# Olypmicube #6a

#### Scott

##### Member
Ok, the "comming in the next couple of months" thing has been up there for like 6 months.

Does anyone know how close we are to acualy having this great puzzle?

#### dougreed

##### Member
No clue. From the videos, the thing looks hella awesome. Whenever it comes out, I hope to be able to buy from the first batch. I can't wait to get my hands on one of those things.

-Doug

#### Richard

##### Member
What exactly is it?

#### gillesvdp

##### Member
Anybody going to Greece soon ?
If so, please try to contact those OlympicCube guys !

#### Richard

##### Member
Oh man! That thing is amazing. Good luck solving that thing...I think i'll stick to my 3X3X3 a while longer....

#### CraigBouchard

##### Member
I'm gunna go with a no comment on this...Lemme just say I'm hoping in the summer???

#### Scott

##### Member
Originally posted by CraigBouchard@Mar 19 2006, 09:01 PM
I'm gunna go with a no comment on this...Lemme just say I'm hoping in the summer???
Hmmm, you just commented after saying you wouldn't

lol

#### Richard

##### Member
When new cubes like this come up, how do find the new algorthims for it? Do you make a computer program for it or what?

#### pjgat09

##### Member
When I solve any cube above 4 I really do it off the top of my head. The only alg i really use cycles 3 edges around. The rest is like a 4x4 solve: centers, edges, 3x3. That was my method for the 40x40.

#### dougreed

##### Member
Originally posted by Richard@Mar 20 2006, 09:04 PM
When new cubes like this come up, how do find the new algorthims for it?
You can use commutators to come up with algorithms for any number of puzzles, not just cubes. You can learn more about their 3x3x3 applications at Joel's Speedcubing Page, and then scale them up to work with nxnxn cubes using the same techniques.

HTH,
Doug

#### Ravi

##### Member
I wonder who'll be the first to BLD one of those...

-Ravi

#### Alexander

##### Member
my bet is Stefan would be the first one

#### pjgat09

##### Member
That would be scary to BLD a +6 order cube!! I can't even do 4!

#### Scott

##### Member
Haha, i cant even do 3!

#### pjk

Staff member
Do you guys know the potential costs of these?

#### CraigBouchard

##### Member
Yes I did comment...but w/e...No clue on potential cost, I'm still hoping around summer...Maybe they'll be sold for the first time at US Nationals...hehehe and not sold anywhere else for a month That would be intense...

Craig

#### Scott

##### Member
Well, if it follows the pattern, it'll cost around 42 USD.

x = cube layers.
p = price.

p = 2x + x

Aproximate Prices
2x2x2 cube, 2(2) + 2 = $6 USD 3x3x3 cube, 3(3) + 3 =$12 USD
4x4x4 cube, 4(4) + 4 = $20 USD 5x5x5 cube, 5(5) + 5 =$30 USD
6x6x6 cube, 6(6) + 6 = $42 USD 7x7x7 cube, 7(7) + 7 =$49 USD
8x8x8 cube, 8(8) + 8 = $74 USD 9x9x9 cube, 9(9) + 9 =$90 USD
10x10x10 cube, 10(10) + 10 = 110 USD Yes i made that formula, no it's not exact, and no... you dont bash it. #### BillT ##### Member Originally posted by Scott+Mar 21 2006, 10:02 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Scott @ Mar 21 2006, 10:02 PM)</td></tr><tr><td id='QUOTE'>you dont bash it. [/b] I'm afraid I must. <!--QuoteBegin-Scott @Mar 21 2006, 10:02 PM x = cube layers. p = price. p = 2x + x 3x3x3 cube, 3(3) + 3 =12 USD
4x4x4 cube, 4(4) + 4 = $20 USD 5x5x5 cube, 5(5) + 5 =$30 USD
[/quote]
If we go with your original formula, these would be the prices:

3x3x3 cube, 2(3) + 3 = $9 USD 4x4x4 cube, 2(4) + 4 =$12 USD
5x5x5 cube, 2(5) + 5 = \$15 USD :blink:

It should be: p = x^2 + x

And yes, I realise I'm being rather picky. (I was bored )

-Bill