8.1 There is a set of weights, where the heaviest weight is 5 times the average weight of all weights. How many weights can be in the set? Provide all possible answers and prove that there are no others.
Solution. Let the average weight of $n$ weights be $x$. Then the heaviest weight is $5x$, and the weight of the remaining $n-1$ weights is $nx - 5x$. This number must be positive, so $n > 5$. For any such $n$, the situation is possible, for example, if all the other weights are equal, each weighing $\frac{x(n-5)}{n-1}$.

Answer: Any number greater than 5.

| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| Correct proof that the number of weights in the set is greater than five | 4 points |
| Examples proving that any number of weights greater than five is possible (or other proof of this fact) | 3 points |
| Examples of such sets of weights (in any number) | 1 point |
| Answer without justification | 0 points |