Its sufficient to prove the thing for a_i cuz q_i=5a_i mod 9. So summibg mod 9 u just get a 5 in front which u can divide by. Also the remainders are periodic of sequence 2 4 6 8 1 3 5 7 0 Summing these will give the only sum attainaible as it has period 9 which is 36 which 9 divides. I am not sure what you mean by remainder, so i assume you simply mean a_i/11

Observe that all the $a_n$ can be written as $\frac{10^{2n}-1}{9}$ and $$10^{2n}-1 \equiv -1^{2n}-1 \equiv 1-1 \equiv 0 \mod 11$$ Basically all the $a_n$ is divisible by 11, Hence $q_i=q_{i+1}=...=q_{i+8}=0$ , $\sum^{i+8}_{i=1}q_i=0$ Which is a multiple of 9

Iora wrote: Observe that all the $a_n$ can be written as $\frac{10^{2n}-1}{9}$ and $$10^{2n}-1 \equiv -1^{2n}-1 \equiv 1-1 \equiv 0 \mod 11$$ Basically all the $a_n$ is divisible by 11, Hence $q_i=q_{i+1}=...=q_{i+8}=0$ , $\sum^{i+8}_{i=1}q_i=0$ Which is a multiple of 9 I think the problem means q_i=a_i/11, and i think we disregard the part "the remainder of the division of a_i by 11", and sum of 8 of those. Note 11/11=1=10^0, 1111/11=101=10^2+10^0, in general 1...1 with 2k digits is 10^0+10^2+...+10^(2k-2). Modulo 9, the numbers 11, 1111, ..., give 1, 1x2, 1x3, ..., 1x8, 1x9=0, 1, 1x2, ..., because the number of powers of 10 summed up defines the number of 1s there are mod 9, which is also defined by simply 1xk mod 9 for 2k digits. This means any consecutive 9 numbers will be 0+1+2+...+8 (these numbers in some cyclic ordering,) = 8(9)/2=0 mod 9, which is indeed a multiple of 9, as desired.