http://olympiads.mccme.ru/vmo/

There are 3 multiples of 6, so there has to be two numbers of the form 3x2^k. However this doesn't leave that many possibilities, the string of 18 must include either 6 and 12 or it must include 12 and 24. Both these possibilities are ruled out because 17 is prime.

In fact, the set of the first two multiples of $6$ can only be $\{6,12\}, \{12,18\}, \{18,24\}, \{48,54\}$, therefore the three multiples of $6$ can only be $\{6,12,18\}, \{12,18,24\}$. In both cases the string of $18$ includes $17$.