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

• Last Updated : 11 Jul, 2022

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

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:

## C++

 // C++ program to find the sum of all// multiples of 3 and 7 below N #include using namespace std; // Function to find sum of AP serieslong 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 Nlong 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 codeint main(){    long long n = 24;     cout << sumMultiples(n);     return 0;}

## Java

 // Java program to find the sum of all// multiples of 3 and 7 below Nimport java.util.*; class solution{ // Function to find sum of AP seriesstatic 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 Nstatic 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 codepublic static void main(String args[]){    long n = 24;     System.out.println(sumMultiples(n));  }} //This code is contributed by Surendra_Gangwar

## Python3

 # Python3 program to find the sum of# all multiples of 3 and 7 below N # Function to find sum of AP seriesdef 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 Ndef 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 coden = 24; print(sumMultiples(n)); # This code is contributed# by mits

## C#

 // C# program to find the sum of all// multiples of 3 and 7 below Nusing System; class GFG{ // Function to find sum of AP seriesstatic 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 Nstatic 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 codestatic public void Main(String []args){    long n = 24;     Console.WriteLine(sumMultiples(n));}} // This code is contributed// by Arnab Kundu



## Javascript



Output:

105

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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