55. Find four such numbers that the sums of the numbers, taken three at a time, are $22, 24, 27, 20$. (This problem was given by the Greek mathematician Diophantus, who lived in the second century AD).
55. Let the required numbers be $x, y, z, t$. Then

$$
\begin{cases}x+y+z=22 \\ x+y+t=24 & \text { Hence } x+y+z+t=31 \\ x+z+t=27 \\ y+z+t=20 & \text { Answer: } x=11, \quad y=4, \quad z=7, \quad t=9\end{cases}
$$