Using the digits 0, 1, 2, 3, 4 and 5 with no repetition, how many 4-digit numbers divisible by 3 can be formed?

The 4 digits must have a digit sum divisible by 3. Possible sets: {0,1,2,3}, {0,1,3,5}, {0,2,3,4}, {0,3,4,5} and {1,2,4,5}. Each set with 0 gives $4!-3!=18$ numbers (0 cannot lead): $4\times 18=72$. The set {1,2,4,5} gives $4!=24$. Total $72+24=96$.