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I saw an interesting calendar in a shop. It is composed of two cubes with numbers written on their 6 sides. By placing these cubes side by side one can make any day of the month from 1 to 31 (even 32). This tickled my mathematical curiosity and made me wonder: what is the largest contiguous range of numbers you can make with 3 cubes? Bonus question: what happens if you allow cubes to be flipped, so 6 can become 9 and vice versa?

This problem is similar to this Counting numbers with 3 dice but here we don't require 0-padding, so the answer is different. For example, here we can use a single die to represent single-digit numbers.

Good luck!

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I really enjoy puzzles like these.

The largest number with 3 dice would have to be 98 (without flipping), since we would need two occurrences of every digit from 1-9 to get past the multiples of 11, plus a 0 somewhere for the tens.
This adds up to 19 (2 x 9 + 1), but we only have 18 (3 x 6) faces. So my solution for without flipping is:
[ 0, 1, 3, 4, 6, 7 ]
[ 1, 2, 4, 5, 7, 8 ]
[ 2, 3, 5, 6, 8, 9 ]
Which can count to 98. The procedure for building up the dice is simply counting up and making sure to have two of each number (except 0) on difference dice.

Bonus question:

As for flipping, we can remove the 9, since 6s can be used.
Now we can get up to 99, but for 100 we need another 0.
So I tried swapping the 9 for 0:
[ 0, 1, 3, 4, 6, 7 ]
[ 1, 2, 4, 5, 7, 8 ]
[ 2, 3, 5, 6, 8, 0 ]
This allows counting up to 110, but it can not make 111, since that requires three 1s, which we don't have.

Curiosity:

I played around with 2 dice, and without flipping, the maximum is indeed 32:
[ 0, 1, 2, 4, 6, 8 ]
[ 1, 2, 3, 5, 7, 9 ]
But with flipping, the maximum is 43 (since 44 is unreachable):
[ 0, 1, 2, 3, 5, 7]
[ 1, 2, 3, 4, 6, 8]

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  • $\begingroup$ Perfect answer, thank you! $\endgroup$ – Dmitry Kamenetsky Sep 26 at 3:31
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I think this is different enough to warrant its own answer for the bonus question. Starting with Matthew Jensen's configuration, swap the last cube to get

[ 0, 1, 3, 4, 6, 7 ]
[ 1, 2, 4, 5, 7, 8 ]
[ 2, 3, 5, 6, 8, 10 ]

This should be fine, since nobody ever said the "number" on each face had to be a digit. Moreover, the 10 can be flipped over to yield 01. This gives us 111 (01 1 1) up through 119. We don't lose 100-110, since we can just use (10 0-9) and (1 10) to make them. We can get 120-139 using the cubes regularly. 140-149 use the 01 again: (01 4 0-9). 150-169 are normal. 170-179 use the 01, 180-199 are normal.

200 is not possible, so we stop here.

Edit: I mentioned this in a comment below, and then decided maybe it was worth adding. The question doesn't specify that numbers need to be positive, so we could certainly rotate a 1 face 90 degrees to get a minus sign. If we allow awkward constructions like -011 (- 01 1), this gets us down to -39 for a total range of 239. If we disallow -011, we can get to -10 for a range of 210.

One last improvement. In the case above where we got stuck at -39, I think we can swap the 8 from Cube 3 with the 4 from Cube 1 without causing any problems. It changes some other constructions (e.g. the 140s are now normal and the 180s use the 01), but this allows us to push much further in the negative direction. We can get to -69 now. So my best solution is
[ 0, 1, 3, 6, 7, 8 ]
[ 1, 2, 4, 5, 7, 8 ]
[ 2, 3, 4, 5, 6, 10 ]

Assuming the most liberal rules (1 can be rotated to a minus sign and "-011" is valid), and assuming I haven't overlooked something, this gets us to 269 consecutive numbers.

