# Euler Problem 26¶

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.

The trick here is to identify a cycle. The easiest way I see to do this is to memorize the remainders. If we have a remainder that we occurred previously there is a cycle.

Let's consider 1/3. The initial remainder is 10. For ten divided by three the new remainder is again ten (or one times ten). So we have a one-cycle of 3.

In [1]:

    def get_cycle_count(nominator, denominator):
        from itertools import count
        assert(nominator == 1)
        remainders = {}
        remainder = nominator
        results = []
        for i in count():
            result = remainder // denominator
            remainder = remainder % denominator
            results.append(result)
            if remainder in remainders:
                return i - remainders[remainder]
            else:
                remainders[remainder] = i
            if remainder == 0:
                return 0
            remainder *= 10

    assert(get_cycle_count(1, 7) == 6)
    assert(get_cycle_count(1, 10) == 0)
    assert(get_cycle_count(1, 6) == 1)

This is a simple divison algorithm. The only special thing is the remainder and that we remember when it occurs the first time. If a remainder occurrs for the second time we substract the position and thus have the lenght of the cycle. With this solution we should be efficient enough to brute force.

In [2]:

    s = max([(get_cycle_count(1, i), i)for i in range(1, 1000)])
    s = s[1]
    assert(s == 983)
    print(s)

    983