9.6. Waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). In the process, the seller wrote down on a piece of paper the weights he used. What is the smallest number of different numbers that could have appeared in his notes, if the weight of each weight is an integer number of kilograms?
Answer: 6 numbers.

Solution. Let's check that with weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, one can weigh any of the given watermelons. Indeed, $2=1+1 ; 4=3+1 ; 6=5+1 ; 8=7+1 ; 11=10+1$; $12=9+3 ; 13=10+3 ; 14=9+5 ; 15=10+5 ; 16=9+7 ; 17=10+7 ; 18=9+9 ; 19=10+9 ;$ $20=10+10$. Thus, 6 different numbers could have been recorded.

We will show that five types of weights are insufficient for the required weighings. If there are five weights, then, generally speaking, twenty watermelons can be weighed. Specifically: five watermelons can be balanced with single weights, five with double weights, and the remaining $\frac{5 \cdot 4}{2}=10$ watermelons with pairs of different weights. However, each combination of weights must be used exactly once.

Notice that half of the watermelons have an odd mass. Let $k$ be the number of weights with an odd mass, and $(5-k)$ be the number of weights with an even mass. Then the number of ways to weigh a watermelon with an odd mass is exactly $k+k(5-k)=6k-k^2$. However, for no $k=0 ; 1 ; 2 ; 3 ; 4 ; 5$ does this expression equal 10 (this can be verified either by substitution or by solving the quadratic equation $6k-k^2=10$).

Therefore, the seller could not have recorded fewer than 6 numbers.

## Grading criteria:

+ a complete and well-reasoned solution is provided

$\pm$ a generally correct solution is provided, containing minor gaps or inaccuracies

$\pm$ it is proven that fewer than six numbers could not have been recorded, a correct set of six weights is indicated, but it is not shown how exactly to weigh the watermelons

dry it is only proven that five weights are insufficient

Ғ it is not proven that five weights are insufficient, but a correct set of six weights is provided (regardless of whether it is shown how exactly to weigh the watermelons)

- only the answer is provided
- the problem is not solved or is solved incorrectly