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x and y are integers, is x even 1. x(y + 1) is even 2. (x +

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Bhai
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x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   29 Jan 2006, 22:55
x and y are integers, is x even

1. x(y + 1) is even
2. (x + 2)(y + 4) is even



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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   29 Jan 2006, 23:46
This is a headache Bhai! I think C.

1. x(y+1) = even

If (x,y) = (1,1) the above can be even.
if (x,y) = (2,2) the above is definitely even. So INSUFF!

2. (x+2)(y+2) = even
if (x,y) = (1,3) the eqn can be even
if (x,y) = (2,3) the eqn is still even. So INSUFF!

Combing both: Here comes the trouble (I don't have a neat approach)
O= odd
E = even
I came up with a truth table type apparoach

x y x(y+1) (x+2)(y+4)
O O True False
O E False True
E O True True
E E True True

So when x is odd, both (1) and (2) cannot be true at the same time. Hence x must be even. Hence C.

Sorry about my vague explanation, but this the best I could come up with.
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 00:21
E.

stmt 1,

take x=2 y=3
2(3+1)=8 conditon satisfied, X is even

take x=y=3
3(3+1)- 12 condition satisfied X isNOT even insuff

stmt2,

x=3
y=2
(3+2) (2+4) =30, condition satisfied, X NOT even

x=2
y=3
4.7=28 , condition satisfied, X NOT even insuff

combining,
x=2
y=0
both conditions satisfied, X even

x=y=3
both conditions satisfied, X NOT even

Hence E
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 02:21
Bhai wrote:
x and y are integers, is x even

1. x(y + 1) is even
2. (x + 2)(y + 4) is even


1) x(y+1) even, so either x, y+1, or both are even => insuff

2) Either x, y or both are even or 0 =>insuff

1+2)

If y is odd, then

- acc. to 1., x can but must not be even (can't be 0)
- combined with 2. it must be even

If y is even, then

- acc. to 1. x must be even
- acc. to 2. x can be even, odd or 0

C it is, as far as I can see
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 02:28
oops..sorry, realized I made a mistake towards the end (x=y=3) this doesnt satifythe equation..answer should be C
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 15:04
my answer is C too..

Good questions Bhai.
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 19:22
Should be "C"

V good Q Bhai :good
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 19:59
1. x(y+1) is even -> x can be even, y+1 can be even, or both can be even. Insufficient.
2 (x+2)(y+4) -> insufficient for the same reason.

Using both, (y+1) = even and (y+4) is even

Note that (y+1) and (y+4) are opposites.

If y is odd, then (y+1) = even and (y+4) is odd.
If (y+4) is odd -> (x+2) has to be even -> x is even (which makes st1 okay as well)

Now, if y is even, then (y+1) = odd, (y+4) = even
(y+1) is odd -> x has to be even -> x+2 is even but it doesn't matter since (y+4) is even.

Both cases, x is even.

Ans C
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 20:01
giddi77 wrote:
This is a headache Bhai! I think C.

1. x(y+1) = even

If (x,y) = (1,1) the above can be even.
if (x,y) = (2,2) the above is definitely even. So INSUFF!

2. (x+2)(y+2) = even
if (x,y) = (1,3) the eqn can be even
if (x,y) = (2,3) the eqn is still even. So INSUFF!

Combing both: Here comes the trouble (I don't have a neat approach)
O= odd
E = even
I came up with a truth table type apparoach

x y x(y+1) (x+2)(y+4)
O O True False
O E False True
E O True True
E E True True

So when x is odd, both (1) and (2) cannot be true at the same time. Hence x must be even. Hence C.

Sorry about my vague explanation, but this the best I could come up with.


The truth table way is interesting. It's amazing how you transform digital electronics to math
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   30 Jan 2006, 20:27
ywilfred wrote:
giddi77 wrote:
This is a headache Bhai! I think C.

1. x(y+1) = even

If (x,y) = (1,1) the above can be even.
if (x,y) = (2,2) the above is definitely even. So INSUFF!

2. (x+2)(y+2) = even
if (x,y) = (1,3) the eqn can be even
if (x,y) = (2,3) the eqn is still even. So INSUFF!

Combing both: Here comes the trouble (I don't have a neat approach)
O= odd
E = even
I came up with a truth table type apparoach

x y x(y+1) (x+2)(y+4)
O O True False
O E False True
E O True True
E E True True

So when x is odd, both (1) and (2) cannot be true at the same time. Hence x must be even. Hence C.

Sorry about my vague explanation, but this the best I could come up with.


The truth table way is interesting. It's amazing how you transform digital electronics to math


You got me here ywilfred! I am a Chip Design Engineer! I am gald you liked it.
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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink] New post   01 Feb 2006, 14:57
C.....no gate logic involved...:)

1. x(y + 1) is even........................INSUFF (y+1) could be even.
2. (x + 2)(y + 4) is even....................INSUFF (y+4) could be even.

2. Simplifying, we get (x+2)(y+4) = xy + 4x + 2y + 8 is EVEN.
From I, xy + y is EVEN.

xy + 4x + y + y + 8 = EVEN
Substituting, xy + y as even, we get
Even + 4x + y + 8 = EVEN.
=> y = EVEN - EVEN - 4X -8
=> y = EVEN.

Since y is EVEN, y+1 is ODD.

From statement I,
x(y+1) = even and if y+1 is ODD, then x is EVEN...SUFF.



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Re: x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink]
01 Feb 2006, 14:57
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