Divide and Conquer

Since all numbers, which are:

• divisible by 8 are also divisible by 2 and 4
• divisible by 9 are also divisible by 3
• divisible by 2 and 3 are also divisible by 6,

therefore we are looking for numbers, which are divisible by:

5 × 8 × 9 = 40 × 9 = 360

(In other words, the LCM of all specified divisors is 360.)

A number is divisible by 360, if it is divisible by all its (prime) factors (to their highest powers, either separately or in a grouped form (where no 2 groups have numbers with a same prime factor, e.g. 9 and 40).

A number is:

a) divisible by 9 (according to the well-known rule), if the sum of the digits of the number itself is divisible by 9.

Another consequence of this, is that if we rearrange the digits of a number, which is divisible by 9, then our new number will also be divisible by 9 (as the sum of the digits remained unchanged).

Since the digits of the 9 digit numbers we are looking for, are all different and there are 10 digits (from 0 to 9) available in the decimal system, therefore we have to leave out exactly one digit in each case.

As the sum of all 10 digits (0 + 1 + ... + 8 + 9 = 45) is divisible by 9, the digit we leave out has to be also divisible by 9 (either 0 or 9), otherwise the sum of the remaining digits (and consequently, our number) won't be divisible by 9.

However, we cannot leave out the 0. This is due to the fact, that 10 | 360, therefore our number has to be divisible by 10, meaning that the last digit always has to be 0 and none of the numbers have the digit 9.

b) divisible by 40, if the number formed of the last 3 digits is divisible by 40 (this is because 40 | 1000)

The possible values of these last 3 digit numbers are:

120, 160, 240, 280, 320, 360, 480, 520, 560, 640, 680, 720, 760 and 840

(14 relevant endings altogether. We had to leave out 9 values (e.g. 000, 080, 880) because they have (at least two) identical digits and 2 other endings (920 and 960) as they have the digit 9).

Since we can put the first 6 digits in any order, our answer should be:

$14 × 6! = \boxed {10080}$

Let $N$ be such a number

• it is divisible by 2 and 5 both.So at end there must be a $0$

• it is divisible by 4 and 8 both.So, N has last two digits will be divisible by 4 and last three digits will be divisible by 8. Such combinations are $120,320,520,720,920,160,360,560,760,960,280,480,680,240,640,840$

• if it is divisible by 9 (as it is also even because ends with zero) then it will be divisible by 3,6 and 9 also.

As it is a nine digit number so from (0-9) one number should be excluded.

The sum of (0-9) is 45 which is divisible by 9.if we exclude one digit except 9 it will be never divisible by 9.But if we remove 9 then its ok.

So we have to select the $14$ numbers from above combinations where 9 is not present i.e.

$120,320,520,720,160,360,560,760,280,480,680,240,640,840$

The rest six numbers will have their own permutations i.e. $6!=720$

So total numbers are $720 \times 14=10080$

No digits appear more than once : the number consists of all except one of the digits 0 through 9.

Divisible by 9 : this requires that the sum of digits is a multiple of 9. Since $0 + 1 + \cdots + 9 = 45$ , this implies that the missing digit is either 0 or 9.

Divisible by 2 and 5 : this requires that the last digit is zero. Thus the missing digit is 9; the number has the form $10\cdot a$ , where $a$ is a permutation of digits 1 through 8.

Divisible by 8 : this is true iff $a$ is a multiple of 4. Divisibility by four depends only on the last two digits. There are 14 possible combinations: $12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96$ .

Divisible by 3, 4, 6 is implied by the previous choices.

First six digits can be any permutation of the six unused digits. This gives $6! = 720$ possible choices. Thus we have $(\text{choices for first six digits})\times(\text{choices for next two digits}) = 720 \cdot 14 = \boxed{10\,080}$ as our final answer.

Excellent Problem