What is the remainder when the 400-digit number formed by repeating the…

A number built by repeating a fixed block of digits many times over is really that block's value added to itself at every place shift — once for each repetition — so its size can be written as the block's value multiplied by a sum of successive powers of ten.

To find the remainder such a number leaves on division by some fixed modulus, first find what a single one of those repeating powers of ten itself leaves as a remainder. If that basic power already leaves a remainder of exactly one, every one of the equally-spaced powers in the sum behaves the same way, so the whole sum simply reduces to the number of repetitions.

1. The number has 400 digits and repeats the 4-digit block 1234, so it is made of exactly 100 copies of that block (400 ÷ 4 = 100).

2. Writing the number as a sum: N = 1234 × (1 + 104 + 108 + … + 10396) — one power of ten for each of the 100 positions where a copy of the block begins.

3. First reduce the basic power: 104 = 10000, and 10000 = 909 × 11 + 1, so 104 leaves remainder 1 on division by 909.

4. Because the basic power already leaves remainder 1, every other power in the sum (108, 1012, and so on, all the way to 10396) also leaves remainder 1 — so the sum of all 100 such powers leaves remainder 100 on division by 909.

5. So N leaves the same remainder as 1234 × 100 = 123400 on division by 909.

6. Dividing 123400 by 909: 909 × 135 = 122715, and 123400 − 122715 = 685 — the remainder.

As an independent check, reduce 1234 by 909 first: 1234 − 909 = 325. Then 325 × 100 = 32500, and 32500 − (909 × 35 = 31815) = 685 — the same remainder.

So the remainder is 685.