Numbers divisible by 12

We know that $12=3x4$ , so if a number is divisible by $12$ it must be divisible by both $3$ and $4$ .

So that a number is divisible by $3$ , the sum of its digits must be divisible by $3$ . $3+4+5+6+7+8=33$ . Since $33$ is divisible by $3$ , all 6-digit numbers composed of the digits $3, 4, 5, 6, 7$ and $8$ will be divisible by $3$ .

So that a number is divisible by $4$ , the last two digits must be divisible by $4$ . In our case, the number must end in: $36,48,56,64,68,76$ and $84$ . If the number ends in $36$ , the first four digits will be $4,5,7,$ and $8$ . And there are exactly $4!=24$ ways of arranging these numbers. Similarly if the number ends in $48,56, 64,68,76,$ and $84$ , the first four digits of each can be arranged in exactly $4!=24$ ways. As there are $24$ ways of arranging the first four digits, there are $24x7=168$ different 6-digit numbers that are divisible by $12$ .