Let's solve the problem step by step.

Problem:

We need to determine how many 4-digit numbers can be formed using the digits $1, 5, 3, 7, 8$ without repetition such that the numbers are divisible by 5.

Key Concepts:

1. Divisibility by 5: A number is divisible by 5 if its last digit is either $0$ or $5$. In this case, since $0$ is not in the set of digits, the last digit must be 5.

2. Formation of 4-digit numbers:

   • Since the last digit is fixed as $5$, we are left with the digits $1, 3, 7, 8$ to arrange in the first three places.
   • These remaining digits can be arranged in different orders to form valid numbers.

3. Permutation Calculation:

   • The number of ways to arrange $n$ distinct objects is given by $n!$ (factorial of $n$).
   • Here, we have $4$ digits (excluding $5$) to arrange in $4$ positions, so the number of arrangements is $4!$.

Solution:

1. Fix $5$ as the last digit.
2. Arrange the remaining $4$ digits ($1, 3, 7, 8$) in the first $3$ positions: $\text{Number of ways} = 4! = 24$

Thus, there are 24 such numbers.

Final Answer:

The correct option is: (1) 24

Would you like a detailed explanation of permutations or help with similar problems? Here are some related questions to expand on:

1. How many 4-digit numbers can be formed using the same digits if repetition is allowed?
2. How many 4-digit numbers divisible by $2$ can be formed from the digits $1, 5, 3, 7, 8$?
3. How many 5-digit numbers can be formed without repetition using these digits?
4. How does the condition of divisibility by $5$ change if $0$ is included in the digits?
5. How many numbers are divisible by $3$ when formed using these digits?

Tip: Always identify constraints (like divisibility rules or digit repetition) before solving permutation problems.