6. Two brothers sold a flock of sheep, receiving as many rubles for each sheep as there were sheep in the flock. Wishing to divide the proceeds equally, they began to take 10 rubles from the total sum in turns, starting with the older brother. After the older brother took 10 rubles again, the younger brother was left with less than 10 rubles. To ensure an equal division, the older brother gave the younger brother his knife. For how many rubles was the knife valued?
6. Answer: 2 rubles. Solution. Let there be $n$ sheep in the flock. Then the brothers earned $n^{2}$ rubles. From the condition, it follows that the number of tens in the number $n^{2}$ is odd. Represent the number $n$ as $10 k+m$, where $k-$ is the number of tens, and $m-$ is the number of units in it. Then $n^{2}=100 k^{2}+20 k m+m^{2}$. Thus, the oddness of the number of tens in the number $n^{2}$ is equivalent to the oddness of the number of tens in the number $m^{2}$. By checking the squares of single-digit numbers, we ensure that the number of tens is odd only for $4^{2}$ and $6^{2}$. In both of these cases, the number $n^{2}$ ends in 6, meaning the younger brother received 4 rubles less than the older brother. To ensure an equal division in this situation, the older brother should give the younger brother 2 rubles.

Grading criteria. Correct solution - 7 points. The fact about the oddness of the number of tens in $m^{2}$ is established - 2 points. In other cases - 0 points.