The sequence $\mathrm{Az}\left(a_{n}\right)$ is defined as follows:

$$
a_{1}=k \text { (positive integer), }
$$

$$
a_{n+1}= \begin{cases}a_{n} / 2, & \text { if } a_{n} \text { is even } \\ a_{n}+5, & \text { if } a_{n} \text { is odd. }\end{cases}
$$

For which positive integers $k$ is it true that 1 appears among the terms of the sequence $\left(a_{n}\right)$?
If $x$ is a positive integer, then $x$ is divisible by 5 if and only if $x / 2$, and $x+5$ are. This means that if $k$ is divisible by 5, then the elements of the sequence are also, so 1 does not appear among them.

Now, suppose $k$ is not divisible by 5; thus, no other element of the sequence is. The elements of the sequence are positive integers, so there is a smallest one among them: let this be $m$. $m$ is now a positive integer not divisible by 5, and it is clearly odd, otherwise $m / 2$ would be its successor and would be smaller than $m$.

Thus, the successor of $m$ is the even $m+5$, and then $(m+5) / 2$ follows, and thus by the choice of $m$,

$$
m \leqq \frac{m+5}{2}
$$

from which $m \leqq 5$.

The condition found is satisfied for two odd numbers not divisible by 5: these are 3 and 1. However, 3 cannot be the smallest element of this sequence, because if $a_{n}=3$, then $a_{n+3}=2$. This means that if $5 \nmid k$, then $m=1$, that is, 1 appears among the elements of the sequence if and only if $k$ is not divisible by 5.

Remark The solution also implies that if $5 \mid k$, then 5 appears among the elements of the sequence, in which case this is the smallest element of the sequence.