19. In a $3 \times 3$ grid, each $1 \times 1$ small square must be filled with a number, with the number in the top-right corner already filled as 30 (as shown in the figure). The numbers to be filled must satisfy the following conditions:
(1) Each number must divide the number in the square directly above it (if there is no square directly above it, this condition does not apply); (2) Each number must divide the number in the square directly to its right (if there is no square directly to its right, this condition does not apply). The number of different ways to fill the grid is $\qquad$.
【Analysis】Considering the prime factor 2: Since a cell has 2, the cell above it and the cell to its right must also have 2. Let the three columns from top to bottom have $a, b, c$ twos, where $0 \leq a \leq b \leq c \leq 3$.
When $a=0$, $b=0$, $c$ can be $0-3$, a total of 4 cases;
When $b=1$, $c$ can be $1-3$, a total of 3 cases;
When $b=2$, $c$ can be $2-3$, a total of 2 cases;
When $b=3$, $c$ can be $3$, a total of 1 case;
Therefore, there are $4+3+2+1=10$ cases in total.

Similarly, when $a=1$, $b=1,2,3$, $c$ has a total of $1+2+3=6$ cases;
When $a=2$, $b=2,3$, $c$ has a total of $1+2=3$ cases;
When $a=3$, $b=3$, $c$ has a total of $1=6$ cases;
Therefore, there are $10+6+3+1=20$ cases in total.

Similarly, considering the prime factors 3 and 5, there are also 20 cases each.
Therefore, there are $20 \times 20 \times 20=8000$ cases in total.