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; }

## 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)

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