9.3. Vasya must write one digit on each face of several dice so that any ordered combination of three digits from 000 to 999 inclusive can be obtained by selecting some three different dice and placing them with the appropriate sides up in the correct order. At the same time, the digits 6 and 9 do not transform into each other when rotated 180 degrees. What is the minimum number of dice Vasya should use?
Answer: 5.

Solution. Since among the possible combinations there must be $000, 111, 222, \ldots, 999$, each digit must appear on at least three different faces (of different cubes), so there must be at least 30 digits on the faces in total, hence there must be at least $30: 6=5$ cubes.

On the other hand, we will show that if the digits are written such that each digit from 0 to 9 appears on the faces of three different cubes, then the requirement of the problem is satisfied. Consider any combination $\mathrm{ABC}$ of three digits. First, choose any cube that contains the digit A. The digit B appears on at least two other cubes besides, possibly, the first one, so we take any of them as the second. Finally, the digit C appears on at least one more cube besides the first and second cubes, so we take it as the third. By placing them in order with the corresponding faces up, we get the required combination ABC. Note that the reasoning does not depend on whether the digits A, B, and C are distinct or not.

Let's provide a possible way to write the digits on the faces of five cubes such that each digit appears on the faces of three different cubes. Write all the digits from 0 to 9 in a row with each digit repeated three times: $000111222 . . .888999$. The first five digits of this row will be written on the first faces of five different cubes, the second five digits will be written on the second faces of five different cubes, and so on.

Grading criteria. Proof of the minimality of 5 cubes: 3 points. Providing a correct example for 5 cubes with justification: 4 points. Lack of justification: minus 2 points.