On a farm - where there are more horses than ducks - the number of cows is one third of the total number of horses and ducks. The sum of the heads and legs of ducks and horses is 100. How many cows are on the farm?
I. Solution: Let the number of horses be $x$, the number of ducks $y$, and the number of cows $z(x>z)$.

According to the problem,

$$
z=\frac{x+y}{3}
$$

and

$$
x+4 x+y+2 y=100
$$

Rearranging equation (2) and using the value of (1), we get

$$
2 x+3(x+y)=2 x+9 z=100
$$

from which

$$
x=\frac{100-9 z}{2}=50-4 z-\frac{z}{2}
$$

Since $x$ must be an integer, $z$ must be divisible by 2, so with the notation $z=2 u$ ($u$ is an integer),

$$
x=50-9 u
$$

From equation (1),

$$
y=15 u-50
$$

$y$ must be positive, so

$$
\begin{gathered}
15 u-50>0 \\
u>\frac{50}{15}>3
\end{gathered}
$$

Since $x$ is also positive, from equation (3),

$$
\begin{gathered}
9 z < 50 \\
z < \frac{50}{9} < 6
\end{gathered}
$$

Since $z$ is an even number, $z$ can be 2, 4, or 6. However, $z$ must be less than 6, so $z$ can only be 2 or 4. If $z=2$, then $u=1$, and from equation (3), $x=41$, which does not satisfy the condition $x>z$. If $z=4$, then $u=2$, and from equation (3), $x=32$, which also does not satisfy the condition $x>z$. Therefore, the only possible value for $z$ is 6, which gives $u=3$, and from equation (3), $x=14$. Substituting $z=6$ and $x=14$ into equation (1), we get $y=10$. This satisfies the condition $x>z$ and the equation $2x-1=27, x=14$, and thus we have the same solution as before: 14 horses, 10 ducks, and 8 cows on the farm.

Kálmán Györy (Ózd, József A. g. III. o. t.)