## Task B-1.2.

If one of the numbers is subtracted from the sum of 10 consecutive odd natural numbers, the result is 2021. What are these numbers and which number was subtracted?
## Solution.

The sum of 10 consecutive odd natural numbers is

$$
\begin{aligned}
(2 x+1)+(2 x+3)+ & (2 x+5)+(2 x+7)+(2 x+9)+(2 x+11)+(2 x+13) \\
& +(2 x+15)+(2 x+17)+(2 x+19)=2021+(2 x+a)
\end{aligned}
$$

where $a \in\{1,3,5,7,9,11,13,15,17,19\}$.

It follows that

$$
20 x+(1+3+5+7+9+11+13+15+17+19)=2021+2 x+a
$$

or

$$
\begin{gathered}
20 x+5 \cdot 20=2021+2 x+a \\
18 x=1921+a
\end{gathered}
$$

so

$$
x=\frac{1921+a}{18}
$$

which can be written in the form

$$
x=106+\frac{13+a}{18}
$$

The number 18 must be a divisor of the number (13+a), and since $a \in\{1,3,5,7,9,11,13,15,17,19\}$, this is only possible for $a=5$, so $x=107$.

The desired numbers are 215, 217, 219, 221, 223, 225, 227, 229, 231, and 233.

We subtracted the number 219.

Note: A student can arrive at the correct solution by estimating and adjusting the numbers around the arithmetic mean of all 10 or 9 given numbers. The arithmetic mean can be estimated in different ways. For example, they might conclude that the sum of all ten numbers is approximately 2223 (2021 : 10 - 11), so the arithmetic mean is an even number around 222, and then adjust the numbers accordingly. For such a procedure with the correct solution, but without an argument for why it is the only solution, the student gets 4 points. If there is no procedure at all, but the correct solution is given, the student gets 2 points.