Shiny Side of the Knife

In counting out $10$ 's of rupees, the elder brother's will be the 'odd' $10$ 's, i.e., the first, third, etc., and the younger brother's will be the 'even' ones, i.e., second, fourth, etc.. So since it is on one of the younger brother's 'turns' that the rupees run out, the 'tens' digit of the total number of rupees must be odd.

Now with $N$ sheep at $N$ rupees apiece the brothers will receive $N^{2}$ rupees to distribute between themselves. Let the last two digits of $N$ be $ab$ . When we calculate $N^{2}$ , the 'ones' digit will be $b^{2} \mod 10$ , and the 'tens' digit will be $(2ab + m) \mod 10$ where $m$ is the 'tens' digit of $b^{2}$ . So to have a 'tens' digit of $N^{2}$ be odd we need to have the 'tens' digit of $b^{2}$ be odd. This is only the case when either $b = 4$ or $b = 6$ , and in both cases $b^{2} \mod 10$ is equal to $6$ . Thus the younger brother took $6$ rupees on his last turn, leaving him with $4$ rupees less than his elder brother.

So to even things out, the older brother must transfer $2$ rupees of his "wealth" to his younger brother to ensure that each of their profits from the sale of the flock are equal. Now assuming that the knife in question belongs solely to the older brother, and given that the transfer of the knife results in the equalizing of profits realized by the two brothers, we can conclude that the knife was worth $\boxed{2}$ rupees.

@Shivam S. Gour The initial ownership of the knife is not clear. As worded, it could either belong to both brothers equally or solely to the older brother. It is only when the latter is assumed that the posted answer of $2$ is in fact correct, so it may be an idea to change the wording near the end to ".... and the elder brother gave him his knife for the sharing to be fair."

@calvin lin Sorry to bother you, but I notice that Shivam is not currently active and that this question was actually posted eight months ago. It's a good question that I believe can be un-flagged if the one word change suggested above is made.