6. A $100 \times 100$ board is filled with numbers. In the bottom-left cell, there is a 0. From any other cell $V$, we consider a path from the bottom-left cell to cell $V$, where you can only move right or up, not diagonally. If you add up all the numbers you encounter along the way and then add 1 for each step you have taken, you get the number that is in cell $V$.

![](https://cdn.mathpix.com/cropped/2024_04_17_01cb006263f00c7722c1g-2.jpg?height=245&width=319&top_left_y=637&top_left_x=1528)
In the partial board shown above, the number 15 is obtained from $(0+1+3+7)+4=15$.

What is the last digit of the number in the top-right cell of the $100 \times 100$ board?
A) 1
B) 3
C) 5
D) 7
E) 9
A6. $\quad$ B) 3 We consider a route from the bottom-left square to the top-right square. This route consists of $99+99=198$ steps. The numbers on the squares along this route we call $a_{0}, a_{1}, a_{2}, \ldots, a_{198}$.

We know that $a_{0}=0, a_{1}=1$ and $a_{2}=3$. To calculate the next number, $a_{3}$, we need to add the numbers $a_{0}, a_{1}$ and $a_{2}$, and add 1 for each step we take. We find that $a_{3}=\left(a_{0}+1\right)+\left(a_{1}+1\right)+\left(a_{2}+1\right)=7$. To then calculate the next number, which is $a_{4}$, we need to take the same sum and add $a_{3}+1$. Thus, we see that $a_{4}=a_{3}+\left(a_{3}+1\right)=15$. In general, we see that

$$
a_{k+1}=\left(a_{1}+1\right)+\ldots+\left(a_{k-1}+1\right)+\left(a_{k}+1\right)=a_{k}+\left(a_{k}+1\right)=2 a_{k}+1
$$

If we only look at the last digits, we will notice that they repeat; after the first 0, we get: $1,3,7,5,1,3,7,5$, and so on. It is easy to explain why this regularity occurs. The last digit of $a_{k+1}$ depends only on the last digit of $a_{k}$. For this reason, this regularity continues up to and including $a_{198}$. Therefore, the numbers $a_{2}, a_{6}, a_{10}, a_{14}, \ldots, a_{198}$ all end in 3.