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M12-31
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16 Sep 2014, 00:47
16 Sep 2014, 00:47
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Difficulty:
45%
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Question Stats:
64% (01:43) correct
36%
(01:37)
wrong
based on 83
sessions
If \(a\) and \(b\) are positive integers, is \(a^2 - b^2\) divisible by 4?
(1) \(a = b + 2\)
(2) Both \(a\) and \(b\) are odd integers
(1) \(a = b + 2\)
(2) Both \(a\) and \(b\) are odd integers
Re M12-31
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16 Sep 2014, 00:48
16 Sep 2014, 00:48
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Official Solution:
If \(a\) and \(b\) are positive integers, is \(a^2 - b^2\) divisible by 4?
(1) \(a = b + 2\)
Substituting the value of \(a\), we get \(a^2 - b^2 = (b + 2)^2 - b^2 = 4b + 4\), which is divisible by 4. Thus, this is sufficient.
(2) Both \(a\) and \(b\) are odd integers
If \(a\) and \(b\) are odd, they can be represented as \(2n + 1\) and \(2k + 1\) respectively, where \(n\) and \(k\) are integers. Therefore, \(a^2 - b^2 = (2n + 1)^2 - (2k + 1)^2 = 4n^2 + 4n - 4k^2 - 4k\), which is divisible by 4. This is sufficient.
Answer: D
If \(a\) and \(b\) are positive integers, is \(a^2 - b^2\) divisible by 4?
(1) \(a = b + 2\)
Substituting the value of \(a\), we get \(a^2 - b^2 = (b + 2)^2 - b^2 = 4b + 4\), which is divisible by 4. Thus, this is sufficient.
(2) Both \(a\) and \(b\) are odd integers
If \(a\) and \(b\) are odd, they can be represented as \(2n + 1\) and \(2k + 1\) respectively, where \(n\) and \(k\) are integers. Therefore, \(a^2 - b^2 = (2n + 1)^2 - (2k + 1)^2 = 4n^2 + 4n - 4k^2 - 4k\), which is divisible by 4. This is sufficient.
Answer: D
Re: M12-31
[#permalink]
11 Oct 2023, 01:32
11 Oct 2023, 01:32
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
