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If x and y are non-zero integers, and \((-x)^x = y^y\), then what is the value of x?

(1) x + y = 0

(2) x is odd

Target question: What is the value of x? Given: x and y are non-zero integers, and \((-x)^x = y^y\) When we see this given information, we should be thinking of a few different ways for the equation to hold true.

For example, if x = 1, then (-x)^x = (-1)^1 = -1. What value of y is necessary for y^y to equal -1? Well, if y = -1, then y^y = (-1)^(-1) = 1/(-1) = -1

So, one possible solution to the given equation is

x = 1 and y = -1Using very similar logic, we can see that

x = -1 and y = 1 is another possible solution to the given equation

Now let's check the statements....

Statement 1: x + y = 0 As we can see from our earlier work, there are at least two possible solutions that satisfy statement 1:

Case a:

x = 1 and y = -1. In this case, the answer to the target question is

x = 1Case b:

x = -1 and y = 1. In this case, the answer to the target question is

x = -1Since we cannot answer the

target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x is odd We should recognize that we can RE-USE the same values we used to show that statement 1 is not sufficient (since 1 and -1 are both ODD)

That is....

Case a:

x = 1 and y = -1. In this case, the answer to the target question is

x = 1Case b:

x = -1 and y = 1. In this case, the answer to the target question is

x = -1Since we cannot answer the

target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined IMPORTANT: Notice that I was able to use the

same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.

In other words:

Case a:

x = 1 and y = -1. In this case, the answer to the target question is

x = 1Case b:

x = -1 and y = 1. In this case, the answer to the target question is

x = -1Since we cannot answer the

target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,

Brent

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