PyPE solution to PE problem 71

    ProjectEuler.net problem 71
    ===========================

    Consider the fraction, n/d, where n and d are positive integers. If nd and
    HCF(n,d)=1, it is called a reduced proper fraction.

    If we list the set of reduced proper fractions for d 8 in ascending order
    of size, we get:

        1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8,
        2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

    It can be seen that 2/5 is the fraction immediately to the left of 3/7.

    By listing the set of reduced proper fractions for d 1,000,000 in
    ascending order of size, find the numerator of the fraction immediately
    to the left of 3/7.

    Solution 1
    ----------

    The first solution is a mostly brute-force approach with one small insight
    applied that greatly reduces the number of calculations required. For each
    possible denominator d, we can easily determine the numerator n that will
    be closest to -- without going over -- the desired value of 3/7:

        n = floor( (3/7) * d )

    We then iterate through the possible denominators, and if the particular
    closest match for a given denominator is a better solution than what we
    had previously, store the new one. The code for this approach is saved
    below for posterity.

    Solution 2
    ----------

    The second approach is based off the notes provided by Project Euler
    (http://projecteuler.net/project/resources/071_87a7f398bb9bd738501cbbc46ec437f5/071_overview.pdf)

    Answer
    ------

    428570