How many five-digit natural numbers have the product of their digits equal to 2000?
Initially, observe that $2000=2^{4} \times 5^{3}$. Since the digits of the number are less than 10, each factor of 5 must be a digit of this number. Additionally, the product of the other digits must be $2^{4}=16$. Thus, we have two cases:

- The missing digits are 2 and 8. In this case, there are five possibilities for placing the 2, four possibilities for placing the 8, and a single possibility for placing each remaining 5. Therefore, we can form $5 \times 4=20$ numbers.
- The missing digits are 4 and 4. In this case, we can choose two places for the digits 4 in $\binom{5}{2}=10$ ways (see Counting Subsets on page 118) and one way to place each remaining 5. Therefore, we can form 10 numbers.
Suggestion: Divide the stones into pairs and perform the weighings, eliminating the lighter stones. Notice that the second heaviest stone can only be eliminated by the heaviest stone.

Thus, a total of $20+10=30$ numbers can be formed.

Suggestion: (a) Divide into two cases according to the color of the central house. (b) Determine the number of $3 \times 3$ boards that can be placed on the $123 \times 123$ board.