<PRE><B>Question 190074</B>
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You are 100% on the right track.  *[tex \LARGE  sqrt{2n}] must be divisible by 3 because that is one of the conditions of the problem.  But it also must be divisible by 2 because for *[tex \LARGE  sqrt{2n}] to be a rational number *[tex \LARGE  2n] must have an even number of factors of 2.  *[tex \LARGE  2n] has one factor of 2 that you can see, so there must be another one buried in the n.  So since *[tex \LARGE  sqrt{2n}] must have a factor of 2 and a factor of 3, *[tex \LARGE  2n] must have at least &lt;i&gt;&lt;b&gt;2&lt;/b&gt;&lt;/i&gt; factors of 2 and at least &lt;i&gt;&lt;b&gt;2&lt;/b&gt;&lt;/i&gt; factors of 3.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2n\ =\ 2\ \times\ 2\ \times\ 3\ \times\ 3\ = \ 36 \ \ \Rightarrow\ \ n = 18] 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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