How many four-digit numbers whose digits add up to 9 are divisible by 11?

I'm not too sure about this, but I don't believe any such numbers exist......here's my reasoning :

Let the 4 digits  be a , b, c and d    

And assume that each digit, except a, can only have a possible range of 0 - 9, inclusive  [a only ranges from 1 - 9, inclusive]

So we have  that

a + b + c + d   = 9    

Rearrange this as   (a + c) + ( b + d)  = 9      (1)

And it can be shown that if  the alternating sums of a 4 digit number equal some multiple of 11 (or, 0)...then that number is divisible by 11

For instance   1331.....we have  1 - 3 + 3 - 1  =  -2 + 2 =  0     and 1331  =  11^3

Or, for instance,  9141    .....we have   9 -1 + 4 - 1  =  8 + 3  = 11   and 9141  = 831 * 11

So.... our 4 digit number is divisible by 11  if    ( a - b) + (c - d ) = 11n    where n is an integer

Rearranging this, we have     (a + c) - (b + d)  = 11n        (2)

Adding (1)  and (2)   we have that

2(a + c)  =  11n + 9 

Now......the left side is even, so n must be odd...now, let n = 1.....then

2(a + c)  = 11(1) + 9

2 (a + c)  = 20

a + c  = 10

But this means that, by  (1),  10 + ( b + d ) = 9  → ( b + d) = -1.....but, since b, d can only range from 0-9, this is impossible

And we can clearly see that if n is any odd integer > 1, then a + c > 10  which by (1), makes (b + d) negative. And this violates our assumption that b, d can only range for 0-9, inclusive.

And n cannot be any negative odd, either....for instance, if n = -1

2(a + c)  = 11(-1) + 9

2(a + c)  = -2

a + c = -1    and this is also impossible, by our assumption

And for any negative odd > 1, (a + c) < -1 which also violates our assumption

Thus, no "n" exists that makes (1) and (2) both true, so there is no 4 digit number that will sum to 9 and also be divisible by 11

P.S. - I'd like some other mathematician(s) to look at this.......!!!!