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Is the positive integer x odd? (1) x = y2 + 4y + 6, where y

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Is the positive integer x odd? (1) x = y2 + 4y + 6, where y [#permalink]

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New post 01 Aug 2006, 02:54
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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.

----------------------------------------------------------

My version of solution:

(1) x=y(y+4)+6 => if y is even, then x would be even; if y is odd, then x is odd;Could be both, hence I think (1) insufficient;

(2) x=y(9z+7)-10 => sma thing again here i think could be both...

(E)

P.S. I'm not sure thats why I posted:)
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Re: Manhattan GMAT's - DS - Is the positive integer x odd? [#permalink]

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New post 01 Aug 2006, 03:09
koSTARica wrote:
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.

----------------------------------------------------------

My version of solution:

(1) x=y(y+4)+6 => if y is even, then x would be even; if y is odd, then x is odd;Could be both, hence I think (1) insufficient;

(2) x=y(9z+7)-10 => sma thing again here i think could be both...

(E)

P.S. I'm not sure thats why I posted:)


There is no y in the second equation,

B

1) when y = even

x = even + even +even = even

When y = odd
x = odd + even + even = odd

Hence not suff.

2) When z = even
x = even + even - even = even

when z = odd
x = odd + odd -even = even

Hence x will be even in all cases. Suff
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Re: Manhattan GMAT's - DS - Is the positive integer x odd? [#permalink]

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New post 01 Aug 2006, 04:37
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jaynayak wrote:
koSTARica wrote:
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.

----------------------------------------------------------

My version of solution:

(1) x=y(y+4)+6 => if y is even, then x would be even; if y is odd, then x is odd;Could be both, hence I think (1) insufficient;

(2) x=y(9z+7)-10 => sma thing again here i think could be both...

(E)

P.S. I'm not sure thats why I posted:)


There is no y in the second equation,

B

1) when y = even

x = even + even +even = even

When y = odd
x = odd + even + even = odd

Hence not suff.

2) When z = even
x = even + even - even = even

when z = odd
x = odd + odd -even = even

Hence x will be even in all cases. Suff


sorry bout 2nd statement misstyped...

your correct thanks...
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Re: Is the positive integer x odd? (1) x = y2 + 4y + 6, where y [#permalink]

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New post 01 Dec 2015, 03:26
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Re: Is the positive integer x odd? (1) x = y2 + 4y + 6, where y   [#permalink] 01 Dec 2015, 03:26
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