code to return divisors and compare common ones

We can make the solution a bit more generic. Let us imagine that we can define an unlimited number of values to determine the common divisors between them. We can do this through the function:

def divisores(*numeros):
    ...

So, if we want the common divisors of 5 and 10, we can call divisores(5, 10) , and if we want the divisors of 42, 100 and 999, we just call divisores(42, 100, 999) . If we consider, for example, input 42, 100 and 999, it is easy to see that all common divisors of the three numbers will be divisor of each number individually, implying that all divisors should be less than or equal to the smallest of the input values , since it does not have a value greater than 42 being divisor of 42. Thus, we know that it is enough to go through all the values between 1 and the smallest value of the entry:

def divisores(*numeros):
    menor = min(numeros)
    for i in range(1, menor+1):
        ...

And to set whether i is a divisor of all input numbers, we can use the all function and check that the remainder of all input numbers by i is 0.

def divisores(*numeros):
    menor = min(numeros)
    for i in range(1, menor+1):
        if all(numero % i == 0 for numero in numeros):
            yield i

With yield we return a generator that iterates over the divisors of all input numbers.

print('Divisores de 5 e 10:', list(divisores(5, 10)))  # [1, 5]
print('Divisores de 42, 100 e 999:', list(divisores(42, 100, 999)))  # [1]
print('Divisores de 14, 38 e 74:', list(divisores(14, 38, 74)))  # [1, 2]

See working at Repl.it | Ideone

    

Taking advantage of the function you have to calculate dividers, you have little time to complete the problem. In order to be able to re-use the function that has the best option is to transform print into return :

def divisores(num):
    n = np.arange(1,num)
    d = num % n
    zeros = d == 0
    return n[zeros]

In fact, it is always better to do this because it makes the function work anywhere without having to force writing.

Now just build a function that gets two numbers, call divisores for each of these numbers and get the common numbers for each set. There are many ways to get the common numbers of each set and I will opt for the set :

def divisores_comuns(num1, num2):
    divs1 = set(divisores(num1)) # obtendo os divisores e transformando em set
    divs2 = set(divisores(num2)) # obtendo os divisores e transformando em set
    return divs1 & divs2 # & é a intereseção de dois sets

See this code working on Ideone

    

Ana,

To receive two values, one must either consider N as a list of integers, or receive two arguments in the function signature, such as:

def divisores(num1, num2):

Although your procedure works, note the following:

You test the division of all numbers between 0 - number received, even if the smallest possible division (2) happens in num / 2;

You again test every numpy array if the rest of the division is 0;

Although not a very charming algorithm, one way you perform fewer iterations is as follows:

div_1 = set() #Cria um set vazio
for i in range(1, int(num1/2+1)): #itera de 1 até metade do valor numérico
    if num1%i == 0: #testa se mod é zero.
        div_1.add(i) #anexa o valor de i ao set de divisores do primeiro numero

Note that this function, like its code, does not return the value itself as a possible divisor (in this case, when you enter 30, the values obtained will be: {1, 2, 3, 5, 6, 10 , 15) (30 excluded). In addition, you can consider using the "set" object instead of lists, since the numbers should not be repeated.

In case, if the final function were to just find the intersection of the common division values, you could include everything in the same loop, as it is not, a simplistic approach is to run another iteration in to identify the same divisors of the other number ( num2).

For the intersection, you could accomplish the following:

lista_divisores_comuns = div_1.intersection(div_2)

The final code would look like:

def divisores(num1, num2):    
    div_1 = set()
    div_2 = set()
    for i in range(1, int(num1/2+1)):
        if num1%i == 0:
            div_1.add(i)

    for i in range(1, int(num2/2+1)):
        if num2%i == 0:
            div_2.add(i)


    lista_divisores_comuns = div_1.intersection(div_2)

    return div_1, div_2, lista_divisores_comuns