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

Given a number n. The task is to find the sum of numbers up to n, that are divisible 2 or 5.

Examples:

Input: n = 2
Output: 2

Input: n = 5
Output: 11


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

A naive approach is to just iterate over the numbers upto 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 complexity of O(n).

Efficient Approach:

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

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

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-

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-

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

sum = sum_2 + sum_5 – sum_10

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach  #include  #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

 

Output:

11


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