6. How many six-digit numbers exist in which four consecutive digits form the number $2021?$
Solution. Consider three types of six-digit numbers: $\overline{a b 2021}, \overline{a 2021 b}, \overline{2021 a b}$ (the bar denotes the decimal representation of the number, where $a, b$ - are digits).

In the first case, $\overline{a b 2021}$, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. These digits are chosen independently. Therefore, we get a total of $9 \cdot 10 = 90$ options.

In the second case, $\overline{a 2021 b}$, the reasoning is repeated, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. These digits are chosen independently. Therefore, we again get $9 \cdot 10 = 90$ options. We check that no number from the first group coincides with a number from the second group. They, in particular, differ by the digit in the hundreds place.

In the third case, $\overline{2021 a b}$, the digits $a$ and $b$ are chosen independently and arbitrarily from 0 to 9. In this case, we have 100 options. No number from the third group coincides with the previous options, they differ by the digit in the hundreds place.

The total number of options is $90 + 90 + 100 = 280$.

Answer. 280 numbers.

Leningrad Region

## All-Russian School Olympiad in Mathematics  Municipal Stage  2021-2022 academic year

7th grade

Grading Criteria

| Problem 1 | Score | For what it is given |
| :---: | :---: | :--- |
|  | 7 | Fully correct solution. A correct diagram is provided. Checking the conditions of the problem for the correct diagram is not required. |
|  |  |  |  |
|  | Incorrect solution. |  |


| Score | For what it is given |  |
| :---: | :---: | :--- |
| 7 | Complete solution. Correct groups of exchanges are obtained. |  |
| 5 | Correct solution. It is not specified how the condition that there are no whistles left after all exchanges is used. As a result, the division of exchanges into groups is not justified. |  |
|  | 3 | A sequence of exchanges is described. The uniqueness of the answer does not follow from the reasoning. |
| 1 | It is written that 50 exchanges were made. Further, it is not written or is incorrect. |  |
| 0 | Incorrect solution and (or) incorrect answer. |  |


| Problem 3 | Score | For what it is given |
| :---: | :---: | :---: |
|  | 7 | Complete solution, the correct answer is obtained. |
|  | 3 | It is noted that for a simple odd $p$, the number $n-1$ will be odd, and $p-1$ - even. Further progress is absent. |
|  | 1 | A check is provided that for $n$, being a power of two, the condition of the problem is satisfied. Proof of the statement formulated in the condition is absent. |
|  | 0 | Incorrect solution and (or) incorrect answer. |


| Score | For what it is given |
| :---: | :--- | :--- |
| 7 | Fully correct solution. A correct example is provided and it is written that all conditions are met. |
| 3 | A correct example is provided, but it is not shown that all conditions are met. Possibly, a partial check is performed (for example, the possibility of walking all paths twice). It is not proven that it is impossible to walk all paths once. |
| 1 | A suitable diagram is provided, the conditions are not checked. |
| 0 | Incorrect solution and (or) incorrect answer. |


| Score | For what it is given |
| :---: | :---: | :--- |
|  | Fully correct solution. It is proven that Petya will win with the correct strategy. The possibility of each necessary move is shown. |
|  | A winning strategy is described. There is no proof of why Petya can always make the necessary move. |
|  | A description is provided of the moves Petya and Vasya will make. It is not assumed that Vasya can choose the number of pebbles at his discretion, and Petya can adapt to his choice. |
| 0 | Incorrect solution and (or) incorrect answer. |


| Problem 6 | Score | For what it is given |
| :---: | :---: | :---: |
|  | 7 | Fully correct solution. Correct answer. |
|  | 6 | Correct solution. Correct answer. The difference from 7 points is that it does not provide a check that no two options for a six-digit number are counted twice. |
|  | 5 | Correct solution. Incorrect answer due to placing the digit 0 in the first position. |
|  | 1 | Partially correct solution, in which only one placement of the pair of unknown digits is allowed (for example, they are placed only at the end). |
|  | 0 | Incorrect solution and (or) incorrect answer that does not meet the criteria for 1 and 5 points. |