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Multiples of 3 or 7

  • Difficulty Level : Easy
  • Last Updated : 16 Aug, 2021

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.




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