\section*{Problem 4 - 121014}

In an old textbook, a problem reports on the following trade:

A farmer wanted to buy several animals from a livestock dealer. The dealer demanded the same price for each animal. The farmer managed to negotiate this price down by exactly the same percentage of the demanded price as it was supposed to be (in groschen).

He now paid 21 groschen per animal. At the original price, his money would have been enough for exactly 3 animals. Now he could buy more animals, spending all his money.

How many animals could the farmer buy in total?
}

The original price \(p_{1}\) is reduced by \(p_{1} \%=\frac{p_{1}}{100}\). Thus, the new price \(p_{2}\) is \(p_{2}=p_{1}-p_{1} \cdot \frac{p_{1}}{100}=\) \(p_{1}-\frac{p_{1}^{2}}{100}\). Since \(p_{2}\) is 21 groschen, we have:

\[
\begin{aligned}
-\frac{p_{1}^{2}}{100}+p_{1} & =21 \\
p_{1}^{2}-100 \cdot p_{1}+2100 & =0 \\
p_{1}^{2}-100 \cdot p_{1}+2500 & =400 \\
\left(p_{1}-50\right)^{2} & =400 \\
p_{1} & =50 \pm 20
\end{aligned}
\]

Thus, the original price \(p_{1}\) was 30 or 70 groschen. With this original price, the farmer could buy exactly three animals. At 21 groschen, he can buy \(n\) animals, spending all his money each time. Therefore, \(3 \cdot p_{1}=21 \cdot n \Leftrightarrow n=\frac{p_{1}}{7}\). Since \(n\) is a natural number, \(p_{1}\) must be divisible by 7. This is true for 70, but not for 30 groschen. Therefore, the original price was 70 groschen. The farmer can now buy 10 animals.

Solutions of the I. Round 1972 adopted from [5]

\subsection*{7.14.2 II. Round 1972, Class 10}