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. 

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 <bits/stdc++.h>
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

Javascript

<script>

// Javascript 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
function sumDivisibles(A, B, M)
{

    // Variable to store the sum
    var sum = 0;

    // Running a loop from A to B and check
    // if a number is divisible by i.
    for(var 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

// A and B define the range
// M is the dividend
var A = 6, B = 15, M = 3;

// Printing the result
document.write(sumDivisibles(A, B, M));

// This code is contributed by rrrtnx

</script>

Output:

42

Time Complexity: O(B-A).
Auxiliary Space: O(1)

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 <bits/stdc++.h>
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

Python3

# Python3 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

Javascript

<script>

// Javascript 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
function findSmallNum(N, K)
{

    // Finding the remainder when N is
    // divided by K
    var 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
function findLargeNum(N, K)
{

    // Finding the remainder when N is
    // divided by K
    var 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
function sumDivisibles(A, B, M)
{

    // Variable to store the sum
    var sum = 0;
    var first = findSmallNum(A, M);
    var 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
    var n = (parseInt(B / M) -
             parseInt((A - 1) / M));

    // Sum of n terms of an AP
    return n * (first + last) / 2;
}

// Driver code

// A and B define the range,
// M is the dividend
var A = 6, B = 15, M = 3;

// Printing the result
document.write( sumDivisibles(A, B, M));

// This code is contributed by rutvik_56

</script>

Output:

42

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