Dividing by 999

Divide $109$ by $999$ and you'll get $109$ as remainder. Divide $109109$ by $999$ and you'll get $218$ as remainder. Divide $109109109$ by $999$ and you'll get $327$ as remainder.

Notice that for every $109$ present in the number, you'll have $109n$ as its remainder when divided by $999$ where $n$ is the number of $109's$ .

Since there are $8$ $109's$ in the number, then

$109n = 109 \times 8$ = $\boxed{872}$

We can rewrite 109109109109109109109109 as 109000000000000000000000+109000000000000000000+...109 and then notice these are each 109 mod 999 because they are 109*1,000,000,000,000,000,000,000 and notice 1,000,000,000,000,000,000,000=1000^7=1^7=1 (mod 999) and this is true for all so we reduce it to 109+109+109+109+109+109+109+109=109x8=872

First you can write 109109109.... as 109.10^21 + 109.10^18 + 109.10^15 + ... + 109.

Dividing each of those terms by 999 we get 109 as reamainder, because we get this remainder 8 times 8.109 = 872.

Define $109_1 = 109, 109_2 = 109109, ...$ . We will prove a recursive relation for $109_n$ . Notice that $109_{n+1} = 1000 \times 109_{n} + 109 \implies 109_{n+1} = 999 \times 109_{n} + 109_n + 109 \implies 109_{n+1} \equiv 109_n + 109 \mod 999$ . Applying this identity recursively:

$109_8 \equiv 109_7 + 109 \equiv 109_6 + 2 \times 109 \equiv ... \equiv 109_1 + 7 \times 109 = 8 \times 109 = 872 \mod 999$