Repeating Decimal Decimals

Copyright 2000 by Paul Niquette. All rights reserved.

The puzzle poses a rather complex pattern:
What ratio of integers produced it?

A proper rational fraction is one that can be expressed as the ratio of two integers, p and q, where p is not greater than q. Rational fractions are expressible as infinitely repeated pattern of decimal ciphers, each of which may be considered to be an integer divided by some power of ten.

Let the repeating pattern be represented by the symbol s having a length of n decimal digits. We reason as follows:

    p / q = infinitum, from which we write
    p / q = s / 10n + s / 102n + s / 103n +... ad infinitum
Sophisticated solvers are delighted with an unending series like this, for they will proceed to perform a set of operations that lead to a self-referent expression. Thus, by multiplying both sides of the equation by 10n, we get...
    10n p / q = s + s / 10n + s / 102n + s / 103n +... ad infinitum,
      wherein the expression s / 10n + s / 102n + s / 103n+... ad infinitum
      can be replaced by p / q with no loss in accuracy to produce...

    10n p / q = s + p / q and all that ad infinitum business goes away. Thus...
    (10n - 1 ) ( p / q ) = s, which can be rearranged to read...
    p = s q / (10n - 1), where 10n - 1 = 999...9 (an integer comprising n 9s).
Since p, q, and s are all integers, it is apparent that the product of s and q must be divisible by 10n - 1; however, s would be given -- not under our control -- and s may or may not be divisible by 10n - 1. In this puzzle, s = 645732, n = 6, and 645732/999999 is definitely not an integer, so we need q to help out here.

You surely noticed that if q = 999999, the conditions are satisfied no matter what the value of s may be, since q carries the entire burden of divisibility. The previous sentence deserves an exclamation point, for we have discovered a general solution for all Repeating Decimal Decimals. Thus, for q = 999999, p = s. That solution always exists. Almost always.

But then you might have also noticed that, since 645732 happens to be divisible by 3, we can let q = 333333, and the conditions are satisfied, making p = 215244, which gives us a second solution to the puzzle. What about q = 111111? That works too, in this case, since 645732 also happens to be divisible by 9, giving us a value of p = 71748.

Accordingly, there are three solutions to our original puzzle.

645732 / 999999
215244 / 333333
71748 / 111111
Things are not quite perfect in the land of Repeating Decimal Decimals. There is one pattern of digits s which cannot be generated as an unending sequence by the ratio of two finite integers p / q, where q is not less than p. Can you find it? {Return}

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