Let a number $N=12124124612468124680124680212460824...$ ¿What is the digit $2018$?

Question

Let a number $N=12124124612468124680124680212460824...$

What is the digit $2018$?

My try

I divided the number into groups, and the last digit of the group coincides with the last digit of $2 (p-1)$ if $p$ is the number of the group.

Example

$N=\underbrace{12}_2|\underbrace{124}_3|\underbrace{1246}_4|\underbrace{12468}_5|\underbrace{124680}_6|\underbrace{1246802}_7|\underbrace{12460824}_8|\dots$

Group $2= 2(2-1)=2$

Group $3= 2(3-1)=4$

Group $4=2(4-1)=6$

Group $5=2(5-1)=8$

Group $6=2(6-1)=10 \implies 0$

But I don't see how to continue, any hints are appreciated.

Answer 1

Group $n \ge 2$ ends on index $n(n+1)/2 - 1$, so you have to pick the largest whole number to solve $n(n+1)/2 -1 = 2018$ then you know the group, and compute the offset in the group similarly. Once you know the offset in the group, note it repeats with a cycle of 6...

Answer 2

From my simple perspective We just need to use triangle numbers $$((n+1)(n+2)/2)-1$$. Using trial and error one can work out that $2016$ is a triangle number but one should take that as the first digit of a sequence because of the $-1$ therefore it is a 2 number across which gives us 4.