**3**
First, notice that 111111 is divisible by 13 (13×8547). So, 11111100 is also divisible by 13 (13×854700), and so on. Also, the sum of any multiples of 13 will also be divisible by 13, which includes any string of 6n 1s (where n is any integer), and obviously those numbers followed by any number of 0s.
Thus, the number formed by 24 1s is divisible by 13:
111111111111111111111111
as is the number formed by 24 1s followed by 26 0s:
11111111111111111111111100000000000000000000000000.
Finally, their sum is a multiple of three:
11111111111111111111111100111111111111111111111111
and if you replace the middle two digits with 13 (literally, add 13000000000000000000000000 to the number), the result will, again, still be a multiple of 13.
Thus, 11111111111111111111111113111111111111111111111111 is divisible by 13, and the digit in question is 3. |