For non-negative integers x, y, and z, is x^z odd?

Question

(1) The product xz is odd.
(2) x = 2^y

Kudos for a correct  solution .

issamL · Accepted Answer

Statement 1 : The product xz is odd implies that x is odd and z is odd, hence x^z is odd. Sufficient

Statement 2 : x = 2^y and we are asked to check whether 2^(yz) is odd.

Here y can be 0 and thus X^Z = 1(odd) or Y and Z can be any other positive integer and then X^Z is even.( Insuff)

So, Answer (A)

camlan1990 · Answer

(1) The product xz is odd.

=> x, z are odd => x^z odd => SUF
(2) x = 2^y

=> x is even => x^z even => SUF

Answer: D

BrainLab · Answer

For non-negative integers x, y, and z, is x^z odd? --> this expression is odd only if x is odd (z is integer and doesn't matter in this case)

(1) The product xz is odd. -> if the product of 2 numbers is odd then we have Odd*Odd. SUFFICIENT 
(2) x = 2^y -> x=even, means x^z is also even SUFFICIENT
Answer D

BrainLab · Answer

Actually Issaml is right... just missed the point about non negative can be 0 (it's not said that it's positive)

Bowtye · Answer

A

stmt 1 says that both x and z are odd. suff

stmt 2 x could be even or odd depending on if y>0. insuff

Aves · Answer

I marked A but when I saw the first 2 comments, I was like WHAT???

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

In this problem, you hopefully found statement (1) to be sufficient, without too much effort. If the product xz is odd, then both integers must be odd. And if x, the base of x^z, is odd, then when raised to any integer it will be odd. Statement (1) guarantees that xz is odd.

But statement (2) seems to guarantee that x is even, as 2 to any exponent is even. Right? There’s one glaring exception: 2^0 = 1 — an important property for many exponent problems. So while most potential values of x, given statement (2), are even, one does exist where it’s odd. Statement (2) is not sufficient, and  the correct answer is A .

Here’s where you can use some higher-level strategy. If you were certain (as you should be) that statement (1) is sufficient with the answer “odd,” then as soon as you see a potential “even” with statement (2) you know that statement (2) is not sufficient! Why? Because the rules of the game dictate that if statement (1) gives you the answer “x must be odd,” then statement (2) cannot say “x must be even.” The only options are “x must be odd” or “x could be either odd or even.” Once you’ve found that “even” answer, then you know the “maybe” part must be coming, based on the rules of the game.

stonecold · Answer

Great Question
It really Tests our Knowledge on the concepts on evens/odds

Here we are given three non negative integers x,y,z
and we need to check if x^z is odd or not
There are two ways in which x^z can be odd here
Case 1=> If z=0, Then the value of x won't even matter as anything^0=1
Case 2=> If x=odd number. Then the value of z wont matter as odd^any non-negative integer is odd

Statement 1
Here as xz=odd
so x and z both must be odd
hence sufficient to say that x^z all be odd
Statement 2
x=2^y
so x can be either 1 (for y=0)
or an even number 
If x=1 => x^z will be odd
but if x=even , then unless z is zero => x^z will be even
And we are given no info on z
hence this statement is insufficient

Hence A

KanishkM · Answer

non-negative integers ->  this means it can be 0 or positive integers

from 1) product xz is odd, here x and y will always be odd

3*1 or 1*5
Sufficient

from 2) x = 2^y
y can be 0 or any other positive integer
so when y =0, 2^0 will make x = 1, which gives a Yes to our question 
Apart from this we can test any other value for y any other positive integer, x will become even, answering no to our question

A

Basshead · Answer

For non-negative integers x, y, and z, is  x^z  odd?

(1) The product xz is odd.

x = odd, z = odd.  x^{odd} = odd . SUFFICIENT.

(2)  x = 2^y

y is a non negative integer, meaning y can be 0. x = even or odd. INSUFFICIENT.

Answer is A.

For non-negative integers x, y, and z, is x^z odd?

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GMAT Data Sufficiency (DS) Questions

For non-negative integers x, y, and z, is x^z odd?

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