Equations (Consec Integers): If q, r, and s are consecutive

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If q, r, and s are consecutive even integers and q < r < s, which of the following CANNOT be the value of s^2 – r^2 – q^2 ?

(A) -20 (B) 0 (C) 8 (D) 12 (E) 16

Would be interested in seeing the various approaches to solving this question.

Thanks in advance.

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by VP_Jim » Tue May 20, 2008 5:39 pm
The first thing that comes to mind for me is to plug in numbers, since all of our answer choices are pretty close together and nice, low numbers to work with, it shouldn't be that bad.

I'd start with q=-2, r=0, and s=2. Plugging that in to the given equation gives us 0, so that's not the answer.

Then I'd change the numbers: q=0, r=2, s=4, which gives us 12, so that's out.

q=2, r=4, s=6 gives us 16. Now since we've hit the highest answer choices, I'd start lowering the numbers.

q=-4, r=-2, s=0 gives us -20, so that's out.

One more thought: if I were crunched for time, I might've stopped once I showed it could be 0 and 12, since we skipped 8, I would've guessed that's the answer.

The only one we haven't used is 8, so that's the answer. A little time consuming but at least we're sure we're correct - the numbers don't lie!
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by luvaduva » Tue May 20, 2008 6:25 pm
Jim's approach is probably the quickest, but if you are just dying for a quadratic then...

q, q+2, q+4

(q+4)^2 - (q+2)^2 - q^2 => q^2 + 8q + 16 - q^2 - 4q - 4 - q^2 => -q^2 + 4q + 12

Set this equal to each answer and see if it produces an acceptable answer. Multiply it through by a negative and remember to do the same with each answer..q^2 -4q -12 = -( ):

a) q^2 -4q -32 = 0 => (q-8)(q+4)=0, q = 8 or -4 OKAY
b) q^2 -4q -12 = 0 => (q+2)(q-6)=0, q =-2 or 6 OKAY
c) q^2 -4q -4 = 0 => NOT OKAY. This is your answer.

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by AleksandrM » Wed May 21, 2008 9:15 am
I think more than anything this number is testing whether you know that 0 is an even integer.