Which player will hold the ball, passing it clockwise as per square number series There are 10 persons, having jersey number marked as 0, 1, 2,.....,9. They are standing in a circular arrangement, as shown in figure:

In a game, they are passing ball clockwise in the following way:
The game starts with the 0 marked player holding the ball.
After the end of 1st round, the ball is passed $1^2 = 1$ times,
 i.e. the ball ends up at the 1 marked player. 
At the end of 2nd round the ball is further passed $2^2 = 4$ times, so the ball ends up at the 5 marked player
At the end of 3rd round the ball is further passed $3^2 = 9$ times, i.e. the ball ends up at the 4 marked player.
Question is:- 
which player will hold the ball, at the end of 2018th round ?  
In solution, 
I expressed the square of every natural number, according to divisibility by 10.
Thus, summing them up, I found that the answer should be the 9 marked player
Isn't it correct?
 A: $9$ looks right.
You can get by a little quicker than summing 2018 squares by using one (or both) of two tricks:
First, find or derive a formula for the sum of the $n$ first squares, and plug $2018$ into that. (I don't go around remembering the formula, but I know it's a cubic polynomial, and I can derive the coefficients by fitting a cubic to the known values $f(0)=0$, $f(1)=1$, $f(2)=5$, $f(3)=14$).
Second, each group of $100$ successive rounds leave the ball at its original position -- because only the last digit of each round number matters for its movement, and after $100$ round each digit will have appeared $10$ times, so its sums cancel each other out modulo $10$.
Therefore you actually only need to sum the squares for the first $18$ rounds -- the remaining $2000$ rounds split into groups of $100$ and therefore have no net effect.
A: Maybe a typo in your answer but your approach is perfectly fine.
We have $$\sum_{i=1}^{2018} i^2 \pmod {10} \equiv 9$$ not $0$. Try to prove the fact. Hope it helps. 
