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If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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04 Mar 2014, 00:14
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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SOLUTIONIf 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. Answer: C.
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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04 Mar 2014, 06:05
Option C. From S1:if y=1,2,3,4,... x=1,4,9,16,... If x=y=1 Ansno If x=9 & y=3 Ansyes Insuff.
From S2:if y>2 Y=3,4,5,6... We don't have any infor for x.So ans will change depending on x. Insuff.
Combine both statements. y=3,4,5,6,... x=9,16,25,36,... Ans will always be yes.
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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08 Mar 2014, 13:51
SOLUTIONIf 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. Answer: C.
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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21 Aug 2014, 22:17
But for this question, we dont need to know the value of X.
If Y > 2, say Y = 3, then the question becomes X^3 < 3^X.
Put any integer value, either positive or negative, the answer will come to yes.
Please correct if I am wrong and thanks in advance for correcting my concepts.



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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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30 Oct 2014, 08:15
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nnitingarg wrote: But for this question, we dont need to know the value of X.
If Y > 2, say Y = 3, then the question becomes X^3 < 3^X.
Put any integer value, either positive or negative, the answer will come to yes.
Please correct if I am wrong and thanks in advance for correcting my concepts. If you put y=4 and x=2, then the condition fails... It becomes 16 on both sides  leading to x^y = y^x. Hence it fails and not sufficient..



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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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30 Oct 2014, 09:02
Thanks for your explanation. Looks like I was not fully devoted when I wrote my previous comments.
This time I solved it before reading any comments. and afterwards, I tried to understand what I had understood last time to fill the gaps.
Thanks and appreciate all your help.



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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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20 Apr 2015, 17:18
Bunuel wrote: 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.
Answer: C. My approach was: Is: \(x^y < y^x\) implying: Is \(x < y^{(x/y)}\) Statement 1: \(x = y^2\) substituting in Is \(x < y^{(x/y)}\); i get: \(y^2 < y^{(y^2/y)}\) equal to Is: \(y^2 < y^y\) insufficient Statement 2: \(y > 2\) substituting in Is \(x < y^{(x/y)}\); i get: is \(x > 2^{(x/2)}\) insufficient Statement 1 + 2: \(y^2 < y^y\) (from statement 1) and since y > 2, then \(y^2\) must be less than \(y^y\) Sufficient.



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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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01 Aug 2015, 07:05
See GMATquantum... has a good solution https://www.youtube.com/watch?v=80aapMWgM3o
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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12 Aug 2015, 07:22
I am stuck on why second statement is not suffeceint. Y>2 tells me that y is positive. X can either be positive or negative integer but not zero. I picked (x=1 and Y=3), (X=1 and Y=3), (X=2 and Y=3), (X=2 and Y=3) even (X= 10 and y=3) All the above combinations give only one answer i.e, "yes". Tell me where I am going wrong.



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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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12 Aug 2015, 07:35
Shreeya24 wrote: I am stuck on why second statement is not suffeceint. Y>2 tells me that y is positive. X can either be positive or negative integer but not zero. I picked (x=1 and Y=3), (X=1 and Y=3), (X=2 and Y=3), (X=2 and Y=3) even (X= 10 and y=3) All the above combinations give only one answer i.e, "yes". Tell me where I am going wrong. The cases you have mentioned are correct and they will give you a "Yes" but we can also take x=y and in this case the answer will be a "no". The question stem or even this statement does not stop us from picking x=y=3.



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If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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18 Dec 2016, 04:14
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf x and y are nonzero integers, is x^y < y^x ? (1) x = y^2 (2) y > 2 Should not first stmt be sufficient substituting 1) (y^2)^y < y^(y^2) y^2y < y^2y; since both are same then no always. Is my manipulation of exponents incorrect?



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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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18 Dec 2016, 07:43
[/quote] Should not first stmt be sufficient
substituting
1) (y^2)^y < y^(y^2) y^2y < y^2y; since both are same then no always. Is my manipulation of exponents incorrect?[/quote]
Your interpretation of exponents is not correct. (y^2)^y is correctly mentioned as y^(2y) while y^(y^2) is not equal to y^(2y), it can not be written in any other way but y^(y^2). You can check this by assuming y = 2.
Hope this helps.
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If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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27 Jan 2018, 21:40
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. Therefore, C is the answer. 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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