# Problem solving approach in this case (about series and combinatorics maybe)

Suppose there is $10$ persons sitting around a round table consisting $10$ seats ofc. They have a ball.

• In the first run the ball ends up to the $1^{1} = 1$st position of the table.
• In the second run the ball travels a more $2^{2} = 4$ persons and thus ends up in the $5$th position of the table.
• In the third run the ball travels another $3^{3}$ persons and ends up in the $2$nd position of the table.

What will be the position in which the ball ends up if it continues to travel for n time thus what will be it's position after completing the nth run ? ( The main question says $n = 2017$ ).

My approches:

1. If I can find the summation of the series $$\sum_{1}^{n}n^{n}$$ then simply write it as $\left(10n + x\right)$ where n is any integer number and $x$ is a single digit, then $x$ should be my answer. But I was unable to find any equation that gives me the summation of the above series. ( not even in the internet, though I didn't search rigorously ).
2. Maybe I should find some pattern here and then solve it. But I failed. ( though I believe that this may be the way to solving this problem ).
• I don't know. I have not been able to solve this and didn't search for the answer. Though I really want to know how to solve it. – Sami Jan 10 '18 at 13:46

Note that the last digits of the number $n^n$ for $n =1$ to $10$ are: $$1,4,7,6,5,6,3,6,9,0$$ with a sum of $47$.
Note that this pattern repeats again from $11-20$, $21-30$ and so on.
Now, the answer is the sum of this sequence upto $2017$ terms modulo $10$. Why? This is because we want the final position of the ball among the people in the round table who are numbered from $1$ to $10$.
The answer is thus: $$\left(47 \times 201 +32 \right) \bmod 10 =9$$ as the ten terms with the sum of $47$ is added $201$ times and the remaining seven terms have a sum of $32$.