## PROBLEM 44. COWS AND SHEEP

Two men jointly owned $x$ cows, which they sold for $x$ dollars a head. With the money they received, they bought sheep at 12 dollars a head. Since the proceeds from the sale of the cows did not divide by 12, they used the remaining money to buy a lamb. Then they divided the flock so that each of them had the same number of animals. The man with the lamb was therefore somewhat shortchanged. To make up for it, the second man gave him his harmonica. How much is the harmonica worth?
Solution. The revenue from sales equals $x^{2}$ dollars. If $x$ were divisible by 6, then $x^{2}$ would be divisible by 36, and consequently, by 12. Since this is not the case, $x$ is not divisible by 6. In this case, $x=12 k+r$, where $|r|=1,2,3,4$ or 5. Accordingly,

$$
\frac{x^{2}}{12}=\frac{(12 k+r)^{2}}{12}=\frac{144 k^{2}+24 k r+r^{2}}{12}=12 k^{2}+2 k r+\frac{r^{2}}{12}
$$

Now, since both people have the same number of animals, the total number of animals is even, which means there is an odd number of sheep and one lamb. Therefore, the quotient of dividing the number $x^{2}$ by 12, which equals the number of sheep bought for $x^{2}$ dollars, must be an odd number. But $12 k^{2}+2 k r$ is an even number. Therefore, the number $r^{2} / 12$ must contribute an odd value to the quotient. This implies that $r^{2}$ must exceed 12, leading to $|r|=4$ or 5. For $|r|=5$ we have

$$
\frac{r^{2}}{12}=\frac{25}{12}=2+\frac{1}{12}
$$

80
giving an even number as the contribution to the quotient. Therefore, $|r|$ must be 4 and $r^{2}=16$. Thus,

$$
\frac{r^{2}}{12}=\frac{16}{12}=1+\frac{4}{12}
$$

and the remainder is 4 (not -4). Therefore, the lamb costs 4 dollars. Consequently, one person had a 4-dollar lamb, and the other had a 12-dollar sheep, and the transfer of a 4-dollar harmonica would bring each of them to a level of 8 dollars.