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

• Last Updated : 03 May, 2021

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

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 using namespace std; // Function to count the number within a range// whose prime factors are only 2 and 3int 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 codeint 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 3static 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 3def 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 codeif __name__ == "__main__" :     l = 1    r = 10         print(findTwoThreePrime(l, r))     # This code is contributed by Ryuga

## C#

 // C# program to count the numbers// within a range whose prime factors// are only 2 and 3using System; class GFG{         // Function to count the number// within a range whose prime// factors are only 2 and 3static 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 codestatic public void Main (){    int l = 1, r = 10;    Console.WriteLine(findTwoThreePrime(l, r));}} // This code is contributed by akt_mit



## Javascript


Output:
6

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