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M70-19

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M70-19  [#permalink]

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New post 03 Sep 2018, 03:51
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

67% (00:51) correct 33% (00:38) wrong based on 3 sessions

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Re M70-19  [#permalink]

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New post 03 Sep 2018, 03:52
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.


Answer: B
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