Which player will hold the ball, passing it clockwise as per square number series

Question

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?

Answer 1

$9$ 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.

Answer 2

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.