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My cousin just learned about numbers and dates in school and I would like to give him a little present: A set of (regular, six-sided) dice with which he can display the current date if he turns them the right way.

When he places the dice, they should be able to display the date of every day and month of the year in the following format:

[M][M] [D][D]

So, if it's the third of September, I want him to be able to place the dice in a way that they show:

[0][9] [0][3]

Now my question:

What is the minimum amount of dice I need to make, and what numbers should I put on each die?

I am going to be honest: He is a great kid, but if this turns out to be too much work - he will get coals for Christmas.

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  • $\begingroup$ A weakness of this puzzle is that four dice are obviously necessary, and there is no reason the normal two-die date display can't simply be doubled to show four digits total. The puzzle might be more interesting if dates are supposed to be shown using as [m][d][d] on 8-sided dice, where [m] is a single face showing 1-12. $\endgroup$ – supercat Nov 18 '16 at 22:27
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My answer:

4 dice. (6 faced).

Numbers on each die:

2 sets of the following 2 dice.
0, 1, 2, 3, 4, 6
0, 1, 2, 5, 7, 8

Reasoning:

1 set will be used for the month and one set for the day.
I will explain how you can get all the numbers from 1 to 31 using one set of dice described above. If you can do it for the day (up to 31) you can do it for the month also with the other set (up to 12).
Let's name the dice A and B in the order listed above.

.

01 - A0, B1
02 - A0, B2
03 - A0, B3
04 - B0, A4
05 - A0, B5
06 - B0, A6
07 - A0, B7
08 - A0, B8
09 - B0, A6 reversed
10 - A1, B0
11 - A1, B1
12 - A1, B2
13 - A1, B3
14 - B1, A4
15 - A1, B5
16 - B1, A6
17 - A1, B7
18 - A1, B8
19 - B1, A6 reversed
20 - A2, B0
21 - A2, B1
22 - A2, B2
23 - A2, B3
24 - B2, A4
25 - A2, B5
26 - B2, A6
27 - A2, B7
28 - A2, B8
29 - B2, A6 reversed
30 - A3, B0
31 - A3, B1

Additional reasoning:

Because you have the numbers 11 and 22 it means 1 and 2 need to go on every die.
You need one die to have a 0, but you cannot form 9 numbers (1 to 9) from a single die, so you need 0 on both of them.
Summarize: We now have 2 dice with the number 0,1 & 2.
Since we need all the digits we need somehow to add the remaining 7 digits in the remaining 6 available faces.
Since 6 can double as a 9 and we never have to create the numbers 66, 69,96 or 99 we get to have 6 number on 6 faces. Just split them 3 on one die and 3 on the other. I don't think the order is important.

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    $\begingroup$ Four dice, that's doable. You saved that little boys christmas! $\endgroup$ – Timme Nov 18 '16 at 14:01
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    $\begingroup$ Fa la la la la, la la, la la!!! $\endgroup$ – Marius Nov 18 '16 at 14:03
  • $\begingroup$ What about Feb 29, 2069? $\endgroup$ – Mad Physicist Nov 18 '16 at 14:38
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    $\begingroup$ @MadPhysicist. What about it? First of all, the year is not supposed to be on the dice. Second, there will be no Feb 29, 2069. There will be Feb 29, 2068 and 2072. $\endgroup$ – Marius Nov 18 '16 at 15:46
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    $\begingroup$ @SE_User the if you want to be "politically correct", regular dice don't have a 9 on them. So if we allow a 9 then we should allow numbers instead of dots. That was the point of the question. Don't try to find loopholes in a nice puzzle and a good answer. You are taking the fun out of it. $\endgroup$ – Marius Nov 28 '16 at 8:25
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The answer is

4

if you note:

You never specified what type of dice. I could use 10 sided dice in this instance with the digits 0-9 on them. And, if you allow more than one digit per dice (unclear), we could go lower.

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  • $\begingroup$ Fair enough.. i will edit the question though. $\endgroup$ – Timme Nov 18 '16 at 13:45
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    $\begingroup$ By the same reasoning yon can say that you can use a 366 faced die with every day-month combination possible. $\endgroup$ – Marius Nov 18 '16 at 14:05
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    $\begingroup$ @Marius Definitely! If you're allowed more than one digit per dice, as I mentioned. This was mostly a joke answer, but at least we're using that lateral-thinking tag to its fullest! $\endgroup$ – TwoBitOperation Nov 18 '16 at 15:08
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    $\begingroup$ I understood that. I wrote my comment as a joke also. we need a special markup for jokes. $\endgroup$ – Marius Nov 18 '16 at 15:45
  • $\begingroup$ @Marius - no need for a markup, just use one of the classic smileys... in particular, this one... :P $\endgroup$ – Glen O Nov 19 '16 at 7:31
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It will not be the answer as it doesn't follow the formatting but you can use only 3 dices.

You only use one type of dice: 0 I II III following each other and X and V on opposite sides.

The trick is, for the month, you show two faces at the same time. Thus you can get the following combination IV VI VII VIII IX.

Same reasoning for the second numeral of the day.

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