greatest possible value of |x-y|

This topic has expert replies
Legendary Member
Posts: 510
Joined: Thu Aug 07, 2014 2:24 am
Thanked: 3 times
Followed by:5 members

greatest possible value of |x-y|

by j_shreyans » Sat Sep 13, 2014 7:57 am
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

OA D

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sat Sep 13, 2014 10:51 am
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
|x-y| = the DISTANCE between x and y.

To MAXIMIZE this distance, try to MAXIMIZE x and MINIMIZE y.
Note the word in red, which implies that y can be equal to 0.
If y=0, we get:
2^x + 2� = x² - 0²
2^x + 1 = x²
x² - 2^x = 1.

The answer choices imply that the distance between x and y cannot be greater than 4.
If y=0, then x must be equal to one of the following values: 0, 1, 2, 3, 4.
Only x=3 satisfies the equation x² - 2^x = 1:
3² - 2³ = 1.

Thus, x=3 and y=0 satisfy the equation 2^x + 2^y = x^2 + y^2.
In this case, |x-y| = |3-0| = 3.

Given that 2^x + 2^y = x^2 + y^2, if the value of y INCREASES -- if y is equal to an integer GREATER THAN 0 -- then the value of x will have to DECREASE.
The result is that x and y will be brought closer together, DECREASING the distance between them.
Thus, the maximum possible distance between x and y is 3.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Sat Sep 13, 2014 10:54 am
Hi j_shreyans,

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 NON-NEGATIVE 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 |X-Y|.....

From the answers, we know that X and Y have to be relatively "close" on the number line. Next, the phrase "non-negative" 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: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image

User avatar
Legendary Member
Posts: 1100
Joined: Sat May 10, 2014 11:34 pm
Location: New Delhi, India
Thanked: 205 times
Followed by:24 members

by GMATinsight » Sun Sep 14, 2014 9:43 pm
j_shreyans 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

OA D
2^x+2^y=x^2+y^2

Since x and y are non-negative integers therefore I would like to use the values
The range worth trying is always set of integers ranging from -3 to +3
Here 0 and 3 satisfy the equation, therefore
Answer: Option D

However x = 2 and y = 4 (or vice-versa) is an automatic solution set to the given equation if x and y were strictly positive (because for this solution set x^y = y^x)
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour