Task 1. The set $M$ consists of all prime divisors of the number 2310. How many numbers are the product of exactly two elements of the set $M$?
Solution. The prime divisors of the number 2310 are 2, 3, 5, 7, and 11, and their product is exactly 2310. Therefore, $M=\{2,3,5,7,11\}$, and the set of pairwise products of elements from $M$ is the set

$$
A=\{6,10,14,22,15,21,33,35,55,77\}
$$

and it has 10 elements. Therefore, the desired number is 10.