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!

1,2,4,8,16,32,64
3,6,12,24,48,96
5,10,20,40,80
7,14,28,56
11,22,44,88
13,26,52; 17,34,68; 19,38,76; 21,42,84; 23,46,92
9,15,25; 36,60,100; 49,63,81
27,45,75; 50,70,98
15 sequences with 56 numbers. 44 remains, so 15 + 22 = 37. 
Not the only 37 solution, some replacements available (36,54,81; 49,70,100; 18,30,50, 50,60,72), so I believe computer program would find better solution quickly if it exists.
Why not 9,27,81 included? Not to spend three "squares" for a single triplet.
