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

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++

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// 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;
}

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Java

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//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    
}

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Python3

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# 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

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C#

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// 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 

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PHP

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<?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
?>

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Output:

6


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Improved By : jit_t, AnkitRai01