<PRE><B>Question 722758</B>
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313 is odd, so not divisible by 2.


The digits of 313 do not sum to a number divisible by 3, so not divisible by 3


Does not end in 0 or 5, so not divisible by 5


Not divisible by 7, 11, 13, or 17 by trial and *[tex \LARGE 19^2\ =\ 361\ &gt;\ 313].  Therefore 313 is prime.


Hence, the best you can do is say *[tex \LARGE c\ =\ \sqrt{313}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
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My calculator said it, I believe it, that settles it
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