SOLUTION: From 1,2,3,4,5 and 6 form a four digit number (where repetition is not allowed). How many such numbers are divisible by 3?

In order to be divisible by 3, the sum of its digits must be divisible by 3.

The sum of the digits of any such 4-digit number cannot be less than
1+2+3+4=10 nor greater than 3+4+5+6=18.

Therefore the sum of the digits must be 12, 15 or 18, the only sums of digits
divisible by 3 in that range.

Case 1.  The number has sum of digits 12.
To have sum of digits 12, the digits must be an arrangement of 1,2,4,5 or
1,2,3,6

Here's why:

  a. If 1 were omitted, the smallest possible sum would be 2+3+4+5=14.  Too
     large
  b. Since you must have a 1, if you leave out 2 then the smallest possible sum
     would be 1+3+4+5=13.  Still too large.
  c. Since 1+2=3, the other two digits must have sum 9, which is only possible
     with 4+5 or 6+3 

So there are 4! permutations of the digits 1,2,4,5 and also
   there are 4! permutations of the digits 1,2,3,6.
So that 2*4! = 2*24 = 48 with sum of digits 12. 

Case 2.  The number has sum of digits 15.
To have sum of digits 15, the digits must be an arrangement of 2,3,4,6 or
1,3,5,6.
  
Here's why:

  a. It must have digit 6, for if 6 were omitted then the greatest possible sum
     of digits would be only 2+3+4+5=14. So the number must have a 6 as a digit.
  
  b. Since it contains a digit 6, the other 3 digits must have sum 9 and
     selected from 1,2,3,4,5.
     
  c. A quick inspection shows that 1+3+5=9 and 2+3+4=9 are the only
     possibilities for a sum of the remaining 3 digits to be 9. 

So there are 4! permutations of the digits 2,3,4,6 and also
there are 4! permutations of the digits 1,3,5,6.
So that 2*4! = 2*24 = 48 with sum of digits 15. 

Case 3.  The number has sum of digits 18. 
Only one combination of 4 digits sums to 18, 3+4+5+6

So there are 4! permutations of the digits 2,3,4,6.
That's 4! = 24 for case 3.

Answer: 48+48+24 = 120

Edwin