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Consecutive integers- faster method anyone? and key idea

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Consecutive integers- faster method anyone? and key idea

by kaps786 » Sat Aug 13, 2011 11:22 am
Hi

An MGMAT qn, the explanations for which I have seen are fairly elaborate algebra - wondering if anyone has a better approximate method or approach to solve this on real question.

I also did not grasp completely what is the underlying concept/key idea being tested.

Kindly help



If q, r, and s are consecutive even integers and q < r < s, which of the following CANNOT be the value of s2 - r2 - q2?


-20
0
8
12
16

OA follows....

OA 8

Thanks
kaps
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by GMATGuruNY » Sat Aug 13, 2011 12:43 pm
kaps786 wrote:Hi

An MGMAT qn, the explanations for which I have seen are fairly elaborate algebra - wondering if anyone has a better approximate method or approach to solve this on real question.

I also did not grasp completely what is the underlying concept/key idea being tested.

Kindly help



If q, r, and s are consecutive even integers and q < r < s, which of the following CANNOT be the value of s2 - r2 - q2?


-20
0
8
12
16

OA follows....

OA 8

Thanks
kaps
I was able to solve very quickly by plugging in values.

-4,-2,0:
0^2 - (-2)^2 - (-4)^2 = -20.
Eliminate A.

-2,0,2:
2^2 - 0^2 - (-2)^2 = 0.
Eliminate B.

0,2,4:
4^2 - 2^2 - 0^2 = 12.
Eliminate D.

Plugging in 2,4,6 -- or a set of greater consecutive even integers -- will yield a result greater than 12.
No way to yield a result of 8.

The correct answer is C.

For the skeptical:

2,4,6:
6^2 - 4^2 - 2^2 = 16.
Eliminate E.
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by saketk » Sun Aug 14, 2011 11:52 am
you can do this question like this as well -
let
q= 2a
r= 2a+2
s= 2a+4

where a is an integer number both positive and negative ( -1,0,1,2,3 etc )

now as per the equation given in the question

s2-r2-q2 -- we get,

4a2+16+16a - (4a2+4+8a) - 4a2

or, -4a2+8a+12 --

now, start plugging values,

a=1, gives 16
a=2, gives 12
a=3, gives 0
a=4, gives -20

Therefore, the correct answer is 8.
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