Cover $\{1,2,...,100\}$ with minimum number of geometric progressions? In another question, posted here by jordan, we are asked whether it is possible to cover the numbers $\{1,2,\ldots,100\}$ with $20$ geometric sequences of real numbers. Naturally, we would like to extend the question:

Problem: What is the minimum number $n$ of geometric progressions $A_1, A_2,\ldots,A_{n}$ of rational* numbers such that
$$
\{1,2,\ldots,100\} \subseteq A_1 \cup A_2\cup \ldots\cup A_{n}?
$$


In the other question, 6005 obtained a lower bound of $31 \leq n$ with an argument about square free integers. We can also obtain an upper bound of $43$ as follows. Consider these $5$ sequences: $[1, 2, 4, 8, 16, 32, 64]$ and $[6, 12, 24, 48, 96]$ and $[5, 10, 20, 40, 80]$ and $[3, 9, 27, 81]$ and $[7, 21, 63]$. Together, these cover $7 + 5 + 5 + 4 + 3 = 24$ terms. The remaining $76$ terms can be covered in at most $38$ sequences, by an argument made here. So we have the bound:
$$31 \leq n \leq 43$$
Can anyone do better?
*We need only consider rational ratios by arguments made in answers to the original question.

(Update) We have a winner!! Thanks to the cumulative efforts of the answerers below, we have arrived at $n = 36$. The upper bound is thanks to jpvee, and the lower bound is due to san. Hooray!

Switching from proving lower bounds to finding upper bounds, I played around with the list of progressions a bit and came up with another small improvement:
The $16$ progressions
$$(1, 2, 4, 8, 16, 32, 64)\\
(3, 6, 12, 24, 48, 96)\\
(5, 10, 20, 40, 80)\\
(7, 14, 28, 56)\\
(9, 18, 36, 72)\\
(11, 22, 44, 88)\\
(13, 26, 52)\\
(15, 30, 60)\\
(17, 34, 68)\\
(19, 38, 76)\\
(21, 42, 84)\\
(23, 46, 92)\\
(25, 35, 49)\\
(27, 45, 75)\\
(50, 70, 98)\\
(81, 90, 100)$$
are completely disjoint and thus cover $7+6+5+3\cdot4+10\cdot3=60$ integers. Using $20$ additional progressions that each cover $2$ of the remaining $100-60=40$ integers yields a cover of $16+20=36$ geometric progressions.
