5. While waiting for customers, the 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). The seller recorded on a piece of paper the weights of the weights he used. What is the smallest number of different numbers that could have appeared in his records, if the weight of each weight is an integer number of kilograms?
Answer: 6.

Solution. With one or two weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, any of the given watermelons can be weighed. 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, six 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 $5 \cdot 4: 2=10$ 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 $k$ of the five weights have an odd mass, and $5-k$ have 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 $6k-k^2=10$).