Yesterday I saw a pair of calendar dice that can be rotated and swapped to show all days of a month between 01 and 31.

enter image description here

The dice have a digit painted on each face.

  • First die: [0, 1, 2, 3, 4, 5]
  • Second die: [0, 1, 2, 6, 7, 8].

The trick is that one of the faces can be rotated to show a 6 or a 9 as required.

enter image description here

With those dice one can also display the number 32. This is the last number that can be reached if we start counting from 01. Even though larger numbers like 42 are possible, there are gaps between 32 and 42 that cannot be represented with those two dice, so we can only count up to 32.


If we were given three dice to count numbers in zero-padded three-digit format starting from 001 and counting up by 1, what would be the largest countable number? How would the dice have to look like?

  • $\begingroup$ What numbers are in these dice? $\endgroup$
    – warspyking
    Nov 23, 2014 at 12:20
  • $\begingroup$ @warspyking One digit per face, like in the calendar example. There's basically no difference between 6 and 9. The puzzle is about finding out an optimal arrangement. $\endgroup$
    – GOTO 0
    Nov 23, 2014 at 12:29
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    $\begingroup$ I know it's late and you've accepted the first answer posted, but you did not restrict the number of faces on the dice... you can count much higher on 20 sided dice than on 6 sided dice... $\endgroup$
    – atk
    Nov 23, 2014 at 15:49

2 Answers 2


First, each die must have a 0 face. We can see this by noting that we need two 0s to make 001, and a third die with {1,2,3,4,5,6} only gets us to 006.

Having this fact, we need 1-8 at least once on some face to reach 010. It doesn't matter (yet) where these are - there are always two other 0s to allow 00x, no matter where x is. If we want to get to 011, we need a second 1 on a different die to the first 1. Similarly for all other numbers, we find that we cannot have a full set of two of each number 1-8 (requires 16 faces) because 3 dice with one face already assigned, leaving us only 15 faces free.

Dropping the highest number, we need 15 faces to reach 087 if we can arrange them suitably.

{0,1,2,3,4,5} {0,1,2,3,6,7} {0,4,5,6,7,8}

Will suffice, though there should be many other suitable arrangements.

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    $\begingroup$ +1 I brute-forced it. 87 is correct and I found the same dice pattern. $\endgroup$
    – d'alar'cop
    Nov 23, 2014 at 12:56
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    $\begingroup$ Fun fact, there were 84 possible dice and 102340 ways we could combine them $\endgroup$
    – d'alar'cop
    Nov 23, 2014 at 13:01

AMBIGRAMS. Ambigrams are dope. The 6 on this die is a simple one - upside down, it looks like a 9, as a six is wont to do. To get higher than 87, we need more of these.


enter image description hereenter image description here

If we add more ambigrams, we can fit more numbers. The first is either a 4 or a 7, and the second a 2 or 3, depending on orientation. So, consider these dice:

[0, 1, 2/3, 4/7, 5, 6/9] [0, 1, 2/3, 4/7, 5, 8] [1, 2/3, 4/7, 5, 6/9, 8]

This set will get you all the way up to 665.

Alternately, you could introduce one more ambigram. I'm a frankly terrible artist, but if one could render one more pair of numbers as one, then all ten digits will fit on each die, allowing you to reach 999!


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