Sum of all numbers divisible by 6 in a given range

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

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

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'. 

C#

// C# program to find sum of numbers 
// divisible by 6 in a given range. 
using System; 
  
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() 
    { 
        int L = 1, R = 20; 
          
        Console.WriteLine(sum(L, R)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

PHP

<?php 
// PHP program to find sum of numbers 
// divisible by 6 in a given range. 
  
// function to calculate the sum of 
// all numbers divisible by 6 in range L-R.. 
function sum($L, $R) 
{ 
      
    // no of multiples of 6 upto r 
    $p = intval($R / 6); 
  
    // no of multiples of 6 upto l-1 
    $q = intval(($L - 1) / 6); 
  
    // summation of all multiples  
    // of 6 upto r 
    $sumR = intval(3 * ($p * ($p + 1))); 
  
    // summation of all multiples  
    // of 6 upto l-1 
    $sumL = intval(($q * ($q + 1)) * 3); 
  
    // returns the answer 
    return $sumR - $sumL; 
} 
  
// Driver Code 
$L = 1; 
$R = 20; 
echo sum($L, $R); 
  
// This code is contributed by Sam007 
?> 

Output:

Time Complexity: O(1)

My Personal Notes

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

C++

Java

Python

C#

PHP

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