[ Examples and counterexamples. Constructions ]  [ Partitions into pairs and groups; bijections ]  [ Decimal number system ]

A subset $X$ of the set of "two-digit" numbers $00,01, \ldots, 98,99$ is such that in any infinite sequence of digits, there are two adjacent digits that form a number in $X$. What is the smallest number of elements that $X$ can contain?
In the sequence of digits $n, m, n, m, \ldots, n, m, \ldots$, there will be two adjacent digits forming a number from $X$. Therefore, for any two digits $n$ and $m$, either the number $\overline{m n}$ is contained in $X$, or the number $\overline{n m}$ is contained in $X$. In particular, when $n=m$, we get that all numbers of the form $\overline{n n}$ are contained in $X$. Consequently, $X$ contains at least $10 +$ $(100-10) / 2 = 55$ numbers.

It remains to prove that there exists a set of 55 elements satisfying the condition of the problem. For example, the set $X=\{\overline{m n} \mid m \leq n\}$ works. Indeed, if in an infinite sequence of digits there are no two consecutive digits where the first is not greater than the second, then the digits in this sequence strictly decrease, which is impossible.

## Answer

55.00