If non- negative integers n & p are not both odd,

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If non- negative integers n & p are not both odd, Which of the following must be odd?
A) np
B) np +2
C) 2n+p
D)2(n +p)
E)2( n+p) +1

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by Jay@ManhattanReview » Thu Sep 27, 2018 10:47 pm
AbhishekRyu wrote:If non- negative integers n & p are not both odd, Which of the following must be odd?

A) np
B) np +2
C) 2n+p
D)2(n +p)
E)2( n+p) +1
Given that n & p are not both odd, and non-negative integers, we have one of n and p even and the other is odd. With this information in mind, let's see the options.

A) np: This is a product of an even and an odd integer; note that EVEN * ODD = EVEN

B) np + 2: EVEN + EVEN = EVEN

C) 2n + p: EVEN * ? + ? => note that product of two evens and product of even and odd is even, thus, EVEN * ? + ? = EVEN + ?.

If ? or p is EVEN, 2n + p is EVEN

D) 2(n + p): EVEN * (n + p) => Product of EVEN and EVEN or ODD is EVEN. Thus, 2(n + p) = EVEN

E) 2(n + p) + 1: EVEN + ODD = ODD. Correct answer.

The correct answer: E

Hope this helps!

-Jay
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by GMATGuruNY » Fri Sep 28, 2018 3:05 am
AbhishekRyu wrote:If non- negative integers n & p are not both odd, Which of the following must be odd?
A) np
B) np +2
C) 2n+p
D)2(n +p)
E)2( n+p) +1
If n=p=0, options A, B, C and D each yield a even value, indicating that these options do NOT have to be odd.
Eliminate A, B, C and D.

The correct answer is E.
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by Scott@TargetTestPrep » Sun Oct 28, 2018 5:20 pm
AbhishekRyu wrote:If non- negative integers n & p are not both odd, Which of the following must be odd?
A) np
B) np +2
C) 2n+p
D)2(n +p)
E)2( n+p) +1
So we see that n and p could be odd/even, even/odd, or both even.

Thus, since 2 times anything is even, and even + 1 is odd, 2(n + p) + 1 is odd.

Answer: E

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