and B define the range, ` `    ``// M is the dividend ` `    ``int` `A = ``6``, B = ``15``, M = ``3``; ` `  `  `    ``// Printing the result ` `    ``System.out.print(sumDivisibles(A, B, M)); ` `  `  `} ` `} ` ` `  `// This code contributed by Princi Singh `

## Python 3

 `# Python 3 program to find the sum of numbers ` `# divisible by M in the given range ` ` `  `# Function to find the largest number ` `# smaller than or equal to N ` `# that is divisible by K ` `def` `findSmallNum(N, K): ` `     `  `    ``# Finding the remainder when N is ` `    ``# divided by K ` `    ``rem ``=` `N ``%` `K ` ` `  `    ``# If the remainder is 0, then the ` `    ``# number itself is divisible by K ` `    ``if` `(rem ``=``=` `0``): ` `        ``return` `N ` `    ``else``: ` `        ``# Else, then the difference between ` `        ``# N and remainder is the largest number ` `        ``# which is divisible by K ` `        ``return` `N ``-` `rem ` ` `  `# Function to find the smallest number ` `# greater than or equal to N ` `# that is divisible by K ` `def` `findLargeNum(N, K): ` `     `  `    ``# Finding the remainder when N is ` `    ``# divided by K ` `    ``rem ``=` `(N ``+` `K) ``%` `K ` ` `  `    ``# If the remainder is 0, then the ` `    ``# number itself is divisible by K ` `    ``if` `(rem ``=``=` `0``): ` `        ``return` `N ` `    ``else``: ` `        ``# Else, then the difference between ` `        ``# N and remainder is the largest number ` `        ``# which is divisible by K ` `        ``return` `N ``+` `K ``-` `rem ` ` `  `# Function to find the sum of numbers ` `# divisible by M in the given range ` `def` `sumDivisibles(A, B, M): ` `     `  `    ``# Variable to store the sum ` `    ``sum` `=` `0` `    ``first ``=` `findSmallNum(A, M) ` `    ``last ``=` `findLargeNum(B, M) ` ` `  `    ``# To bring the smallest and largest ` `    ``# numbers in the range [A, B] ` `    ``if` `(first < A): ` `        ``first ``+``=` `M ` ` `  `    ``if` `(last > B): ` `        ``first ``-``=` `M ` ` `  `    ``# To count the number of terms in the AP ` `    ``n ``=` `(B ``/``/` `M) ``-` `(A ``-` `1``) ``/``/` `M ` ` `  `    ``# Sum of n terms of an AP ` `    ``return` `n ``*` `(first ``+` `last) ``/``/` `2` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``# A and B define the range, ` `    ``# M is the dividend ` `    ``A ``=` `6` `    ``B ``=` `15` `    ``M ``=` `3` ` `  `    ``# Printing the result ` `    ``print``(sumDivisibles(A, B, M)) ` ` `  `# This code is contributed by Surendra_Gangwar `

## C#

 `// C# program to find the sum of numbers ` `// divisible by M in the given range ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG{ ` `   `  `// Function to find the largest number ` `// smaller than or equal to N ` `// that is divisible by K ` `static` `int` `findSmallNum(``int` `N, ``int` `K) ` `{ ` `    ``// Finding the remainder when N is ` `    ``// divided by K ` `    ``int` `rem = N % K; ` `   `  `    ``// If the remainder is 0, then the ` `    ``// number itself is divisible by K ` `    ``if` `(rem == 0) ` `        ``return` `N; ` `    ``else` `   `  `        ``// Else, then the difference between ` `        ``// N and remainder is the largest number ` `        ``// which is divisible by K ` `        ``return` `N - rem; ` `} ` `   `  `// Function to find the smallest number ` `// greater than or equal to N ` `// that is divisible by K ` `static` `int` `findLargeNum(``int` `N, ``int` `K) ` `{ ` `    ``// Finding the remainder when N is ` `    ``// divided by K ` `    ``int` `rem = (N + K) % K; ` `   `  `    ``// If the remainder is 0, then the ` `    ``// number itself is divisible by K ` `    ``if` `(rem == 0) ` `        ``return` `N; ` `    ``else` `   `  `        ``// Else, then the difference between ` `        ``// N and remainder is the largest number ` `        ``// which is divisible by K ` `        ``return` `N + K - rem; ` `} ` `   `  `// Function to find the sum of numbers ` `// divisible by M in the given range ` `static` `int` `sumDivisibles(``int` `A, ``int` `B, ``int` `M) ` `{ ` `    ``// Variable to store the sum ` `    ``int` `first = findSmallNum(A, M); ` `    ``int` `last = findLargeNum(B, M); ` `   `  `    ``// To bring the smallest and largest ` `    ``// numbers in the range [A, B] ` `    ``if` `(first < A) ` `        ``first += M; ` `   `  `    ``if` `(last > B) ` `        ``first -= M; ` `   `  `    ``// To count the number of terms in the AP ` `    ``int` `n = (B / M) - (A - 1) / M; ` `   `  `    ``// Sum of n terms of an AP ` `    ``return` `n * (first + last) / 2; ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// A and B define the range, ` `    ``// M is the dividend ` `    ``int` `A = 6, B = 15, M = 3; ` `   `  `    ``// Printing the result ` `    ``Console.Write(sumDivisibles(A, B, M)); ` `   `  `} ` `} ` `  `  `// This code is contributed by Rajput-Ji `

Output:

```42
```

Time Complexity: O(1).

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