Task 3. (12 points)

The arithmetic mean of ten different natural numbers is 20, and the arithmetic mean of any nine of these numbers is not less than 17. Find the maximum possible value of the largest of these numbers.

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# Solution.

We will order the 10 given numbers in ascending order.

According to the problem, the sum of the first 9 numbers cannot be less than $9 \cdot 17=153$.

Therefore, the 10th largest number cannot be greater than $20 \cdot 10-153=47$.

At the same time, the set 17, 17, $, 17, 47$ satisfies the condition of the problem.

Answer. 47

| Content of the criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| The answer is correct. It is shown that the maximum possible value cannot be greater than 47. The set containing 47 is not specified. | $+/ 2$ | 6 |
| All main logical steps of the solution are provided. Due to a computational error or typo, an incorrect answer is obtained. |  |  |
| The answer is correct. It is not shown that 47 is the maximum possible value. | $\mp$ | 2 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |