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Algebra

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by [email protected] » Sat Oct 26, 2013 5:04 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


While one solution is testing the answer values, is their any other approach too?
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by ani781 » Sat Oct 26, 2013 5:31 am
Lest not taking up a lot of time, I would plug in numbers as that would seem fastest ... sadly I don't have a more mathematical approach within the 2 min time frame.

Remember 0 is a +ve Integer.
So I would start with D. Diff 3. Let's say x=3, y =0. Then 2^x+2^y=8+1=9 ; x^2+y^2=3^2+0=9. Others don't satisfy.

IMO D.
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by GMATGuruNY » Sat Oct 26, 2013 6:11 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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by theCodeToGMAT » Sat Oct 26, 2013 7:09 am
ani781 wrote:Lest not taking up a lot of time, I would plug in numbers as that would seem fastest ... sadly I don't have a more mathematical approach within the 2 min time frame.

Remember 0 is a +ve Integer.
So I would start with D. Diff 3. Let's say x=3, y =0. Then 2^x+2^y=8+1=9 ; x^2+y^2=3^2+0=9. Others don't satisfy.

IMO D.
Ani your solution is nice. But, I would like to correct you on one statement in your solution.

"0" is NOT positive.. here, we selected "0" because question says that numbers are "non-negative"
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by Matt@VeritasPrep » Sun Oct 27, 2013 10:48 pm
This is NOT an easy one! Here's an attempt at a definitive explanation, but it's time-consuming and technical (the number plugging strategies recommended above would at least give you a 50/50 shot between D and E, which is probably better on test day).

Rewrite the equation as 2ˣ - x² = y² - 2ʸ

There are three possibilities for the right hand side of the equation (the easier one to work with):

y² - 2ʸ = 0
y² - 2ʸ < 0
y² - 2ʸ > 0

Let's deal with each one.

Suppose y² - 2ʸ = 0. In this case, y = 2 or y = 4. 2ˣ - x² = 0 also implies x = 2 or x = 4, so in this case our maximum value of |x - y| is |4 - 2| or |2 - 4|, either of which = 2.

Suppose y² - 2ʸ < 0. Then we also have 2ˣ - x² < 0, which only happens if x = 3 and which gives 2ˣ - x² = -1. That means y² - 2ʸ = -1, which happens only if y = 0 or y = 1. So we could have |x - y| = |3 - 0| = 3.

Suppose y² - 2ʸ > 0. This happens if y = 3, and gives y² - 2ʸ = 1, which implies that 2ˣ - x² = 1. This happens only if x = 0 or x = 1, so again we could have |x - y| = |0 - 3| = 3.

We've exhausted all our options, so the maximum is 3.
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by [email protected] » Sun Oct 27, 2013 11:23 pm
Hi ani781,

You shouldn't be upset that you didn't have a more mathematical approach to this question. GMAT questions can usually be solved using a number of different approaches, so on any given question, you're looking for an approach that:

1) Gets you the correct answer
2) Is easy to use
3) Is efficient, so you don't waste time

You'd be amazed at how often the "math approach" is neither easiest nor fastest. Stay flexible and organized and you'll see how easy most GMAT questions are.

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