Gmat_mission wrote:If \(q, r\), and \(s\) are consecutive even integers and \(q < r < s\), which of the following CANNOT be the value of \(s^2 - r^2 - q^2?\)

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

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

Let's take convenient values of q, r and s. Say q = 2; thus, r = 4 and s = 6

Thus, \(s^2 - r^2 - q^2=6^2-4^2-2^2=16\). So, Option E is ruled out.

Looking at the options, we find that 16 is the highest value among all. Thus, we must keep trying with smaller values of q, r and s.

"¢ q = 0, r = 2, and s = 4: \(s^2 - r^2 - q^2 = 4^2-2^2-0^2=12\): Option D is ruled out.
"¢ q = -2, r = 0, and s = 2: \(s^2 - r^2 - q^2 = 2^2-0^2-(-2)^2=0\): Option B is ruled out.

You would observe that the no value of \(s^2 - r^2 - q^2\) would lie between 0 and 12; thus, Option C = 8 cannot be the value of \(s^2 - r^2 - q^2\).

The correct answer: C

Hope this helps!

-Jay
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Gmat_mission wrote:If \(q, r\), and \(s\) are consecutive even integers and \(q < r < s\), which of the following CANNOT be the value of \(s^2 - r^2 - q^2?\)

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

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

If we let q = 0, r = 2 and s = 4, then s^2 - r^2 - q^2 = 16 - 4 - 0 = 12.

If we let q = -2, r = 0 and s = 2, then s^2 - r^2 - q^2 = 4 - 0 - 4 = 0.

If we let q = -4, r = -2 and s = 0, then s^2 - r^2 - q^2 = 0 - 4 - 16 = -20.

If we let q = 2, r = 4 and s = 6, then s^2 - r^2 - q^2 = 36 - 16 - 4 = 16.

At this point, we see that the value of s^2 - r^2 - q^2 can't be 8 (since it can be any of the other 4 choices).

Answer: C