## Task A-1.5.

How many numbers at least need to be removed from the set $\{1,2,3, \ldots, 2020\}$ so that the resulting set does not contain the product of any two of its distinct elements?
## Solution.

The correct answer is 44.

1 point

We need to prove that at least 44 numbers must be removed from the given set. First, the number 1 must be removed from the set, because otherwise, with any other element $a$ from the set, it would form the product $1 \cdot a=a$.

Let's consider the three-element sets

$$
\{2,87,174\},\{3,86,258\}, \ldots,\{44,45,1980\}
$$

or sets of the form $\{k, 89-k, k \cdot(89-k)\}$, for $k=2,3, \ldots, 44$.

From each of these 43 sets, we must remove at least one number, so we conclude that together with the number 1, we must remove at least 44 numbers in total.

An example of a set from which we have removed 44 numbers is the one where we have removed the numbers $1,2,3, \ldots, 44$, i.e., the set $\{45,46, \ldots, 2020\}$.

The product of any two elements of this set is at least $45 \cdot 46=2070>2020$, so it definitely does not belong to the set.

1 point

## COUNTY COMPETITION IN MATHEMATICS

## 2nd grade - high school - A variant

4th March 2020.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.