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Find the sum of all multiples of 2 and 5 below N
• Difficulty Level : Basic
• Last Updated : 09 Apr, 2021

Given a number N. 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 is the number of terms, is the starting term, and 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 ``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;``}`

## 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

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## Javascript

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Output:
`110`

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