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Oddly Enough - July 24, 2006 Challenge Problem

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marc.gagnon
 Post subject: Oddly Enough - July 24, 2006 Challenge Problem
 Post Posted: Fri Jun 03, 2011 9:36 pm 
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Students


Posts: 17
For the following question - can the answer be determined by number properties, or is this a problem that requires plugging in odd and even values of y and z?

Question Stem
Is the positive integer x odd?

(1) x = y2 + 4y + 6, where y is a positive integer.

(2) x = 9z2 + 7z - 10, where z is a positive integer.
Solution:
(2) SUFFICIENT: If z is even, then 9z2 + 7z – 10 will be even. For example, if z = 2, then 9z2 + 7z – 10 = 36 + 14 – 10 = 40. If z is odd, then 9z2 + 7z – 10 will still be even. For example, if z = 3, then 9z2 + 7z – 10 = 81 + 21 – 10 = 92. So no matter what the value of z, x will be even and we can answer "no" to the original question.

The correct answer is B.
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j.william.bell
 Post subject: Re: Oddly Enough - July 24, 2006 Challenge Problem
 Post Posted: Sat Jun 04, 2011 1:45 am 
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Students


Posts: 7
(1) y(y+4) + 6
Look at cases where y is even and then odd.
If y is even, it's product with the term inside the parenthesis will be even so E+E = E
If y is odd, y+4 is odd, so O*O+E = O
Insufficient

(2) z(9z+7) - 10
Same as before, look at cases where z is even and then odd
If z is even, it doesn't matter what the term in parenthesis is since it's multiplying an even number, so E-E=E
If z is odd, 9z+7 is even, so O*E-E=E
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jnelson0612
 Post subject: Re: Oddly Enough - July 24, 2006 Challenge Problem
 Post Posted: Sun Jun 05, 2011 8:49 am 
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ManhattanGMAT Staff


Posts: 1857
Nice job william.

I think this one illustrates the advantages of both testing numbers (we can rule out statement 1 pretty quickly using this method) and of using number properties rules (as illustrated in the solution to statement 2).

_________________
Jamie Nelson
ManhattanGMAT Instructor
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