Difference between revisions of "12" - Rhea

Revision as of 19:48, 21 January 2009

So, you will need to go a little further with explanations of why but the way to go about this one is:

$ \lfloor \sqrt{1000} \rfloor + \lfloor \sqrt[3]{1000} \rfloor - \lfloor \sqrt[6]{1000} \rfloor $

To see why this is correct, draw a Venn Diagram, and take out the common terms.

I didn't think of the 6th root approach, so i just when through and counted the cubes (since there would only be 10) and I found that only 1 and 64 are both squares and cubes.

You missed one. 729 is both a square and a cube. I believe the correct equation would be;

$ \lfloor \sqrt{1000} \rfloor + \lfloor \sqrt[3]{1000} \rfloor - \lfloor \sqrt[5]{1000} \rfloor $

This yields the same answer, but I believe the intersection is the set $ x^2*x^3=x^5 $ meaning you would need the 5th root, not the sixth.

--Jberlako 21:27, 21 January 2009 (UTC)

I have been playing around with this one, and the original equation is right, but I can't figure out why. Can you explain why it is the 6th root rather than the 5th?

--Jberlako 22:34, 21 January 2009 (UTC)

I'm so lost. Is this question "How many bit strings are there of length six or less?"? It's in section 5.1 of my book for some reason!

I guess I don't get why you say the original equation is correct, I thought the second one looked right.

--Rhollowe 00:48, 22 January 2009 (UTC)