vitaliyGMAT wrote:
\(\frac{1}{y} = \frac{bc}{100}\)
\(y = \frac{100}{bc}\)
\(bc\) should be a two digit factor of 100, where b and c are distinct "non zero" digits. The only factor that satisfies above criteria is 25. So our b=2 and c=5.
Back to options:
I. Yes.
II. Not always ----> 1/4 = 0.25
III. Yes, because y should be a factor of \(100 = 2^2*5^2\), which corresponds to the fact that we have terminating decimal (\(y = 2^a*5^b\)).
Hence D.
This is a nice solution -- much cleaner than the one in the spoiler tag in the OP -- but it's not quite right. In the original version of the problem, we were not told that the digits were nonzero. So the number "bc" needs to be a factor of 100, but if our digits can be zero, then "bc" can be "04", for example; it's not the case that "25" is the only possible value of "bc". Here we might have x = 26 and y = 25, say, and x/y = 1.04. In the question's original wording, the only digit we can be sure is nonzero is c, because if c were zero, the decimal would be "a.b", and not "a.bc".
I gather the question meant to declare a, b and c to all be nonzero, and if that's true, then the solution to item I becomes correct, but then the solution for item II becomes incorrect, since 1/4 = 0.25 is no longer possible (the digit a is zero in that case).
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