Well I decided to work it out myself, so here's my solution.

So we know:

A+B+C+D is evenB&D is a multiple of 11
A&B is divisible by 9
A&C is the square of an integer


So, from thi we can infer:

A+B+C+D mod 2 = 0
B=D since any multiple of 11 less than 100 has two digits the same
A&B mod 9 = 0
B&D mod 11 = 0
sqr(A&C) is in the range 4<=A&C<=9 since A&C is a two digit number

Now, since B=D and A+B+C+D is even

B+D is even, therefore
If B is odd, D is odd
If B is even, D is even

This implies that:

A+C is even, therefore
If A is odd, C is odd
If A is even, C is even

since the sum of an odd and an even number will always be odd.
The sum of A+C and B+D must be even according to the stated conditions
so

(A+C) mod 2 = (B+D) mod 2


Now, since A&C is in the set {16, 25, 36, 49, 64, 81} and we know that A+C is even,
we can infer that A&C cannot be in the set {25, 36, 49, 81} since for those values of
A&C, A+C is odd. This contradicts the logic we established just now.

This means that A&C must be in the set {16, 64}
So A=1, C=6
OR
A=6, C=4

Now, we know that A&B is a factor of 9.

Therefore A&B must be in the range
10<=A&B<=19 if A=1
OR
60<=A&B<=69 if A=6

If A=1 and (A&B mod 9)=0 then
A&B must be 18

If A=6 and (A&B mod 9)=0 then
A&B must be 63

So we have either:

(1) A=1, B=8, C=6

OR

(2) A=6, B=3, C=4


Now, we know B=D and (A+B+C+D mod 2)=0 (A+B+C+D is even)

CASE (1):
A=1, B=8, C=6, D=8
A+B+C+D=23 ODD
So we reject case (1)

CASE (2):
A=6, B=3, C=4, D=3
A+B+C+D=16 EVEN
So we accept case (2)

Therefore the only possible 4 digit number to satisfy the stated conditions is:

6343

[Edited by HarryW on 04-13-2000 at 08:26 PM]