11. Let $S=\{1,2,3, \cdots, 500\}$, choose any 4 different numbers from $S$, and arrange them in ascending order to form a geometric sequence with a positive integer ratio. Find the number of such geometric sequences.
11. Solution 1 Let the geometric sequence be $a_{1}, a_{1} q, a_{1} q^{2}, a_{1} q^{3}(a, q$ are positive integers, $q \geqslant 2)$,

Thus, $a_{1} q^{3} \leqslant 500, q^{3} \leqslant 500 / a_{1} \leqslant 500$, then $q \leqslant \sqrt[3]{500}$.
Therefore, $2 \leqslant q \leqslant 7$.
That is, $q=2,3,4,5,6,7$.
When $q=2$, $1 \leqslant a_{1} \leqslant\left[\frac{500}{2^{3}}\right]=62$, there are 62 geometric sequences.
When $q=3$, $1 \leqslant a_{1} \leqslant\left[\frac{500}{3^{3}}\right]=18$, there are 18 geometric sequences.
When $q=4$, $1 \leqslant a_{1} \leqslant\left[\frac{500}{4^{3}}\right]=7$, there are 7 geometric sequences.
When $q=5$, $1 \leqslant a_{1} \leqslant\left[\frac{500}{5^{3}}\right]=4$, there are 4 geometric sequences.
When $q=6$, $1 \leqslant a_{1} \leqslant\left[\frac{500}{6^{3}}\right]=2$, there are 2 geometric sequences.
When $q=7$, $1 \leqslant a_{1} \leqslant\left[\frac{500}{7^{3}}\right]=1$, there is 1 geometric sequence.
In total, there are $62+18+7+4+2+1=94$ geometric sequences.
Solution 2 Since $7^{3}<500<8^{3}$, there are 6 geometric sequences with 1 as the first term (with $2,3,4,5,6,7$ as the common ratio), $2 \cdot 6^{3}<500<2 \cdot 7^{3}$, there are 5 geometric sequences with 2 as the first term (with $2,3,4,5,6$ as the common ratio), $3 \cdot 5^{3}<500<3 \cdot 6^{3}$, there are 4 geometric sequences with 3 as the first term (with $2,3,4,5$ as the common ratio), $4 \cdot 5^{3}<500<4 \cdot 6^{3}$, there are 4 geometric sequences with 4 as the first term (with $2,3,4,5$ as the common ratio), there are 3 geometric sequences with $5,6,7$ as the first term (with $2,3,4$ as the common ratio) each,
there are 2 geometric sequences with $8,9, \cdots, 18$ as the first term (with 2,3 as the common ratio) each, there is 1 geometric sequence with $19,20, \cdots, 62$ as the first term (with 2 as the common ratio) each.
Thus, $6+5+4 \times 2+3 \times 3+11 \times 2+44=94$ (sequences)