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If x and y are both integers, which is larger, x^x or y^y? [#permalink]
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18 Feb 2012, 09:33
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If x and y are both integers, which is larger, x^x or y^y? (1) x = y + 1 (2) x^y > x and x is positive.
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Re: Which is larger, x^x or y^y? [#permalink]
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Answer should be C. Here is how: If \(x\) and \(y\) are both integers, which is larger, \(x^x\) or \(y^y\) ? Statement A: \(x=y+1\) So \(x\) and \(y\) are consecutive integers. Remember they can be positive, negative or \(0\) (from the question stem). Suppose \(y=1\) and \(x=2\) , then \(x^x\) is larger , but suppose \(y=2\) and \(x =1\) then \(y^y\) is larger. Hence Insufficient. Statement B: \(x^y>x\) and \(x\) is positive. Knowing that \(x>0\), \(x^y>x\) is only possible if \(y>1\) . Please note even when \(y=0\) , \(x>0\) and \(x\) is an integer. So now we know that \(y\) is positive and \(x\) is positive but we do not know which is larger. Hence InsufficientCombined: From Statement 2 we know that \(x\) and \(y\) are both \(>0\) and from Statement 1 we know that \(x\) is bigger. So YES. \(x^x\) is bigger than \(y^y\) Sufficient. Hence Answer C . Seriously Kudos Hungry
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Re: Which is larger, x^x or y^y? [#permalink]
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18 Feb 2012, 12:08
omerrauf wrote: Answer should be C. Here is how: If \(x\) and \(y\) are both integers, which is larger, \(x^x\) or \(y^y\) ? Statement A: \(x=y+1\) So \(x\) and \(y\) are consecutive integers. Remember they can be positive, negative or \(0\) (from the question stem). Suppose \(y=1\) and \(x=2\) , then \(x^x\) is larger , but suppose \(y=2\) and \(x =1\) then \(y^y\) is larger. Hence Insufficient. Statement B: \(x^y>x\) and \(x\) is positive. Let's rearrage this a bit. \(x^y>x\) so \(x^yx>0\) so x*(y1)>0 and we know that x is +ve Now in order for \(x*(y1)>0\) the factor \((y1)\) has to be \(>0\) so we know that \(y>1\) but we do not know which is bigger, \(x\) or \(y\). Hence InsufficientCombined: From Statement 2 we know that \(x\) and \(y\) are both \(>0\) and from Statement 1 we know that x is bigger. So YES. \(x^x\) is bigger than \(y^y\) Sufficient. Hence Answer C Seriously Kudos Hungry The red part is not correct. If x and y are both integers, which is larger, x^x or y^y? (1) x = y + 1 > if \(y\) is positive integer then \(x^x=(y+1)^{y+1}>y^y\) but if \(y=2\) then \(x=1\) and \(x^x=1<\frac{1}{4}=y^y\) (2) x^y > x and x is positive > since \(x\) is positive then \(x^{y1}>1\) > since \(x\) and \(y\) are integers then \(y>1\). If \(x=1\) and \(y=2\) then \(x^x<y^y\) but if \(x=3\) and \(y=2\) then \(x^x>y^y\). Not sufficient. (1)+(2) From (2) \(y>1\), so it's a positive integer then from (1) \(x^x=(y+1)^{y+1}>y^y\). Sufficient. Answer: C. P.S. Not a GMAT style question.
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Re: If x and y are both integers, which is larger, x^x or y^y? [#permalink]
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18 Feb 2012, 21:05
Had already corrected that before you wrote bunuel. While writing the solution I mistakenly took \(x^yx\) for \(xyx\) but corrected it when i was reading my solution after I had posted. Thankyou anyways!
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Exponents [#permalink]
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04 Mar 2012, 10:15
If x and y are both integers, which is larger, x^x or y^y?
x = y + 1 x^y > x and x is positive.



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Re: Exponents [#permalink]
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04 Mar 2012, 10:43
St 1: x=y+1 eg; y=2; then x= 2+1 = 3 then 3^3 > 2^2 y=3 then x= 3+1 = 2 then 2^2 > 3^3 Sufficient St 2: x^y>x eg; x = 2 y = 3 then x^y>x Thus, x^x < y^y However, if x = 4 and y = 3 then also x^y> x But, x^x > y^y Not sufficient ANS: A I dont know if this is the best approach.
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Re: Exponents [#permalink]
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04 Mar 2012, 13:34
http://www.platinumgmat.com/practice_gm ... on_id=2197
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Re: Which is larger, x^x or y^y? [#permalink]
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27 Dec 2013, 10:09
Bunuel wrote: omerrauf wrote: Answer should be C. Here is how: If \(x\) and \(y\) are both integers, which is larger, \(x^x\) or \(y^y\) ? Statement A: \(x=y+1\) So \(x\) and \(y\) are consecutive integers. Remember they can be positive, negative or \(0\) (from the question stem). Suppose \(y=1\) and \(x=2\) , then \(x^x\) is larger , but suppose \(y=2\) and \(x =1\) then \(y^y\) is larger. Hence Insufficient. Statement B: \(x^y>x\) and \(x\) is positive. Let's rearrage this a bit. \(x^y>x\) so \(x^yx>0\) so x*(y1)>0 and we know that x is +ve Now in order for \(x*(y1)>0\) the factor \((y1)\) has to be \(>0\) so we know that \(y>1\) but we do not know which is bigger, \(x\) or \(y\). Hence InsufficientCombined: From Statement 2 we know that \(x\) and \(y\) are both \(>0\) and from Statement 1 we know that x is bigger. So YES. \(x^x\) is bigger than \(y^y\) Sufficient. Hence Answer C Seriously Kudos Hungry The red part is not correct. If x and y are both integers, which is larger, x^x or y^y? (1) x = y + 1 > if \(y\) is positive integer then \(x^x=(y+1)^{y+1}>y^y\) but if \(y=2\) then \(x=1\) and \(x^x=1<\frac{1}{4}=y^y\) (2) x^y > x and x is positive > since \(x\) is positive then \(x^{y1}>1\) > since \(x\) and \(y\) are integers then \(y>1\). If \(x=1\) and \(y=2\) then \(x^x<y^y\) but if \(x=3\) and \(y=2\) then \(x^x>y^y\). Not sufficient. (1)+(2) From (2) \(y>1\), so it's a positive integer then from (1) \(x^x=(y+1)^{y+1}>y^y\). Sufficient. Answer: C. P.S. Not a GMAT style question. Just curious, why not a GMAT style question? Thanks Cheers! J




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