If x and y are integers, is xy odd ? (1) x^y is odd (2) x + y is even

Question

If x and y are integers, is xy odd?

(1) x^y is odd.

(2) x + y is even.

BrentGMATPrepNow · Accepted Answer

Great question Bunuel!!

Given : x and y are integers

Target question :    Is xy odd?

Statement 1 : x^y is odd  
There are several values of x and y that satisfy statement 1. Here are two: 
 Case a : x = 1 and y = 1. Notice that 1^1 = 1, and 1 is odd. In this case, xy = (1)(1) = 1, which means the answer to the target question is  YES, xy is odd 
 Case b : x = 2 and y = 0. Notice that 2^0 = 1, and 1 is odd. In this case, xy = (2)(0) = 0, which means the answer to the target question is  NO, xy is not odd 
Since we can’t answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : x + y is even 
  Pro tip:   When testing values, always check to see whether you can save some time and reuse values from the previous statement. In this case, we can definitely use them since both pairs of values also satisfy statement 2: 
 Case a : x = 1 and y = 1. Notice that 1+1 = 2, and 2 is even. In this case, xy = (1)(1) = 1, which means the answer to the target question is  YES, xy is odd 
 Case b : x = 2 and y = 0. Notice that 2+0 = 2, and 2 is even. In this case, xy = (2)(0) = 0, which means the answer to the target question is  NO, xy is not odd 
Since we can’t answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined    
Notice that I was able to use the  same counter-examples  to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED. 
In other words, 
 Case a : x = 1 and y = 1. In this case, xy = (1)(1) = 1, which means the answer to the target question is  YES, xy is odd 
 Case b : x = 2 and y = 0. In this case, xy = (2)(0) = 0, which means the answer to the target question is  NO, xy is not odd 
Since we can’t answer the  target question  with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

G770 · Answer

For xy to be odd, x and y both should be odd.

Consider statement 1 alone, if x^y is odd, it implies that x is odd. We cannot say anything about y, so statement 1 is not sufficient.

Consider statement 2 alone, if x+y is even then x,y pair can be odd,odd or even,even 
Therefore statement 2 is not sufficient.

Consider both statements together, statement 1 implies that x is odd and if x is odd, then according to the statement 2, y must be odd too.
Therefore, xy is odd.
Both the statements together are sufficient.

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abuc0112 · Answer

If x and y are integers, is xy odd?

(1) x^y is odd. insufficient. this information tells us that X is odd but Y could be odd or evenn

(2) x + y is even. Insufficient. X and Y are both even or both odd

1 & 2 Togther is sufficient. both X and Y are odd, there the product XY will be odd.

SohGMAT2020 · Answer

If x and y are integers, is xy odd?

(1) x^y is odd.

(2) x + y is even.

For xy to be odd, both x and y have to be odd. ( Since X and Y are integers)

From Statement 1

x^y is odd; From this we can conclude that x is odd. However we can't conclude anything about y.

Not sufficient ( Eliminate A and D)

From Statement 2

x + y is even ; From this we can conclude that either both x and y are odd or both x and y are even. ie they are of the same sign. However we cannot conclude what sign they are.

Not sufficient ( Eliminate B) 
 
Combining

From 1 we got x is odd, and from 2 we got both x and y are of same sign. So combining we can get both x and y are odd. And hence we can sufficiently answer the question - is xy odd?

So, C is the correct answer IMO

BrentGMATPrepNow · Answer

Be careful. x doesn't have to be odd.

neetib · Answer

For xy to be odd -> both x and y should be odd
St1.x^y is odd 
odd raise to power any no is odd no.
so by st1 , we know x is odd but there is no info abt y -->>Not sufficient

st2.x+y=even
so either both x and y are even or both x and y are odd .-->>not sufficient

st1 & st 2
by st1 we know x is odd .therefore by st2 y is also odd

OA :C

abuc0112 · Answer

Thank you very much, Brent for pointing that out to me. After reading your analysis, I see where I made my mistake.

laurars · Answer

Many here are saying C, the correct answer is E.

We know from the statement that x and y must be odd

(1) x^y is odd.

If y>0 ---> x must be odd
If x=0 ---> x could be even or odd, since either would result in 1 = odd
Therefore statement 1 is not enough to know about x, and less so about y

(2) x + y is even.
Either both are odd or both are even, not enough

(1+2)
From 1 we know that x could be either odd or even, and from two we know the same information.
Insufficient info.

If x and y are integers, is xy odd ? (1) x^y is odd (2) x + y is even

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If x and y are integers, is xy odd ? (1) x^y is odd (2) x + y is even

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