What is the remainder when 1212121212............ (upto 300 digits) is divided…

Concept: When a k-digit block is repeated to build a larger number, and 10 to the power k leaves remainder 1 when divided by m, each repetition of the block contributes the same remainder modulo m as a single block. So the remainder of the full number modulo m equals (value of one block times the number of repetitions), reduced modulo m.

Application: This number is formed by repeating the two-digit block '12' to reach 300 digits, so it consists of 150 repetitions of '12'.

1. Since a two-digit block is being repeated, check 10 squared mod 99: 100 mod 99 = 1, so shifting by two digits leaves the remainder unchanged modulo 99.

2. Because 10 squared is congruent to 1 (mod 99), the remainder contributed by each repetition of '12' is just 12 itself, so the combined contribution of all 150 repetitions is 12 times 150 = 1800.

3. Reduce 1800 modulo 99: 99 times 18 = 1782, and 1800 minus 1782 = 18, so the number leaves remainder 18 when divided by 99.

Cross-check: Verify the pattern on a smaller case -- '1212' (two repetitions of '12') divided by 99 gives 99 times 12 = 1188, remainder 24, which matches 12 times 2 = 24 mod 99. Scaling the same relationship up to 150 repetitions confirms the remainder for the full 300-digit number.

Reference working (handwritten):