3. Misha calculated the products $1 \times 2, 2 \times 3$, $3 \times 4, \ldots, 2017 \times 2018$. For how many of them is the last digit zero?
Solution. The last digit of the product depends on the last digits of the factors. In the sequence of natural numbers, the last digits repeat every ten. In each ten, in the sequence of products, four products end in zero: ... $4 \times \ldots 5, \ldots 5 \times \ldots 6, \ldots 9 \times \ldots 0, \ldots 0 \times \ldots 1$. ... There are 201 complete tens. And at the end, there are seven more products, two of which end in zero.

Thus, a total of $4 \times 201 + 2 = 806$ products end in zero.

Criteria. Any correct solution: 7 points.

In the solution, two zeros from the last incomplete ten are lost: 4 points.

Only the correct answer is provided: 1 point.