Let's determine the highest power of 3 by which the product of integers from 1 to 2000 is divisible.
In the sequence of numbers ranging from 1 to 2000, there is the number 666, which is divisible by 3; 222, which is divisible by 9; 74, which is divisible by 27; 24, which is divisible by 81; 8, which is divisible by 243; 2, which is divisible by 729. Therefore, the highest power of 3 by which the given product is divisible is:

$$
3^{666} \cdot 3^{222} \cdot 3^{74} \cdot 3^{24} \cdot 3^{8} \cdot 3^{2}=3^{996}
$$

(Manheim Emil.)

The problem was also solved by: Beck F., Détshy K., Erdős A., Freibauer E., Horovitz E., Kertész L., Petrogalli G., Schwartz E., Szabó I., Szabó K., Weisz J.