 GeeksforGeeks App
Open App Browser
Continue

# Multiples of 3 or 7

Given a positive integer n, find count of all multiples of 3 or 7 less than or equal to n.
Examples :

Input: n = 10
Output: Count = 4
The multiples are 3, 6, 7 and 9

Input: n = 25
Output: Count = 10
The multiples are 3, 6, 7, 9, 12, 14, 15, 18, 21 and 24

Recommended Practice

A Simple Solution is to iterate over all numbers from 1 to n and increment count whenever a number is a multiple of 3 or 7 or both.

## C++

 `// A Simple C++ program to find count of all``// numbers that multiples``#include``using` `namespace` `std;` `// Returns count of all numbers smaller than``// or equal to n and multiples of 3 or 7 or both``int` `countMultiples(``int` `n)``{``    ``int` `res = 0;``    ``for` `(``int` `i=1; i<=n; i++)``       ``if` `(i%3==0 || i%7 == 0)``           ``res++;` `    ``return` `res;``}` `// Driver code``int` `main()``{``   ``cout << ``"Count = "` `<< countMultiples(25);``}`

## Java

 `// A Simple Java program to``// find count of all numbers``// that multiples``import` `java.io.*;` `class` `GFG``{``    ` `// Returns count of all numbers``// smaller than or equal to n``// and multiples of 3 or 7 or both``static` `int` `countMultiples(``int` `n)``{``    ``int` `res = ``0``;``    ``for` `(``int` `i = ``1``; i <= n; i++)``    ``if` `(i % ``3` `== ``0` `|| i % ``7` `== ``0``)``        ``res++;` `    ``return` `res;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``System.out.print(``"Count = "``);``    ``System.out.println(countMultiples(``25``));``}``}` `// This code is contributed by m_kit`

## Python3

 `# A Simple Python3 program to``# find count of all numbers``# that multiples` `# Returns count of all numbers``# smaller than or equal to n``# and multiples of 3 or 7 or both``def` `countMultiples(n):``    ``res ``=` `0``;``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``if` `(i ``%` `3` `=``=` `0` `or` `i ``%` `7` `=``=` `0``):``            ``res ``+``=` `1``;`` ` `    ``return` `res;` `# Driver code``print``(``"Count ="``, countMultiples(``25``));` `# This code is contributed by mits`

## C#

 `// A Simple C# program to``// find count of all numbers``// that are multiples of 3 or 7``using` `System;` `class` `GFG``{``    ` `// Returns count of all ``// numbers smaller than``// or equal to n  and``// are multiples of 3 or``// 7 or both``static` `int` `countMultiples(``int` `n)``{``    ``int` `res = 0;``    ``for` `(``int` `i = 1; i <= n; i++)``    ``if` `(i % 3 == 0 || i % 7 == 0)``        ``res++;` `    ``return` `res;``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``Console.Write(``"Count = "``);``    ``Console.WriteLine(countMultiples(25));``}``}` `// This code is contributed by ajit`

## PHP

 ``

## Javascript

 ``

Output

`Count = 10`

Time Complexity : O(n)
Auxiliary Space: O(1)

An efficient solution can solve the above problem in O(1) time. The idea is to count multiples of 3 and add multiples of 7, then subtract multiples of 21 because these are counted twice.

`count = n/3 + n/7 - n/21 `

## C++

 `// A better C++ program to find count of all``// numbers that multiples``#include``using` `namespace` `std;` `// Returns count of all numbers smaller than``// or equal to n and multiples of 3 or 7 or both``int` `countMultiples(``int` `n)``{``   ``return` `n/3 + n/7 -n/21;``}` `// Driver code``int` `main()``{``   ``cout << ``"Count = "` `<< countMultiples(25);``}`

## Java

 `// A better Java program to``// find count of all numbers``// that multiples``import` `java.io.*;` `class` `GFG``{``    ` `// Returns count of all numbers``// smaller than or equal to n``// and multiples of 3 or 7 or both``static` `int` `countMultiples(``int` `n)``{``    ``return` `n / ``3` `+ n / ``7` `- n / ``21``;``}` `// Driver code``public` `static` `void` `main (String args [] )``{``    ``System.out.println(``"Count = "` `+``                        ``countMultiples(``25``));``}``}` `// This code is contributed by aj_36`

## Python 3

 `# Python 3 program to find count of``# all numbers that multiples` `# Returns count of all numbers``# smaller than or equal to n and``# multiples of 3 or 7 or both``def` `countMultiples(n):``    ``return` `n ``/` `3` `+` `n ``/` `7` `-` `n ``/` `21``;` `# Driver code``n ``=` `((``int``)(countMultiples(``25``)));``print``(``"Count ="``, n);` `# This code is contributed``# by Shivi_Aggarwal`

## C#

 `// A better Java program to``// find count of all numbers``// that multiples``using` `System;` `class` `GFG``{``    ` `// Returns count of all numbers``// smaller than or equal to n``// and multiples of 3 or 7 or both``static` `int` `countMultiples(``int` `n)``{``    ``return` `n / 3 + n / 7 - n / 21;``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``Console.WriteLine(``"Count = "` `+``                       ``countMultiples(25));``}``}` `// This code is contributed by m_kit`

## PHP

 ``

## Javascript

 ``

Output

`Count = 10`

Time Complexity : O(1)
Auxiliary Space: O(1)
Exercise:
Now try the problem of finding sum of all numbers less than or equal to n and multiples of 3 or 7 or both in O(1) time.
This article is contributed by Saurabh Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.