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

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
```

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

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.

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

 ` `

Output:

```110
```

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