A farmer sold his house geese. By the end of the market, he sold exactly one-tenth as many pieces as the price per piece he set. He divided the revenue between his two sons. Starting with the elder, the boys alternately received one hundred forints each, until finally, only a few ten-forint coins were left for the younger one. Then the father gave him his pocket knife and said that this way both of them received the same value. How much was the knife worth?
The number of hares sold - as a quantity - is an integer, let it be $x$. Then the price of one hare was $10 x$ Ft, and the total revenue was $10 x^{2} \mathrm{Ft}$. In the story, there is no mention of values smaller than $10 \mathrm{Ft}$, only multiples of this (integer) value, so - considering $10 \mathrm{Ft}$ as 1 unit of currency - the amount of money collected is $x^{2}$, a square number, and the value of the knife is the amount of money that rounds up the last digit of the square number to 10 (at most 8 units of currency, because the younger boy received at least 2 units of currency last).

The number of hundred-forint notes (ten-unit notes) distributed is odd, because the last one was received by the same person who received the first one, so the digit in the tens place of $x^{2}$ is odd. From this, we can determine the last digit of $x^{2}$. Let $x=10 y+z$, where $y, z$ are integers and $z<10$, then $x^{2}=100 y^{2}+20 y z+z^{2}=20 y(5 y+z)+z^{2}$. The first term is an even multiple of 10, so the tens digit of $z^{2}$ is odd, and the digit in the units place is the same as that of $x^{2}$. Taking the square of natural numbers less than 10, we only get an odd tens digit twice: in 16 and 36, but the last digit in both cases is 6, so the knife was worth 4 units of currency, $40 \mathrm{Ft}$.