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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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15 Dec 2009, 15:48
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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right)
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Re: nonnegative integers  MGMAT Challenge [#permalink]
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axl_oz wrote: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right) Suppose you got the answer of 2 for the values of \(x\) and \(y\) as 4 and 2. \(2^4+2^2=4^2+2^2\) > \(42=2\) But if we check for \(y=0\), we'll get: \(2^x+2^0=x^2+0^2\) > \(2^x+1=x^2\) > \(2^x=(x1)(x+1)\) > \(x=3\) \(2^3+2^0=9=3^2+0^2\) \(xy=30=3\) 4 can not be the greatest value as when you increase \(x\) so as \(xy\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\).
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Re: nonnegative integers  MGMAT Challenge [#permalink]
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19 Dec 2009, 07:20
Great Problem! Since your trying to find the greatest value of XY, you just have to assume that Y=0, like Bunel said and then use the "hit and trial" approach like xcusem... Said. The algebratic approach is great too, but I know for me personally it opens up the opportunity for me to make silly mistakes. So I try to not use it unless necessary.
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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08 Feb 2013, 09:27
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axl_oz wrote: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right) Since we need to maximize the value of x – y, we can do that in two ways...1)make y negative, which is not possible as per the question...2)make y= 0..putting y=0 you will get an equation in x and on hit and trial method u will get the value of x as 3, which will satisfy the equation.... putting x=3 and y=0, we will get the value of x – y as 3.



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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20 Mar 2013, 12:16
I have a question for Bunuel: we know that x−y = ((xy)*2)*1/2 = (x*2+y*22xy)*1/2
so substituting the value of x*2 +y*2 as 2*x + 2*y in the above equation
I got x−y = (2*x + 2*y 2xy)*1/2
then, substituting values for x (taking y=0)...the max value for x can be anything more than 0, coz if you take (x=4) then u'l end up with x−y = 4
please can you help me with this problem, where did I go wrong???



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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21 Mar 2013, 03:39



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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21 Mar 2013, 07:35
sorry...what i meant to say was, if you took the eq: x−y = (2*x + 2*y 2xy)*1/2
and substitute the values (x = 4 & y = 0), then we'll end up with x−y = (17)*1/2 (which is almost equal to 4)
but before u mentioned the max value of x−y = 3.
did i do something wrong??



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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21 Mar 2013, 09:34
Brunel and all, Is it a rule to apply one value as zero whenever it is given: 1) Both x and y are nonnegative integers 2) we need to find the max value of xy What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise?
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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21 Mar 2013, 21:56
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surya167 wrote: Brunel and all,
Is it a rule to apply one value as zero whenever it is given:
1) Both x and y are nonnegative integers 2) we need to find the max value of xy
What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise? Usually, when you are checking for numbers, you do check for 0. It's often a transition point for patterns. Secondly, the question used the term 'nonnegative integers' instead of 'positive integers'  this means 0 would probably have a role to play. There are no such rules but common sense says that we must not ignore 0. Also, this question doesn't really test max min concepts. It is a direct application of your understanding of exponential relations discussed in the post: http://www.veritasprep.com/blog/2013/01 ... cognition/Now, when we look at the equation, 2^x + 2^y = x^2 + y^2, some things come to mind: 1. It is not very easy to find values that satisfy this equation. 2. But there must be some values which satisfy since we are looking for a value of x – y 3. If x = y = 2, the equation is satisfied since all terms become equal and x – y = 0 which is the minimum value of x – y. Usually, the left hand side will be greater than the right hand side (as discussed in the post, 2^n will usually be greater than x^2 except in very few cases). So we must focus on those 'very few cases'. Also, we need to make x and y unequal. We know (from the post) that 2^4 = 4^2 is one solution so we could put x = 4 while keeping y = 2. The equation will be satisfied and x – y = 2 Now, we also know that 2^x < x^2 when x = 3. So is there a solution there as well? The difference between 2^3 and 3^2 is of 1 so can we create a difference of 1 between the other two terms? Sure! If y = 0, then 2^0 = 1 but 0^2 = 0. So another solution is 2^3 + 2^0 = 3^2 + 0^2. Here, x – y = 3 which is the maximum difference. The reason we can be sure that there are no other values is that as you go ahead of 4 on the number line, 2^n will be greater than n^2 (again, discussed in the post). So both left hand side terms will be greater than the right hand side terms i.e. 2^x > x^2 and 2^y > y^2. So, for no other values can we satisfy this equation. Answer (D)
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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the expression xy to reach its maximum we need y to be 0. Hence, we need to find X.
therefore, 2^x+2^y=x^2+^2 > 2^x+1=x^2 what means that x is odd. Only 3 satisfies this equation: 2^3+1=3^2. Hence, x must be equal 3



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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Bunuel wrote: misanguyen2010 wrote: axl_oz wrote: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right) my answer: x – y for x and y 2^x + 2^y = x^2 + y^2 A. x= y x=y=1 2 + 2 = 1 + 1 wrong B. 1 x=2, y=1 4 + 2 = 4 + 1 wrong C. 2 x=3, y=1 8 + 2 = 9 + 1 rightD. 3 x=4, y =1 16 + 2 = 16 + 1 wrong E. 4 x=5, y =1 32 + 2 = 25 + 1 wrong My answer is C. Please note that the correct answer is D, not C. Hi thank you for your reply. I explained what i confused. Of course I read previous answers and all chose D. However, from what i found, i chose C. That s why i posted here. I dont know which is wrong in my answer. Please help!



