The problem asks for the smallest 4-digit even number that can be formed using the digits 0, 2, 5, 7, and 8. Each digit can only be used once.

Solution Approach:

1. Constraints: The number must be even, so the last digit must be either 0, 2, or 8.
2. Minimization: To minimize the number, the first digit (thousands place) should be the smallest non-zero digit, as the number cannot start with 0.
3. Step-by-step Construction:
   • For the first digit, choose the smallest non-zero digit, which is 2.
   • For the last digit (because the number must be even), choose the smallest available even digit. Since 2 is already used in the first place, the next smallest available even digit is 0.
   • Now, the remaining digits to place in the second and third positions are 5, 7, and 8. To minimize the number, choose the smallest remaining digits for the next positions.

Forming the Number:

• First digit: 2
• Last digit: 0
• Second and third digits: Choose the smallest remaining digits (5 and 7).

Thus, the smallest number is 2570.

Would you like more details or explanations?

Here are five related questions to consider:

1. How would the solution change if you were forming the largest 4-digit even number?
2. What if each digit could be repeated? How would that impact the smallest number?
3. Can you list all possible 4-digit even numbers formed from these digits?
4. What strategies can be used to minimize or maximize numbers given specific digit constraints?
5. How does this problem relate to permutation and combination principles?

Tip: When forming the smallest or largest number, always focus on placing the largest (or smallest) digits in the highest value positions (thousands, hundreds, etc.).