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a,b, and c are consecutive integers where a<b<c. which

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VP
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a,b, and c are consecutive integers where a<b<c. which [#permalink]

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29 Dec 2007, 10:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

a,b, and c are consecutive integers where a<b<c. which of the following could not be the value of c^2-b^2-a^2?

a. -21
b. -12
c. -6
d. 0
e. 3
Senior Manager
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29 Dec 2007, 11:21
[quote="marcodonzelli"]a,b, and c are consecutive integers where a<b<c>B(4-B)

B=0 ===> 0 possible
B=1 ===> 3 possible
B=5==> -5
B=6==> -12
B=7=>-21

-6 is not possible

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29 Dec 2007, 11:38
Its C.
Let a =x, then b = x+1, c= x+2

c^2-b^2-a^2
=>(x+2)^2 - (x+1)^2 -x^2
=> -x^2 +2x + 3

x can't be an integer when -x^2 +2x + 3 = -6
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29 Dec 2007, 11:50
marcodonzelli wrote:
nope OA is 3

Hmm OA is 3 ?? Not convinced ..
c^2-b^2-a^2
(0, 1,2) gives a value of 3 and 2,3,4 also gives a value of 3..
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29 Dec 2007, 12:24
Manager
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29 Dec 2007, 12:37
OA cant be 3. Damager is right.

If A=2, B=3, C=4, then the answer is 3.
If 3,4,5, then answer is 0.
If 5,6,7, then -12.
If 6,7,8, then -21.

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29 Dec 2007, 18:05
mkl_in_2001 wrote:
Its C.
Let a =x, then b = x+1, c= x+2

c^2-b^2-a^2
=>(x+2)^2 - (x+1)^2 -x^2
=> -x^2 +2x + 3

x can't be an integer when -x^2 +2x + 3 = -6

thats a good way of solving it.
good job!
Re: consecutive integers   [#permalink] 29 Dec 2007, 18:05
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