Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only
a finite number of nonzero digits?
B. This question is really asking about terminating decimales.
In order to have a terminating decimal 1 of the 3 scenarios MUST be in place.
X, Y, and Z are any number. (I believe they can be any number, not sure if negatives would affect Z, I tested a few negative vales for Z make 2^Z --> 1/2^Z, seems ok though no forever decimals) Correct me on this if Im wrong please.
Anyway 3 scenarios are:
X/2^z X/5^z or X/2^z*5^y
There cannot be any other numbers in the bottom such as 3,7, etc... or it does not neccesarily consitute a terminating decimal!