# 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

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.

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## C++

`// A Simple C++ program to find count of all` `// numbers that multiples` `#include<iostream>` `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

`<?php` `// A Simple PHP 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` `function` `countMultiples(` `$n` `)` `{` ` ` `$res` `= 0;` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `if` `(` `$i` `% 3 == 0 || ` `$i` `% 7 == 0)` ` ` `$res` `++;` ` ` `return` `$res` `;` `}` `// Driver code` `echo` `"Count = "` `,countMultiples(25);` `// This code is contributed by aj_36` `?>` |

## Javascript

`<script>` `// A Simple JavaScript 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` `function` `countMultiples(n)` `{` ` ` `let res = 0;` ` ` `for` `(let i = 1; i <= n; i++)` ` ` `if` `(i % 3 == 0 || i % 7 == 0)` ` ` `res++;` ` ` `return` `res;` `}` `// Driver code` `document.write( ` `"Count = "` `+countMultiples(25));` `// This code is contributed by bobby` `</script>` |

**Output :**

Count = 10

**Time Complexity :** O(n)

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<iostream>` `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

`<?php` `// A better PHP 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` `function` `countMultiples(` `$n` `)` `{` `return` `floor` `(` `$n` `/ 3 + ` `$n` `/ 7 - ` `$n` `/ 21);` `}` `// Driver code` `echo` `"Count = "` `, countMultiples(25);` `// This code is contributed by ajit` `?>` |

## Javascript

`<script>` `// JavaScript 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` `function` `countMultiples(n)` `{` `return` `Math.floor(n / 3 + n / 7 - n / 21);` `}` `// Driver code` `document.write( ` `"Count = "` `+countMultiples(25));` `// This code is contributed by bobby` `</script>` |

**Output :**

Count = 10

**Time Complexity : **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.

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