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Re: If negative integers k and p are NOT both even, which of the [#permalink]
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!
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Re: If negative integers k and p are NOT both even, which of the [#permalink]
marcovg4 wrote:
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!

Actually irrespective of whether they are odd or even, choice E gives an odd number. It's 2 times some number -1 or simply even number - 1, hence odd.
You don't need to look beyond.
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Re: If negative integers k and p are NOT both even, which of the [#permalink]
Bunuel,

Does "NOT both even" mean both k and p cant be even simultaneously?
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Re: If negative integers k and p are NOT both even, which of the [#permalink]
marcovg4 wrote:
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


There are 2 possibilities -

1. Both k & p are odd

Plug in 2 values k = -1 & p = -3

(A) kp = 3
(B) 4(k + p) = -16
(C) k – p = 2
(D) k + 1 – p = -3
(E) 2(k + p) – 1 = -9

2. Either k or p is odd and the other is even

Plug in 2 values k = -1 & p = -2

(A) kp = 2
(B) 4(k + p) = -12
(C) k – p = 1
(D) k + 1 – p = -2
(E) 2(k + p) – 1 = -7

Thus, the answer must be (E) -7
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Re: If negative integers k and p are NOT both even, which of the [#permalink]
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Re: If negative integers k and p are NOT both even, which of the [#permalink]
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