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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07 Dec 2013, 05:52
misanguyen2010 wrote: Bunuel wrote: misanguyen2010 wrote: my answer: x – y for x and y 2^x + 2^y = x^2 + y^2 A. x= y x=y=1 2 + 2 = 1 + 1 wrong B. 1 x=2, y=1 4 + 2 = 4 + 1 wrong C. 2 x=3, y=1 8 + 2 = 9 + 1 right D. 3 x=4, y =1 16 + 2 = 16 + 1 wrong E. 4 x=5, y =1 32 + 2 = 25 + 1 wrong
My answer is C. Please note that the correct answer is D, not C. Hi thank you for your reply. I explained what i confused. Of course I read previous answers and all chose D. However, from what i found, i chose C. That s why i posted here. I dont know which is wrong in my answer. Please help! To get the greatest value of xy as 3 consider x=3 and y=0. Notice that these values satisfy \(2^x + 2^y = x^2 + y^2\) > \(2^3 + 2^0 =9= 3^2 + 0^2\). Hope it helps.
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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12 Dec 2013, 02:02
For x>4,y>4 2^x > x^2  2^y > y^2 Hence maximum we need to check till (x,y) = {0,1,2,3}
By hit and trial .
For x=3,y=1 and x=4,y=2 xy = 2
Hence the answer is 2. Draw the graph for 2^x and x^2. The solution becomes simpler .



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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28 Dec 2013, 04:25
Here is how I done it: 1) If xy needs to be max then Y=0, because Y² is only positive 2) Check the answers, those are only integers, you are therefore looking for an integer 3) You have the equation 2^x +1 = X² 4) Use the different choices and you will see that only 3 matches. Answer D Hope it helps!
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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29 Jul 2015, 21:07
Maximize xy by either making y negative or y = 0. y cannot be negative as given, so make y = 0.
Plug 0 in for y to get x^2  1 = 2^x x^2 1^2=2^x If two integers have a median value, then they have a difference of squares. We already know that they have a difference of squares, so we need to find the median value. (x+1)(x1)=2^x x can only be odd numbers for there to be a median integer. Plug 3 and you get 4*2 = 2^3. That works, so x=3.



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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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01 Sep 2015, 04:33
Bunuel wrote: axl_oz wrote: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right) Suppose you got the answer of 2 for the values of \(x\) and \(y\) as 4 and 2. \(2^4+2^2=4^2+2^2\) > \(42=2\) But if we check for \(y=0\), we'll get: \(2^x+2^0=x^2+0^2\) > \(2^x+1=x^2\) > \(2^x=(x1)(x+1)\) > \(x=3\) \(2^3+2^0=9=3^2+0^2\) \(xy=30=3\) 4 can not be the greatest value as when you increase \(x\) so as \(xy\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\). how to simplify this withlout calculator 2^x = x^21



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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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anik19890 wrote: Bunuel wrote: axl_oz wrote: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 This is a challenge problem on MGMAT. I do not have the answer to the question... But on solving the problem I got the answer of 2 (not sure if it is right) Suppose you got the answer of 2 for the values of \(x\) and \(y\) as 4 and 2. \(2^4+2^2=4^2+2^2\) > \(42=2\) But if we check for \(y=0\), we'll get: \(2^x+2^0=x^2+0^2\) > \(2^x+1=x^2\) > \(2^x=(x1)(x+1)\) > \(x=3\) \(2^3+2^0=9=3^2+0^2\) \(xy=30=3\) 4 can not be the greatest value as when you increase \(x\) so as \(xy\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\). how to simplify this withlout calculator 2^x = x^21You can simplify via : 1. Graphical method. Plot graphs for \(2^x\) and \(x^21\) and see where these intersect and these points of intersections will give you possible values of x satisfying the given equation or 2. Iterative process wherein you find what value of x satisfy the equation \(2^x\) = \(x^2  1\) , in this case as \(2^x\) > 0 (for all x), I will not use x<0 values, x=3 satisfies the value. Graphical method is a bit more 'complex' if you are not comfortable with graphs and coordinate geometry. If given as a part of GMAT question, rest assured you can follow method 2 and you will be able to find a small enough value satisfying the given equation. In the case above, I saw that \(2^x\)> 0 for all x, thus started with x =1,2,3... etc Hope this helps.



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If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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01 Sep 2015, 11:05
[quote="dharam831"]If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative integers, what is the greatest possible value of x – y? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 hi, It is given that xy should be maximize and both x and y can not be negative (they are nonnegative integers). This means to maximize xy y value should be zero. Now if y=0, then the equation will be : 2^x + 2^0 = x^2 + 0^2 => 2^x +1 = x^2 => x^2  2^x = 1 This is possible in only one condition when 3^2  2^3. and hence x = 3. So, maximum value of xy is 30 = 3 Ans(D).
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Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]
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06 Mar 2018, 15:09
Hi All, This question is layered with clues to help us solve it, but the work that we'll do is more "brute force" than anything else. We're given 3 pieces of information: 1: 2^x+2^y=x^2+y^2 2: X and Y are NONNEGATIVE integers (this means that they're either 0 or positive) 3: The answers are small integers (0  4, inclusive) We're asked for the GREATEST possible value of XY….. From the answers, we know that X and Y have to be relatively "close" on the number line. Next, the phrase "nonnegative" is interesting  it gets me thinking that one of the values is probably going to be 0. Now, let's do a few brute force calculations so we can see the results: 2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 0^2 = 0 1^2 = 1 2^2 = 4 3^2 = 9 4^2 = 16 According to the given equation, we need to sum two values from the first list and sum the two corresponding values from the second list (and have the results equal one another). There are a couple of ways to do that, but we want the GREATEST possible difference between X and Y…. If X = 3 and Y = 0, then we'd have 2^x+2^y=x^2+y^2 8 + 1 = 9 + 0 9 = 9 3  0 = 3 Final Answer: GMAT assassins aren't born, they're made, Rich
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