The Mystery of the number 11

Well, $987652413$ is divisible by $11.$ A larger $9$ -digit number with distinct digits would look like $98765ABCD$ with $A,B,C,D \le 4.$ Divisibility rules say that $B-A+D-C$ should be congruent to $4$ mod $11.$ But $(B+D)-(A+C)$ is between $1-7 = -6$ and $7-1 = 6,$ so it must equal $4.$ This narrows down the possibilities for the last four digits; here's the complete list: $0213 \\ 0312 \\ 1203 \\ 1302 \\ 1324 \\ 1423 \\ 2314 \\ 2413$ Since $2413$ is the largest of these, the answer is $\fbox{987652413}.$