Posted by Muffy on Saturday, October 31, 2009 at 1:57pm.
My question is I^793
I think the answer is just I because it is the only number that can go into it. Is that correct?
Pre Calc - bobpursley, Saturday, October 31, 2009 at 2:06pm
are you looking for factors of x^793?
or i^793, where i is the sqrt(-1)?
factors of x^793 are very numerous. x^793 is not a prime number.
i^793=i^792 * 1
so the conclusion is that i^(4n-2) is -1
otherwise, i to even is 1
WEll, is 792=4n-2
4n=794, or n is not an integer, so
finally, i^792*i= i and it has only one factor, i.
Pre Calc - Muffy, Saturday, October 31, 2009 at 2:17pm
So I was correct the answer is i right?
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