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26 Jul 2006, 12:05
Kudos
Bookmarks
https://www.manhattangmat.com
Is the positive integer x odd?
(1) x = y^2 + 4y + 6, where y is a positive integer.
(2) x = 9z^2 + 7z - 10, where z is a positive integer.
Ok, this seems really straight forward (OA will not be posted until next week), but how can it be this straight forward if it's listed as a 700+ problem?? That's why I have to assume there is a trick.
I get B.
Thanks!
Is the positive integer x odd?
(1) x = y^2 + 4y + 6, where y is a positive integer.
(2) x = 9z^2 + 7z - 10, where z is a positive integer.
Ok, this seems really straight forward (OA will not be posted until next week), but how can it be this straight forward if it's listed as a 700+ problem?? That's why I have to assume there is a trick.
I get B.
Thanks!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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26 Jul 2006, 12:15
Kudos
Bookmarks
B
St1: x = (ODD or EVEN) + EVEN + EVEN = ODD or EVEN : INSUFF
St2:
Case1: If z is odd
x = ODD + ODD -EVEN = EVEN
Case2: If z is even
x = EVEN + EVEN - EVEN = EVEN
SUFF
St1: x = (ODD or EVEN) + EVEN + EVEN = ODD or EVEN : INSUFF
St2:
Case1: If z is odd
x = ODD + ODD -EVEN = EVEN
Case2: If z is even
x = EVEN + EVEN - EVEN = EVEN
SUFF
26 Jul 2006, 12:16
Kudos
Bookmarks
S1: x =y^2 + 4y +6
If y is odd => x = odd + even + even = odd
If y is even =>x= even + even + even = even
Not sufficient.
S2: x = 9z^2 + 7z - 10
If z is odd => x = odd * odd + odd -even = even
If z is even =>x = even + even-even = even
Sufficient.
Answer: B.
Not sure why its a 700+ problem.
If y is odd => x = odd + even + even = odd
If y is even =>x= even + even + even = even
Not sufficient.
S2: x = 9z^2 + 7z - 10
If z is odd => x = odd * odd + odd -even = even
If z is even =>x = even + even-even = even
Sufficient.
Answer: B.
Not sure why its a 700+ problem.
