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# If x and y are integers, is xy divisible by 3 ? 1. (x+y)^2

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If x and y are integers, is xy divisible by 3 ? 1. (x+y)^2 [#permalink]

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27 Apr 2009, 15:51
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If x and y are integers, is xy divisible by 3 ?

1. (x+y)^2 is divisible by 9

2. (x-y)^2 is divisible by 9.

Is there a way not to plug numbers?

Kudos [?]: 303 [0], given: 5

Director
Joined: 23 May 2008
Posts: 800

Kudos [?]: 86 [0], given: 0

Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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27 Apr 2009, 21:46
If x and y are integers, is xy divisible by 3 ?

1. (x+y)^2 is divisible by 9

2. (x-y)^2 is divisible by 9.

Is there a way not to plug numbers?

C, i think

every thing divisible by 9 is divisible by 3
never mind the squares, prime factors will be the same

1) only x and y numbers that add up to 9 and are multiples of 3 will be divisible by 3 if you took xy
5+4 doesnt work, but 6+3 works, so insuff

2) similar to statement 1
11-2 doesnt work but 12-3 works, so insuff

together,
both x+y and x-y have to be divisible by 9 so you have to use multiples of three, so sufficient

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Manager
Joined: 05 Jul 2008
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GMAT 2: 740 Q51 V38
Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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29 Apr 2009, 03:26
1
KUDOS
If x and y are integers, is xy divisible by 3 ?

1. (x+y)^2 is divisible by 9

2. (x-y)^2 is divisible by 9.

Is there a way not to plug numbers?

(1)=> (x+y) is divisible by 3
(2)=> (x-y) is divisible by 3

(1) or (2) is not enough

(1) and (2): (x+y) -(x-y) is divisible by 3 so 2y is divisible by 3 so y is divisible by 3 => x is divisible by 3 too.
So xy is divisible by 9....
C for me

Kudos [?]: 124 [1], given: 40

Retired Moderator
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Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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29 Apr 2009, 06:02
Can you break this down for me? I did not understand that part.

(1) and (2): (x+y) -(x-y) is divisible by 3 so 2y is divisible by 3 so y is divisible by 3 => x is divisible by 3 too.
So xy is divisible by 9....
DavidArchuleta wrote:
If x and y are integers, is xy divisible by 3 ?

1. (x+y)^2 is divisible by 9

2. (x-y)^2 is divisible by 9.

Is there a way not to plug numbers?

(1)=> (x+y) is divisible by 3
(2)=> (x-y) is divisible by 3

(1) or (2) is not enough

(1) and (2): (x+y) -(x-y) is divisible by 3 so 2y is divisible by 3 so y is divisible by 3 => x is divisible by 3 too.
So xy is divisible by 9....
C for me

Kudos [?]: 303 [0], given: 5

Manager
Joined: 05 Jul 2008
Posts: 136

Kudos [?]: 124 [0], given: 40

GMAT 2: 740 Q51 V38
Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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29 Apr 2009, 07:47
(1) and (2): (x+y) -(x-y) is divisible by 3
so 2y is divisible by 3 => y is divisible by 3
x+y is divisible by 3 while y is divisible by 3 => x is divisible by 3.

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Manager
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Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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29 Apr 2009, 07:53
bandit wrote:
DavidArchuleta wrote:
If x and y are integers, is xy divisible by 3 ?

1. (x+y)^2 is divisible by 9

2. (x-y)^2 is divisible by 9.

Is there a way not to plug numbers?

(1)=> (x+y) is divisible by 3
(2)=> (x-y) is divisible by 3

(1) or (2) is not enough

(1) and (2): (x+y) -(x-y) is divisible by 3 so 2y is divisible by 3 so y is divisible by 3 => x is divisible by 3 too.
So xy is divisible by 9....
C for me

Good Explanation

I still think D is the answer here, and we don't need both statements to answer the question.
However, everyone seems to think it's C, and therefore could anyone please explain why the following reasoning is faulty:

stmnt1 - (x+y)^2 is divisible by 9 => (x^2+2xy+y^2) is div. by 9 => x^2, 2xy and y^2 are all standalone divisible by 9.
Since 2xy is divisible by 9, then xy must be divisible by 9 (because 2 is not). Therefore, if xy is divisible by 9, then it is also divisible by 3.

Same logic for statement 2.

Anyone?

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Manager
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Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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29 Apr 2009, 07:56
stmnt1 - (x+y)^2 is divisible by 9 =>
(x^2+2xy+y^2) is div. by 9 (right statement)
=> x^2, 2xy and y^2 are all standalone divisible by 9.(a very very wrong statement)
1+2+6 is divisible by 9

Since 2xy is divisible by 9, then xy must be divisible by 9 (because 2 is not). Therefore, if xy is divisible by 9, then it is also divisible by 3.

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Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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29 Apr 2009, 17:23
DavidArchuleta wrote:
stmnt1 - (x+y)^2 is divisible by 9 =>
(x^2+2xy+y^2) is div. by 9 (right statement)
=> x^2, 2xy and y^2 are all standalone divisible by 9.(a very very wrong statement)
1+2+6 is divisible by 9

Since 2xy is divisible by 9, then xy must be divisible by 9 (because 2 is not). Therefore, if xy is divisible by 9, then it is also divisible by 3.

I later came to realize the mistake myself - should have read the other people' solutions above.

thanks

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Retired Moderator
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Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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30 Apr 2009, 06:20
David, what I don't get is how did you get 2y from (x+y) - (x-y)? Not sure of the rationale behind this... thx.

DavidArchuleta wrote:
(1) and (2): (x+y) -(x-y) is divisible by 3
so 2y is divisible by 3 => y is divisible by 3
x+y is divisible by 3 while y is divisible by 3 => x is divisible by 3.

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Schools: Chicago Booth 2011
Re: If x and y are integers, is xy divisible by 3 ? [#permalink]

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30 Apr 2009, 18:20
Remember, both x and y are integers, so (x+y) and (x-y) are also integers.
If (x+y)^2 is divisible by 9, then (x+y) is divisible by 3, so (x+y) = 3A (where A is some unknown integer)
If (x-y)^2 is divisible by 9, then (x-y) is divisible by 3, so (x-y) = 3B (where B is some unknown integer)

Therefore, (x+y) - (x-y) = 2y = 3(A-B), so y = 3(A-B)/2. (A-B) must be even because y is an integer. Therefore y is divisible by 3.
Similarly, (x+y) + (x-y) = 2x = 3(A+B), so x = 3(A+B)/2. (A+B) must be even because x is an integer. Therefore x is divisible by 3.

So, xy must be divisible by 3.

Kudos [?]: 29 [0], given: 6

Re: If x and y are integers, is xy divisible by 3 ?   [#permalink] 30 Apr 2009, 18:20
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# If x and y are integers, is xy divisible by 3 ? 1. (x+y)^2

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