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
Analytical Puzzle Solution
