9.24 How many four-digit numbers divisible by 5 can be formed from the digits $0,1,3,5,7$, if each number must not contain identical digits?
9.24 The desired numbers end with five or zero. If the last digit is 0, then there are $A_{4}^{3}=4 \cdot 3 \cdot 2=24$ such numbers. If the last digit is 5, then the first digit cannot be 0 (the desired numbers are four-digit); in this case, there are $A_{4}^{3}-A_{3}^{2}=18$ such numbers. Therefore, 42 numbers satisfy the condition of the problem.

Answer: 42 numbers.