Snow White has a row of 101 plaster dwarfs arranged by weight from the heaviest to the lightest in her garden, with the weight difference between any two adjacent dwarfs being the same. One day, Snow White weighed the dwarfs and found that the first, heaviest, dwarf weighed exactly $5 \mathrm{~kg}$. What surprised Snow White the most was that when she placed the 76th to 80th dwarfs on the scale, they weighed as much as the 96th to 101st dwarfs. What is the weight of the lightest dwarf?

(M. Mach)
Let's denote the difference in weights of two adjacent dwarfs as $x$. The first dwarf weighs $5 \mathrm{~kg}$, the second $5-x$, the third $5-2 x, \ldots$, the $n$-th dwarf weighs $5-(n-1) \cdot x(\mathrm{~kg})$. The sum of the weights of the 76th to 80th dwarfs is

$$
(5-75 x)+(5-76 x)+(5-77 x)+(5-78 x)+(5-79 x)=25-385 x .
$$

The sum of the weights of the 96th to 101st dwarfs is

$$
(5-95 x)+(5-96 x)+(5-97 x)+(5-98 x)+(5-99 x)+(5-100 x)=30-585 x .
$$

We thus get the equation from which we can solve for $x$:

$$
\begin{aligned}
25-385 x & =30-585 x, \\
200 x & =5, \\
x & =0.025(\mathrm{~kg}) .
\end{aligned}
$$

The lightest dwarf weighs $5-100 \cdot 0.025=2.5(\mathrm{~kg})$.

Another solution. Let's denote the difference in weights of two adjacent dwarfs as $x$. The first dwarf weighs $5 \mathrm{~kg}$, the second $5-x$, the third $5-2 x, \ldots$, the 101st dwarf weighs $5-100 x(\mathrm{~kg})$.

The difference in weight between the 76th and 96th dwarfs is $20x$. The same difference applies between the 77th and 97th, 78th and 98th, 79th and 99th, and 80th and 100th dwarfs. Therefore, the total weight of the 76th to 80th dwarfs is $100 x$ greater than the total weight of the 96th to 100th dwarfs. For the 76th to 80th dwarfs to weigh the same as the 96th to 101st dwarfs, the weight of the 101st dwarf must be equal to $100 x$.

We thus get the equation from which we can easily solve for $x$:

$$
\begin{aligned}
5-100 x & =100 x, \\
200 x & =5, \\
x & =0.025(\mathrm{~kg}) .
\end{aligned}
$$

The lightest dwarf weighs $100 \cdot 0.025=2.5(\mathrm{~kg})$.

Evaluation. 4 points for setting up the equation that allows the calculation of the difference in weights of two adjacent dwarfs; 2 points for the equation manipulation and the result of the problem.

Note. In neither of the given solutions does the contestant necessarily have to calculate that $x=0.025 \mathrm{~kg}$. It suffices, for example, to determine that $100 x=2.5 \mathrm{~kg}$.