Remainder of repdigit 99...99 divided by 13

Analyze the behavior of repeated digits (repmultuples) under division by 13. A well-known number theory pattern shows that any single digit repeated exactly 6 times forms a number that is perfectly divisible by 7, 11, and 13 (e.g., 999999 \equiv 0 ± od{13}).

Since every block of six 9s leaves a remainder of 0, group the 99 occurrences of the digit 9 into blocks of 6: 99 \div 6 = 16 complete blocks, with a remainder of 3 individual digits.

This means the massive remainder calculation collapses down to evaluating just the last three digits: 999 ± od{13}. Execute long division: 999 = 13 × 76 + 11. The remainder left behind is exactly 11.

Answer: (a).