2018 again

2018 divided by 4 gives remainder of 2. Second number is 0.

First, let's factor 2018 into $2\times1009$ . Since 2018 is a multiple of 2, we know that the $2018^{\text{th}}$ number in the sequence must be either 0 or 8. It is either 0 or 8 because they are always the $2n^{\text{th}}$ number in the sequence.

Because we know it is either 0 or 8, we can simplify this problem to:

What is the $1009^{\text{th}}$ number in this sequence: 080808080808...?

Since 1009 is odd, that means it has to be $\boxed{0}$ , because 0 is always the $2n+1^{\text{th}}$ number in the sequence.

In the sequence 20182018201820182018... the pattern continues by writing 2,0,1,8 in order. That means after 4 numbers the pattern will repeat. To find the 2018th number, we must group every pattern 2, 0, 1, 8 until the 2018th number. That means we must divide 2018 by 4, which gives an answer of 504 remainder 2. That means there are 504 patterns and 2 numbers until the 2018th number. The two numbers are the first two numbers in pattern:2 and 0. The 2018th digit is the last digit, so it should be 0.