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!
 A: 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.
A: 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.
A: Nice problem!
Using help from the computer, I found that within the positive integers below 100, there are
66 geometric progressions of length 3,
6 geometric progressions of length 4,
3 geometric progressions of length 5,
1 geometric progression of length 6 and
1 geometric progression of length 7
(after eliminating those that are proper subsequences of longer ones).
These progressions of length 3 or longer cover a total of 65 integers $\le 100$; denote the set of these 65 integers by $\mathbb{M}$. If the $6+3+1+1=11$ progressions of lengths $\ge4$ were all disjoint (which they are not), they together would cover $6\cdot4+3\cdot5+6+7=52$ elements of $\mathbb{M}$; for the remaining ones, at least $\lceil(65-52)/3\rceil=5$ additional progressions of length 3 are necessary.
The remaining 35 integers outside of $\mathbb{M}$ can only be contained in progressions of length 2 and thus need at least $\lceil35/2\rceil=18$ of those progressions to cover them.
Therefore, a lower bound of the progressions needed to cover all positive integers up to 100 is $11+5+18=34$.

Update 2015-08-08: I just saw that my reasoning is a bit flawed; see my comment below.
A: 100, 60, 36. 99, 66, 44. 98, 84, 72. 96, 48, 24, 12, 6, 3. 92, 46, 23. 90, 30, 10. 88, (44,) 22, 11. 84, 42, 21. 81, 54, (36,) (24,) 16. 80, 40, 20, (10,) 5. 76, 38, 19. 75, 45, 27. 68, 34, 17. 64, 32, (16,) 8, 4, 2, 1. That's 49 numbers, with 14 progressions. So we can certainly do it with 40 sequences. Or we can keep going. 56, 28, 14, 7. 52, 26, 13. 49, 35, 25. So, 38 sequences. 
EDIT: turns out this question was discussed last year on MathOverflow, and I even contributed (a little bit) to the discussion. I recommend having a look at that discussion before continuing it here. 
