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How many six digit integers can be formed so that the number formed by the first, second and fourth digits (counting from left) as well as the other number formed by the third, fifth and sixth digits is divisible by 11? It is required that the third and fourth digits are different.

We assume that the 6-digit integer is: abcdef 11|abd and 11|cef where, c is not equal to d.
Number of such 3-digit integer divisible by 11 is: (90-10+1)=81
If abd={110, 220, 330, 440, 550, 660, 770, 880, 990} where d=0, c must be different from d each time. For each of this abd, there are 81 choices for cef. Total=81×9
Now, if abd={121, 231, 341, 451, 561, 671, 781, 891} where d=1, all integers starting with 1 (where c=1) will be rejected. 9 integers divisible by 11 start with 1. For each of these integers, there are (81-9) or, 72 choices for cef. Total=72×8
For {132, 242, 352, 462, 572, 682, 792, 902}, integers started with 2 are rejected. Total=72×8
For {143, 253, 363,..., 913}, integers started with 3 are rejected. Total=72×8
For {154, 264, 374,..., 924}, integers started with 4 are rejected. Total=72×8
For {209, 319, 429,..., 979}, integers started with 9 are rejected. Total=72×8.
So, total number of such integers=81×9+72×8×9 =9 (81+8×72)=9 (81+576)=9×657 = 5913