5. A little squirrel collected 15 nuts in the forest weighing $50, 51, \ldots, 64$ grams. He knows the weight of each nut. Using a balance scale, the little squirrel tries to prove to his friends that the first nut weighs 50 g, the second 51 g, the third 52 g, and so on (initially, the friends know nothing about the weights of the nuts). What is the minimum number of weights the little squirrel will need, if both the weights and the nuts can be placed on either pan of the balance, and the number of weighings is unlimited? (The weights of the weights are known to both the little squirrel and his friends. There is an unlimited supply of weights weighing $1, 2, \ldots, 1000$ g.)
Answer. One weight.

Solution. Note that it is impossible to do without weights altogether, since if the weight of each nut is halved, the results of all weighings will not change. We will show that the baby squirrel can manage with the help of one 1 g weight. First, he will convince his friends that the nuts weigh $n, (n+1), \ldots, (n+14)$ g. For this, he will place a nut weighing 51 g on one pan of the scales, and a nut weighing 50 g and the weight on the other pan. The scales will show equality, and his friends will learn that the weighed nuts differ by exactly one gram. Then the same needs to be done with nuts weighing 52 and 51 g, and so on. Finally, with the last weighing, the baby squirrel will show his friends what $n$ is. For this, he will place nuts weighing from 50 to 57 g on one pan of the scales, and the remaining nuts and the weight on the other pan. The scales will again show equality. Thus, the friends will know that

$$
n+(n+1)+\cdots+(n+7)=(n+8)+(n+9)+\cdots+(n+14)+1
$$

Therefore, $8 n+28=7 n+78$, i.e., $n=50$.

Comment. Correct solution - 20 points. A solution with 2 weights - 2 points. A solution with more than 2 weights - 1 point. Incorrect or no solution - 0 points.