PyPE solution to PE problem 4

    ProjectEuler.net problem 4
    ==========================

    A palindromic number reads the same both ways. The largest palindrome made
    from the product of two 2-digit numbers is 9009 = 91 * 99.

    Find the largest palindrome made from the product of two 3-digit numbers.

    Solution
    --------

    If we call the two numbers i and j, then we first find the upper and lower
    bounds for i and j. For 3-digit numbers, these are 999 and 100 respectively.

    We choose to iterate between these bounds reversely, such that once we reach
    a value of p=i*j that is less than the maximum palindrome so far, we break
    out of the loop as all subsequent iterations will be smaller again.

    We also choose our minor iteration of j to begin at i; this way we prevent
    double-checking values such as i=132, j=528 vs i=528, j=132.

    Lastly, we note that the largest possible multiple (999*999=998001) is six
    digits long. We also note that 111111 is a palindrome, and thus the solution
    to the puzzle must lie between these two values (i.e. that it is a six-digit
    number. In order for a 6-digit number to be palindromic, it must have at most
    three distinct digital values, which we will label a, b, and c. By analysis,
    we know that any solution will be of the form

        p = 100000*a + 10000*b + 1000*c + 100*c + 10*b + a
          = 100001*a + 10010*b + 1100*c
          = 11 * (9091*a + 910*b + 100*c)

    We now know that at least one of i or j must have a factor of 11, and so we
    can use this knowledge to check only such combinations (i.e. if i does not
    have a factor of 11, then we set j to the largest multiple of 11 less than
    or equal to i and reverse-iterate j by steps of 11).

    Answer
    ------

    906609