SOLUTION: In how many ways can you arrange the digits 8, 7, 6, and 3 to form a four-digit number divisible by 9, using each digit once and only once?

There is no way! Why? Because the sum of the 
digits of any multiple of 9 must itself be a 
multiple of 9. However the sum of these digits 
is 8+7+6+3=24 which is not a multiple of 9, 
regardless of how you arrange them.

Proof that the sum of the digits of a multiple of
9 is a multiple of 9:

Suppose the number which is a multiple of 9 
has the digits A,B,C, and D.  Then the number is

1000A + 100B + 10C + D

Suppose this equals to a multiple of 9, 
say, 9 times an integer N, or 9N

1000A + 100B + 10C + D = 9N

We write 1000 as 999+1, 100 as 99+1, 
and 10 as 9+1:

(999+1)A + (99+1)B + (9+1)C + D = 9N

999A + A + 99B + B + 9C + C + D = 9N

A + B + C + D = 9N - 999A - 99B - 9C

Factor out 9 on the right:

A + B + C + D = 9(N - 111A - 11B - C)

The right side is a multiple of 9 and the 
left side is the sum of the digits.

That proves that the sum of the digits of 
a multiple of 9 is also a multiple of 9.

Therefore no matter how you arrange the 
digits 8, 7, 6, and 3, it can never be a 
multiple of 9, because their sum is 24, 
which is not a multiple of 9.

On the other hand, if you could trade the 
6 for a 9,  then the sum of the digits 
would be 27, which is a multiple of 9,
and therefore all 24 ways to arrange 
the digits 8, 7, 9, and 3 will be a 
multiple of 9, as you see below:

 1.  3789 = 9×421
 2.  3798 = 9×422
 3.  3879 = 9×431
 4.  3897 = 9×433
 5.  3978 = 9×442
 6.  3987 = 9×443
 7.  7389 = 9×821
 8.  7398 = 9×822
 9.  7839 = 9×871
10.  7893 = 9×877
11.  7938 = 9×882
12.  7983 = 9×887
13.  8379 = 9×931
14.  8397 = 9×933
15.  8739 = 9×971
16.  8793 = 9×977
17.  8937 = 9×993
18.  8973 = 9×997
19.  9378 = 9×1042
20.  9387 = 9×1043
21.  9738 = 9×1082
22.  9783 = 9×1087
23.  9837 = 9×1093
24.  9873 = 9×1097

Edwin