Sum of all numbers in the given range which are divisible by M
Last Updated :
12 Jul, 2025
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 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
# 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# 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
<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>
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.
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 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 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# 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
<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>
Time Complexity: O(1)
Auxiliary Space: O(1)
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