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
greatest possible value of |x-y|
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|x-y| = the DISTANCE between x and y.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
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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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
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
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2^x+2^y=x^2+y^2j_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
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)
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