what do you mean when you say "assemble"
As in taking 9 3x3x3 cubes to make a single 9x9x9 cube?
Dear forum members,
Firstly, I am new to this forum and a complete non-math person, so please forgive me if I posted in the wrong area.
I am an artist and am working on a sculpture that requires some math that is way above my paygrade (I know my 7 times tables, up to 7x7!), so I am hoping you can help.
I have read that the number of atoms contained in the visible universe is estimated at 10^{82 }I would like to build a sculpture that expresses this number visually, so I thought of assembling a number (?) of Rubik's Cubes which would contain approximately 10^{82} possible permutations.
According to Wikipedia, a 3x3 Rubiks cube has approximately forty-three quintillion permutations. (I don't even know how to write that as scientific notation )
What I would like to figure out is:
How many Rubiks Cubes do I need to assemble to display (more than) 10^{82} possible permutations.
I would very much appreciate your help. Have a nice day!
You will need a LOT of Rubik's cubes if you just sum the permutation possibilities. If you count ALL the possible permutations on all your Rubik's cubes as if it was one object, you can multiply the permutations, in which case 5 will suffice (4 is not quite enough).
In case you were wondering, 1 quintillion is 10^{18}.
Hello and thanks for replying.
Romsek: I wouldn't worry too much about the assembling part. But wouldn't making a 9x9x9 cube take 27 3x3x3 cubes?
Deveno: Thank you - I've done some research and I reckon it would only take maybe 3 cubes...
My maths is hopeless of course, so I am probably wrong, but wouldn't 43^{18 }x 43^{18 }=1849^{36 }?
If so, I think a third 43^{18} would get us over the line for 10^{78}
Am I wrong?
I used wolfram|alpha for some comparison
http://www.wolframalpha.com/input/?i=10%5E82 = 10 sexvigintillion
You'll have to do it the way Deveno advised cause seems like you will need a very big amount of rubiks cubes otherwise--so much that I don't want to think about it.
http://www.wolframalpha.com/input/?i...%5E%2818%29%29
43 quintillion is not $43^{18}$, it's $4.3 \times 10^{19}$
The base being exponentiated is 10, not 43 (which results in a smaller number), the result of said exponentiation is then multiplied by 4.3.
$(4.3 \times 10^{19})^4 = (4.3)^4(10^{19})^4 \approx 3.4 \times 10^{78}$