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If x and y are positive integers, is x an even int? 1)

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If x and y are positive integers, is x an even int? 1) [#permalink] New post 29 Oct 2007, 10:20
If x and y are positive integers, is x an even int?

1) x(y+5) is an even int
2) 6y^2 + 41y + 25 is even int


I can't seem to find a way for (2) to be an even integer no matter what I plug in and I think it is a waste of time to factor (2)

any ideas?
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 [#permalink] New post 29 Oct 2007, 11:46
C

Start with stmt 2:

Plug in 1 in (2) and it yields 6+41+25 = 72 which is even
Plug in 2 in (2) and it yields 24+82+25 = 131 which is odd
Plug in 3 in (2) and it yields 54+123+25 = 202 which is even
Plug in 4 in (2) and it yields 96+164+25 = 285 which is odd

It tells us that y is odd but doesn't tell anything about x

Now come to stmt1
Plug in 1 for y in (1) and it yeilds = 6x which is even for any values of x
Plug in 3 for y in (1) and it yeilds = 8x which is even for any values of x

So, both statements are required.
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 [#permalink] New post 29 Oct 2007, 15:00
Careless mistake. Even after solving the whole thing picked wrong one. Yes, x can be either even or odd as per these two statements. So no definitve answer. E
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Re: gmatprep 2 even and odd [#permalink] New post 29 Oct 2007, 20:55
combres wrote:
If x and y are positive integers, is x an even int?

1) x(y+5) is an even int
2) 6y^2 + 41y + 25 is even int


I can't seem to find a way for (2) to be an even integer no matter what I plug in and I think it is a waste of time to factor (2)

any ideas?


1) x can be ever or odd. Try x = 2 ,y=3 and x=3 y=3

Insuff.

2) nothing about X. but it does establish that y must be odd. here is why:
6y^2+41y+25

O=odd
E=Even

6(O)+41(O) +25 ---> E+O +O --> O +O--> Even

If y is even then 6(E) +41(E) +25---> E+E+O which is NOT even.

1&2:

X can still be even or odd. If y is odd then y+5 must be even. however, if x is odd O(E)=E or E(E)=E.

Insuff.


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 [#permalink] New post 30 Oct 2007, 02:32
St1:
x(y+5) could be x = even or (y+5) = even or both even. Insufficient.

St2:
6y^2 + 41y + 25 = even

25 = odd

Can be 6y^2 = odd, 41y = even, 25 =odd then odd+even+odd = even.
But if 41y = even, then y must be even and y^2 = even and 6y^2 cannot be odd.

So it must be 6y^2 = even, 41y = odd, 25 = odd then even+odd+odd= even.
Now we know y = odd. But we do not know x. Insufficient.

St1 and St2:
Since we know y is odd, y+5 = even. So x(y+5) = even but x can be odd or even. Insufficient.

Ans E
  [#permalink] 30 Oct 2007, 02:32
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