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15 Nov 2021, 01:43
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If \(x\) is a positive integer, what is the remainder when \(x\) is divided by 2?
(1) \((-1)^{(x^2)} = -1\)
(2) \(n^x = n^{(2x - 1)}\)
(1) \((-1)^{(x^2)} = -1\)
(2) \(n^x = n^{(2x - 1)}\)
15 Nov 2021, 01:43
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Bookmarks
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
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
