Count numbers from range whose prime factors are only 2 and 3

Last Updated : 08 Jul, 2022

Given two positive integers L and R, the task is to count the elements from the range [L, R] whose prime factors are only 2 and 3.
Examples:

Input: L = 1, R = 10
Output: 6
2 = 2
3 = 3
4 = 2 * 2
6 = 2 * 3
8 = 2 * 2 * 2
9 = 3 * 3
Input: L = 100, R = 200
Output: 5

Approach: Start a loop from L to R and for every element num:

While num is divisible by 2, divide it by 2.
While num is divisible by 3, divide it by 3.
If num = 1 then increment the count as num has only 2 and 3 as its prime factors.

Print the count in the end.
Below is the implementation of the above approach:

C++

// C++ program to count the numbers within a range
// whose prime factors are only 2 and 3
#include <bits/stdc++.h>
using namespace std;

// Function to count the number within a range
// whose prime factors are only 2 and 3
int findTwoThreePrime(int l, int r)
{
    // Start with 2 so that 1 doesn't get counted
    if (l == 1)
        l++;

    int count = 0;

    for (int i = l; i <= r; i++) {
        int num = i;

        // While num is divisible by 2, divide it by 2
        while (num % 2 == 0)
            num /= 2;

        // While num is divisible by 3, divide it by 3
        while (num % 3 == 0)
            num /= 3;

        // If num got reduced to 1 then it has
        // only 2 and 3 as prime factors
        if (num == 1)
            count++;
    }

    return count;
}

// Driver code
int main()
{
    int l = 1, r = 10;
    cout << findTwoThreePrime(l, r);
    return 0;
}

Java

//Java program to count the numbers within a range 
// whose prime factors are only 2 and 3 

import java.io.*;

class GFG {
    
// Function to count the number within a range 
// whose prime factors are only 2 and 3 
static int findTwoThreePrime(int l, int r) 
{ 
    // Start with 2 so that 1 doesn't get counted 
    if (l == 1) 
        l++; 

    int count = 0; 

    for (int i = l; i <= r; i++) { 
        int num = i; 

        // While num is divisible by 2, divide it by 2 
        while (num % 2 == 0) 
            num /= 2; 

        // While num is divisible by 3, divide it by 3 
        while (num % 3 == 0) 
            num /= 3; 

        // If num got reduced to 1 then it has 
        // only 2 and 3 as prime factors 
        if (num == 1) 
            count++; 
    } 

    return count; 
} 

// Driver code 
    public static void main (String[] args) {

        int l = 1, r = 10; 
        System.out.println (findTwoThreePrime(l, r)); 
    }
//This code is contributed by ajit    
}

Python3

# Python3 program to count the numbers 
# within a range whose prime factors 
# are only 2 and 3

# Function to count the number within 
# a range whose prime factors are only 
# 2 and 3
def findTwoThreePrime(l, r) :

    # Start with 2 so that 1 
    # doesn't get counted
    if (l == 1) :
        l += 1

    count = 0

    for i in range(l, r + 1) :
        num = i

        # While num is divisible by 2, 
        # divide it by 2
        while (num % 2 == 0) :
            num //= 2;

        # While num is divisible by 3, 
        # divide it by 3
        while (num % 3 == 0) :
            num //= 3

        # If num got reduced to 1 then it has
        # only 2 and 3 as prime factors
        if (num == 1) :
            count += 1

    return count

# Driver code
if __name__ == "__main__" :

    l = 1
    r = 10
    
    print(findTwoThreePrime(l, r))
    
# This code is contributed by Ryuga

// C# program to count the numbers 
// within a range whose prime factors 
// are only 2 and 3 
using System;

class GFG
{
        
// Function to count the number 
// within a range whose prime
// factors are only 2 and 3 
static int findTwoThreePrime(int l, int r) 
{ 
    // Start with 2 so that 1 
    // doesn't get counted 
    if (l == 1) 
        l++; 

    int count = 0; 

    for (int i = l; i <= r; i++) 
    { 
        int num = i; 

        // While num is divisible by 2,
        // divide it by 2 
        while (num % 2 == 0) 
            num /= 2; 

        // While num is divisible by 3, 
        // divide it by 3 
        while (num % 3 == 0) 
            num /= 3; 

        // If num got reduced to 1 then it  
        // has only 2 and 3 as prime factors 
        if (num == 1) 
            count++; 
    } 
    return count; 
} 

// Driver code 
static public void Main ()
{
    int l = 1, r = 10; 
    Console.WriteLine(findTwoThreePrime(l, r)); 
}
}

// This code is contributed by akt_mit

PHP

<?php
// PHP program to count the numbers 
// within a range whose prime factors 
// are only 2 and 3

// Function to count the number 
// within a range whose prime 
// factors are only 2 and 3
function findTwoThreePrime($l, $r)
{
    // Start with 2 so that 1 
    // doesn't get counted
    if ($l == 1)
        $l++;

    $count = 0;

    for ($i = $l; $i <= $r; $i++) 
    {
        $num = $i;

        // While num is divisible by 2,
        // divide it by 2
        while ($num % 2 == 0)
            $num /= 2;

        // While num is divisible by 3, 
        // divide it by 3
        while ($num % 3 == 0)
            $num /= 3;

        // If num got reduced to 1 then it has
        // only 2 and 3 as prime factors
        if ($num == 1)
            $count++;
    }
    return $count;
}

// Driver code
$l = 1;
$r = 10;
echo findTwoThreePrime($l, $r);
    
// This code is contributed by ajit
?>

JavaScript

<script>

    // Javascript program to count the numbers 
    // within a range whose prime factors 
    // are only 2 and 3 
    
    // Function to count the number 
    // within a range whose prime
    // factors are only 2 and 3 
    function findTwoThreePrime(l, r) 
    { 
        // Start with 2 so that 1 
        // doesn't get counted 
        if (l == 1) 
            l++; 

        let count = 0; 

        for (let i = l; i <= r; i++) 
        { 
            let num = i; 

            // While num is divisible by 2,
            // divide it by 2 
            while (num % 2 == 0) 
                num = parseInt(num / 2, 10); 

            // While num is divisible by 3, 
            // divide it by 3 
            while (num % 3 == 0) 
                num = parseInt(num / 3, 10); 

            // If num got reduced to 1 then it  
            // has only 2 and 3 as prime factors 
            if (num == 1) 
                count++; 
        } 
        return count; 
    } 
    
    let l = 1, r = 10; 
    document.write(findTwoThreePrime(l, r)); 
    
</script>