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# Calendar dice

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Rigil Kent
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Three men make a tiger.

 « on: July 18, 2016, 14:35:04 PM »

This desk toy made me wonder what the largest number, M, is which can be displayed by arranging two six sided dice of which any given side may contain a number from 0 to 9 such that any whole number from 01 to M can also be built.

Obviously the maker of the toy has solved the problem up to 31 - a date that comes along every two months or so - but can you go bigger?

(Hint: It won't be 99 because there will be some numbers smaller than 99 that cannot be built.)

Rigil

ETA: Just noticed that six and nine are one and the same symbol .... just to make it infernally interesting.

 « Last Edit: July 18, 2016, 16:41:37 PM by Rigil Kent » Logged
Rigil Kent
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Three men make a tiger.

 « Reply #1 on: July 18, 2016, 17:04:00 PM »

Die A: 0 1 2 3 6 7
Die B: 0 4 5 1 2 8

Takes it up to a lackluster 32.
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brianvds
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 « Reply #2 on: July 19, 2016, 04:45:01 AM »

Die A: 0 1 2 3 6 7
Die B: 0 4 5 1 2 8

Takes it up to a lackluster 32.

And how can you tell there aren't other solutions that may give us higher values for M? This is an interesting math problem, but I have not the vaguest clue how to solve it in a rigorous sort of way so that I can prove M must be whatever number.

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Rigil Kent
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Three men make a tiger.

 « Reply #3 on: July 19, 2016, 08:05:57 AM »

And how can you tell there aren't other solutions that may give us higher values for M?
That's the million dollar question! Anyway, I don't know if it's particularly rigorous, but I reasoned like this:

We are required to paint at least the numbers 0 1 2 3 4 5 6 7 8 to potentially arrive at 10. (Recall the six can be used as a nine).
Also, the number 0 must appear on both dice otherwise, counting up,  we will get stuck on 06.
To go bigger than 10, some numbers will have to appear on both dice to allow us to make 11, 22, 33, 44, 55, 66 etc.
If 0 1 and 2 are painted on both dice, it leaves enough space to include the other required numbers 3 4 5 6 7 8.
If 0 1 2 and 3 are painted on both dice, it leaves insufficient space to include the other required numbers 4 5 6 7 8.
Therefore, we can only duplicate 0 1 and 2.
Since 3 cannot be duplicated, we know that 33 will be impossible to build, so the next highest candidate for M must be 32.
Then all that is left to do is to find an arrangement that will allow building all the numbers 01 to 32, and at least one arrangement,

Die A: 0 1 2 3 4 5
Die B: 0 1 2 6 7 8

allows us to do just that.

Rigil

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brianvds
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 « Reply #4 on: July 19, 2016, 12:16:38 PM »

Don't know if that's rigorous (I'm not even sure how to tell whether a math proof is really really rigorous), but it looks clever enough to me. Perhaps someone else will come up with a different idea...
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Rigil Kent
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Three men make a tiger.

 « Reply #5 on: July 20, 2016, 08:54:55 AM »

An extension to the fun :

We know 1 die  will give M=6.
and 2 dice will give M=32.
So following the same rules, what value for M will be possible using 3 dice to allow us to build 001 through M?

Rigil
 « Last Edit: July 20, 2016, 09:54:05 AM by Rigil Kent, Reason: One die will give M=6, of course. Duh. » Logged
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