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

$\lfloor \sqrt{1000} \rfloor + \lfloor \sqrt{1000} \rfloor - \lfloor \sqrt{1000} \rfloor$

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;

$\lfloor \sqrt{1000} \rfloor + \lfloor \sqrt{1000} \rfloor - \lfloor \sqrt{1000} \rfloor$

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 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)

If you increase the problem to numbers less than 5000, you include one more integer that is a square and cube, namely 4096.

$\sqrt{5000}=5.49$

$\sqrt{5000}=4.13$

This shows that the sixth root is correct. I think I understand what is going on here now. If you have a number that has both a square and cube root it must be the result of the cube of a square or the square of a cube. Look at the 4 numbers we have in the intersection of the sets for this problem

$1=(1^2)^3$

$64=4^3=8^2$ where $4=2^2$ and $8=2^3$ meaning $64=(2^2)^3$

$729=27^2=9^3$ where $27=3^3$ and $9=3^2$ meaning $729=(3^2)^3$

$4096=16^3=64^2$ where $16=4^2$ and $64=4^3$ meaning $4096=(4^2)^3$

As we all know, $(x^a)^b=x^{b*a}$ which in our case gives us $x^6$.

--Jberlako 11:05, 22 January 2009 (UTC)

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale. Dr. Paul Garrett