# Sum of all the multiples of 3 and 7 below N

Given a number N, the task is to find the sum of all the multiples of 3 and 7 below N.
Note: A number must not repeat itself in the sum.

Examples:

Input: N = 10
Output: 25
3 + 6 + 7 + 9 = 25

Input: N = 24
Output: 105
3 + 6 + 7 + 9 + 12 + 14 + 15 + 18 + 21 = 105

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

Approach:

• We know that multiples of 3 form an AP as S3 = 3 + 6 + 9 + 12 + 15 + 18 + 21 + …
• And the multiples of 7 form an AP as S7 = 7 + 14 + 21 + 28 + …
• Now, Sum = S3 + S7 i.e. 3 + 6 + 7 + 9 + 12 + 14 + 15 + 18 + 21 + 21 + …
• From the previous step, 21 is repeated twice. In fact, all of the multiples of 21 (or 3*7) will be repeated as they are counted twice, once in the series S3 and again in the series S7. So, the multiples of 21 need to be discarded from the result.
• So, the final result will be S3 + S7 – S21

The formula for the sum of an AP series is :
n * ( a + l ) / 2
Where n is the number of terms, a is the starting term, and l is the last term.

Below is the implementation of the above approach:

 // CPP program to find the sum of all // multiples of 3 and 7 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 3 and 7 below N long long sumMultiples(long long n) {     // Since, we need the sum of     // multiples less than N     n--;        return sumAP(n, 3) + sumAP(n, 7) - sumAP(n, 21); }    // Driver code int main() {     long long n = 24;        cout << sumMultiples(n);        return 0; }

 // Java program to find the sum of all // multiples of 3 and 7 below N import java.util.*;    class solution {    // 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 3 and 7 below N static long sumMultiples(long n) {     // Since, we need the sum of     // multiples less than N     n--;        return sumAP(n, 3) + sumAP(n, 7) - sumAP(n, 21); }    // Driver code public static void main(String args[]) {     long n = 24;        System.out.println(sumMultiples(n));     } }    //This code is contributed by Surendra_Gangwar

 # Python3 program to find the sum of  # all multiples of 3 and 7 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 3 and 7 below N def sumMultiples(n):        # Since, we need the sum of     # multiples less than N     n -= 1;        return int(sumAP(n, 3) +                 sumAP(n, 7) -                 sumAP(n, 21));    # Driver code n = 24;    print(sumMultiples(n));    # This code is contributed  # by mits

 // C# program to find the sum of all  // multiples of 3 and 7 below N  using System;     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 3 and 7 below N  static long sumMultiples(long n)  {      // Since, we need the sum of      // multiples less than N      n--;         return sumAP(n, 3) + sumAP(n, 7) -                          sumAP(n, 21);  }     // Driver code  static public void Main(String []args)  {      long n = 24;         Console.WriteLine(sumMultiples(n));  }  }     // This code is contributed  // by Arnab Kundu



Output:
105

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