5-digit numbers

To be divisible by $9$ , the sum of the digits of the number must be divisible by $9$ . Ignoring the digit $9$ , there is exactly one digit of each remainder modulo $9$ . Thus, whatever the first four digits of the number are, there is exactly one possible final digit that will leave the final number $9$ -free yet divisible by $9$ . Thus there are as many five-digit numbers that are $9$ -free, yet divisible by $9$ as there are ways of choosing $9$ -free four-digits numbers. Thus $a = 8\times9\times9\times9 = 5832$ .

To be divisible by $5$ , the final digit of the number must be $0$ or $5$ . Thus, whatever the first four digits of the number are (provided that they do not contain $5$ ), there is exactly one possible final digit ( $0$ ) that will leave the number $5$ -free yet divisible by $5$ . Thus $b = 8\times9\times9\times9 = 5832$ .

To be divisible by $4$ , the final two digits of the number must form a number divisible by $4$ . Since we cannot use $4$ , there are exactly two final digits that can be chosen, whatever the penultimate digit is and provided that the first four digits are $4$ -free, to make the final number $4$ -free yet divisible by $4$ (if the penultimate digit is even, the final digit can be $0$ or $8$ , while the final digit can be either $2$ or $6$ when the penultimate digit is odd). Thus $c = 8 \times9\times9\times9\times2 = 11664$ .

To be divisible by $3$ , the sum of the digits of the number must be divisible by $3$ . Ignoring the digit $3$ , there is exactly three available digits of each remainder modulo $93$ (we have $0,6,9$ giving remainder $0$ , then $1,4,7$ giving remainder $1$ , and finally $2,5,8$ giving remainder $2$ ). Thus, whatever the first four digits of the number are, provided they are $3$ -free, there is exactly three possible final digits that will leave the final number $3$ -free, yet divisible by $3$ . Thus $a = 8\times9\times9\times9\times3 = 17496$ .

This makes $a+b+c+d = 8\times9\times9\times9(1 + 1 + 2 +3) = \boxed{40824}$ .