# 8. Solution.

a) Suppose there are 9 numbers in the set. Then five of them do not exceed the median, which is the number 2. Another four numbers do not exceed the number 13. Therefore, the sum of all numbers in the set does not exceed

$$
5 \cdot 2 + 4 \cdot 13 = 62
$$

Since the arithmetic mean is 7, the sum of the numbers in the set is $7 \cdot 9 = 63$. This is a contradiction. The set cannot consist of 9 numbers.

b) Let the number of numbers in the set be $2n + 1$ ($n$ is a natural number). In the set, there will be exactly $n + 1$ numbers, each of which does not exceed the median, which is the number 2. The remaining $n$ numbers do not exceed the number 13. Therefore, the sum of all numbers in the set does not exceed

$$
13n + 2(n + 1) = 15n + 2
$$

On the other hand, this sum is equal to $7(2n + 1) = 14n + 7$. From the inequality $14n + 7 \leq 15n + 2$, we get that $n \geq 5$. Therefore, the set contains no fewer than $2 \cdot 5 + 1 = 11$ numbers. To show that this is possible, consider the following example. The set

$$
2; 2; 2; 2; 2; 2; 13; 13; 13; 13; 13
$$

consists of 11 numbers and satisfies conditions 1-4.
Answer: a) no; b) 11.

Scoring criteria

| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| The correct estimate of the number of numbers in part (b) is found, but no example is given | 2 points |
| Part (a) is solved correctly | 1 point |
| The solution is incorrect or missing (in particular, only the answer is given) | 0 points |