## Task 19/63

A farmer leaves his two sons, among other things, a flock of sheep. The brothers have this flock sold by an intermediary, instructing him to sell each sheep for as many marks as there are sheep in the flock.

The intermediary brings the proceeds in ten-mark notes and a remainder of small change that does not amount to a full ten-mark note. The brothers divide the money so that both receive the same number of ten-mark notes. One ten-mark note and the small change remainder are left over. Then the older brother says to the younger: "I'll take the note, and you get the remainder and a pocket knife I just bought, then we will both have received the same amount."

How much did the pocket knife cost?
If the herd consisted of $n=10 x+y$ sheep, then the revenue was $n^{2}=(10 x+y)^{2}=100 x^{2}+20 x y+y^{2}$ Marks, where $y$ is a single-digit number. Since, apart from the small change, a 10-Mark note was left over when dividing the money, the number of 10-Mark notes must be odd.

Since the number $10 x^{2}+2 x y$ is always even, the only possible values for $y^{2}$ are 16 or 36; for $y^{2}=1 ; 4 ; 9 ; 25 ; 49 ; 64 ; 81$, the number of 10-Mark notes would be even. This means that the small change $r=6$ Marks.

If we denote the price of the pocket knife in Marks by $t$, then it must obviously hold that

$$
10-t=r+t \rightarrow 10-r=2 t=4 \rightarrow t=2
$$

Thus, the pocket knife costs 2 Marks.