Is there a 9 digit number that fits these requirements?

Question

Ok, the requirements are that
this 9 digit number has to use each digit
1-9 exactly once, for example it could be 192837465

ok, here's the trick. the number has to be divisible
by the amount of digits it has.
so if we remove the last digit on the right and are left with 8
digits, it has to be divisible by the number 8.

if we remove the last two digits on the right , it has to be
divisible by the number 7
and so on..

divisible meaning it'll divide and result in a whole number
without any values to the right of the decimal place

Answer 1

There is exactly one such number, and it is 381654729.

Edit: actually, on second thought, I'll just give you the link to the proof, as I'm too tired to type it out right now.

http://www.everything2.com/index.pl?node_id=1489175

Answer 2

Let the number be abcdefghi

The first one, a, can be any. Any digit is divisible by 1.

The fifth one, e, must be 5. The second one must be an even number.

b, d, f, g and h must be even, so, since there are 5 odds, a, c, e and h are odd, but not 5

The sum of the 9 digits is 9*10 / 2, which is divisible by 9, so the last one, i, can be any of the odds numbers, but not 5. The same happens to a.

The 8th number, h, has to be one so that fgh is divisible by 8

The 3rd one, c, must be one that makes that abc is divisible by 3, so, a + b + c is a multiple of 3.

The 4th number, d, must be so that the third and the fourth (cd) is divisible by 4. d must be even too because its a multiple of 4, so its a multiple of 2.

The 6th, f, must be even and the the sum of the 5 first must be a multiple of 3. If d is 4 or 8, then c must even too, and this is not possible (remember that cd is a multiple of 4).

a + b + c + d + 5 + f must be multiple of 3

But a + b + c is a multiple of 3, so d + 5 + f is a multiple of 3 too. d can be 2 or 6

If d = 2, then 2 + 5 + f = 7 + f, a multiple of 3, so that f is even. So f can be 2 (but I would be repeating the 2), or 8, so f must be 8.

If d = 6, then 6 + 5 + f = 11 + f and f must be 4

Until now our number can be

1)abc258ghi or
2) abc654ghi

1) b and h can be 4 and 6

So, we have

1.1- a4c258g6i or 1.2- a6c258g4i

2) b and h can be 2 or 8.

So, we have

2.1 a2c654g8i or a8c654g2i

Lets study these possibilities

1.1- 8g6 must be a multiple of 8, so g can be 1 or 9

1.1.1- a4c25816i
1.1.2- a4c25896i

a4c is a multiple of 3, where a and c are odd numbers.

1.1.1.1- a, c and i arent 5 or 1, so they are 3, 7 or 9
1.1.1.2- a, c and i arent 5 or 9, so they are 1, 3 or 7

You go on from here.

Ana

Answer 3

Start with the five.. you know wher it must go because for a number to be divisible by 5 it must end in 0 or 5... the digits must alternate odd, even,odd, even..by divisibility tests.

Answer 4

NO i gave up. tried several combinations.

Answer 5

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