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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. | 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. | ||
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| + | You missed one. 729 is both a square and a cube. I believe the correct equation would be; | ||
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| + | <math>\lfloor \sqrt{1000} \rfloor + \lfloor \sqrt[3]{1000} \rfloor - \lfloor \sqrt[5]{1000} \rfloor</math> | ||
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| + | This yields the same answer, but I believe the union is the set <math>x^2*x^3=x^5</math> meaning you would need the 5th root, not the sixth. | ||
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| + | --[[User:Jberlako|Jberlako]] 21:27, 21 January 2009 (UTC) | ||
Revision as of 17:27, 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 union 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)
