Brainstorm

Let's begin.

For ease of understanding, let the 200 digits on the required number be termed as 200 boxes to be occupied by the numbers 1,2, 3.. where one box can accommodate only one digit.

Let's consider the case for the numbers 1 to 9:

1 occupies the 1st box from left.

The last box that 2 occupies is 3(=1+2)rd from left.

The last box that 3 occupies is 6(=1+2+3)th from the left. And so on.

Thus, from the pattern we can conclude that the last box occupied by the number n will be the 0.5 x n(n+1)-th box from left. [Since, sum of n natural numbers = 0.5 x n(n+1)]

Hence, 0.5 x n(n+1)-th box from left will have n as the digit. From this we get 45th box from left to be occupied by 9. The 46th and 47th boxes will have 1 and 0 respectively.

Now comes the tricky part with 2-digit numbers.

Last box occupied by is 2-digit number is actually occupied only by the digit in it's unit's place. Similarly first box occupied by a 2-digit number is occupied by the ten's place of the number and not the whole number itself. I will be using these words from now on, try not to get confused.

Since we know that every 2-digit number takes up 2 boxes on the required number, we modify the formula to be: n(n+1)-th box from left is the last box occupied by n and has unit's place of n in it.

But this formula is not consistent with the first 9 numbers as they are 1-digit numbers. These numbers are being filled in the boxes twice as many times as is required. So, we subtract 45 [Since 45=0.5 x 9(9+1)].

So, for a 2-digit number, the last box occupied by it will be n(n+1)-45 th box and will contain the unit's place of n.

Now, we need to find n such that the last box occupied by it comes after the 200th box and the first box occupied by it comes before the 200th box,i.e.,

last box occupied by n: n(n+1)-45>200 and

first box occupied by n: (n-1)n-45 +1<200

Solving these, we get n=16. and the first box occupied by the number 16 as the 196th box. Therefore, the 196th box has the digit 1. The 197th has 6.. the 200th has the digit 1.

So our number is 122333.......1516161. Now for checking divisibility by 8, we check the divisibility of the last 3 digits, i.e., 161. 161=8(20)+1

Hence, the remainder is 1.

I know it's long and might be confusing but that's the best I can do to explain this.

The number on expanding will be like

1223334444..........15...1516161 so the last three digit which is required for the divisibility of 8 is 161

Which on division gives 1 s remainder

Here are 200 digit😃: 12233344445555566666677777778888888899999999910101010101010101010111111111111111111111112121212121212121212121213131313131313131313131313141414141414141414141414141415151515151515151515151515151516161