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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
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 01:51
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Official Solution:

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


We can rearrange the equation, putting all the \(x\)’s on one side and all the \(y\)’s on the other side:
\(2^x - x^2 = y^2 - 2^y\)

Now, list the values of \(2^n\) and \(n^2\) for the first several nonnegative integers \(n\). In fact, go ahead and compute the differences both ways (both \(2^n - n^2\) and \(n^2 - 2^n\)).
\(n\) \(2^n\) \(n^2\) \(2^n - n^2\) \(n^2 - 2^n\) 0 1 0 1 -1 1 2 1 1 -1 2 4 4 0 0 3 8 9 -1 1 4 16 16 0 0 5 32 25 7 -7 6 64 36 28 -28
From this point on, \(2^n\) grows much faster than \(n^2\), so the differences explode. This means that in order to have a valid equation \((2^x - x^2 = y^2 - 2^y)\), we will have to use small values of the integers. We want values in the \(2^n - n^2\) column to match values in the \(n^2 - 2^n\) column, and to maximize the value of \(|x - y|\), we want to pick values from different rows - as far apart as possible.

If we pick \(x = 0\) and \(y = 3\) (or vice versa), then we get a valid equation:
\(2^0 - 0^2 = 3^2 - 2^3\)
\(1 - 0 = 9 - 8\)

These values of \(x\) and \(y\) are as far apart as possible, so we get \(|x - y| = 3\).


Answer: D
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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New post 28 Jan 2016, 05:17
Bunuel wrote:
Official Solution:

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


We can rearrange the equation, putting all the \(x\)’s on one side and all the \(y\)’s on the other side:
\(2^x - x^2 = y^2 - 2^y\)

Now, list the values of \(2^n\) and \(n^2\) for the first several nonnegative integers \(n\). In fact, go ahead and compute the differences both ways (both \(2^n - n^2\) and \(n^2 - 2^n\)).
\(n\) \(2^n\) \(n^2\) \(2^n - n^2\) \(n^2 - 2^n\) 0 1 0 1 -1 1 2 1 1 -1 2 4 4 0 0 3 8 9 -1 1 4 16 16 0 0 5 32 25 7 -7 6 64 36 28 -28
From this point on, \(2^n\) grows much faster than \(n^2\), so the differences explode. This means that in order to have a valid equation \((2^x - x^2 = y^2 - 2^y)\), we will have to use small values of the integers. We want values in the \(2^n - n^2\) column to match values in the \(n^2 - 2^n\) column, and to maximize the value of \(|x - y|\), we want to pick values from different rows - as far apart as possible.

If we pick \(x = 0\) and \(y = 3\) (or vice versa), then we get a valid equation:
\(2^0 - 0^2 = 3^2 - 2^3\)
\(1 - 0 = 9 - 8\)

These values of \(x\) and \(y\) are as far apart as possible, so we get \(|x - y| = 3\).


Answer: D



Hello there Bunuel.

I'm new here!

Do we have any other method other than this one to solve this question?

Thanks in advance.

P.S. You're doing great work! Kudos to you!

Kudos [?]: 0 [0], given: 1

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D
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Re: S97-03 [#permalink]

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New post 28 Jan 2016, 05:25
alishandhanani17 wrote:
Bunuel wrote:
Official Solution:

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


We can rearrange the equation, putting all the \(x\)’s on one side and all the \(y\)’s on the other side:
\(2^x - x^2 = y^2 - 2^y\)

Now, list the values of \(2^n\) and \(n^2\) for the first several nonnegative integers \(n\). In fact, go ahead and compute the differences both ways (both \(2^n - n^2\) and \(n^2 - 2^n\)).
\(n\) \(2^n\) \(n^2\) \(2^n - n^2\) \(n^2 - 2^n\) 0 1 0 1 -1 1 2 1 1 -1 2 4 4 0 0 3 8 9 -1 1 4 16 16 0 0 5 32 25 7 -7 6 64 36 28 -28
From this point on, \(2^n\) grows much faster than \(n^2\), so the differences explode. This means that in order to have a valid equation \((2^x - x^2 = y^2 - 2^y)\), we will have to use small values of the integers. We want values in the \(2^n - n^2\) column to match values in the \(n^2 - 2^n\) column, and to maximize the value of \(|x - y|\), we want to pick values from different rows - as far apart as possible.

