If q, r, and s are consecutive even integers and q

Question

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

ytsejam · Accepted Answer

Let the possible outcome of the equation be c.

Assuming the three numbers to be (r-2), r & (r+2), and substituting in the equation we get the equation as

8r-r^2 = c

Now multiplying by -1 we get

r^2-8r=-(c)

which becomes

r^2-8r+c=0

The determinant of this quadratic equation is thus:

64-4*c

now we compute the determinant for each of the options. The value of c which does not give a perfect square as the determinant will make the roots irrational and hence is not a valid answer:
a. c=-20 gives D=144 -> Rational roots
b. c=0 gives D=64 -> Rational roots
c. c=8 gives D=32 -> Irrational roots
d. c=12 gives D=16 -> Rational roots
e. c=16 gives D=0 -> Rational roots

Hence 8 cannot be the value of the expression.

Answer is C.

PareshGmat · Answer

s > r > q are consecutive even integers

Testing for  s^2 - r^2 - q^2

I: For s = 0, r = -2, q = -4,

s^2 - r^2 - q^2 = -20

II: For s = 2, r = 0, q = -2

s^2 - r^2 - q^2 = 0

III: For s = 4, r = 2, q = 0

s^2 - r^2 - q^2 = 12

Step II & Step III have consecutive representation of values of s, r, q. No other combination is possible.

8 cannot be the answer as for value of s = 6 & above, the resultant would be greater than 12

Answer = C

pacifist85 · Answer

This is fine, but is there an approach that is not based on trial and error? Because, on first site, I don't see why 8 would be the value.. And I guess there should be, otherwise there is not much point in this question. Anyone can plug in numbers and test them.. Am I missing anything?

mehrdadtaheri92 · Answer

If we changed the all 3 variables to one variable ( for example r ) we see that : s= r+2 and q= r-2 . so we should replace these equations in the

main equation and we get this equation: s^2 - r^2- q^2= (r+2)^2 - R^2- (r-2)^2 = 8r-r^2

so If we replaced any even integer in this new equation we see that ONLY 8 can not obtain.

If we replace 10 we get 8 (10) - (10)^2 = -20 so option A is ruled out

If we replace 4 we get 32- 16 = 16 etc So the only 8 can not the answer so option c.....

sunita123 · Answer

my approach was:
 take s=r-2 and q=r+2
=>s^2-r^2-q^2=(r-2)^2-r^2-(r+2)^2
by solving we will have -r(r+8)
when r =2, value is -20
when r= -2, value is 12
when r= 0, value is 0
when r=-4, value is 16

so remaining option is C:8

kplusprinja1988 · Answer

a more simple method would be using the quadratic equation.Since the numbers are consecutive

s=n+2
q=n
r=n-2

after substituting the values we get

n^2 +4+4n-n^2-n^2-4n+4n
which will be 
8n-n^2

equate the values to this equation to get value of n.

vitaliyGMAT · Answer

Let’s denote consecutive even integers as follows: 2x, 2x+2, 2x+4
As to the question we have following:

(2x+4)^2-(2x+2)^2-(2x)^2  by simplifying this expression we’ll get:
 -4x^2+8x+12

Now we need to put above polynomial into correspondence to our question choices and find out if resultant expression has integer solutions or, read it other way, can it be factorized or not.

a)  -4x^2+8x+12=-20 
 -4x^2+8x+32=0 
 -4(x-4)(x+2)=0  Yes.

b)  -4x^2+8x+12=0 
 -4(x-3)(x+1)=0  Yes.

c)  -4x^2+8x+12=8 
  -4(x2-2x-1)=0  This polynomial does not have integer roots.

d)  -4x^2+8x+12=12 
  -4x(x+2)=0  Yes.

e)  -4x^2+8x+12=16 
  -4(x-1)^2=0  Yes.

Answer C.

EMPOWERgmatRichC · Answer

Hi All,

We're told that Q, R and S are consecutive EVEN integers, so we're restricted there (notice that the prompt does NOT say anything about the 3 values being positive though). While this question might look a bit complex, you can "brute force" this question quickly by following the instructions and TESTing values. Using these values, here are the results (note that we're looking for what CANNOT be the value):

0, 2, 4 = 16 - 4 - 0 = 12 Eliminate D 
2, 4, 6 = 36 - 16 - 4 = 16 Eliminate E 
-2, 0, 2 = 4 - 0 - 4 = 0 Eliminate B 
-4, -2, 0 = 0 - 4 - 16 = -20 Eliminate A

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

Solution:

Since q, r, and s are consecutive even integers and q < r < s, then in terms of q, we see that r = q + 2 and s = q + 4. Therefore, we have:

s^2 – r^2 – q^2 
= (q + 4)^2 - (q + 2)^2 - q^2

= q^2 + 8q + 16 - q^2 - 4q - 4 - q^2

= 4q + 12 - q^2

Now, let’s set the expression to each of the given choices and solve for q. If an even integer value of q can’t be found, then that choice is the correct answer.

A) -20

4q + 12 - q^2 = -20

q^2 - 4q - 32 = 0

(q - 8)(q + 4) = 0

q = 8 or q = -4

B) 0

4q + 12 - q^2 = 0

q^2 - 4q - 12 = 0

(q - 6)(q + 2) = 0

q = 6 or q = -2

C) 8

4q + 12 - q^2 = 8

q^2 - 4q - 4 = 0

Since this can’t be factored over the integers, we can’t find an even integer value of q. Thus, C is the correct answer.

Answer: C

AbhiS101 · Answer

Let 2X, 2(X+1) & 2(X+2) be 3 consecutive even numbers
If we put these numbers in given eqn & take 4 common
as square of 2 will have 4 common in all
4  or you can say
4*

A. 4*  = 4* 3 consecutive integers(0,-1,-2)
B. 4* , -1, 0, 1
C. 4* , no such integers combination present, where sum of 3 consecutive integers is 2
D. 4 , 4,3,2
E. 4 , 3,2,1

IMO C

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If q, r, and s are consecutive even integers and q < r < s, which of

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