Find the sum of all multiples of 2 and 5 below N

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Given a number N. and the task is to find the sum of all multiples of 2 and 5 below N ( N may be up to 10^10).

Examples:

Input : N = 10
Output : 25
Explanation : 2 + 4 + 6 + 8 + 5

Input : N = 20
Output : 110

Approach :

We know that multiples of 2 form an AP as:

2, 4, 6, 8, 10, 12, 14….(1)

Similarly, multiples of 5 form an AP as:

5, 10, 15……(2)

Now, Sum(1) + Sum(2) = 2, 4, 5, 6, 8, 10, 10, 12, 14, 15….
Here, 10 is repeated. In fact, all of the multiples of 10 or 2*5 are repeated because it is counted twice, once in the series of 2 and again in the series of 5. Hence we’ll subtract the sum of the series of 10 from Sum(1) + Sum(2).

The formula for the sum of an A.P is :

n * ( a + l ) / 2
Where $n$ is the number of terms, $a$ is the starting term, and $l$ is the last term.

So, the final answer is:

S₂ + S₅ – S₁₀

Below is the implementation of the above approach:

C++

// CPP program to find the sum of all
// multiples of 2 and 5 below N
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of AP series
long long sumAP(long long n, long long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
long long sumMultiples(long long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
int main()
{
    long long n = 20;
 
    cout << sumMultiples(n);
 
    return 0;
}

C

// C program to find the sum of all
// multiples of 2 and 5 below N
#include <stdio.h>
 
// Function to find sum of AP series
long long sumAP(long long n, long long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
long long sumMultiples(long long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
int main()
{
    long long n = 20;
 
    printf("%lld",sumMultiples(n));
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.

Java

// Java program to find the sum of all
// multiples of 2 and 5 below N
 
class GFG{
// Function to find sum of AP series
static long sumAP(long n, long d)
{
    // Number of terms
    n /= d;
 
    return (n) * (1 + n) * d / 2;
}
 
// Function to find the sum of all
// multiples of 2 and 5 below N
static long sumMultiples(long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);
}
 
// Driver code
public static void main(String[] args)
{
    long n = 20;
 
    System.out.println(sumMultiples(n));
}
}
// This code is contributed by mits

Python3

# Python3 program to find the sum of 
# all multiples of 2 and 5 below N 
 
# Function to find sum of AP series 
def sumAP(n, d):
 
    # Number of terms 
    n = int(n / d); 
 
    return (n) * (1 + n) * (d / 2); 
 
# Function to find the sum of all 
# multiples of 2 and 5 below N 
def sumMultiples(n):
 
    # Since, we need the sum of 
    # multiples less than N 
    n -= 1; 
 
    return (int(sumAP(n, 2) + sumAP(n, 5) -
                              sumAP(n, 10))); 
 
# Driver code 
n = 20; 
 
print(sumMultiples(n)); 
     
# This code is contributed by mits

C#

// C# program to find the sum of all 
// multiples of 2 and 5 below N 
 
using System;
 
public class GFG{
     
    // Function to find sum of AP series 
static long sumAP(long n, long d) 
{ 
    // Number of terms 
    n /= d; 
 
    return (n) * (1 + n) * d / 2; 
} 
 
// Function to find the sum of all 
// multiples of 2 and 5 below N 
static long sumMultiples(long n) 
{ 
    // Since, we need the sum of 
    // multiples less than N 
    n--; 
 
    return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10); 
} 
 
// Driver code 
     
    static public void Main (){
            long n = 20; 
 
        Console.WriteLine(sumMultiples(n)); 
    } 
} 

PHP

<?php
// PHP program to find the sum of all 
// multiples of 2 and 5 below N 
// Function to find sum of AP series 
function sumAP($n, $d) 
{ 
    // Number of terms 
    $n = (int)($n /$d); 
 
    return ($n) * ((1 + $n) * 
                   (int)$d / 2); 
} 
 
// Function to find the sum of all 
// multiples of 2 and 5 below N 
function sumMultiples($n) 
{ 
    // Since, we need the sum of 
    // multiples less than N 
    $n--; 
 
    return sumAP($n, 2) + sumAP($n, 5) - 
                          sumAP($n, 10); 
} 
 
// Driver code 
$n = 20; 
 
echo sumMultiples($n); 
 
// This code is contributed 
// by Sach_Code
?>

Javascript

<script>
 
// Java script program to find the sum of all 
// multiples of 2 and 5 below N 
 
// Function to find sum of AP series 
function sumAP(n, d) 
{ 
     
    // Number of terms 
    n = parseInt(n / d); 
 
    return (n) * ((1 + n) * 
         parseInt(d) / 2); 
} 
 
// Function to find the sum of all 
// multiples of 2 and 5 below N 
function sumMultiples(n) 
{ 
     
    // Since, we need the sum of 
    // multiples less than N 
    n--; 
 
    return sumAP(n, 2) + sumAP(n, 5) - 
           sumAP(n, 10); 
} 
 
// Driver code 
n = 20; 
 
document.write( sumMultiples(n)); 
 
// This code is contributed by bobby 
 
</script>

Output:

Time complexity: O(1), since there is no loop or recursion.

Auxiliary Space: O(1), since no extra space has been taken.

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