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Multiples of 3 or 7

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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<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)
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<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)
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.



Last Updated : 30 Aug, 2022
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