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Sum of the numbers upto N that are divisible by 2 or 5

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  • Last Updated : 31 Jul, 2022
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Given a number n. The task is to find the sum of numbers up to n, that are divisible by 2 or 5.
Examples: 

Input: n = 2
Output: 2

Input: n = 5
Output: 11

A naive approach is to just iterate over the numbers up to n and check if it divisible by 2 or 5. If it is divisible then just add this number to our required sum. And finally, we got our total sum with a complexity of O(n).
Efficient Approach: 

 1. First find the numbers that are divisible by 2. So, these numbers for an AP, having 

first term = 2, difference = 2, Number of terms = n/2 
So, sum given by-

    \begin{align*} Sum=\frac{n*\left(2*a+(n-1)*d\right)}{2}\\ \end{align*} put the value, we got \begin{align*} Sum_2=\frac{\frac{n}{2}*\left(4+\left(\frac{n}{2}-1\right)*2\right)}{2}\\ \end{align*}

2. Secondly we find the numbers that are divisible by 5. So, these number for an AP, having 
 

first term = 5, difference = 5, Number of terms = n/5 
So, sum given by-

    \begin{align*} Sum=\frac{n*\left(2*a+(n-1)*d\right)}{2}\\ \end{align*} put the value, we got \begin{align*} Sum_5=\frac{\frac{n}{5}*\left(10+\left(\frac{n}{5}-1\right)*5\right)}{2}\\ \end{align*}

3. First we find the numbers that are divisible by 2 and 5.so, these number for an AP, having 
 

first term =10, difference = 10, Number of terms = n / 10 
So, sum given by-

    \begin{align*} Sum=\frac{n*\left(2*a+(n-1)*d\right)}{2}\\ \end{align*} put the value, we got \begin{align*} Sum_{10}=\frac{\frac{n}{10}*\left(20+\left(\frac{n}{10}-1\right)*10\right)}{2}\\ \end{align*}

4. As we have to find the sum of numbers divisible by 2 or 5. So, the required sum is given by-
 

sum = sum_2 + sum_5 – sum_10
 

Below is the implementation of the above approach: 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
 
// Function to find the sum
ll findSum(int n)
{
 
    ll sum2, sum5, sum10;
 
    // sum2 is sum of numbers divisible by 2
    sum2 = ((n / 2) * (4 + (n / 2 - 1) * 2)) / 2;
 
    // sum5 is sum of number divisible by 5
    sum5 = ((n / 5) * (10 + (n / 5 - 1) * 5)) / 2;
 
    // sum10 of numbers divisible by 2 and 5
    sum10 = ((n / 10) * (20 + (n / 10 - 1) * 10)) / 2;
 
    return sum2 + sum5 - sum10;
}
 
// Driver code
int main()
{
    int n = 5;
 
    cout << findSum(n) << endl;
    return 0;
}

Java




// Java implementation of
// above approach
import java.lang.*;
import java.util.*;
 
class GFG
{
 
// Function to find the sum
static long findSum(int n)
{
    long sum2, sum5, sum10;
     
    // sum2 is sum of numbers
    // divisible by 2
    sum2 = ((n / 2) * (4 +
            (n / 2 - 1) * 2)) / 2;
     
    // sum5 is sum of number
    // divisible by 5
    sum5 = ((n / 5) * (10 +
            (n / 5 - 1) * 5)) / 2;
     
    // sum10 of numbers divisible
    // by 2 and 5
    sum10 = ((n / 10) * (20 +
             (n / 10 - 1) * 10)) / 2;
     
    return sum2 + sum5 - sum10;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 5;
    System.out.println(findSum(n));
}
}
 
// This code is contributed by Raj

Python3




# Python3 implementation of
# above approach
 
# Function to find the sum
def findSum(n):
 
     
    # sum2 is sum of numbers divisible by 2
    sum2 = ((n // 2) * (4 + (n // 2 - 1) * 2)) // 2
 
    # sum5 is sum of number divisible by 5
    sum5 = ((n // 5) * (10 + (n // 5 - 1) * 5)) // 2
 
    # sum10 of numbers divisible by 2 and 5
    sum10 = ((n // 10) * (20 + (n // 10 - 1) * 10)) // 2
 
    return sum2 + sum5 - sum10;
 
 
# Driver code
if __name__=='__main__':
    n = 5
    print (int(findSum(n)))
     
 
# this code is contributed by Shivi_Aggarwal

C#




// C# implementation of
// above approach
using System;
 
class GFG
{
 
// Function to find the sum
static long findSum(int n)
{
    long sum2, sum5, sum10;
     
    // sum2 is sum of numbers
    // divisible by 2
    sum2 = ((n / 2) * (4 +
            (n / 2 - 1) * 2)) / 2;
     
    // sum5 is sum of number
    // divisible by 5
    sum5 = ((n / 5) * (10 +
            (n / 5 - 1) * 5)) / 2;
     
    // sum10 of numbers divisible
    // by 2 and 5
    sum10 = ((n / 10) * (20 +
             (n / 10 - 1) * 10)) / 2;
     
    return sum2 + sum5 - sum10;
}
 
// Driver code
public static void Main ()
{
    int n = 5;
    Console.WriteLine(findSum(n));
}
}
 
// This code is contributed by inder_verma

PHP




<?php
// PHP implementation of above approach
 
// Function to find the sum
function findSum($n)
{
 
    // sum2 is sum of numbers
    // divisible by 2
    $sum2 = ((int)($n / 2) * (4 +
            ((int)($n / 2) - 1) * 2)) / 2;
     
    // sum5 is sum of number
    // divisible by 5
    $sum5 = ((int)($n / 5) * (10 +
                  ($n / 5 - 1) * 5)) / 2;
     
    // sum10 of numbers divisible
    // by 2 and 5
    $sum10 = ((int)($n / 10) * (20 +
                   ($n / 10 - 1) * 10)) / 2;
     
    return $sum2 + $sum5 - $sum10;
}
 
// Driver Code
$n = 5;
echo findSum($n);
 
// This code is contributed by Raj
?>

Javascript




<script>
 
// Javascript implementation of above approach
 
// Function to find the sum
function findSum(n)
{
 
    var sum2, sum5, sum10;
 
    // sum2 is sum of numbers divisible by 2
    sum2 = parseInt((parseInt(n / 2) *
    (4 + (parseInt(n / 2) - 1) * 2)) / 2);
 
    // sum5 is sum of number divisible by 5
    sum5 = parseInt((parseInt(n / 5) *
    (10 + (parseInt(n / 5) - 1) * 5)) / 2);
 
    // sum10 of numbers divisible by 2 and 5
    sum10 = parseInt((parseInt(n / 10) *
    (20 + (parseInt(n / 10) - 1) * 10)) / 2);
 
    return sum2 + sum5 - sum10;
}
 
// Driver code
var n = 5;
document.write( findSum(n));
 
</script>

Output

11

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


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