If x, y, and z are positive integers such that x = z^2 + 6z + 3

Question

If x, y, and z are positive integers such that x = z^2 + 6z + 3 and x + y is even, which of the following cannot be the value of y?

A. y = z^2 + z + 1
B. y = z^2 + 2z + 1
C. y = 2z^2 + z + 2 
D. y = 2z^2 + z + 3
E. y = 2z^2 + 2z + 1

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Kinshook · Answer

ExpertsGlobal5

If x, y, and z are positive integers such that x = z^2 + 6z + 3 and x + y is even, which of the following cannot be the value of y?

x = (z+3)^2 - 6; Since x + y is even ; Either both x & y are even or both are odd

A. y = z^2 + z + 1; x+y = z^2 + 6z + 3 + z^2 + z + 1 = 2z^2 + 7z + 4; If z is even, x+y is even
B. y = z^2 + 2z + 1; x+y = z^2 + 6z + 3 + z^2 + 2z + 1 = 2z^2 + 8z + 4; Always true
C. y = 2z^2 + z + 2; x+y = z^2 + 6z + 3 + 2z^2 + z + 2 = 3z^2 + 7z + 5; If z is odd, x+y is odd; If z is even; x+y is odd; NOT FEASIBLE
D. y = 2z^2 + z + 3; x+y = z^2 + 6z + 3 + 2z^2 + z + 3 = 3z^2 + 7z + 6; Always true
E. y = 2z^2 + 2z + 1; x+y = z^2 + 6z + 3 + 2z^2 + 2z + 1 = 3z^2 + 8z + 4; If z is even, x+y is even

IMO C

Dereno · Answer

Given that x,y,z are all positive integers.

Such that x,y,z > 0

x+y = even

This can occur in two cases:

1. Even + even = even

2. odd + odd = Even.

So, x = y = either odd or even.

Given that:  x = z^2 + 6z + 3

if x = Odd ,  then 6z + 3 , irrespective of z, 6z is even and 3 is odd.

6z + 3 = even + odd = Odd

x = z^2 + Odd

if x = Even, it can occur only if z^2 is Odd.

if x = Odd, it can occur only if z^2 is Even.

So, the Combination of (x,y,z) is  (Odd, Odd , Even )  Or  (Even, Even, Odd).

Option A : let’s use the case 1 :  (Odd, Odd , Even )

Odd = even + even + odd = even + odd = odd. Valid

Option B:

Odd = Even + Even + 1 = Even + 1 = Odd. Valid.

Option C:

Odd = Even + Even + Even = Even (  Invalid  )

let’s check case 2, for the same option

Even = Even + Odd + Even = Odd ( Invalid  ).

Hence,  Option C

ExpertsGlobal5 · Answer

Explanation:

x = z^2 + 6z + 3
  We need to select the value of y for which (x + y)  cannot  be even  .

A.
x + y = (z^2 + 6z + 3) + (z^2 + z + 1) = 2z^2 + 7z + 4
Since 2z^2 and 4 are even, 2z^2 + 4 must be even.
However, since 7z can either be even or odd,   x + y can either be even or odd . 
Thus, this answer choice is incorrect .

B.
x + y = (z^2 + 6z + 3) + (z^2 + 2z + 1) = 2z^2 + 8z + 4
Since 2z^2, 8z, and 4 are even,   x + y must be even.  
Thus, this answer choice is incorrect .

C.
x + y = (z^2 + 6z + 3) + (2z^2 + z + 2) = 3z^2 + 7z + 5
 Possibility 1 :   If z is even  , then 3z*2 and 7z will be even. This implies that 3z^2 + 7z will be even, and   x + y will be odd  . 
 Possibility 2 :   If z is odd  , then 3z*2 and 7z will be odd. This implies that 3z^2 + 7z will be even, and   x + y will be odd  . 
  x + y cannot be even.  
Thus, this answer choice is correct .

D.
x + y = (z^2 + 6z + 3) + (2z^2 + z + 3) = 3z^2 + 7z + 6
 Possibility 1 :   If z is even  , then 3z^2 and 7 z  will be even. This implies that 3z^2 + 7z will be even, and   x + y will be even  . 
 Possibility 2 :   If z is odd  , then 3z^2 and 7 z  will be odd. This implies that 3z^2 + 7z will be even, and   x + y will be even  . 
  x + y must be even.  
Thus, this answer choice is incorrect .

E.
x + y = (z^2 + 6z + 3) + (2z^2 + 2z + 1) = 3z^2 + 8z + 4
Since 8 z  and 4 are even, 8 z  + 4 must be even.
However, since 3z^2 can either be even or odd,   x + y can either be even or odd . 
Thus, this answer choice is incorrect .

C is the correct answer choice.

If x, y, and z are positive integers such that x = z^2 + 6z + 3

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GMAT Problem Solving (PS) Questions

If x, y, and z are positive integers such that x = z^2 + 6z + 3

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