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03 Sep 2018, 04:51
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
57% (01:01) correct
43%
(01:11)
wrong
based on 28
sessions
History
Date
Time
Result
Not Attempted Yet
For positive integers \(x\) and \(y\), is \(\sqrt{x^y}\) an integer?
(1) \(y\) is an odd number.
(2) \(x\) is a perfect square.
(1) \(y\) is an odd number.
(2) \(x\) is a perfect square.
03 Sep 2018, 04:52
Kudos
Bookmarks
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
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
Moderator:
