Divisors Count

Problem

Return the number of positive divisors of n.

Loop i up to the square root and count each divisor, adding two for its partner unless i is exactly the square root. Assume n >= 1.

Hint

Divisors pair up as i and n // i, except for a perfect square root.

case 1

in: [1]

expected: 1

case 2

in: [12]

expected: 6

case 3

in: [36]

expected: 9

case 4

in: [97]

expected: 2

case 5

in: [1000]

expected: 16

Starter codePython

def divisor_count(n):
    # Your code here
    pass

5 test cases · run your code against them in the editor