## Task 1 - 190511

(A historical problem, 2000 years B.C.)

In a cage, rabbits and pheasants are locked up. These animals together have 40 heads and 104 feet.

State the number of all rabbits and the number of all pheasants that are in the cage!
Let the number of rabbits be $k$, and the number of pheasants be $f$.

Then $k+f=40$ and $4 \cdot k+2 \cdot f=104$ since they have 4 and 2 legs respectively.

Subtracting the first equation twice from the second gives $2 k=24$, so $k=12$ and thus $f=28$.