If we pick \(x = 0\) and \(y = 3\) (or vice versa), then we get a valid equation:
\(2^0 - 0^2 = 3^2 - 2^3\)
\(1 - 0 = 9 - 8\)

These values of \(x\) and \(y\) are as far apart as possible, so we get \(|x - y| = 3\).


Answer: D



Hello there Bunuel.

I'm new here!

Do we have any other method other than this one to solve this question?

Thanks in advance.

P.S. You're doing great work! Kudos to you!


Check alternative solutions here: if-2-x-2-y-x-2-y-2-where-x-and-y-are-nonnegative-inte-88016.html

Hope it helps.
_________________

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re S97-03 [#permalink]

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New post 15 Feb 2016, 10:20
I don't agree with the explanation. Take X=3 Y =1 X-Y = 2 , equality holds true.
Take X=4 Y=1 X-Y =3 , equality doesn't hold true..

True answer should be C not D.. Am i missing something ?

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New post 15 Feb 2016, 10:26
vyomesh wrote:
I don't agree with the explanation. Take X=3 Y =1 X-Y = 2 , equality holds true.
Take X=4 Y=1 X-Y =3 , equality doesn't hold true..

True answer should be C not D.. Am i missing something ?


Please re-read the solution. Where does it say that the greatest possible value of |x-y| is for x=4 and y=1? It's for x=0 and y=3 (or vice versa).
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Collection of Questions:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S97-03 [#permalink]

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New post 17 Feb 2016, 00:44
X and Y are nonnegative integers and we need values of X and Y such that mod (X-Y) is maximum.
In order to get maximum difference, one value must be zero, so we can assume Y = 0.
Then,looking at all options only X=3 satisfies the equation.

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Re: S97-03 [#permalink]

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New post 17 Feb 2016, 00:44
X and Y are nonnegative integers and we need values of X and Y such that mod (X-Y) is maximum.
In order to get maximum difference, one value must be zero, so we can assume Y = 0.
Then,looking at all options only X=3 satisfies the equation.

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New post 14 Apr 2017, 01:27
I didn't get this
could anyone explain deeply please ?

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New post 14 Apr 2017, 01:29

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New post 14 Apr 2017, 01:35
got it thank you bunuel !

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New post 14 Apr 2017, 02:03
x=3 and y=1, satisfy the equation.
C can be the correct answer

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New post 14 Apr 2017, 02:48
arsschan wrote:
x=3 and y=1, satisfy the equation.
C can be the correct answer


Please read the discussion above and follow the links provided. The correct answer is D, not C.

x = 3 and y = 0 also satisfy the equation, which gives \(|x-y|=|3-0|=3\). Answer: D.
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Collection of Questions:
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New post 01 May 2017, 00:22
Hi,

So, if x=0,y=4 then equation becomes 2^0 + 2^4 = 0^2 + 4^2 (Each side equals 16) now,the difference between x and y becomes 4. So,should the answer not be 4?

Please help!

Thanks! :)

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New post 01 May 2017, 02:18
stressed wrote:
Hi,

So, if x=0,y=4 then equation becomes 2^0 + 2^4 = 0^2 + 4^2 (Each side equals 16) now,the difference between x and y becomes 4. So,should the answer not be 4?

Please help!

Thanks! :)


\((non-zero \ integer)^0 = 1\).

So, \(2^0 + 2^4 = 1 + 16 = 17\), not 16.

Hope it helps.
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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S97-03   [#permalink] 01 May 2017, 02:18
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