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Topic: HARD: Calendar Cubes I (Read 2765 times) |
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ddavis
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HARD: Calendar Cubes I
« on: Jul 25th, 2002, 7:47pm » |
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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
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| « Last Edit: Sep 2nd, 2003, 8:34pm by Icarus » |
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kevin
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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.
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James Jones
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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
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ddavis
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Re: HARD: Calendar Cubes - spoiler
« Reply #3 on: Jul 26th, 2002, 7:40am » |
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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
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Neil Sedaka
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This works (using the 9 as a 6 also).
0 1 2 3 4 5
0 1 2 7 8 9
Any difficult riddles?
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James Jones
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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.
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Neil Sedaka
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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 !!!
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jmlyle
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Re: HARD: Calendar Cubes
« Reply #7 on: Jul 30th, 2002, 1:32pm » |
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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
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william wu
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Re: HARD: Calendar Cubes
« Reply #8 on: Jul 30th, 2002, 7:18pm » |
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on Jul 26th, 2002, 8:52am, Neil Sedaka wrote: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 !!! |
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pwhahaha 
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| « Last Edit: Jul 30th, 2002, 7:21pm by william wu » |
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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thetorpedodog
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Re: HARD: Calendar Cubes I
« Reply #10 on: Jun 27th, 2004, 10:45am » |
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Additional solution:
Cube 1: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , 1 1 1 1 1 1 1 1 1 1 1 1 1 1 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 , 4 4 4 4 4 4 4 4 4 4 4 4 4 4 , 5 5 5 5 5 5 5 5 5 5 5 5 5 5 , 7 7 7 7 7 7 7 7 7 7 7 7 7 7
Cube 2: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , 1 1 1 1 1 1 1 1 1 1 1 1 1 1 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 , 3 3 3 3 3 3 3 3 3 3 3 3 3 3 , 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 6/9 , 8 8 8 8 8 8 8 8 8 8 8 8 8 8
Added irrelevant note: whoever can should really replace the <table> tags in the "glow" output with <span>s.
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JavaMan
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Re: HARD: Calendar Cubes I
« Reply #11 on: May 26th, 2006, 8:49am » |
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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 "u" for "n" and "p" for "d" in Calendar Cube II.
The answer that I found for calendar cube II is:
cube 1: j g p m o f
cube 2: u a s c v b
cube 3: n l e r y t
I feel that the solution is not uique, but I'm just too lazy to find another.
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astaroth
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Re: HARD: Calendar Cubes I
« Reply #12 on: Sep 7th, 2006, 5:25pm » |
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I tried using "U" for "C" and came up with:
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
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| « Last Edit: Sep 7th, 2006, 5:26pm by astaroth » |
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rmsgrey
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Re: HARD: Calendar Cubes I
« Reply #13 on: Sep 8th, 2006, 5:43am » |
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on Sep 7th, 2006, 5:25pm, astaroth wrote:I tried using "U" for "C" and came up with:
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 |
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I don't see how you're getting OCT?
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astaroth
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Re: HARD: Calendar Cubes I
« Reply #14 on: Sep 8th, 2006, 10:20am » |
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Right, I lost 'T' :/
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TheNumberScott
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Re: HARD: Calendar Cubes I
« Reply #15 on: Oct 26th, 2006, 12:22pm » |
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| hidden: | | 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. |
It would look like:
| hidden: | 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 |
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