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Math Expert V
Joined: 02 Sep 2009
Posts: 58445

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Difficulty:   45% (medium)

Question Stats: 40% (00:50) correct 60% (01:10) wrong based on 10 sessions

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For positive integers $$x$$ and $$y$$, is $$\sqrt{x^y}$$ an integer?

(1) $$y$$ is an odd number.

(2) $$x$$ is a perfect square.

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Math Expert V
Joined: 02 Sep 2009
Posts: 58445

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Official Solution:

We’ll go for ALTERNATIVE because we can use numbers to eliminate contradictory answers.

(1) Looking at the smallest odd powers, we can see that if $$x = y = 1$$, then $$\sqrt{1^1}=1$$, which is an integer, but if $$x = y = 3$$, then $$\sqrt{3^3}=\sqrt{27}$$, which is not an integer. Two contradictory results means that there’s insufficient information. (A) and (D) are eliminated.

(2) We can try different perfect squares as the value of $$x$$: 1, 4, 9, 16… No matter which value of $$y$$ we try, the result is always an integer. For example: $$\sqrt{4^1}=2$$, $$\sqrt{4^2}=4$$, $$\sqrt{4^3}=8$$. Even if we don’t see why this happens, if we don’t find a contradictory example – we should mark answer choice (B). The reason is that when we have both a power and a square, it doesn’t matter what we do first. So $$\sqrt{x^y}=(\sqrt{x})^y$$, and since $$x$$ is a perfect square, then $$\sqrt{x}$$ is an integer. Raised to the power of $$y$$ (an integer), the result must be an integer.

_________________ Re M70-19   [#permalink] 03 Sep 2018, 04:52
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