B1. Peter is bored during the lockdown. Therefore, he decides to write down numbers for a whole day. He creates a sequence of numbers that starts with 0, 1, and -1 and continues infinitely. On the line below, he writes the same sequence of numbers, but shifted one position to the right. On the third line, he writes the same sequence of numbers again, now shifted another position to the right. He then adds the three numbers vertically. (Thus, he skips the first two positions and starts with $-1 + 1 + 0$). This results in a multiple of three each time. Peter's sheet looks like this:

![](https://cdn.mathpix.com/cropped/2024_04_17_88f9acd303931df7c94dg-1.jpg?height=200&width=600&top_left_y=1553&top_left_x=731)

The first number in the top row is 0, the second number is 1, the third number is -1, and so on. Determine the 2021st number in the top row.
B1. 2020

We look at a small piece of the three rows of numbers placed one below the other.

$$
\begin{array}{cccccccc}
\ldots & a & b & c & d & \ldots & \ldots & \ldots \\
\ldots & \ldots & a & b & c & d & \ldots & \ldots \\
\ldots & \ldots & \ldots & a & b & c & d & \ldots
\end{array}
$$

In the two middle columns, we see that when we add the numbers in the column with $d$, $c$, and $b$, we get 3 more than when we add the numbers in the column with $c$, $b$, and $a$. This gives us $(d+c+b)-(c+b+a)=3$, or $d-a=3$. This means that if we move three places further in the top row, the number becomes 3 larger.

We are looking for the 2021st number in the row. Since 2021 gives a remainder of 2 when divided by 3, we only need to look at what happens to the numbers that are in positions that give a remainder of 2 when divided by 3. The second number is 1, the fifth number (three places further) is then $1+3=4$, and so on. This row looks like this:

$$
\ldots, 1, \ldots, \ldots, 4, \ldots, \ldots, 7, \ldots, \ldots, 10, \ldots, \ldots, 13, \ldots
$$

In general, we see that for a number $n$ that gives a remainder of 2 when divided by 3, the $n$-th number in the row is equal to $n-1$. The 2021st number is thus 2020.