ProjectEuler#26 Reciprocal cycles

#A unit fraction contains 1 in the numerator. The decimal representation of
#the unit fractions with denominators 2 to 10 are given:

#    1/2    =   0.5
#    1/3    =   0.(3)
#    1/4    =   0.25
#    1/5    =   0.2
#    1/6    =   0.1(6)
#    1/7    =   0.(142857)
#    1/8    =   0.125
#    1/9    =   0.(1)
#    1/10   =   0.1

#Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
#It can be seen that 1/7 has a 6-digit recurring cycle.

#Find the value of d < 1000 for which 1/d contains the longest recurring
#cycle in its decimal fraction part.

#Problem: Floating point representation limitation.
#Examble: 1/19.0 = 0.05263157894736842 while
#1/19 = 0.052631578 947368421  ; 18 repeating digits

#Analysis:
#Using long division to convert the fraction (rational number) into decimal form.
#if any point the reminder is 0 its a terminating decimal.
#if we arrive to a remainder as the one we began with, the decimal will start to repeat.
#Source: http://en.wikipedia.org/wiki/Repeating_decimal
#Section: Decimal expansion and recurrence sequence

max = 1000
maxCount = 0
maxDivisor = 0
for divisor in range(2,max):
    divdend = 10
    quotient = []
    reminder = []
    while True:
        if divdend < divisor:
            quotient.append(0)
            divdend *= 10
        else:
            quotient.append(divdend/divisor)
            reminder.append((divdend%divisor)*10)
            if reminder[-1] == 0:
                break
            elif len(reminder) > 1 and reminder[-1] in\
                 reminder[:len(reminder)-1]:
                break
            divdend = reminder[-1]

    if reminder[-1] != 0:
        while quotient[0] == 0:
            quotient.remove(0)
        if len(quotient)-1 >= maxCount:
            maxCount = len(quotient)-1
            maxDivisor = divisor

print maxDivisor, maxCount