If x and y are integers, is x^y

Question

If x and y are integers, is x^y<y^x ?

(1) x^y= 16
(2) x and y are consecutive even integers.

jaizen · Answer

I would say that the asnwer is A.

From 1, we can say have the following combinations.
x=2, y=4. Ans: Not greater
x=4, y=2. Ans: Not greater
x=16, y=1. Ans: Not greater

All three possibilities of combination gives us a consistent answer as 'No'. Therefore A is sufficient.

What is the OA?

fluke · Answer

What if:
x=-2, y=4
x=-4, y=2

jaizen · Answer

I forgot to include signs. Well, since (1) says that the x^y = 16, this means that y has to be a positive, even number. So we can add two more combinations. 
x=-4, y=2
x=-2, y=4

The answer still holds.

jaizen · Answer

My very first kudo! This is special... thank you fluke. :-D

dreambeliever · Answer

guys, i was looking at this again.. and the OA seems debatable.

the question does not say if x<y or not.

so for 1, if x=16 and y = 1 then  x^y  = 16 and  y^x  = 1.
then  x^y  >  y^x 
but if y=16 and x =1 then  x^y  = 1 and  y^x  = 16. and  x^y  <  y^x .

so A is not sufficient.

combining the 2 statements, the only possibility will be (x,y) = (4,2) and (2,4). in both cases however,  x^y  =  y^x  and the ans to the question will be a No.
So i think the ans to this question should be C.

please let me know if i'm missing something.

nikhil007 · Answer

mmm..even i was confused, a bit..

X^Y can be -4^2, -2^4, 2^4 or 16^1 but in all cases above y^X is smaller than 16...so i guess its A In DS problem any concrete answer YES or NO can be an answer so i guess we do have an answer here..

mydreammba · Answer

Guys initially marked but now confused.....

1. it will satisfy for both

x=4, y=2 and x=2 y=4 ....the value is not greater But for x=16 y=1 X^Y>Y^X so we are getting two answers....so then how can the answer be A?????????

Can anyone please explain????

Harrypotterforever · Answer

I feel it is c. Because different values of X and Y are giving different relations.

Lets se, X=2, Y=4 then x^y=y^x

if we have x=16, y=1, then x^y > y^x

So, I feel C is correct answer.

gurpreetsingh · Answer

you dont need to know whether x>y.

there are 3 possibilities for statement 1.
x=2 y =4
x=4 y=2
x=16 y=1
for all three the answer to the question is NO. so it is sufficient.

shinbhu · Answer

1. x^y = 16. Only possible relationship is x=4,y=2. In this case x^y=y^x. Suff.
2. Say x=2, y=3, we get 2^3<3^2. Say x=3, y=4, we get 3^4>4^3. Insuff.

A

Apex231 · Answer

If x and y are integers, is x^y<y^x ? 
(1) x^y = 16
(2) x and y are consecutive even integers.

(1) x^y = 16
x y can be 
2 4
4 2 
16 1

not sufficient.

(2) x and y are consecutive even integers.
we don't know whether x > y or y > x.

not sufficient

1 + 2

x y can be
2 4 => x^y = 16 and y^x = 16
4 2 => x^y = 16 and y^x = 16
0 2 => x^y = 0 and y^x = 1
-2 -4 => x^y = 1/16 and y^x = 1/16

not sufficient.

Bunuel · Answer

If x and y are integers, is x^y x^y=16>\frac{1}{16}=y^x --> the answer to the question is NO; x=-2 and y=4 --> x^y=16>\frac{1}{16}=y^x --> the answer to the question is NO; x=2 and y=4 --> x^y=16=y^x --> the answer to the question is NO; x=4 and y=2 --> x^y=16=y^x --> the answer to the question is NO; x=16 and y=1 --> x^y=16>1=y^x --> the answer to the question is NO. As you can see in ALL 5 possible cases the answer to the question "is x^y if x=2 and y=4 the answer will be NO but if x=0 and y=2 the answer will be YES. Not sufficient. Answer: A.

alchemist009 · Answer

If x and y are integers, is x^y < y^x ?
(1) x^y = 16
(2) x and y are consecutive even integers.

according to statement 1 the possible pair of x and y would be (16, 1), (1, 16), (2,4), (4,2)

2^4=4^2 in this case the answer is no

if x= 1 then 1^16<16^1 in this case the answer is yes
if x=16 then 16^1>1^16 in this case the answer is No

so statement 1 is insufficient . am i right or am i missing something?

thuylinh · Answer

You are wrong: For the statement 1, the possible is (2,4) or (4,2) only.
1^16=1 not 16.

alchemist009 · Answer

yep i got now. i am just missing the x^y= 16 part.

cyberjadugar · Answer

Hi,

Since it is mentioned that x and y are integers.
 x^y = 16  gives following solution,
(x,y) = (-4,2), (-2,4), (2,4), (4,2) & (16,1)

to check  x^y < y^x  using above values of (x,y)
 16 = (-4)^2 
 16 = (-2)^4 
 16 = (2)^4 
 16 = (4)^2 
 16 = (16)^1

Although the answer is same (i.e, (A)), but   I want to emphasis on the fact that all the cases should be considered  .
This would help in other DS questions.

Jashan79 · Answer

The answer must be C because
the two of the conditions derived from the first equation is sufficient when x=2 or 4. But what about the value of x as 16.
Please correct the answer.

Bunuel · Answer

It's not clear what you mean at all. The answer is A, not C. Please check the discussion. For example, this post: https://gmatclub.com/forum/if-x-and-y-a ... l#p1030984

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x and y are integers, is x^y<y^x ?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

If x and y are integers, is x^y<y^x ?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards