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        riddles >> hard >> HARD: Calendar Cubes I
        
(Message started by: ddavis on Jul 25^th, 2002, 7:47pm)

Title: HARD: Calendar Cubes I
Post by ddavis on Jul 25^th, 2002, 7:47pm

I really liked this riddle. A good hint for those of you who haven't yet solved it is to recognize that if we write the number 6 on one side of a cube, that character can also be used as 9...just turn it upside down.

Once you figure this out, the rest of the riddle is pretty trivial.

Don

Title: Re: HARD: Calendar Cubes
Post by kevin on Jul 26^th, 2002, 7:18am

Well.... almost. So, 6=9, but with only 6 faces on a cube and 10 numbers required (including 0) you still need to make an additional 3 numbers.... and you can't make 2 look like 5.

Although it would be a bit of an impractical calender,
I think you need to use transparent cubes and calculator style numbers. This way, for instance, you can have 7 opposite 4 to make 6 or 9, and 0 opposite 3 to make 8. But even using just 0,1,2,3,4,7 you still can't make 5. In which case, I think you need to add another bit of complexity by having different coloured numbers (white/black) opposite each other so that some bits cancel each other out - at which point I decide to stop thinking about it.

Title: Re: HARD: Calendar Cubes
Post by James Jones on Jul 26^th, 2002, 7:34am

I'm not sure if my answer is acceptable, but the solution is trivial if we don't restrict ourselves to base 10. Using base 6, I write 0, 1, 2, 3, 4, 5 on each cube and can represent everything from 0 to 35 (base 10) easily.

- James

Title: Re: HARD: Calendar Cubes - spoiler
Post by ddavis on Jul 26^th, 2002, 7:40am

Nope, you can use base 10. Think about what numbers you have to have on both cubes...1 (to represent the 11th) and 2 (to represent the 22nd). From there you can work out the rest. For prectical resons you will also need the 0 on both cubes. Remember, no months go above 31 days.

So the answer is:
Cube 1 - 0, 1, 2, 3, 4, 5
Cube 2 - 0, 1, 2, 6, 7, 8

The 6 does double duty as a 9 also.

Don

Title: Re: HARD: Calendar Cubes
Post by Neil Sedaka on Jul 26^th, 2002, 7:45am

This works (using the 9 as a 6 also).

0 1 2 3 4 5
0 1 2 7 8 9

Any difficult riddles?

Title: Re: HARD: Calendar Cubes
Post by James Jones on Jul 26^th, 2002, 8:33am

Using 6 for 9 is clever, but hey, using base 6 is clever, too. ;)

If you want a "difficult riddle", Neil Sedaka, try this one:

Prove that every even number greater than 2 is the sum of two primes.

Title: Re: HARD: Calendar Cubes
Post by Neil Sedaka on Jul 26^th, 2002, 8:52am

You want me to prove Goldbach's conjecture ?

Don't you think I would have taken the $1million from Faber and Faber instead of mere kudos from you?

I'm far too busy working on a Grand Unifying theory of World Peace and an end to Hunger and Disease - in Base 6 !!!

Title: Re: HARD: Calendar Cubes
Post by jmlyle on Jul 30^th, 2002, 1:32pm

Before I read the answer, I came up with a "sly" way to do this that could actually count from 00 to 5B in base 13. It involves putting the cubes "next to each other", but not side by side from the point of view of the viewer. That would be 76, I think, without mulling over it.

Having this many consecutive numbers available easily fits the 00 to 31 digit requirment (in base 10) requirment.

Using the real answer, you could probably vastly increase the high end, but I don't want to bother.

On the off chance that any one wants to figure out how to do this, I won't post the solution just yet. Post if you want the answer to how I did it....

--jmlyle

Title: Re: HARD: Calendar Cubes
Post by william wu on Jul 30^th, 2002, 7:18pm

on 07/26/02 at 08:52:15, Neil Sedaka wrote:

pwhahaha ;D

Title: Re: HARD: Calendar Cubes
Post by william wu on Aug 4^th, 2002, 3:14pm

Calendar Cubes II is now available.
http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml#calendarCubes2

Title: Re: HARD: Calendar Cubes I
Post by thetorpedodog on Jun 27^th, 2004, 10:45am

Additional solution:

Cube 1: 0, 1, 2, 4, 5, 7
Cube 2: 0, 1, 2, 3, 6/9, 8

Added irrelevant note: whoever can should really replace the <table> tags in the "glow" output with <span>s.

Title: Re: HARD: Calendar Cubes I
Post by JavaMan on May 26^th, 2006, 8:49am

Once you know that you can use 6 as 9 in calendar cube I, I'm sure soon enough you'll find that you can use "[hide]u[/hide]" for "[hide]n[/hide]" and "[hide]p[/hide]" for "[hide]d[/hide]" in Calendar Cube II.

The answer that I found for calendar cube II is:

[hide]cube 1: j g p m o f
cube 2: u a s c v b
cube 3: n l e r y t[/hide]

I feel that the solution is not uique, but I'm just too lazy to find another.

Title: Re: HARD: Calendar Cubes I
Post by astaroth on Sep 7^th, 2006, 5:25pm

I tried using "[hide]U[/hide]" for "[hide]C[/hide]" and came up with:
[hide]cube 1: E G J O R Y
cube 2: A C/U D F S V
cube 3: B C/U L M N P[/hide]

Title: Re: HARD: Calendar Cubes I
Post by rmsgrey on Sep 8^th, 2006, 5:43am

on 09/07/06 at 17:25:55, astaroth wrote:

I tried using "[hide]U[/hide]" for "[hide]C[/hide]" and came up with:
[hide]cube 1: E G J O R Y
cube 2: A C/U D F S V
cube 3: B C/U L M N P[/hide]

I don't see how you're getting OCT?

Title: Re: HARD: Calendar Cubes I
Post by astaroth on Sep 8^th, 2006, 10:20am

Right, I lost 'T' :/

Title: Re: HARD: Calendar Cubes I
Post by TheNumberScott on Oct 26^th, 2006, 12:22pm

[hideb]if you write the letters correctly, an upside-down e can look like an a. And I see no reason that a u can't be used as an n and a c. In Fact, I think you could use a J as an upside down P too.[/hideb]

It would look like:
[hideb]cube 1: J/P, F, M, G, O, D
cube 2: a/e, T, u/c/n, L, V
cube 3: n/c/u, B, R, Y, S[/hideb]