Integer ?s

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by rattanas » Wed May 21, 2008 7:57 am
Yes the answer should be c..as 1 and 2 alone are not sufficient...to solve it would be like y^3=27..so y =3

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by VP_RedSoxFan » Wed May 21, 2008 8:02 am
(1) This statement narrows the value of y to {1, 3, 9, 27} but obviously not a single value like we're looking for so its insufficient.

(2) x = y^2. The most tempting thing about this one is if you let knowledge and information from (1) bleed onto it. This actually doesn't limit y to anything other than y^2 being an integer or, rather, that y is the square root of x.

Combining the two, however gets us there because (1) limits our choices to {1, 3, 9, 27} and (2) limits our possibilities to perfect square roots. In that set of available choices, only 3 is the square root of another factor of 27.

Ans: [C]
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by imadummy » Wed May 21, 2008 8:04 am
While I'm a Yankees fan, I appreciate the clear solution you gave RedSox man.