11.2. In a row, 21 numbers are written sequentially: from 2000 to 2020 inclusive. Enthusiastic numerologists Vova and Dima performed the following ritual: first, Vova erased several consecutive numbers, then Dima erased several consecutive numbers, and finally, Vova erased several consecutive numbers (at each step, they erased consecutive natural numbers, without skipping over the gaps that formed). In the end, the sum of the numbers erased by Vova turned out to be exactly four times the sum of the numbers erased by Dima, and only one number remained from the sequence. Which numbers can be definitely said to have been erased by Dima?
Solution. Let $n$ be the sum of the numbers erased by Dima, then $4n$ is the sum of the numbers erased by Vova, $5n$ is the sum of the numbers erased by both, and $42210 - 5n$ is the number that remained unerased (42210 is the sum of all numbers from 2000 to 2020). According to the condition, $2000 \leq 42210 - 5n \leq 2020$, from which $8038 \leq n \leq 8042$ - it is clear that Dima erased 4 numbers, which are consecutive by condition; at the same time, $8038 = 2008 + 2009 + 2010 + 2011, \quad 8042 = 2009 + 2010 + 2011 + 2012$ and the intermediate values of $n$ are not sums of four consecutive numbers. When $n = 8038$, we get $5n = 40190, 42210 - 5n = 2020$, and when $n = 8042$, we get $5n = 40210, 42210 - 5n = 2000$ - both options are consistent, but in any case, the numbers 2009, 2010, 2011 were erased by Dima.

Answer: 2009, 2010, 2011.

Solutions to the mathematics olympiad problems for the 2019-2020 academic year

## 11th Grade