GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jul 2018, 09:44

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# S97-03

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

16 Sep 2014, 01:51
00:00

Difficulty:

95% (hard)

Question Stats:

28% (01:28) correct 72% (01:26) wrong based on 98 sessions

### HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

16 Sep 2014, 01:51
1
1
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 $$|x - y| = 3$$.

_________________
Intern
Joined: 19 Jan 2016
Posts: 3

### Show Tags

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 $$|x - y| = 3$$.

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!
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

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 $$|x - y| = 3$$.

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.
_________________
Intern
Joined: 10 May 2015
Posts: 1

### Show Tags

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

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).
_________________
Manager
Joined: 16 Oct 2015
Posts: 73
Location: India
Concentration: Technology, General Management
Schools: ISB '19, IIMA , IIMB, IIMC
GMAT 1: 520 Q44 V17
GMAT 2: 530 Q44 V20
GMAT 3: 710 Q48 V40
GPA: 3.45
WE: Research (Energy and Utilities)

### Show Tags

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.
Manager
Joined: 16 Oct 2015
Posts: 73
Location: India
Concentration: Technology, General Management
Schools: ISB '19, IIMA , IIMB, IIMC
GMAT 1: 520 Q44 V17
GMAT 2: 530 Q44 V20
GMAT 3: 710 Q48 V40
GPA: 3.45
WE: Research (Energy and Utilities)

### Show Tags

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.
Intern
Joined: 24 Feb 2017
Posts: 37

### Show Tags

14 Apr 2017, 01:27
I didn't get this
could anyone explain deeply please ?
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

14 Apr 2017, 01:29
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
_________________
Intern
Joined: 24 Feb 2017
Posts: 37

### Show Tags

14 Apr 2017, 01:35
got it thank you bunuel !
Intern
Joined: 28 Feb 2017
Posts: 2

### Show Tags

14 Apr 2017, 02:03
x=3 and y=1, satisfy the equation.
C can be the correct answer
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

14 Apr 2017, 02:48
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.
_________________
Intern
Joined: 15 Apr 2017
Posts: 4
Location: India
GMAT 1: 690 Q49 V33
GPA: 3.3

### Show Tags

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?

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

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?

Thanks!

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

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

Hope it helps.
_________________
S97-03 &nbs [#permalink] 01 May 2017, 02:18
Display posts from previous: Sort by

# S97-03

Moderators: chetan2u, Bunuel

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.