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--[[User:Rhollowe|Rhollowe]] 00:48, 22 January 2009 (UTC) | --[[User:Rhollowe|Rhollowe]] 00:48, 22 January 2009 (UTC) | ||
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| + | If you increase the problem to numbers less than 5000, you include one more integer that is a square and cube, namely 4096. | ||
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| + | <math>sqrt[5]{5000}=5.49.... whereas sqrt[6]{5000}=4.13...</math> | ||
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| + | 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 | ||
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| + | <math> 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.</math> | ||
| + | |||
| + | --[[User:Jberlako|Jberlako]] 11:05, 22 January 2009 (UTC) | ||
Revision as of 06:05, 22 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)
If you increase the problem to numbers less than 5000, you include one more integer that is a square and cube, namely 4096.
$ sqrt[5]{5000}=5.49.... whereas sqrt[6]{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)
