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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)

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.

4 can not be the greatest value as when you increase \(x\) so as \(x-y\) to be \(4\), \(2^x+2^y\) will always be more than \(x^2+y^2\).
_________________

Great Problem! Since your trying to find the greatest value of X-Y, 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.

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.

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 non-negative integers 2) we need to find the max value of x-y

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?
_________________

Is it a rule to apply one value as zero whenever it is given:

1) Both x and y are non-negative integers 2) we need to find the max value of x-y

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 'non-negative 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.

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.

Re: If 2^x + 2^y = x^2 + y^2, where x and y are nonnegative inte [#permalink]

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06 Dec 2013, 10:47

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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 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!

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 |x-y| as 3 consider x=3 and y=0. Notice that these values satisfy \(2^x + 2^y = x^2 + y^2\) --> \(2^3 + 2^0 =9= 3^2 + 0^2\).

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 |x-y| 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.

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 |x-y| 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)(x-1)=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.

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.