If the 1000-digit number formed by repeating the block '2357' (235723572357…)…

When a number is built by repeating a fixed-length block, its remainder on division by 101 can be found using the identity 104 ≡ 1 (mod 101) — because 102 ≡ -1 (mod 101), squaring gives 104 ≡ 1 (mod 101). This means every group of 4 digits contributes to the remainder independently: the number's remainder mod 101 equals (value of the repeating block mod 101 × number of repetitions) mod 101, whenever the block itself is exactly 4 digits long.

Here, the block '2357' has 4 digits, and the 1000-digit number is this block repeated 250 times (1000 ÷ 4 = 250). Apply the identity step by step:

1. Find the remainder of the block itself: 2357 ÷ 101 gives 101 × 23 = 2323, so 2357 leaves remainder 34 (2357 − 2323 = 34), i.e., 2357 ≡ 34 (mod 101).

2. Since 104 ≡ 1 (mod 101), each of the 250 repeated blocks contributes its own remainder, so the full number's remainder is (250 × 34) mod 101 = 8500 mod 101.

3. Divide 8500 by 101: 101 × 84 = 8484, so 8500 − 8484 = 16. Hence the number ≡ 16 (mod 101).

Cross-check by reducing 250 modulo 101 first: 250 mod 101 = 48 (250 − 2×101 = 48); then 48 × 34 = 1632, and 1632 mod 101 = 16 (101 × 16 = 1616, 1632 − 1616 = 16) — the same result, confirming the answer.

Therefore, the remainder is 16.