BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Algebra -1

Expert replies
by guerrero » Fri Apr 05, 2013 12:29 pm
If 2^x + 2^y = x^2 + y^2, where x and y are non negative integers, what is the greatest possible value of |x - y|?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


How to Approach this type of questions?

thanks!


OA3
Join the discussion
Source: — Problem Solving |

by rintoo22 » Fri Apr 05, 2013 12:48 pm
Hi,

I did this type of question few days ago. If the question mentions greatest possible value, then we can make one of the 2 variables as 0.

eg. x=0;y=3

lets put the values in 2^x + 2^y = x^2 + y^2
2^0+2^3 = 0^2 + 3^2
9=9

So |x-y| = 3.
Join the discussion

by srcc25anu » Fri Apr 05, 2013 12:55 pm
I would start with some number testing.

if i take 1, its 2 + 2 = 1 + 1 (doesnt work)
if i take 2, its 4 + 4 = 4 + 4 (WORKS) in this case |x-y| = 0

I would also test any combination where x^y = y^x. One I know by solving other similar questions is that 2^4 = 4^2 = 16
so if x = 4 and y = 2, then 2^4 + 2^2 = 4^2 + 2^2 = TRUE

hence |x-y| = 2
hence C
Join the discussion

by rintoo22 » Fri Apr 05, 2013 1:06 pm
Hi srcc25anu,

your method is rt "I would also test any combination". However in this question it asks greatest possible value. So in my opinion, you answer c)2 is correct in an essence but it not the greatest.
Greatest would be 3.
Join the discussion

by GMATGuruNY » Fri Apr 05, 2013 1:08 pm
guerrero 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

OA3
|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
Join the discussion

by srcc25anu » Fri Apr 05, 2013 1:10 pm
rintoo22 wrote:Hi srcc25anu,

your method is rt "I would also test any combination". However in this question it asks greatest possible value. So in my opinion, you answer c)2 is correct in an essence but it not the greatest.
Greatest would be 3.
You are correct. thanks for pointing that out to me.
i missed it.
Join the discussion