Need some help with my program that finds divisors and prime/perfect numbers

[leo255]

So, here's what I have...

As you can see by my code, if a number is prime, I just output 1 and that number. For whatever reason, I'm not getting the divisors, when it is not a prime number.

Also, when I'm finding out if a number is perfect, when trying out the perfect number, 496, I found that I outputted the following numbers: 1, 2, 4, 8, 16, 31, 62, 124, 248.

It should be outputting 16, 32, 64, etc. so I'm not sure what's wrong with my code! Thanks in advance!!

#include <iostream>

using namespace std;

int main()
{
    int userNum = 0, sum = 0;

    cout << "\nThis program will take numbers entered by a user, and will determine if a number is a prime number or a perfect number." << endl;

    cout << "\nPlease enter a number between 1 and 1000: \n";
    cin >> userNum;

    while (userNum < 1 || userNum > 1000){

    cout << "\nPlease enter a number between 1 and 1000: \n";
    cin >> userNum;

    }

    cout << "\nDivisors: \n";

    if (userNum % 2 == 1){

    cout << userNum%2 << " and " << userNum << endl;
    cout << "\nThis number is a prime number. \n";

    }

    else{
    for (int z = 1; z <= userNum/2; z++){
    if (userNum % z == 0)
    cout << z << endl;

    cout << "\nThis number is not a prime number. \n";
    }
    }

    cout << "\nDivisors: \n";

    for (int i = 1; i <= userNum/2; i++){
    if (userNum % i == 0)
    sum += i;
    cout << i << endl;

    }

    if (sum == userNum)
    cout << "\nThe number " << userNum << " is a perfect number. \n";
    else
    cout << "\nThe number " << userNum << " is not a perfect number. \n";
    
    return 0;
}

[Chervil]

496 / 64 = 7.75
496 / 32 = 15.5
That is, neither 32 nor 64 are factors of 496, so these should not be output.

2*2*2*2*31 = 496
1+2+4+8+16+31+62+124+248 = 496

[leo255]

Thanks, you're right. I still don't understand why my program does not state that it is a perfect number.

Also, don't get why the else statement on line 30 does not give me the divisors of the number.

[Chervil]

You can probably adapt this to give the required output. Your code was almost correct, apart from the checking for prime numbers which was seriously flawed. I would normally give hints rather than a full solution, but I borrowed the function isPrime() from a current thread on the front page of this forum.

#include <iostream>
#include <cmath>

using namespace std;


bool isPrime(int num);

int main()
{
    int userNum = 496;
    int sum = 0;
    
    cout << userNum << endl;
    
    if (isPrime(userNum))
        cout << " is prime" << endl;
    
    cout << "\nDivisors: \n";
    
    for (int i = 1; i <= userNum/2; i++)
    {
        if (userNum % i == 0)
        {
            sum += i;
            cout << i << endl;
        }
    }
    
    if (sum == userNum)
        cout << "\nThe number " << userNum << " is a perfect number. \n";
    else
        cout << "\nThe number " << userNum << " is not a perfect number. \n";
    
    return 0;
}

bool isPrime(int num)
{
    if (num == 1)
        return false;

    int root = sqrt((double)num);

    for (int x = 2; x <= root; x++)  
    {
        if (num % x == 0)
            return false;
    }

    return true;
}