23. Given 5 numbers: $a_{1}=1, a_{2}=-1, a_{3}=-1, a_{4}=1, a_{5}=-1$. The following numbers are determined as follows: $a_{6}=a_{1} \cdot a_{2}, a_{7}=a_{2} \cdot a_{3}, a_{8}=a_{3} \cdot a_{4}$ and so on. What is $a_{1988}$?
23. Let's construct a table:

| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | -1 | -1 | 1 | -1 | -1 | 1 | -1 | -1 | 1 | -1 | -1 |

We notice that in the obtained sequence, the group of numbers $1, -1, -1$ repeats periodically. If there are 1988 numbers, then the number of complete periods will be $662 (1988: 3=$

![](https://cdn.mathpix.com/cropped/2024_05_21_00604dc020e3721fc1f4g-168.jpg?height=105&width=1374&top_left_y=964&top_left_x=341)
This means that the number at the 1988th position will be -1.