Given a range L-R, find the sum of all numbers divisible by 6 in range L-R

L and R are very large.

Examples:

Input : 1 20 Output : 36 Explanation: 6 + 12 + 18 = 36 Input : 5 7 Output : 6 Explanation: 6 is the only divisible number in range 5-7

A **naive approach **is be to run a loop from L to R and sum up all the numbers divisible by 6.

An **efficient approach **is to sum all the numbers divisible by 6 up to R in sum, and sum all numbers divisible by 6 up to L-1. And then there subtraction will be the answer.

sum = 6 + 12 + 8 + …….(R/6)terms.

sum = 6(1 + 2 + 3……R/6 terms)

sumR = 3 * (R/6) * (R/6+1)

similarly we get

sumL as 3 * ((L-1)/6) * ((L-1/6)+1)

and the final answer as sumR – sumL.

## C++

// CPP program to find sum of numbers divisible // by 6 in a given range. #include <bits/stdc++.h> using namespace std; // function to calculate the sum of // all numbers divisible by 6 in range L-R.. int sum(int L, int R) { // no of multiples of 6 upto r int p = R / 6; // no of multiples of 6 upto l-1 int q = (L - 1) / 6; // summation of all multiples of 6 upto r int sumR = 3 * (p * (p + 1)); // summation of all multiples of 6 upto l-1 int sumL = (q * (q + 1)) * 3; // returns the answer return sumR - sumL; } // driver program to test the above function int main() { int L = 1, R = 20; cout << sum(L, R); return 0; }

## Java

// Java program to find sum of numbers // divisible by 6 in a given range. import java.io.*; class GFG { // function to calculate the sum // of all numbers divisible by 6 // in range L-R.. static int sum(int L, int R) { // no of multiples of 6 upto r int p = R / 6; // no of multiples of 6 upto l-1 int q = (L - 1) / 6; // summation of all multiples of // 6 upto r int sumR = 3 * (p * (p + 1)); // summation of all multiples of // 6 upto l-1 int sumL = (q * (q + 1)) * 3; // returns the answer return sumR - sumL; } // driver program public static void main(String[] args) { int L = 1, R = 20; System.out.println(sum(L, R)); } } // This code is contributed by Prerna Saini

## Python

# Python3 program to find sum of numbers divisible # by 6 in a given range. def sumDivisible(L, R): # no of multiples of 6 upto r p = int(R/6) # no of multiples of 6 upto l-1 q = int((L-1)/6) # summation of all multiples of 6 upto r sumR = 3 * (p * (p + 1)) # summation of all multiples of 6 upto l-1 sumL = (q * (q + 1)) * 3 # returns the answer return sumR - sumL # driver code L = 1 R = 20 print(sumDivisible(L,R)) # This code is contributed by 'Abhishek Sharma 44'.

Output:

36

Time Complexity: O(1)

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.