## Task B-3.5.

Christmas ornaments are packed in two types of boxes, red and green. In red boxes, the ornaments are packed in five rows with four ornaments each, and in green boxes, in three rows with six ornaments each. In how many different ways can we choose the number of red and green boxes to pack 2018 Christmas ornaments? There should be no unfilled boxes, and no ornament should be left unpacked.
## First solution.

Red boxes are packed with 20 balls each, and green boxes with 18 balls each. Let $x$ be the number of red boxes, and $y$ be the number of green boxes. We need to determine $x, y \in \mathbb{N}_{0}$ such that $20 x + 18 y = 2018$, or solve the equation $10 x + 9 y = 1009, x, y \in \mathbb{N}_{0}$.

One obvious solution is the ordered pair $(100, 1)$.

Therefore, $10(x - 100) + 9(y - 1) = 0$.

From this, $x - 100 = -9k, y - 1 = 10k$, for some $k \in \mathbb{N}_{0}$,

so the solutions are of the form $x = 100 - 9k, y = 10k + 1$.

Since $x$ and $y$ are from the set $\mathbb{N}_{0}$, from $100 - 9k \geqslant 0$ and $10k + 1 \geqslant 0$ it follows that $k < 11.1$ so $k \in \{0, 1, 2, 3, \ldots, 10, 11\}$.

Therefore, there are a total of 12 solutions.