Line 19: Line 19:
  
 
--[[User:Jberlako|Jberlako]] 21:27, 21 January 2009 (UTC)
 
--[[User:Jberlako|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?
 +
 +
--[[User:Jberlako|Jberlako]] 22:34, 21 January 2009 (UTC)

Revision as of 17:34, 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)

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett