4.51 Find the smallest (greater than 1) natural number such that it is at least 600 times larger than each of its prime divisors?
(Leningrad Mathematical Olympiad, 1989)
[Solution] Note that, when the natural number $m$ contains a prime factor greater than 3, then
$$
m \geqslant 5 \cdot 600=3000 \text {. }
$$

Therefore, to find the smallest natural number $m$, we should first consider numbers of the form $2^{a} \cdot 3^{b}$.
Since $2 \cdot 600=1200,3 \cdot 600=1800$, then
$$
m \geqslant 1800 \text {. }
$$

And the smallest natural number $m$ greater than 1800 that contains only the prime factors 2 and 3 is $m=1944=2^{3} \cdot 3^{5}$, thus 1944 is the answer.