12. There are 94 pieces of $4 \times 10 \times 19$ rectangular bricks, to build a tower consisting of 94 bricks (one brick placed on top of another), the number of different possible heights for the tower is $\qquad$
12. 465

Let the edges of the tower with lengths $19 \times 10 \times 4$ be $x, y, 94-x-y$ blocks respectively. The height of the tower is $h$.
$$
\begin{aligned}
h & =19 x + 10 y + 4 \times (94 - x - y) \\
& = 376 + 15 x + 6 y \quad \text{where} \quad x + y \leq 94.
\end{aligned}
$$

Since $19 - 10 = 9$, $10 - 4 = 6$, and $9 \times 2 = 6 \times 3$, we have $19 \times 2 + 4 \times 3 = 10 \times 5$.
Therefore, we can assume $y < 5$.
Now consider whether different $(x, y)$ pairs result in different tower heights $(y < 5)$.
Let $y_1 < 5$, $y_2 < 5$, $x_1 + y_1 \leq 94$, $x_2 + y_2 \leq 94$,
and $376 + 15 x_1 + 6 y_1 = 376 + 15 x_2 + 6 y_2 \Rightarrow 15(x_1 - x_2) = 6(y_2 - y_1)$
$\Rightarrow 5(x_1 - x_2) = 2(y_2 - y_1)$, so $5(y_2 - y_1)$, thus $y_2 = y_1$, $x_1 = x_2$.
Therefore, the number of different tower heights is the number of $(x, y)$ pairs.
If $y = i$, then $x + y \leq 94 \Rightarrow x \leq 94 - i$, for $i = 0, 1, 2, 3, 4$, the number of $(x, i)$ pairs is $94 - i + 1$.
The total number of different tower heights is $95 + 94 + 93 + 92 + 91 = 465$.