$15 \cdot 1$ How to strike out the least number of numbers from $1,2, \cdots, 1982$ so that none of the remaining numbers equals the product of any two other numbers? Which numbers should be struck out?
(16th All-Soviet Union Mathematical Olympiad, 1982)
[Solution] Strike out the 43 numbers $2,3,4, \cdots, 44$.
At this point, except for 1, the product of any two numbers is greater than $45^{2}$, and thus greater than 1982.
If any 42 numbers are struck out, one of the following triplets must remain:
$(2,87,2 \cdot 87),(3,86,3 \cdot 86), \cdots,(t, 89-t, t(89-t)), \cdots,(44, 45,44 \cdot 45), \quad 2 \leqslant t \leqslant 44$.

Since the function $t(89-t)$ is strictly increasing when $2 \leqslant t \leqslant 44$, the 129 numbers in the above triplets are all distinct and do not exceed
$$
44 \cdot 45=1980<1982 \text {. }
$$

Therefore, 43 is the minimum number of numbers that need to be struck out.