Thread: Prime numbers and factors

Okay, so the following code is "supposed" to print numbers 1-1000 whose prime factors sum up to be a prime number. What it DOES print is seemingly random numbers that follow no pattern (primes/not primes). Here's my code:

Code:

#include <iostream>                     //Print numbers 1-1000 whose prime factors added together form a prime number
bool isPrime(int number);
bool isDivisible(int number, int divisor);
int sum_of_pf(int number);


using namespace std;


int main()
{
    for(int i=1;i<=1000;i++)
    {
      int S=sum_of_pf(i);
      if(isPrime(S))
      {
          cout<<i<<endl;
      }
    }
    cin.ignore();
    cin.get();
}


bool isPrime(int number)
{
    for(int i=2; i<number; i++)
    {
        if(isDivisible(number, i))
        {
            return false;
        }
    }
    return true;
}


bool isDivisible(int number, int divisor)
{
    return number%divisor==0;
}


int sum_of_pf(int number)
{
    int sum=0;
    int aux=number;
    for(int i=2;i<=number/2;i++)
    {
        while(aux%i==0)
        {
            sum+=i;
            aux/=i;
        }
    }
    return sum;
}

Can someone point out my error(s)?

Awful lot of crickets around here....

Code:

bool

Code:

 isPrime(int number)

Code:

{
    for(int i=1; i<number; i++)
    {
        if(isDivisible(number, i))
        {
            return false;
        }
    }
returntrue;
}

That was your issue....

I'm afraid I don't understand what you're telling me to do. The function isPrime is working properly.

Shouldn't sum_of_pf() be only adding up the numbers that are prime as indicated by isPrime()?

You are checking whether S is prime, but then printing i.

hornet 07 when i changed int 2 to int 1 everything was working fine...

@rcgldr
sum_of_pf() adds the prime factors of a number, not the prime numbers.

@tabstop
Well, if S(sum of i's prime factors) is a prime number, then I should print the number i. That is what the exercise is saying.

@jocdrew21
I changed 2 into 1. No difference. Besides, when trying to establish if a number is prime, you disregard 1 as a possible divisor.