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# a,b,c are consecutire integers where a < b < c. Which

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Director
Joined: 10 Feb 2006
Posts: 657
a,b,c are consecutire integers where a < b < c. Which [#permalink]

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17 Sep 2007, 13:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

a,b,c are consecutire integers where a < b < c. Which of the following values are not possible for the equation c^2 - b^2 - a^2?

-12
-6
0
3
4

Please provide the most efficient way to answer this and the number theory that applies. Thanks
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Manager
Joined: 08 Oct 2006
Posts: 211

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17 Sep 2007, 14:27
I think the only way is to actually calculate it.

(a+2)^2-(a+1)^2-a^2
---> -a^2+2a+3

Only B does not fit.
You can plug in the values for various integers or actually calculate the eqn witth the answer choices and B is the only one that can not be calculated.
Manager
Joined: 29 Jul 2007
Posts: 182

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17 Sep 2007, 14:57
(n+2)^2-(n+1)^2-n^2 = -n^2+2n+3

The question asks for which of the answer choices makes the above expression -n^2+2n+3) not factorable (i.e. prime).

Answer choice B (-6) is the only one that makes the expression prime.

-n^2+2n+3= -6
-n^2+2n+9=0 is not factorable.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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17 Sep 2007, 23:07
a = n-1
b = n
c = n+1

c^2-b^2-a^2
= (n+1)^2 - n^2 - (n-1)^2
= (n^2 + 2n + 1) - n^2 - (n^2 - 2n + 1)
= 4n - n^2
= n(4-n)

If n = 0, n(4-n) = 0 --> c is out
If n = 1, n(4-n) = 3 --> d is out
If n = 2, n(4-n) = 4 --> e is out
If n = -2, n(4-n) = -12 --> a is out

Ans: B
17 Sep 2007, 23:07
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