Estonija, 2010

a) Note that $3c_{i-1} - 1$ is never divisible by $3$ and if $\frac{1}{2}c_{i-1}$ is divisible by $3$, then also $c_{i-1}$ is divisible by $3$. Thus, if some term $c_k = n$ is divisible by $3$, then, up to it, only dividing by $2$ is used to build the terms (i.e., $c_i = \frac{1}{2}c_{i-1}$ for every $i$ such that $1 \le i \le k$) and, consequently, $c_0 > c_1 > \dots > c_k = n$. But this contradicts the condition $c_0 < n$. Hence every positive multiple of $3$ is pure.

b) If $n$ is not divisible by $3$, then $n = 3k + 1$ or $n = 6k + 2$ or $n = 6k + 5$. If $n = 3k + 1$, then taking $c_0 = 2k + 1$ gives $c_1 = 6k + 2$ and $c_2 = 3k + 1 = n$. Thereby $k > 0$ since $n > 1$, therefore $c_0 < n$. Hence none of such numbers $n$ is pure. If $n = 6k + 2$, then taking $c_0 = 2k + 1$ gives $c_1 = 6k + 2 = n$, whereby $c_0 < n$. Hence also none of such numbers $n$ is pure. Hence, among the positive integers $n > 1$ not divisible by $3$, only those of the form $n = 6k + 5$ can be pure.