If x, y, and z are consecutive integers (x

Question

If x, y, and z are consecutive integers (x < y < z), which of the following cannot be the value of  z^2 - y^2 - x^2? ?

A. -12
B. -6
C. 0
D. 3
E. 4

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Dereno · Answer

Since the values are not too far, we can go for the plug and play approach.

x, y, and z are consecutive integers (x < y < z).

Integers means they can be both positive or negative.

Z 
 Y 
 X 
 
  -1 
 -2 
 -3 
 
  0 
 -1 
 -2 
 
  1 
 0 
 -1 
 
  2 
 1 
 0 
 
  3 
 2 
 1

let row 1 denote case 1, followed by case 2, 3 till 5.

substuiting the values in the equation:   z^2 - y^2 - x^2? .  We get the following cases.

case 1:  1-4-9 = -12

case 2:  0-1-4 = -5

case 3:  1-0-1 = 0

case 4:  4-1-0 = 3

case 5:  9-4-1 = 4.

HENCE, -6 cannot be an answer.  Option B.

Aproach 2:
  
This is a detailed approach and time consuming too.

The three digits are consecutive. Let’s the numbers be z, y= z-1; and x = Z-2.

Substuiting these values in the equation  z^2 - y^2 - x^2? , we get

z^2 - (z-1)^2 -(z-2)^2

solving we get ,  -z^2 + 6z -5

equate it to the answer choices,  -z^2 + 6z -5  = -12 , solving it we get z = -1, 7.

if z = 7, y=6, x = 5. This implies 49-36-25 = -12 .

This process is repeated for z= -1.

//  since solving it was lengthy and time consuming I have not solved it. Pls verify it  //

for each answer choice, we get two values of z, substuiting in the equation we get integer values. Except for option B (-6). Hence the answer,

Ayeka · Answer

x, y, and z are consecutive integers (x < y < z), try to put values for x,y,z 
(Z, Y, X) ..................z^2-y^2-x^2 
(-1, -2, -3) ............. 1-4-9=-12 
(0, -1, -2) ............... 0-1-4=-5 
(1, 0, -1) ................. 1-0-1=0 
(2, 1, 0) ................... 4-1-0=3 
(3, 2, 1) ................... 9-4-1=4 
There is no (-6) in these solutions, so answer is (-6)

Answer: B

ravi1522 · Answer

Hello ,
As x,y,z are consecutive integer 
so we can write them as N-1,N,N+1
z2−y2−x2= (N+1)^2-N^2-(N-1)^2

So we get 4N-N^2
No we equate N we various value 
N=0 we get 0
N=1 we get 3
N=2 we get 4
N=5 we get -5
N=6 we get -12
So we can not get -6

Hence option B is correct

Abhiswarup · Answer

We have to check with different options
x,y,z are consecutive integers, x<y<z, Sets can be= 1,2,3:0,1,2:-1,0,1:-2,-1,0:-3,-2,-1

Z^2-y^2-x^2= 4,3,0,-5,-12

-6 is not possible. B is the answer.

If x, y, and z are consecutive integers (x < y < z), which of the

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