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15 Aug 2019, 01:14
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D
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:
67% (01:04) correct
33%
(01:13)
wrong
based on 67
sessions
History
Date
Time
Result
Not Attempted Yet
If n and m are integers, is nm odd?
(1) n + 1 = m
(2) m/n is an even integer.
(1) n + 1 = m
(2) m/n is an even integer.
15 Aug 2019, 06:51
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Statement 1 tells us n and m are consecutive integers, so one of them is even, and the product mn is even.
Statement 2 tells us m/n = e, where e is some even number, so m = e*n, and m is even. So mn will be even.
So each Statement alone is sufficient to give us a "no" answer to the question, and the answer is D.
Statement 2 tells us m/n = e, where e is some even number, so m = e*n, and m is even. So mn will be even.
So each Statement alone is sufficient to give us a "no" answer to the question, and the answer is D.
17 Sep 2019, 02:35
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Bunuel
Official Explanation
We need to know whether nm is odd. Both are integers, so each will be odd or even. On to the statements, separately first.
Statement (1) will succumb to analysis by cases. n is either even or odd. If n is even, m is odd, and the product nm will be an even times an odd number, which is always even. When two numbers are multiplied, the product contains all of the factors of the two numbers multiplied. That's why any integer multiplied by an even number must be even; the factor of 2 carries over into the product. Moving on: if n is odd, then m is even, and nm is again even. These cases are exhaustive, so nm must be even. Statement (1) is sufficient.
Statement (2) tells us that m/n = even. Hence, m = n*even. That means that m must be even, since the factor of 2 in that number we are calling "even" will be contained within m. Since m is even, and n is an integer, nm must be even. We have answered the question definitively, so Statement (2) is sufficient.
The correct answer is (D).
