If 235723572357...... (1000 digit) is divided by 101 then what will be the…

Concept: For a number formed by repeating a fixed digit-block k times, if 10L ≡ 1 (mod m), where L is the block length, then every repetition of the block contributes the same reduced remainder regardless of its position — the whole number's remainder mod m equals (k × block remainder) mod m.

Application:

1. The number is made of 250 repetitions of the 4-digit block '2357' (1000 digits ÷ 4 digits per block = 250).

2. 100 ≡ -1 (mod 101), so 104 = 1002 ≡ (-1)2 ≡ 1 (mod 101) — every 4-digit block sits at a power-of-10 position that is congruent to 1 mod 101.

3. So every block contributes 2357 mod 101 to the total, independent of where it sits in the number.

4. 2357 mod 101 = 34 (since 101 × 23 = 2323 and 2357 − 2323 = 34).

5. Total ≡ 250 × 34 (mod 101) = 8500 (mod 101).

6. 8500 mod 101 = 16 (since 101 × 84 = 8484 and 8500 − 8484 = 16).

Cross-check: For a smaller case with only 2 repetitions ('23572357', an 8-digit number), the same rule gives 2 × 34 = 68 (mod 101), which matches a direct reduction of that shorter number — confirming the block-independence rule before scaling it up to all 250 repetitions.

Result: the remainder is 16.