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If p^q + r is an even number which of the following m

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akshaytanejaf1
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If p^q + r is an even number which of the following m [#permalink] New post   20 Oct 2019, 00:13
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If \(p^q + r\) is an even number which of the following must also be even?

A. p+q+r

B. pqr

C. p+r

D. q+r

E. None of these
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E
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Re: Number Properies Question [#permalink] New post   20 Oct 2019, 00:24
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Please, read the rules for posting on the board. This is a problem solving and must go under the right sub-forum NOT in the general quant section.

https://gre.myprepclub.com/forum/rules-for ... -1083.html
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Re: If p^q + r is an even number which of the following m [#permalink] New post   20 Oct 2019, 00:39
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p^q must be even and r as well

Even + even = Even

However, q could be odd or even

\(2^1+2=4\)

\(2^2+2=6\)

\(2^3+2=10\)

All the answer choices could be even or odd. But the question is a must be true

E is the answer
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Re: If p^q + r is an even number which of the following m [#permalink] New post   22 Oct 2019, 10:20
Why can't C be the answer?

(p^q) and r can be (even and even) or (odd and odd) since their sum is even.
When even p+r = even
When odd p+r = even
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Re: If p^q + r is an even number which of the following m [#permalink] New post   22 Oct 2019, 10:37
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3152gs wrote:
Why can't C be the answer?

(p^q) and r can be (even and even) or (odd and odd) since their sum is even.
When even p+r = even
When odd p+r = even


Here's a counter-example:

If \(p^q + r\) is even, it COULD be the case that p = 2, q = 0 and r = 3 (since \(2^0 + 3 = 1+3=4\), which is even)

In this case, \(p+r=2+3=5\), which is odd so p + r need not be even.
Eliminate C

Cheers,
Brent
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Re: If p^q + r is an even number which of the following m [#permalink] New post   22 Oct 2019, 10:48
Thanks. i completely ignored the fact that q could also be 0. Fell right into the trap. smh
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Re: If p^q + r is an even number which of the following m [#permalink] New post   23 Nov 2019, 07:11
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pqr can be also odd
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Re: If p^q + r is an even number which of the following m [#permalink] New post   23 Nov 2019, 08:02
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For the term to be even, both p^q and r has to be either odd or even at the same time.
If r is odd, p^q has to be odd --> For p^q has to be odd p has to be odd q can be either odd or even;
This can help us in eliminating all the choices. Hence, E
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