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!
Answer
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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