Bunuel wrote:
OFFICIAL SOLUTION:
If x is a positive integer, what is the remainder when x is divided by 2?
(1) \((-1)^{(x^2)} = -1\).
This implies that x^2 is odd (if it were even, then (-1)^even = 1). x^2 = odd, on the other hand means that x is odd (since given that x is an integer). Any odd number divided by 2 gives the remainder of 1. Sufficient.
(2) \(n^x = n^{(2x - 1)}\).
Be careful not to fall into the trap. Remember we can automatically equate the exponents of equal bases when that base does not equal 0, 1 or -1:
\(1^x = 1^y\), for any values of x and y (they are not necessarily equal);
\((-1)^x = (-1)^y\), for any even values of x and y (they are not necessarily equal);
\(0^x = 0^y\), for any non-zero x and y (they are not necessarily equal).
Thus, for \(n^x = n^{(2x - 1)}\), we could equate the exponents if we knew that n is not 0, 1 or -1. In this case we'd have: x = 2x - 1, which would give x = 1. But if n is 0, 1 or -1, then we cannot equate the exponents. For example, if n = 1, then x can be even as well as odd. Not sufficient.
Answer: A.
Thank you "BUNUEL"...for nice expalination...
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