To solve this problem, let’s analyze the criteria required for the four-digit number, which is presented as $57\_\_$:

1. Divisibility by 4: For a number to be divisible by 4, its last two digits must form a number that is divisible by 4.

2. Divisibility by 5: For a number to be divisible by 5, its last digit must be 0 or 5.

3. Divisibility by 9: For a number to be divisible by 9, the sum of its digits must be a multiple of 9.

Step-by-Step Solution

Let's denote the missing digits as $A$ and $B$, forming the number $57AB$.

1. Checking divisibility by 5:

   • Since the last digit $B$ must be 0 or 5, we’ll test cases where $B = 0$ and $B = 5$.

2. Checking divisibility by 4:

   • If $B = 0$, the number $57A0$ needs the last two digits $A0$ to be divisible by 4.
   • If $B = 5$, the number $57A5$ has last two digits $A5$, which can’t be divisible by 4 (since 05, 15, 25, etc., are not divisible by 4).
   • So, we conclude that $B = 0$ to meet this condition.

3. Checking divisibility by 9:

   • Now, with $B = 0$, we have $57A0$.
   • The sum of the digits must be a multiple of 9: $5 + 7 + A + 0 = 12 + A$.
   • For $12 + A$ to be a multiple of 9, $A$ must be 6 (since $12 + 6 = 18$, and 18 is a multiple of 9).

Conclusion

The number that satisfies all conditions is 5760.

Would you like a detailed breakdown of each divisibility rule, or do you have any questions about this solution?

Here are some related questions to explore further:

1. What other combinations of numbers could create a four-digit number divisible by different sets of numbers?
2. How can you quickly check if a number is divisible by 4, 5, or 9 without doing extensive calculations?
3. How do the divisibility rules apply to numbers with more than four digits?
4. Could this method be extended to check divisibility for more than three conditions?
5. What patterns do numbers divisible by both 4 and 9 tend to follow?

Tip: To quickly determine divisibility by 9, always check if the sum of the digits is a multiple of 9; it’s a helpful shortcut!