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16 Sep 2014, 01:53
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Question Stats:
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56%
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If \(x\) and \(y\) are positive integers and \(( \sqrt{x} )^y = y\), then \(x - y\) could equal which of the following?
A. -2
B. -1
C. 1
D. 2
E. 4
A. -2
B. -1
C. 1
D. 2
E. 4
16 Sep 2014, 01:53
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Official Solution:
If \(x\) and \(y\) are positive integers and \(( \sqrt{x} )^y = y\), then \(x - y\) could equal which of the following?
A. -2
B. -1
C. 1
D. 2
E. 4
Since both \(x\) and \(y\) are restricted to positive integers, we should start with low positive integers.
If \(x = 1\) and \(y = 1\), then we have \(( \sqrt{1} )^1 = 1\), which is true. Thus, \(x - y\) could equal \(1 - 1 = 0\). However, 0 is not among our answer choices, so we must keep looking.
If \(x = 1\), then \(\sqrt{x}\) raised to any power is 1. Thus, no other values of \(y\) (besides 1) could equal that result, and we should look at other possible values of \(x\) and \(y\).
If \(x = 2\), then we have two possible values of \(y\):
\(y = 2\): \(( \sqrt{2} )^2 = 2\)
\(y = 4\): \(( \sqrt{2} )^4 = 4\)
In the first case, we have \(x - y = 2 - 2 = 0\) again. However, in the second case, we have \(x - y = 2 - 4 = -2\).
As it turns out, there are no other possible combinations of \(x\) and \(y\), but we don't need to prove that; we can stop here.
Answer: A
If \(x\) and \(y\) are positive integers and \(( \sqrt{x} )^y = y\), then \(x - y\) could equal which of the following?
A. -2
B. -1
C. 1
D. 2
E. 4
Since both \(x\) and \(y\) are restricted to positive integers, we should start with low positive integers.
If \(x = 1\) and \(y = 1\), then we have \(( \sqrt{1} )^1 = 1\), which is true. Thus, \(x - y\) could equal \(1 - 1 = 0\). However, 0 is not among our answer choices, so we must keep looking.
If \(x = 1\), then \(\sqrt{x}\) raised to any power is 1. Thus, no other values of \(y\) (besides 1) could equal that result, and we should look at other possible values of \(x\) and \(y\).
If \(x = 2\), then we have two possible values of \(y\):
\(y = 2\): \(( \sqrt{2} )^2 = 2\)
\(y = 4\): \(( \sqrt{2} )^4 = 4\)
In the first case, we have \(x - y = 2 - 2 = 0\) again. However, in the second case, we have \(x - y = 2 - 4 = -2\).
As it turns out, there are no other possible combinations of \(x\) and \(y\), but we don't need to prove that; we can stop here.
Answer: A
07 Jul 2015, 11:46
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