Counting How Many Factors An Integer Has

So I have written a function to determine how many factors a number has and list that number. However my function is not outputting the correct information. def num_factors(integer

Solution 1:

With the line for i in range(1,integer+1): you are looping through all the numbers between 1 and your integer, including 1 and your integer, which are of course factors.

So for example if integer = 5, you'd loop over 1, 2, 3, 4, and 5. Out of these 1 and 5 are both of course factors of 5.

You could edit the line to for i in range(2,integer): to fix the error. The resulting code would look like this:

defnum_factors(integer):
    result = 0for i inrange(2,integer):
        if integer%i == 0:
            result +=1return result

print(num_factors(5))
print(num_factors(6))
print(num_factors(97))
print(num_factors(105))
print(num_factors(999))

Though as someone in the comments suggested, you could reduce the search space even more.

Solution 2:

sympy provides this function to find prime factors.

>>>from sympy.ntheory import factorint>>>factorint(6008)   # 6008 = (2**3) * (751**1)
{2: 3, 751: 1}

Solution 3:

The issue is that you are counting division by 1 and the test integer itself.

You either need to subtract 2 or skip 1 and integer to get what you want as the desired output:

def num_factors(integer):
    result=0for i inrange(2,integer): # skips 1and integer...
        if integer%i ==0:
            result+=1returnresult

And even better is to realize that for every x * y that is a factor of integer we only need to find one of them (ie, 16 has 2,4,and 8 as factors. Count 2 twice (2 x 8=16) and 4 once (4 x 4=16)) And since one will be less than or equal to the square root of integer just loop to the square root of integer and increment by 2 instead of by 1 and only make a fraction of the tests (and make the result 1000's of times faster):

def num_factors(integer):
    result=0
    stop=int(float(integer)**0.5)
    for i inrange(2,stop+1):
        if integer%i ==0:
            result+=2
    if stop*stop==integer: result-=1returnresultfor x in (5,6,97,105,999):
    print(f'{x}: {num_factors(x)}')

Prints:

5: 06: 297: 0105: 6999: 6

---

BTW: It is, in fact, customary to count 1 and the integer itself as factors. So all these results should be +2 and your original solution was actually correct. To make the solutions above correct, just start with result=2

Solution 4:

I would do it this way:

def num_factors(integer):
    for i in range(1, integer, 1):
        if integer % i == 0:
            print(i)

print(num_factors(5))
print(num_factors(6))
print(num_factors(97))
print(num_factors(105))
print(num_factors(999))

Solution 5:

try this :

num_factors=lambda integer : len([e  for e in range(1,integer) ifinteger%e==0 and e not in [1,integer]])



print(num_factors(5))
print(num_factors(6))
print(num_factors(97))
print(num_factors(105))
print(num_factors(999))