64. First, two numbers 2 and 3 are written, then they are multiplied and their product, the number 6, is recorded, then the product of the two last single-digit numbers $3 \cdot 6=18$, then $1 \cdot 8=8$, then $8 \cdot 8=64$ and so on. As a result, we get the following sequence of digits:

## $2361886424 \ldots$

Which digits cannot appear in such a sequence? Which digit stands at the thousandth place?
64. As a result of analyzing the sequence of digits $236188642483261224832612248 \ldots$, we notice that starting from the tenth digit, the group of digits 48326122 repeats (as in the previous problem, let's call it the period). Before the first period, there are 9 digits, and the period itself consists of 8 digits. Among these digits, there are no $5, 7, 9$, and 0. Therefore, these digits will be absent in the sequence, no matter how long it is extended. To find the digit in the thousandth place, since there are 9 digits before the period, there will be 991 digits in the first and subsequent periods up to the desired digit $(1000-9=991)$. Since the period consists of 8 digits, there will be 123 complete periods plus 7 more digits of the 124th period $(991=8 \cdot 123+7)$. The seventh digit of the period is 2. Therefore, the digit in the thousandth place in the sequence is 2.