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MBA20
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For integers a and b, is a^b an even or odd integer? 10 Jun 2019, 06:38
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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:

72% (01:29) correct 28% (02:19) wrong based on 18 sessions

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For integers a and b, is \(a^b\) an even or odd integer?

1. \(a^2\) - \(b^2\) is an even integer
2. \(a^2\)/\(b^2\) is an even integer

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B
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IanStewart
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For integers a and b, is a^b an even or odd integer? 10 Jun 2019, 08:02
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MBA20
For integers a and b, is \(a^b\) an even or odd integer?

1. \(a^2\) - \(b^2\) is an even integer
2. \(a^2\)/\(b^2\) is an even integer
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This question is a mess. For one thing, it's open to two perfectly legitimate interpretations (you could think it's just asking if a^b is even, or you could think it's asking "is a^b an integer that is either even or odd?"). For another, the "right" answer is not right.

Using Statement 2 alone, we could have, say, a = 4 and b = 2. Then a^b is an even integer. But since all we know about a and b is that they are integers, we could also have, say, a = -4 and b = -2. Then Statement 2 is true, but a^b is not an integer at all.

The answer is only B here if they tell you that a and b are positive integers. Then Statement 2 ensures that more 2s divide a than b, so a must be even, and a^b is even any time a is even, when b is a positive integer. But this is not a good question.
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MBA20
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For integers a and b, is a^b an even or odd integer? 10 Jun 2019, 09:55
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IanStewart
MBA20
For integers a and b, is \(a^b\) an even or odd integer?

1. \(a^2\) - \(b^2\) is an even integer
2. \(a^2\)/\(b^2\) is an even integer


This question is a mess. For one thing, it's open to two perfectly legitimate interpretations (you could think it's just asking if a^b is even, or you could think it's asking "is a^b an integer that is either even or odd?"). For another, the "right" answer is not right.

Using Statement 2 alone, we could have, say, a = 4 and b = 2. Then a^b is an even integer. But since all we know about a and b is that they are integers, we could also have, say, a = -4 and b = -2. Then Statement 2 is true, but a^b is not an integer at all.

The answer is only B here if they tell you that a and b are positive integers. Then Statement 2 ensures that more 2s divide a than b, so a must be even, and a^b is even any time a is even, when b is a positive integer. But this is not a good question.
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Thank you IanStewart, I got it wrong on the Mock test, hence thought of posting. Assuming someone will understand if I hav't.
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For integers a and b, is a^b an even or odd integer? 28 Jun 2019, 11:49
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Such confusing Q's wont come on the official test

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