36.43 As shown in the figure, a particle moves in the first quadrant. In the first minute, it moves from the origin to $(1,0)$, and then it continues to move back and forth in directions parallel to the $x$-axis and $y$-axis, moving 1 unit length each minute. Therefore, after 1989 minutes, the position of the particle is
(A) $(35,44)$.
(B) $(36,45)$.
(C) $(37,45)$.
(D) $(44,35)$.
(E) $(45,36)$.
(40th American High School Mathematics Examination, 1989)
[Solution] From the problem statement, we can see that when $n$ is odd, the particle reaches the point $(0, n)$ after $(n+1)^{2}-1$ minutes; when $n$ is even, the particle reaches the point $(n, 0)$ after $(n+1)^{2}-1$ minutes.
$\because 44^{2}=1936$, the particle reaches the point $(0,43)$ after 1935 minutes, and then reaches the point $(0,44)$ after 1936 minutes.
$$
1989-1936=53=44+9,
$$

Thus, the particle reaches the point $(44,44)$ after 1980 minutes, and then reaches the point $(44,35)$ after 9 more minutes, i.e., at 1989 minutes.
Therefore, the answer is $(D)$.