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26 May 2017, 04:42
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If \(x\) and \(y\) are positive integers, are \(x\) and \(y\) consecutive?
(1) \(x + y = 3\)
(2) \(x - y = 1\)
(1) \(x + y = 3\)
(2) \(x - y = 1\)
26 May 2017, 04:42
Kudos
Bookmarks
Official Solution:
You should take the 1st step of the variable approach and modify the original condition and the question. In order for \(x\) and \(y\) to be consecutive, the difference between \(x\) and \(y\) should be 1. Thus, in the case of con 2), if \(x - y = 1\), it is always yes, and sufficient. However, this is also an integer question, one of the key questions. Thus, if you apply "CMT 4(B: if you get A or B too easily, consider D)", in the case of con 1), it is stated that x and y are also positive integers in the original condition, and only \((x,y) = (1,2), (2,1)\) is possible because it satisfies \(x + y = 3\), hence yes, and sufficient. As such, you should always be careful of "CMT 4(B: if you get A or B too easily, consider D)". Therefore, the answer is D.
Answer: D
You should take the 1st step of the variable approach and modify the original condition and the question. In order for \(x\) and \(y\) to be consecutive, the difference between \(x\) and \(y\) should be 1. Thus, in the case of con 2), if \(x - y = 1\), it is always yes, and sufficient. However, this is also an integer question, one of the key questions. Thus, if you apply "CMT 4(B: if you get A or B too easily, consider D)", in the case of con 1), it is stated that x and y are also positive integers in the original condition, and only \((x,y) = (1,2), (2,1)\) is possible because it satisfies \(x + y = 3\), hence yes, and sufficient. As such, you should always be careful of "CMT 4(B: if you get A or B too easily, consider D)". Therefore, the answer is D.
Answer: D
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