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If x and y are both integers, which is larger, x^x or y^y? [#permalink]
 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]
18 Feb 2012, 11:38
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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+1So 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^ySufficient. Hence Answer C . Seriously Kudos Hungry 
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Re: Which is larger, x^x or y^y? [#permalink]
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+1So 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 re-arrage this a bit. x^y>x so x^y-x>0 so x*(y-1)>0 and we know that x is +ve Now in order for x*(y-1)>0 the factor (y-1) 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^ySufficient. 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^{y-1}>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]
18 Feb 2012, 21:05
Had already corrected that before you wrote bunuel. While writing the solution I mistakenly took x^y-x for xy-x but corrected it when i was reading my solution after I had posted. Thankyou anyways!
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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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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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