If 235723572357….. (a 1000-digit number formed by repeating 2357) is divided…

Concept: For a number formed by repeating a k-digit block several times, if 10k ≡ 1 (mod m), then shifting the number by one full block does not change its residue class mod m. So the remainder of the whole (very large) number modulo m equals the remainder of (number of repetitions × remainder of one block modulo m), reduced modulo m again — the periodicity of powers of 10 replaces the giant number with one small block computation.

Applying it here:

1. The repeating block is 2357, a 4-digit block, and the full number has 1000 digits, so the block repeats 1000 ÷ 4 = 250 times.

2. 104 = 10000, and 10000 = 101 × 99 + 1, so 104 ≡ 1 (mod 101) — shifting by one block leaves the residue unchanged mod 101.

3. Reduce one block: 2357 = 101 × 23 + 34, so 2357 ≡ 34 (mod 101).

4. Every block contributes the same residue 34, and there are 250 of them, so the whole number ≡ 250 × 34 (mod 101) = 8500 (mod 101).

5. 8500 = 101 × 84 + 16, so the remainder is 16.

Cross-check: The same block-residue rule on fewer repetitions matches direct division: two blocks (23572357) give 34 × 2 = 68 (mod 101), and three blocks (235723572357) give 34 × 3 = 102 ≡ 1 (mod 101) — both match dividing those shorter numbers by 101 directly, confirming the method scales correctly to 250 repetitions.

So the remainder when this 1000-digit repeating number is divided by 101 is 16.