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marcovg4
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E
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If negative integers k and p are NOT both even, which of the following must be odd?
(A) kp
(B) 4(k + p)
(C) k – p
(D) k + 1 – p
(E) 2(k + p) – 1
(A) kp
(B) 4(k + p)
(C) k – p
(D) k + 1 – p
(E) 2(k + p) – 1
26 Jan 2013, 16:00
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marcovg4
If a negative k and p are NOT both even, which of the following must be odd?
A. kp
B. 4(k + p)
C. k - p
D. k + 1 - p
E. 2(k + p) - 1
A. kp
B. 4(k + p)
C. k - p
D. k + 1 - p
E. 2(k + p) - 1
The question should read:
If negative integers k and p are NOT both even, which of the following must be odd?
(A) kp --> can be even if either k or p is even.
(B) 4(k + p) --> always even.
(C) k – p --> can be even as well as odd. For example consider k-p=odd-odd=even and k-p=odd-even=odd.
(D) k + 1 – p --> can be even as well as odd. For example consider, k+1-p=odd+odd-odd=odd and k+1-p=odd+odd-even=even.
(E) 2(k + p) – 1 --> always odd, because: 2(k+p)-1=2*integer-odd=even-odd=odd.
Answer: E.
Hope it's cler.
marcovg4
Joined: 15 Mar 2012
Last visit: 04 Jan 2017
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26 Jan 2013, 16:20
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Totally!
I had a dilemma yesterday while answering the question, whether the answer was A or E, but a closer look shows that "both NOT even" infer that one of them could be even, and that's the part I had trouble with. I thought that "both NOT even = both ARE odd".
Thanks for the help Bunuel!
I had a dilemma yesterday while answering the question, whether the answer was A or E, but a closer look shows that "both NOT even" infer that one of them could be even, and that's the part I had trouble with. I thought that "both NOT even = both ARE odd".
Thanks for the help Bunuel!
