Euler #004 - Largest palindrome product

"""
The palindrome can be written as: abccba

Which then simpifies to:
100000a + 10000b + 1000c + 100c + 10b + a

And then:
100001a + 10010b + 1100c

Factoring out 11, you get:
11(9091a + 910b + 100c)

So the palindrome must be divisible by 11.
Seeing as 11 is prime, at least one of the numbers must be divisible by 11.
So brute force in Python, only with less numbers to be checked

"""

def is_palindrome(n):
    nstr = str(n)
    for idx in range(len(nstr) + 1 // 2):
        if nstr[idx] != nstr[-(idx + 1)]:
            return False
    return True

ans = n1 = n2 = comp = 0
# for num1 in range(100, 1000):
#     for num2 in range(num1, 1000):
for num1 in range(999, 99, -1): # num1 range 999 to 100
    for num2 in range(num1 - num1 % 11, 99, -11): # num2 range from (nearest 11 factor of num1) to (all 11 factors > 99)
        product = num1 * num2
        comp += 1
        if product > ans and is_palindrome(product):
            ans, n1, n2 = product, num1, num2

print(ans, n1, n2, comp)