At the outset, let us consider a positive integer N having at least 4 digits.

Then, we can write N as:

N=1000p+q

Reducing both sides module 999, we have:

N= (p+q) mod 999

Rem: This is the basic structure of the well known divisibility rule of 999 and its divisors 333,111,37,27 etc.

Applying this procedure to the given problem, we have:

[2323.....23 (4018 digits) ] mod 999

= [ 2323...232(4015 digits)+323] mod 999

=[ 2323....23(4012 digits)+232+323] mod 999

=[2323....23(4012 digits)+555] mod 999

So, with each 6-digit of reduction mod 999 of the series, we get an additional term of 555 by replacement at each step.

Now, we observe that:

2018=2014+4=6*669+4

Therefore, we will reduce 6 digits in each step for a total of 669 steps until precisely 4 digits remain.

So, [2323....23(4018 digits)] mod 999

= [2323+ 669*555] mod 999

By divisibility rule of 999, we have:

2323=(2+323) mod 999=325 mod 999

and, 669*555=666*555+3*555

Now, 666*555 is divisible by 999 as:

666 is divisible by 333 and 555 is divisible by 3.

Accordingly:

(669*555) mod 999

= (3*555) mod 999

=1665 mod 999

=(1+665)mod 999

=666 mod 999

Again, we recall that:

2323 = 325 mod 999

Consequently, the required remainder = 666+325 = 991.

*Edited on ***January 5, 2022, 12:30 am**