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# If a and b are positive integers (a > b), is a^2 b^2

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SVP
Joined: 21 Jan 2007
Posts: 1867
Location: New York City
If a and b are positive integers (a > b), is a^2 b^2  [#permalink]

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21 Nov 2007, 02:37
If a and b are positive integers (a > b), is a^2 – b^2 divisible by 4?

1. a = b + 2
2. a and b are odd

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Director
Joined: 13 Nov 2003
Posts: 592
Location: BULGARIA

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21 Nov 2007, 03:01
Hi,
Stmnt 1) by substituting a with b+2 in stem we get a^2-B^2=4b+4 so A is sufficient
Stmnt2 is Insuff
So A
Manager
Joined: 08 Nov 2007
Posts: 73

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21 Nov 2007, 03:48
BG wrote:
Hi,
Stmnt 1) by substituting a with b+2 in stem we get a^2-B^2=4b+4 so A is sufficient
Stmnt2 is Insuff
So A

What if A=10 and B=8?

In that case we end up with 100 - 64 = 46 which is not divisible by 4...

I make it C... but only on trial and error - all odd numbers with a difference of 2 seem to produce a result divisible by 4.
VP
Joined: 28 Dec 2005
Posts: 1042

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21 Nov 2007, 05:14
I get B

Statement 1 isnt enough because a^2-b^2=(a+b)(a-b)=(a+b)2

Not sufficient

Statement two says a and b are odd, therefore a^2 and b^2 are odd as well.

Just start picking numbers and you will see that the result is always divisible by 4
Manager
Joined: 08 Nov 2007
Posts: 73

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21 Nov 2007, 06:03
pmenon wrote:
I get B

Statement 1 isnt enough because a^2-b^2=(a+b)(a-b)=(a+b)2

Not sufficient

Statement two says a and b are odd, therefore a^2 and b^2 are odd as well.

Just start picking numbers and you will see that the result is always divisible by 4

Good spot - I agree - B seems to work regardless of the gap between the numbers - so long as one is larger than the other.
Intern
Joined: 14 Nov 2007
Posts: 10

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21 Nov 2007, 15:10
I think D.

a^2 - b^2 = (a+b)(a-b)

(1) a = b + 2, this implies a^2 - b^2 = 4(b + 1) this is divisible by 4

(2) a and b are odd, this implies a^2 - b^2 = (a+b)(a-b) = even x even. this is divisble by 4.

is that wrong?
Senior Manager
Joined: 11 Jun 2007
Posts: 307

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21 Nov 2007, 16:32
yes, D for the same reason.
Director
Joined: 09 Jul 2007
Posts: 784
Location: London
Re: 12.31 divisibility by 4  [#permalink]

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21 Nov 2007, 16:48
bmwhype2 wrote:
If a and b are positive integers (a > b), is a^2 – b^2 divisible by 4?

1. a = b + 2
2. a and b are odd

1.
a^2=b^2+4b+4
a^2-b^2=4b+4=4(b+1)--suff.

2.
a=5, b=3. then 25-6=19 nope.
a=7, b=3. then 49-9=40 yep.
so not suff.

A
Senior Manager
Joined: 11 Jun 2007
Posts: 307
Re: 12.31 divisibility by 4  [#permalink]

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21 Nov 2007, 16:58
Ravshonbek wrote:
bmwhype2 wrote:
If a and b are positive integers (a > b), is a^2 – b^2 divisible by 4?

1. a = b + 2
2. a and b are odd

1.
a^2=b^2+4b+4
a^2-b^2=4b+4=4(b+1)--suff.

2.
a=5, b=3. then 25-6=19 nope.
a=7, b=3. then 49-9=40 yep.
so not suff.

A

Yo buddy..5^2 - 3^2 = 25 - 9 = 16
Senior Manager
Joined: 11 Jun 2007
Posts: 307

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21 Nov 2007, 16:59
alrussell wrote:
BG wrote:
Hi,
Stmnt 1) by substituting a with b+2 in stem we get a^2-B^2=4b+4 so A is sufficient
Stmnt2 is Insuff
So A

What if A=10 and B=8?

In that case we end up with 100 - 64 = 46 which is not divisible by 4...

I make it C... but only on trial and error - all odd numbers with a difference of 2 seem to produce a result divisible by 4.

Director
Joined: 09 Jul 2007
Posts: 784
Location: London
Re: 12.31 divisibility by 4  [#permalink]

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21 Nov 2007, 17:01
eyunni wrote:
Ravshonbek wrote:
bmwhype2 wrote:
If a and b are positive integers (a > b), is a^2 – b^2 divisible by 4?

1. a = b + 2
2. a and b are odd

1.
a^2=b^2+4b+4
a^2-b^2=4b+4=4(b+1)--suff.

2.
a=5, b=3. then 25-6=19 nope.
a=7, b=3. then 49-9=40 yep.
so not suff.

A

Yo buddy..5^2 - 3^2 = 25 - 9 = 16

hahahaha, thanks mate
D stamped
Manager
Joined: 03 Sep 2006
Posts: 163

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21 Nov 2007, 18:45
Got D:
I. [(a - b) (a + b)] / 4 = [(b + 2 - b)*(b + 2 + b)]/4 = [2*2*(b + 1)]/4 SUFF

II. [(odd - odd)*(odd + odd)]/4 = [ev * ev]/4 <-- by definition, hence SUFF

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
21 Nov 2007, 18:45