What is the remainder when 121212........ (300 digits) is divided by 99 ?

To find the remainder of a number N modulo m, we can express N as a sum of place-value terms and reduce each term modulo m separately. Here, since 102 = 100 = 99 + 1, we get 102 ≡ 1 (mod 99), so every even power of 10 is also ≡ 1 (mod 99).

1. The 300-digit number consists of the two-digit block "12" repeated 150 times (300 ÷ 2 = 150), since each repetition contributes exactly two digits.

2. Write N as a sum of place-value terms: N = 12 + 12×102 + 12×104 + … + 12×10298, i.e. 150 terms, each equal to 12 times an even power of 10.

3. Since 102 ≡ 1 (mod 99), every even power of 10 (100, 102, 104, …) is also ≡ 1 (mod 99).

4. So each term 12×10(2k) ≡ 12×1 = 12 (mod 99), and summing all 150 terms gives N ≡ 150×12 (mod 99).

5. Compute 150×12 = 1800.

6. Divide 1800 by 99: 99×18 = 1782, so 1800 = 99×18 + 18, giving remainder 18.

To verify the method on a smaller case: a 4-digit number formed the same way, "1212" (2 blocks of "12"), gives 12×2 = 24 by the same formula, and indeed 1212 ÷ 99 = 12 remainder 24 (99×12 = 1188, 1212 − 1188 = 24) — confirming the block-sum-mod-99 approach. Applying the same logic to 150 blocks confirms the result above.

Hence, the remainder is 18.