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16 Sep 2014, 00:37
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If \(x\) is a positive odd integer, which of the following numbers must be even?
A. \(x^{\frac{6}{3} }\)
B. \(\frac{x^2 - 1}{2}\)
C. \(xp + 1\), where \(p\) is a prime number
D. \(\frac{(x + 7)(x - 2)}{2}\)
E. \(x^{x - 1}\)
A. \(x^{\frac{6}{3} }\)
B. \(\frac{x^2 - 1}{2}\)
C. \(xp + 1\), where \(p\) is a prime number
D. \(\frac{(x + 7)(x - 2)}{2}\)
E. \(x^{x - 1}\)
16 Sep 2014, 00:37
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Official Solution:
If \(x\) is a positive odd integer, which of the following numbers must be even?
A. \(x^{\frac{6}{3} }\)
B. \(\frac{x^2 - 1}{2}\)
C. \(xp + 1\), where \(p\) is a prime number
D. \(\frac{(x + 7)(x - 2)}{2}\)
E. \(x^{x - 1}\)
Let's evaluate each option:
A. \(x^{\frac{6}{3} }\). This is equivalent to \(x^2\), which will be odd because the square of an odd number is odd (\(odd^2 = odd\)).
B. \(\frac{x^2 - 1}{2}\). This is equivalent to \(\frac{x^2 - 1}{2} = \frac{(x - 1)(x + 1)}{2}\). Since \(x\) is odd, both \((x - 1)\) and \((x + 1)\) are even, so their product is a multiple of 4. Dividing by 2 gives an even result.
C. \(xp + 1\), where \(p\) is a prime number. If \(p = 2\), then \(xp + 1 = odd*even + 1 = odd\).
D. \(\frac{(x + 7)(x - 2)}{2}\). Since \(x\) is odd, the expression becomes \(\frac{even*odd}{2}\). If that even number, \((x + 7)\), isn't a multiple of 4, the overall expression will be odd. For example, consider \(x = 3\).
E. \(x^{x - 1}\). Given that \(x\) is a positive odd number, the expression equals to \(\text{positive odd number}^even\), which is odd.
Answer: B
If \(x\) is a positive odd integer, which of the following numbers must be even?
A. \(x^{\frac{6}{3} }\)
B. \(\frac{x^2 - 1}{2}\)
C. \(xp + 1\), where \(p\) is a prime number
D. \(\frac{(x + 7)(x - 2)}{2}\)
E. \(x^{x - 1}\)
Let's evaluate each option:
A. \(x^{\frac{6}{3} }\). This is equivalent to \(x^2\), which will be odd because the square of an odd number is odd (\(odd^2 = odd\)).
B. \(\frac{x^2 - 1}{2}\). This is equivalent to \(\frac{x^2 - 1}{2} = \frac{(x - 1)(x + 1)}{2}\). Since \(x\) is odd, both \((x - 1)\) and \((x + 1)\) are even, so their product is a multiple of 4. Dividing by 2 gives an even result.
C. \(xp + 1\), where \(p\) is a prime number. If \(p = 2\), then \(xp + 1 = odd*even + 1 = odd\).
D. \(\frac{(x + 7)(x - 2)}{2}\). Since \(x\) is odd, the expression becomes \(\frac{even*odd}{2}\). If that even number, \((x + 7)\), isn't a multiple of 4, the overall expression will be odd. For example, consider \(x = 3\).
E. \(x^{x - 1}\). Given that \(x\) is a positive odd number, the expression equals to \(\text{positive odd number}^even\), which is odd.
Answer: B
mmrigank
Joined: 29 Nov 2013
Last visit: 14 Apr 2024
Posts: 4
Given Kudos: 55
Location: United States (CA)
Schools: IE Sept "18 (A) IIMA (A) Cox '19 (A)
GMAT 1: 690 Q49 V35
GPA: 4
25 Jul 2016, 12:43
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I understand the solution,, but what is wrong with this understanding
(x+7)(x−2)/2 => x+7 = odd+odd = even
x-2 => odd -even = odd
(x+7)(x−2) => even*odd = even
even/2 (even) => even
what is wrong with above ?
(x+7)(x−2)/2 => x+7 = odd+odd = even
x-2 => odd -even = odd
(x+7)(x−2) => even*odd = even
even/2 (even) => even
what is wrong with above ?
