Calendar Cubes: Two Dice That Show Every Date
A desk calendar shows the day of the month using two cubes placed side by side. Each of the 12 faces carries a single digit from $0$ to $9$, and the two cubes can be placed in either order, so the pair must be able to display every two-digit day from $01$ to $31$ (single-digit days are shown with a leading zero, so $01, 02, \ldots, 09$ must all be displayable).
(a) Which six digits go on each cube? Find a labelling that works, and explain the key trick that makes it possible.
(b) Show that without that trick no labelling can display all 31 dates.
Hints
- Days 11 and 22 force both cubes to carry a 1 and a 2. Days 01 through 09 then force a 0 on whichever cube does not hold a given digit, so both cubes need a 0 as well.
- With 0, 1, 2 on both cubes you have used 6 faces and have 6 left, but the digits 3, 4, 5, 6, 7, 8, 9 are seven digits. Something has to give.
- Turn a 6 upside down and it reads as a 9. So you only need to place 3, 4, 5, 6, 7, 8 on the remaining six faces, for example $\{0,1,2,3,4,5\}$ and $\{0,1,2,6,7,8\}$.
Worked Solution
How to Think About It: Start from the dates that impose the tightest constraints (repeated digits and leading zeros), count faces, and see where the count fails. Then look for slack in the physical setup rather than in the arithmetic.
Approach: Derive the forced digits, count, and resolve the deficit with the 6/9 rotation.
Formal Solution:
*Step 1 -- Forced repeated digits.* The dates $11$ and $22$ require a $1$ on both cubes and a $2$ on both cubes.
*Step 2 -- Forced zeros.* The dates $01, 02, \ldots, 09$ require every digit $1$ through $9$ to be paired with a $0$ on the other cube. Suppose only one cube carried a $0$; then the other cube would have to hold all of $1$ through $9$, which is nine digits on six faces. Impossible, so both cubes carry a $0$.
*Step 3 -- Counting the remaining faces.* Each cube now holds $0, 1, 2$, using $6$ of the $12$ faces. The remaining six faces must accommodate the digits $3, 4, 5, 6, 7, 8, 9$, which is seven distinct digits. Since $7 > 6$, no assignment of ten distinct glyphs works. This proves part (b): with each face showing one fixed digit, the calendar is impossible.
*Step 4 -- The trick.* A $6$ rotated by $180^\circ$ reads as a $9$. So a single face can serve as both $6$ and $9$, and only six symbols $\{3, 4, 5, 6, 7, 8\}$ need placing on the six free faces. The count now matches exactly.
*Step 5 -- A working labelling.* Cube A: $\{0, 1, 2, 3, 4, 5\}$. Cube B: $\{0, 1, 2, 6, 7, 8\}$ with the $6$ doubling as $9$. Checks: $01$-$09$ use a $0$ from one cube and the digit from the other (the $9$ is the flipped $6$ on cube B with the $0$ on cube A); $10$-$19$ use the $1$ on one cube and the units digit on the other; $20$-$29$ likewise with the $2$; $30$ and $31$ use the $3$ on cube A with $0$ or $1$ on cube B. A brute-force enumeration of all labellings confirms that solutions exist only when the 6/9 rotation is allowed (there are 20 such unordered pairs of labellings, all of which put $0, 1, 2$ on both cubes), and this one is the standard choice.
Answer: (a) Cube A: $0, 1, 2, 3, 4, 5$; cube B: $0, 1, 2, 6, 7, 8$, where the $6$ is turned upside down to display $9$. (b) Both cubes must carry $0, 1, 2$, leaving six faces for the seven digits $3$ through $9$, so without the 6/9 trick it is impossible.
Intuition
A simple count shows the puzzle is impossible as stated: the three shared digits 0, 1, 2 eat six faces, leaving six faces for seven digits. The way out is to notice that the constraint "12 faces, 10 digits" has hidden slack because a 6 and a 9 are the same glyph rotated. Lateral puzzles like this test whether you check your modelling assumptions before declaring a problem infeasible, a habit that matters when a "no-arbitrage" or "impossible fill" conclusion rests on an assumption that does not actually bind.