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Math Expert V
Joined: 02 Sep 2009
Posts: 56269
If x and y are nonzero integers, is x^y < y^x ?  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 67% (02:05) correct 33% (02:13) wrong based on 1503 sessions

### HideShow timer Statistics The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

(1) x = y^2
(2) y > 2

Data Sufficiency
Question: 121
Category: Arithmetic; Algebra Arithmetic operations; Inequalities
Page: 161
Difficulty: 650

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Math Expert V
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Posts: 56269
Re: If x and y are nonzero integers, is x^y < y^x ?  [#permalink]

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SOLUTION

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

(1) x = y^2 --> if $$x=y=1$$, then $$x^y=1=y^x$$, so the answer would be NO BUT if $$y=3$$ and $$x=9$$, then $$x^y=9^3<y^x=3^9$$, so the answer would be YES. Not sufficient.

(2) y > 2. No info about $$x$$, not sufficient.

(1)+(2) From (1) $$x = y^2$$, thus the question becomes: is $$(y^2)^y<y^{(y^2)}$$? --> is $$y^{2y}<y^{(y^2)}$$? Now, since from (2) $$y=integer>2$$, then $$2y$$ will always be less than $$y^2$$, therefore $$y^{2y}$$ will be less than $$y^{(y^2)}$$. Sufficient.

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Math Expert V
Joined: 02 Sep 2009
Posts: 56269
Re: If x and y are nonzero integers, is x^y < y^x ?  [#permalink]

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SOLUTION

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

(1) x = y^2 --> if $$x=y=1$$, then $$x^y=1=y^x$$, so the answer would be NO BUT if $$y=3$$ and $$x=9$$, then $$x^y=9^3<y^x=3^9$$, so the answer would be YES. Not sufficient.

(2) y > 2. No info about $$x$$, not sufficient.

(1)+(2) From (1) $$x = y^2$$, thus the question becomes: is $$(y^2)^y<y^{(y^2)}$$? --> is $$y^{2y}<y^{(y^2)}$$? Now, since from (2) $$y=integer>2$$, then $$2y$$ will always be less than $$y^2$$, therefore $$y^{2y}$$ will be less than $$y^{(y^2)}$$. Sufficient.

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Re: If x and y are nonzero integers, is x^y < y^x ?  [#permalink]

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables ($$x$$ and $$y$$) and 0 equations, C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
$$x^y = (y^2)^y = y^{2y}$$
$$y^x = y^{y^2}$$
Since $$y > 2$$, we have $$y^2 > 2y$$ and $$y^{2y} < y^{y^2}$$.
Thus both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1)
$$x = 9, y = 3 ⇒ 9^3 = 3^6 < 3^9$$ : Yes
$$x = 1, y = 1 ⇒ 1^1 = 1^1$$ : No
The condition 1) only is not sufficient.

Condition 2)
$$x = 1, y = 4 ⇒ 1^2 < 4^1$$ : Yes
$$x = -1, y = 4 => (-1)^4 = 1 > 4^{-1} = \frac{1}{4}$$ : No
The condition 2) only is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If x and y are nonzero integers, is x^y < y^x ?  [#permalink]

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Bunuel wrote:
If x and y are nonzero integers, is x^y < y^x ?

(1) x = y²
(2) y > 2

Target question: Is x^y < y^x ?

Given: x and y are nonzero integers

Statement 1: x = y²
Take the original target question, and replace x with y² to get: Is (y²)^y < y^(y²)?
Simplify the left side power to get: Is y^(2y) < y^(y²)?

Let's TEST some values of y.
Case a: If y = 1, then y^(2y) = 1 and y^(y²) = 1. In this case, the answer to the target question is NO, it is NOT the case that y^(2y) < y^(y²)
Case b: If y = 3, then y^(2y) = 3^6 and y^(y²) = 3^9. In this case, the answer to the target question is YES, it IS the case that y^(2y) < y^(y²)
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y > 2
Since we have no information about x, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1, we changed the target question to read Is y^(2y) < y^(y²)?
Now let's divide both sides of the inequality by y^(2y) to get: Is 1 < [y^(y²)/y^(2y)]?
Apply the Quotient Law to get: Is 1 < y^(y² - 2y)?
Notice that, if y is an INTEGER greater than 2 (i.e., statement 2), then y² - 2y is greater than 1
So, y^(y² - 2y) MUST be greater than 1
So, the answer to the target question is YES it IS the case that 1 < y^(y² - 2y)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent
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Re: If x and y are nonzero integers, is x^y < y^x ?  [#permalink]

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1
From Statement 1 x=y^2 so x and y both can be equal to 1 , then x^y cannot be less than y^x as both are equal
When y=3 then x=9 then x^y will be less than y^x. So no unique answer. Hence statement 1 alone is not suffice

Statement 2, y>2 does not provides us with any data about x hence not suffice as we cannot compare the expressions

Both statements together are required. When we combine both the statements, only value possible is y=3 and x=9
Hence Option C

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Wont give up till i hit a 700+ Re: If x and y are nonzero integers, is x^y < y^x ?   [#permalink] 16 Oct 2018, 20:15
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