# Problem 1.

A sequence of 100 non-zero numbers is written in a row. Each number, except the first and the last, is equal to the product of the two adjacent numbers. The first number is 2018. Find the last number in such a sequence.

#
# Solution.

Let's denote the hundred numbers as $a_{1}, \ldots, a_{100}$. According to the condition,

$$
a_{1}=\frac{a_{2}}{a_{3}}, \quad a_{2}=\frac{a_{3}}{a_{4}}, \ldots, \quad a_{98}=\frac{a_{99}}{a_{100}}
$$

Multiplying the first and second equations, we get

$$
a_{1} a_{2}=\frac{a_{2} a_{3}}{a_{3} a_{4}} \Rightarrow a_{1}=\frac{1}{a_{4}}
$$

Similarly, $a_{4}=\frac{1}{a_{7}}$. Therefore, $a_{1}=a_{7}$.

Reasoning in the same way further, we get

$$
a_{1}=a_{7}=\ldots=a_{97}=2018
$$

Now $a_{100}=\frac{1}{a_{97}}=\frac{1}{2018}$.

Answer: $\quad a_{100}=\frac{1}{2018}$.