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even/odd qustion

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even/odd qustion

by morbius14 » Wed Oct 24, 2012 10:41 am
Hi, not sure where my logic is wrong here...

Is the positive integer p even? (1) p^2 + p is even. (2) 4p + 2 is even.

The OA is E. But if we combine both statements as (p^2 + p) + (4p+2) = even, since Even + Even = Even.
We get p^2 + 5p + 2 = even. Now since all the terms would have to be even in order get an even outcome, it must be that p^2 = even and p must be even...

Please shed some light if I'm wrong somewhere, thank you!
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by Brent@GMATPrepNow » Wed Oct 24, 2012 10:56 am
morbius14 wrote: Is the positive integer p even?
(1) p^2 + p is even
(2) 4p + 2 is even.
Target question: Is p even?

Statement 1: p^2 + p is even
Factor to get: p(p+1) is even
Notice that p and p+1 are consecutive integers, which means one will be even and the other will be odd. However, it's unclear which one (p or p+1) is even.
To see what I mean, consider these two cases that satisfy statement 1.
case a: p=2, in which case p is even
case b: p=1, in which case p is not even
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 4p + 2 is even.
Factor to get: 2(2p+1) is even
Since 2 times any integer will result in an even number, we can see that 4p + 2 will be even for any integer p.
So statement 2 tells us nothing about p.
If there are any doubts, consider these two cases that satisfy statement 2.
case a: p=2, in which case p is even
case b: p=1, in which case p is not even
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
So we know that p^2 + p is even and we know that 4p + 2 is even.
If both of these are true, there are still some possible cases with conflicting results:
case a: p=2, in which case p is even
case b: p=1, in which case p is not even
Since we still cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

Cheers,
Brent

PS: If anyone is interested, we have a free video on using a table to test possible odd/even cases: https://www.gmatprepnow.com/module/gmat- ... ies?id=839
Last edited by Brent@GMATPrepNow on Wed Oct 24, 2012 11:08 am, edited 1 time in total.
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by Brent@GMATPrepNow » Wed Oct 24, 2012 11:02 am
morbius14 wrote: We get p^2 + 5p + 2 = even. Now since all the terms would have to be even in order get an even outcome, it must be that p^2 = even and p must be even...
p^2 + 5p + 2 = even --- agreed!

However, this does not mean that p must be even.

Try p = 1.
We get 1^2 + 5(1) + 2 = 8 (even)

Try p = 2.
We get 2^2 + 5(2) + 2 = 16 (even)

So, if p^2 + 5p + 2 is even, p can still be even or odd.

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Oct 24, 2012 11:06 am
morbius14 wrote:
But if we combine both statements as (p^2 + p) + (4p+2) = even, since Even + Even = Even.
We get p^2 + 5p + 2 = even. Now since all the terms would have to be even in order get an even outcome, it must be that p^2 = even and p must be even...
The main problem is in green above.

We know that p^2 + 5p + 2 is even
Since 2 is even, we know that the sum (p^2 + 5p) must be even. There are two ways that this can occur:
1) p^2 is even, and 5p is even
2) p^2 is odd, and 5p is odd

Cheers,
Brent
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by morbius14 » Wed Oct 24, 2012 11:12 am
Brent@GMATPrepNow wrote:
morbius14 wrote:
But if we combine both statements as (p^2 + p) + (4p+2) = even, since Even + Even = Even.
We get p^2 + 5p + 2 = even. Now since all the terms would have to be even in order get an even outcome, it must be that p^2 = even and p must be even...
The main problem is in green above.

We know that p^2 + 5p + 2 is even
Since 2 is even, we know that the sum (p^2 + 5p) must be even. There are two ways that this can occur:
1) p^2 is even, and 5p is even
2) p^2 is odd, and 5p is odd

Cheers,
Brent
Oh Thank you so much! Totally get it now.
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