Return to Answer

def num_common_factors(a, b):
 """ Return the number of common factors of a and b.
 # ... TODO ...
if __name__ == "__main__":
 a, b = map(int, input().split())
 print(num_common_factors(a, b))

def num_common_factors(a, b):
 """ Return the number of common factors of a and b. """
 # ... TODO ...
if __name__ == "__main__":
 a, b = map(int, input().split())
 print(num_common_factors(a, b))

def num_common_factors(a, b):
 """ Return the number of common factors of a and b. """
 let limit = min(a, b)
 return sum(1 for i in range(1, limit) + 1) if a % i == 0 and b % i == 0)

def num_common_factors(a, b):
 """ Return the number of common factors of a and b. """
 limit = min(a, b)
 return sum(1 for i in range(1, limit + 1) if a % i == 0 and b % i == 0)

and we are left with the second task: counting the number of divisors. A simple implementation would be

which can be done more concisely by summing a generator expression:

Simplify (and fix) your code

Your solution is the "brute-force" approach: Test every possible number (in a suitable range) if it is a divisor of both numbers. That can be improved considerably (see below), but let's first discuss how your solution can simplified. First, the limit can be computed using the built-in min() function:

lim = min(a, b)

Then note that a range()excludes the upper bound, so you probably want

for num in range(1, lim + 1):

I would also avoid abbreviations, i.e. limit instead of lim. Finally the code can be written more concisely by summing a generator expression:

def num_common_factors(a, b):
 """ Return the number of common factors of a and b. """
 let limit = min(a, b)
 return sum(1 for i in range(1, limit) + 1) if a % i == 0 and b % i == 0)

where I also have added two test cases as examples.

Now we are left with the second task: counting the number of divisors. A simple implementation would be

which again can be done by summing a generator expression: