M21-07

Question

If  x ,  y , and  z  are even positive integers, each of the following must be an even integer EXCEPT:

A.  \frac{xyz}{4}  
B.  x^{(y + z)}  
C.  (x+y)^{z-1}  
D.  x-y-z  
E.  y^{z-x}

Bunuel · Answer

Official Solution:

If  x ,  y , and  z  are even positive integers, each of the following must be an even integer EXCEPT:

A.  \frac{xyz}{4}  
B.  x^{(y + z)}  
C.  (x+y)^{z-1}  
D.  x-y-z  
E.  y^{z-x}

Let's evaluate each option:
 
A.  \frac{xyz}{4} 
 
Given that all the variables are even, the product  xyz  will be a multiple of 8, thus  \frac{xyz}{4}  will certainly be a multiple of 2, hence even.
 
B.  x^{(y + z)} 
 
Since all the variables are even and positive, this expression would equate to  even^{positive \ even \ number} , which will always be even.
 
C.  (x+y)^{z-1} 
 
Similarly, this expression would equate to  even^{positive \ odd \ number} , which will also always be even.
 
D.  x-y-z 
 
This operation will always yield an even result for even numbers.
 
E.  y^{z-x} 
 
This expression could yield an even number. However, it's not necessarily even. Remember that any non-zero number raised to the power of 0 is 1. Therefore, if  z=x , this option becomes  y^{z-x}=y^0=1 , which is odd.

Answer: E

sureshhbs · Answer

Sir Question says "Must be even integer". So either it can be odd or it can be 0 ( as it doesnt say the resultant will be positive)

If we take option D i.e. x-y-z (let x=4 y=2 and z=2), the resultant would be 0 
this also satisfies the equation in my opinion.

Please clarify this doubt. Thanks

Bunuel · Answer

ZERO: 1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself. Check more here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030

b9n920 · Answer

Hi Bunuel, I have a doubt about choice C. In case z=0, it turns out to be the reciprocal of (x+y), which may not be an even integer, isn't it? Pls clarify. I had this Q as my 1st Q in quant section in a GmatClub test. Spent 2 full minutes and got demotivated

Pls help

Thanks, Binit

Bunuel · Answer

You should read questions more carefully: If  x ,  y , and  z  are even  POSITIVE  integers... 0 is not positive.

b9n920 · Answer

Yes. I should be more careful in reading. I unnecessarily got nervous on the first Q and wasted time. Thanks so much for ur suggestion.

Binit.

Klenex · Answer

Hi bunuel,

My question might be ridiculous but pls. reply. If x and z are same integers then why would the author name it differently? In questions like these, i should also consider the fact that all integers might be same?

Bunuel · Answer

Unless it is explicitly stated otherwise, different variables CAN represent the same number.

adymyr2 · Answer

hi  Bunuel ,

I have a question on D. 
If all my number are 2, meaning 2-2-2 = -2, which is not an even integer, as negatives cannot be even or odd, doesn`t D qualify as an answer as well?

Bunuel · Answer

An even number is an   integer   that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, ..., -4, -2, 0, 2, 4, ... are all even integers.

An odd number is an   integer   that is not evenly divisible by 2. So, ..., -3, -1, 1, 3, 5, ... are all odd integers.

jayditya · Answer

Hi

what if x=2 ,y=4, z=6 
they all are positive even integers
and then xyz/4 is not an integer
246/4=123/2. which is not an integer

how option A is wrong

Bunuel · Answer

xyz there means x*y*z, not a three-digit number xyz.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

sssdd1234 · Answer

I think this is a  high-quality  question and I agree with explanation.

Shabaz18gmat · Answer

I don’t quite agree with the solution. For Option 'C', if you take z=0, then no matter what numbers you take for 'x' & 'y', (x+y)^(z-1) will be a fraction, not even an integer, and you can also distinct values for x, y, and z.

brunel , 
On one instance you mention that zero is neither odd nor even integer but you also mentioned that any integer which can be divided by 2 is even integer and zero can be divided by 2.

Just want some clarification in the above statement

One doubt, should we assume in some cases that any two of the variables may be equal if 'distinct' word is not present?

Bunuel · Answer

1. 0 is even.

2. 0 is neither positive nor negative, so since we're given that the unknowns are even positive integers, you cannot consider z = 0.

3. Unless explicitly stated otherwise, different variables can represent the same value.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{0}{x} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

Hope it helps.

commodivero · Answer

I like the solution - it’s helpful.

Fisayofalana · Answer

I like the solution - it’s helpful.

swap29 · Answer

For Option E - if y = 6; z = 2 and x =4 it would mean 6^2-4 = 6^-2 = 1/6^2 which is not an integer. Hence, E is the correct option for this question as it is not an even integer. Is this reasoning correct?

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