SOLUTION: how many four digit numbers that are divisible by 3 can b formed using the digits 0,2,3,5,8, if no digit occurs more than once in each number?

In order to be divisible by 3, the sum of the digits must be
divisible by 3, so we must selct 4 of those 5 digits that have
a sum which is divisible by 3. The sum of all 5 digits is 18,
so we can either leave out the 0 or the 3.  So,

we can select digits 0,2,5,and 8, since 0+2+5+8 = 15, which is divisible by 3.
Or we can select digits 2,3,5,and 8, since 2+3+5+8 = 18, which is divisible by
3.

If we select the digits 0,2,5,8

We can select the first digit any of 3 ways. (it can't be 0).
We can select the second digit any of 3 ways (it can be 0).
We can select the third digit either of 2 ways.
We can select the fourth digit 1 way.

That's 3*3*2*1 = 18 ways

If we select the digits 2,3,5,8

We can select the first digit any of 4 ways. 
We can select the second digit any of 3 ways.
We can select the third digit either of 2 ways.
We can select the fourth digit 1 way. 

That's 4*3*2*1 = 4! = 24

Total: 18+24 = 42 ways.

Edwin