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

p / q = 0.sss...ad infinitum, from which we write
p / q = s / 10ⁿ + s / 10²ⁿ + s / 10³ⁿ +... ad infinitum

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).

ou 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.
 

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}