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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 | 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 | ||
| − | <math>1=(1^2)^3 | + | <math>1=(1^2)^3</math> |
| − | <math>64=4^3=8^2</math> | + | <math>64=4^3=8^2</math> where <math>4=2^2</math> and <math>8=2^3</math> meaning <math>64=(2^2)^3</math> |
| − | <math>4=2^2</math> <math>8=2^3</math> meaning <math>64=(2^2)^3</math> | + | |
| − | <math>729=27^2=9^3</math> where <math>27=3^3</math> <math>9=3^2</math> meaning <math>729=(3^2)^3</math> | + | <math>729=27^2=9^3</math> where <math>27=3^3</math> and <math>9=3^2</math> meaning <math>729=(3^2)^3</math> |
<math>4096=16^3=64^2</math> where <math>16=4^2</math> and <math>64=4^3</math> meaning <math>4096=(4^2)^3</math> | <math>4096=16^3=64^2</math> where <math>16=4^2</math> and <math>64=4^3</math> meaning <math>4096=(4^2)^3</math> | ||
| − | As we all know, <math>(x^a)^b=x^ | + | As we all know, <math>(x^a)^b=x^{b*a}</math> which in our case gives us <math>x^6</math>. |
--[[User:Jberlako|Jberlako]] 11:05, 22 January 2009 (UTC) | --[[User:Jberlako|Jberlako]] 11:05, 22 January 2009 (UTC) | ||
Revision as of 06:14, 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 $ $ 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)
