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by Jay@ManhattanReview » Sat Apr 15, 2017 12:02 am
Mo2men 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

Source: Manhattan
Hi Mo2men,

Since we want the greatest possible value of |x - y| and the greatest possible value in options is 4, let's assume that |x - y| = 4.

We'll try to keep y to the minimum, thus y = 0.

Say x = 4 and y = 0

=> 2^x + 2^y = x^2 + y^2 => 2^4 + 2^0 = 4^2 + 0^2 => 16 + 1 > 16 + 0. This is not the solution.

Say x = 3 and y = 0

=> 2^x + 2^y = x^2 + y^2 => 2^3 + 2^0 = 3^2 + 0^2 => 8 + 1 = 9 + 0. This is the solution.

The greatest possible value |x - y| = 3.

The correct answer: D

Hope this helps!

-Jay
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by GMATGuruNY » Sat Apr 15, 2017 1:37 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.
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