How many ways?

• For an integer $n$ to be $6|n$ , it must be $3|n$ and $2|n$ .
• For $3|n$ , then the digit sum of $n$ must also be divisible by $3$ . Out of the six combinations of $5$ digits selected from the set $\{0,1,2,3,4,8\}$ only two subsets $\{0,1,2,4,8\}$ and $\{1,2,3,4,8\}$ satisfy the condition.
• For $2|n$ , then the last digit of $n$ must be even. So, for the two subset above, it cannot end with $1$ or $3$ .
• Also a five-digit number cannot start with a $0$ .

$\text{Case 1: } \{0,1,2,4,8\}:$

• For first digit $= 1$ , all permutations for the remaining four digits are acceptable. Therefore, $N_a = 4! = 24$ .

• For first digit $= \{2,4,8\}$ , as $1$ cannot be the last digit, we first permute the remaining three digits without $1$ then multiplied by $3$ for the three positions $1$ can assume. Therefore, $N_b = 3\times 3! \times 3 = 54$ .

  $\text{Case 2: } \{1,2,3,4,8\}:$

• For first digit $= \{1,3\}$ , the last digit cannot be $3$ or $1$ repectively. Therefore, $N_c = 2\times 3! \times 3 = 36$ .

• For first digit $= \{2,4,8\}$ , the last digit cannot be $1$ or $3$ . Therefore, $N_d = 3\times 2! \times 3! = 36$ .

Therefore, the total number of five-digit numbers meeting the condition is:

$N = N_a+N_b+N_c+N_d = \boxed{150}$ .

A number is divisible by 6 if this number is divisible by 2 and 3. For this reason, we get two case ,

case [1] = 1,2,3,4,8 ( because ( 1 + 2 + 3 + 4 + 8)%3=0 ) )

case[2] = 0,1,2,4,8 ( because (0 + 1 + 2 + 4 + 8)%3=0 ) )

Now , case[1] ,

5! - ( 4! * 2 ) = 72 [ 1 and 3 does not sit as last digit ]

So, solution of case[1] is = 72.

Again , case[2] ,

Total permutation = 5!

first position of number dose not allow 0 ,

So , 5! - 4! = 96

last position of number dose not allow 1 ,

So , 96 - 4! = 72

But , some number are common , where , 0 is first position and 1 is last position.

So , 72 + 3! = 78

Now , case[1] + case[2] = 72 + 78 = 150 .