Bunuel
If x > 0, is 1/x a prime number ?
(1) \(x^x = \frac{1}{\sqrt{2}}\)
(2) The highest common factor of integers 1/x^2 and 1/x^4 is the square of a prime number
M36-22
Official Solution:If \(x > 0\), is \(\frac{1}{x}\) a prime number ? (1) \(x^x = \frac{1}{\sqrt{2}}\)
Let's see how we can express \(\frac{1}{\sqrt{2}}\) so that the base and the exponent are the same (so that to mimic \(x^x\)).
\(\frac{1}{\sqrt{2}}=\frac{1}{2^{\frac{1}{2}}}=(\frac{1}{2})^{\frac{1}{2}}\);
So, \(x\) can be \(\frac{1}{2}\) We can also express \(\frac{1}{\sqrt{2}}\) as \(\frac{1}{\sqrt{2}}=(\frac{1}{2})^{\frac{1}{2}}=(\frac{1}{2})^{2*\frac{1}{4}}=((\frac{1}{2})^2)^{\frac{1}{4}}=(\frac{1}{4})^{\frac{1}{4}}\)
So, \(x\) can also be \(\frac{1}{4}\) Not sufficient.
(2) The highest common factor of integers \(\frac{1}{x^2}\) and \(\frac{1}{x^4}\) is the square of a prime number.
Say \(\frac{1}{x^2}=k\). Then the highest common factor of \(k=\frac{1}{x^2}\) and \(k^2=\frac{1}{x^4}\) would obviously be \(k\). So, we are told that \(k=\frac{1}{x^2}\) is the square of a prime number:
\(\frac{1}{x^2}=prime^2\);
\(\frac{1}{x}=prime\).
Sufficient.
Answer: D