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  • $\begingroup$ Wow great work in pushing the boundary! $\endgroup$ – Dmitry Kamenetsky Sep 26 at 3:33
  • $\begingroup$ This is also the maximum. 200 itself would be doable with 20 and 50 (flipped giving 05 and 02), but then you would lose 122 and 155. $\endgroup$ – Zizy Archer Sep 26 at 10:52
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    $\begingroup$ Yes, without resorting to behavior that I think is outside of the spirit of the question, that's the best you can do. There are truly silly things like turning an 8 on its side to make two zeros, or turning the 10 on its side to make an underlined zero. But maybe less silly would be to turn a 1 on its side to make a minus sign, since the question doesn't specify that the numbers need to be positive. Doing so would get us down to -39 for a total range of 239. $\endgroup$ – hdsdv Sep 26 at 11:00
  • $\begingroup$ Potentially, with the 2-digit faces, you could 'hide' one digit behind another block. For example: [ 2 ][ 0 ]0 ] with [ 10 ] hiding behind the [ 0 ]. But as hdsdv said, that's far beyond the spirit of the question. $\endgroup$ – Matthew Jensen Sep 26 at 20:36
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To make a sequence of 100 consecutive numbers you need two of each digit (0-8). As this uses all 18 faces, we cannot have a third occurrence of any digit. The largest contiguous range is 111 numbers in the range 0 to 110.

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As long as you have all the digits, you can make all the 1-digit numbers. So the question becomes: how far can you get with the 2-digits, which start at 10

Without allowing 6 and 9 to be flipped:

Start with:
First two dice: 0,1,2,3,4,5,7,8,9,,,_
Third dice: 1,2,3,4,5,6
and you'll get to 65.

So, throw in a 6 in the first/second dice and you can get to 87 provided you separate your first two carefully:
0,1,2,3,4,7
5,6,7,8,9,_
1,2,3,4,5,6

and I haven't even used everything yet. It must be possible to improve on that!

Edit: And it is. There are 560 ways to get to 98. eg
(1, 2, 3, 5, 7, 8), (3, 4, 5, 6, 8, 9), (0, 1, 2, 4, 6, 7)
(2, 4, 5, 6, 8, 9), (0, 1, 3, 4, 6, 7), (1, 2, 3, 5, 7, 8)
This turns out the be the maximum (brute force search).

Allowing 6 and 9 to be flipped opens up a few extra possibilities:

If you repeat the 7,8 on your first two dice:
7,8,0,1,2,3 7,8,4,5,69,0 1,2,3,4,5,69
you can get to 110 (which is the biggest possible as @DanielMathias points out in his answer).

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I find it strange that this answer is missing and I hope it's within what was asked, even if a bit impractical to use at first and rather different from the other answers.

Base question (no flipping):

The idea is to use a base that better fits cubes, like base 6. This solution uses cubes [0, 1, 2, 3, 4, 5] and to convert it back to the usual decimal base you just need to do a A*36 + B*6 + C, where A, B and C are the three cubes. With this it's possible to cover 6³ = 214 combinations.

But as it was said that not all cubes must be present, than it should be allowed to consider numbers differently based on the total number of cubes. Specifically it's possible to consider an offset to avoid duplicates. 1 cube no offset. 2 cubes +6, 3 cubes +42. In this case it's possible to reach 6 + 6² + 6³ = 6 + 36 + 216 = 258 values.

I believe this to be inside the question as were considered valid both multi digit cubes and using a rotated 1 to get a -. Additionally it's not that uncommon to use numbers outside base 10, just think about the dates themselves which can roughly be approximated to a mixed base number (days in base 31 and months in base 12).

Bonus question with flipping.

This is a bit more stretched, but if the base part was fine than I guess this one is fine too. I wrote it mostly in case anyone is curious to the natural extension of my solution to the base question to the bonus one.

If it's possible to rotate a number so that 6 becomes 9 and that 1 becomes -, than each digit has four possible rotations. To keep track of the rotation a line can be wrote under the digit. This means that we can add to the senary digit a +4*Rotation. If it's straight you have +0, if it's turned on the left +6, if it's upside down +12 and if it's turned on the right +18. This makes the cubes able to represent any digit of base 24 and using the math of the base question we have 13824 distinct and consecutive values if adding an offset based on the number of digits is not allowed and 14424 if it is.

While this was a bit far from the other answers, they are still normal numbers just in different base. For example this is not that far from Babylonian numerals as their digit was a compound in the same way my digits are, just in different bases and with a different writings.

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