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24 May 2022, 03:35
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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:
85%
(hard)
Question Stats:
31% (01:23) correct
69%
(01:19)
wrong
based on 154
sessions
History
Date
Time
Result
Not Attempted Yet
Is p an odd integer?
(1) \((p^5)^5\) is an odd integer.
(2) \(\sqrt[5]{\sqrt[5]{p}}\) is an odd integer.
(1) \((p^5)^5\) is an odd integer.
(2) \(\sqrt[5]{\sqrt[5]{p}}\) is an odd integer.
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24 May 2022, 07:36
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Bunuel
Target question: Is p an odd integer?
Statement 1: \((p^5)^5\) is an odd integer.
Since \((p^5)^5 = p^{25}\), statement 1 is really telling us that \(p^{25}\) is an odd integer.
There are several values of p that satisfy statement 1. Here are two:
Case a: If \(p = 1\), then \(p^{25} = 1^{25} = 1\), and \(1\) is an odd integer. Since \(p = 1\), the answer to the target question is YES, p is an odd integer
Case b: If \(p = \sqrt[25]{3}\), then \(p^{25} = (\sqrt[25]{3})^{25} = 3\), and \(3\) is an odd integer. Since \(p = \sqrt[25]{3}\), the answer to the target question is NO, p is not an odd integer
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: \(\sqrt[5]{\sqrt[5]{p}}\) is an odd integer.
First off, \(\sqrt[5]{\sqrt[5]{p}} = \sqrt[25]{p}\), which means statement 2 is telling us that \(\sqrt[25]{p} =\) some odd integer
Let's raise both sides of this equation to the power of \(25\) to get: \((\sqrt[25]{p})^{25} =\) (some odd integer)^25
Simplify: p = (some odd integer)^25
Since any odd integer raised to the power of 25 will always be odd, we can be certain that p is odd
Statement 2 is SUFFICIENT
Answer: B
06 Apr 2023, 04:46
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Bunuel
Official Solution:
Is \(p\) an odd integer?
(1) \((p^5)^5\) is an odd integer.
The above implies that \(p^{25}\) is an odd integer. If \(p\) itself is an odd integer, it would give a YES answer to the question. However, \(p\) can also be an irrational number, like the 25th root of an odd integer, for example, \(\sqrt[25]{2}\), giving a NO answer to the question. Not sufficient.
(2) \(\sqrt[5]{\sqrt[5]{p}}\) is an odd integer.
The above implies that \(\sqrt[25]{p}\) is an odd integer (\(\sqrt[5]{\sqrt[5]{p}}=\sqrt[5]{p^{\frac{1}{5}}}=p^{\frac{1}{25}}=\sqrt[25]{p}\)). Taking it to the 25th power gives \(p = odd^{25}\). Since an odd number raised to a positive integer power remains odd, then \(p\) must also be an odd integer. Sufficient.
Answer: B
