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# S97-03

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Math Expert
Joined: 02 Sep 2009
Posts: 43302

Kudos [?]: 139235 [0], given: 12781

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16 Sep 2014, 00:51
Expert's post
4
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

29% (01:28) correct 71% (01:26) wrong based on 93 sessions

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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

_________________

Kudos [?]: 139235 [0], given: 12781

Math Expert
Joined: 02 Sep 2009
Posts: 43302

Kudos [?]: 139235 [0], given: 12781

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16 Sep 2014, 00:51
Expert's post
1
This post was
BOOKMARKED
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 $$|x - y| = 3$$.

_________________

Kudos [?]: 139235 [0], given: 12781

Intern
Joined: 19 Jan 2016
Posts: 3

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

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28 Jan 2016, 04:17
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 $$|x - y| = 3$$.

Hello there Bunuel.

I'm new here!

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

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

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

Math Expert
Joined: 02 Sep 2009
Posts: 43302

Kudos [?]: 139235 [0], given: 12781

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28 Jan 2016, 04:25
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 $$|x - y| = 3$$.

Hello there Bunuel.

I'm new here!

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

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.
_________________

Kudos [?]: 139235 [0], given: 12781

Intern
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15 Feb 2016, 09:20
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 ?

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

Math Expert
Joined: 02 Sep 2009
Posts: 43302

Kudos [?]: 139235 [0], given: 12781

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15 Feb 2016, 09:26
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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Kudos [?]: 139235 [0], given: 12781

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Joined: 16 Oct 2015
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16 Feb 2016, 23:44
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.

Kudos [?]: 27 [0], given: 12

Manager
Joined: 16 Oct 2015
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Kudos [?]: 27 [0], given: 12

GMAT 1: 520 Q44 V17
GMAT 2: 530 Q44 V20
GMAT 3: 710 Q48 V40

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16 Feb 2016, 23:44
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.

Kudos [?]: 27 [0], given: 12

Intern
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14 Apr 2017, 00:27
I didn't get this
could anyone explain deeply please ?

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Math Expert
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Posts: 43302

Kudos [?]: 139235 [0], given: 12781

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14 Apr 2017, 00:29
mkumar26 wrote:
I didn't get this
could anyone explain deeply please ?

Check alternative solutions here: http://gmatclub.com/forum/if-2-x-2-y-x- ... 88016.html
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Kudos [?]: 139235 [0], given: 12781

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14 Apr 2017, 00:35
got it thank you bunuel !

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14 Apr 2017, 01:03
x=3 and y=1, satisfy the equation.
C can be the correct answer

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Math Expert
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14 Apr 2017, 01:48
x=3 and y=1, satisfy the equation.
C can be the correct answer

x = 3 and y = 0 also satisfy the equation, which gives $$|x-y|=|3-0|=3$$. Answer: D.
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Kudos [?]: 139235 [0], given: 12781

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30 Apr 2017, 23:22
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?

Thanks!

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Math Expert
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01 May 2017, 01:18
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?

Thanks!

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

So, $$2^0 + 2^4 = 1 + 16 = 17$$, not 16.

Hope it helps.
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Kudos [?]: 139235 [0], given: 12781

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

Moderators: chetan2u, Bunuel

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