Divide 283 into two (positive) parts such that one part is divisible by 13 and the other by 17.

(For the latter, see the article "Integer Solutions of Linear Equations" in the current issue.)
The task can also be formulated as finding the positive integer values of $x$ and $y$ that satisfy the equation

$$
13 x + 17 y = 283
$$

$13 x$ and $17 y$ are the two sought parts.

From (1),

$$
x = \frac{283 - 17 y}{13} = 21 - y + \frac{10 - 4 y}{13} = 21 - y - u
$$

from which

$$
y = \frac{10 - 13 u}{4} = 2 - 3 u + \frac{2 - u}{4} = 2 - 3 u + v
$$

from which

$$
u = 2 - 4 v
$$

Substituting this value back into (3) and (2),

$$
y = 2 - 6 + 12 v + v = 13 v - 4
$$

and

$$
x = 21 - 13 v + 4 + 2 - 4 v = 27 - 17 v
$$

According to the problem,

$$
\begin{aligned}
& x = 27 - 17 v > 0, \text{ or } v < \frac{27}{17} \approx 1.59, \text{ and } \\
& y = 13 v - 4 > 0, \text{ or } v > \frac{4}{13} \approx 0.31
\end{aligned}
$$

from which it follows that only

$$
v = 1
$$

satisfies the problem.

Thus, $x = -17 + 27 = 10, y = 13 - 4 = 9$, so the two sought parts are

$$
13 x = 13 \cdot 10 = 130, \text{ and } 17 y = 17 \cdot 9 = 153
$$

László Farkas (Ózd, József Attila g. II. o. t.)