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

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SVP
Joined: 16 Oct 2003
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x and y are integers, is x even 1. x(y + 1) is even 2. (x + [#permalink]

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29 Jan 2006, 22:55
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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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VP
Joined: 21 Sep 2003
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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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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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Director
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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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Director
Joined: 17 Dec 2005
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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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Director
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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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Senior Manager
Joined: 13 Jun 2005
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30 Jan 2006, 15:04

Good questions Bhai.

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SVP
Joined: 14 Dec 2004
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30 Jan 2006, 19:22
Should be "C"

V good Q Bhai

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GMAT Club Legend
Joined: 07 Jul 2004
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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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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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VP
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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.
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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VP
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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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01 Feb 2006, 14:57
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