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16 Sep 2014, 01:05
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Is the integer \(x\) even?
(1) \((x + 1)(x - 2)\) is even
(2) \((x - 3)(x + 5)\) is even
(1) \((x + 1)(x - 2)\) is even
(2) \((x - 3)(x + 5)\) is even
16 Sep 2014, 01:05
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Official Solution:
Is the integer \(x\) even?
(1) \((x + 1)(x - 2)\) is even
If \(x\) is even, then \((x + 1)(x - 2) = odd*even = even\). Conversely, if \(x\) is odd, \((x + 1)(x - 2) = even*odd = even\). Therefore, both even and odd values of \(x\) satisfy this statement. Not sufficient.
(2) \((x - 3)(x + 5)\) is even>
If \(x\) were even, \((x - 3)(x + 5)\) would be the product of two odd integers, thus odd, not even as given in the statement. Therefore, \(x\) must be odd. Sufficient.
Answer: B
Is the integer \(x\) even?
(1) \((x + 1)(x - 2)\) is even
If \(x\) is even, then \((x + 1)(x - 2) = odd*even = even\). Conversely, if \(x\) is odd, \((x + 1)(x - 2) = even*odd = even\). Therefore, both even and odd values of \(x\) satisfy this statement. Not sufficient.
(2) \((x - 3)(x + 5)\) is even>
If \(x\) were even, \((x - 3)(x + 5)\) would be the product of two odd integers, thus odd, not even as given in the statement. Therefore, \(x\) must be odd. Sufficient.
Answer: B
18 Apr 2022, 06:37
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Bunuel
Key Properties:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN
#4. (ODD)(ODD) = ODD
#5. (ODD)(EVEN) = EVEN
#6. (EVEN)(EVEN) = EVEN
Target question: Is integer x even?
Statement 1: (x + 1)(x - 2) is even
There are only two possible cases to consider: x is even or x is odd. So let's examine each case.
Case a: If x is even, then (x + 1)(x - 2) = (EVEN + 1)(EVEN - 2) = (ODD)(EVEN)= EVEN. This case satisfies statement 1, which means the answer to the target question is YES, x is even
Case b: If x is odd, then (x + 1)(x - 2) = (ODD + 1)(ODD - 2) = (EVEN)(ODD) = EVEN. This case satisfies statement 1, which means the answer to the target question is NO, x is not even
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: (x - 3)(x + 5) is even
Once again, we'll test both possible cases...
Case a: If x is even, then (x - 3)(x + 5) = (EVEN - 3)(EVEN + 5) = (ODD)(ODD)= ODD. This case does NOT satisfy statement 2. So, we must ignore it.
Case b: If x is odd, then (x - 3)(x + 5) = (ODD - 3)(ODD + 5) = (EVEN)(EVEN)= EVEN. This case satisfies statement 2, which means the answer to the target question is NO, x is not even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
