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

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.

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

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.

\(\)

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..

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????

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.

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.

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

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.

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

We have an YES/NO data sufficiency question. In a Yes/No Data Sufficiency question, each statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".

(1) \(x^y=16\), since \(x\) and \(y\) are integers then following cases are possible:
\(x=-4\) and \(y=2\) --> \(x^y=16>\frac{1}{16}=y^x\) --> 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 \(x^y<y^x\)?" is NO. Thus this statement is sufficient.

(2) x and y are consecutive even integers --> 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.

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?

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

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

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.

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.

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

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