Need help. Prime Time

[shola]

Can anyone help me figure out what I'm doing wrong in my code. It seems that it only divides entered number by 2. My idea is to divide the entered integer by 2 until it go in anymore then divide that by 3 until it go in anymore and use counters to count 2s and 3s. Check if the integer is divisible by 2 and 3 by % operator. Thank you!

Write a program to determine if a natural number,greater than 1, has only 2 and/or 3 as prime factors (but no other prime factors) and
how many of each factor (2 and 3) it does have.

Example 1: Input is 30
No
Example 2: Input is 18
Yes
Twos=1, Threes=2

#include <iostream>
#include <iomanip> 
#include <cmath> 

using namespace std;

int main() 
{
  int num = 0;      //User entered number
  int two = 0;      //Counter for 2's
  int three = 0;    //Counter for 3's
  int rem;          //Remainder after devision by 2 or 3
  
  cout << "Eneter a positive integer: ";
  cin >> num;
  
  while ( rem == 0 ) 
  {
    rem = num / 2;

       if ( num % 2 == 0 || rem % 2 == 0 ) 
       {
         two++;
       }
       else if ( num % 3 == 0 || rem % 3 == 0)
       {
         three++;
       }
       else 
        cout << "NO" << endl;
  }

  cout << "number 2s " << two << endl;
  cout << "number 3s " << three << endl;

  return 0;
}

[JLBorges]

> while ( rem == 0 )

What is the value of rem when this is evaluated for the first time?

#include <iostream>

int main()
{
    unsigned int number ;
    std::cin >> number ;

    if( number != 0 )
    {
        int n_two = 0 ;
        while( number%2 == 0 ) // keep dividing by two, as many times as possible
        {
            number /= 2 ;
            ++n_two ;
        }

        int n_three = 0 ;
        while( number%3 == 0 ) // keep dividing by three, as many times as possible
        {
            number /= 3 ;
            ++n_three ;
        }

        if( number == 1 ) // only prime factors are two and three
            std::cout << "Yes.\n2s: " << n_two << "  3s: " << n_three << '\n' ;

        else std::cout << "No.\n" ; // there was a residue not divisible by either 2 or 3
    }
}

[beko]

Without even considering whether your approach is correct or not:

Line 12: You don't initialize rem
Line 17: You loop as long as rem is 0. First of all, rem is never zero as you didn't initialize it, second, I think this condition doesn't make much sense. If at all, you'd want to loop as long as rem is not zero.

Also, you never update num's value, so in line 19 you always get the same rem, which will make it loop infinitely eventually - if your condition was appropriate, that is.

Here's how I did it (I won't give out the solution yet so you can try it for yourself):
1.) You don't need rem at all, just work with num.
2.) Make a if/else branch. First check whether it's not divisible by either two or three. That's the trivial case. Otherwise you'll have to compute the 2s and 3s.

Well, since someone else already did it, I might as well share my approach x)

#include <iostream>
#include <iomanip> 
#include <cmath> 

using namespace std;

int main() 
{
  int num = 0;      //User entered number
  int two = 0;      //Counter for 2's
  int three = 0;    //Counter for 3's
  
  cout << "Eneter a positive integer: ";
  cin >> num;

  // if it's not divisible by either, then just say so
  if(num%2 != 0 && num%3 != 0) {
    cout << "NO!" << endl;
  // otherwise do the computation
  } else {
    while(num%2 == 0 || num%3 == 0) { // loop as long as num is divisible by either one
      if(num%2 == 0) {
        two++;
        num = num/2; // don't forget to update accordingly!
      } else {
        three++;
        num = num/3;
      }
    }
    // I forgot this!
    if(num == 1)
        std::cout << "Yes." << std::endl;
    else
        std::cout << "No." << std::endl;
    // (*)   
  } 

  // this will be printed out every time. If you don't want that, put this to (*)
  cout << "number 2s " << two << endl;
  cout << "number 3s " << three << endl;

  return 0;
}

[shola]

Thank you very much!