Questions about counting the number of triples arranged in geometric progression 
Problem:  Three tickets are chosen from a set of $100$ tickets numbered $1,2,3,\ldots,100$. Find the number of choices such that the numbers on the three tickets are in geometric progression.


Solution: Let $k, n \in \mathbb Z_+$ s.t. $n \ge 2, \ kn^2 \le 100.$ Then the number of possibilities for $k$ is $\lfloor \frac{100}{n^2} \rfloor$. Hence total number of $k, kn, kn^2$ with integer common ratio is $\sum_{n = 2}^{10}\lfloor \frac{100}{n^2} \rfloor = 25 + 11 + 6 + 4 + 2 + 2 + 1 + 1 + 1 = 53.$  Now the number of $k, k\frac mn, k\frac{m^2}{n^2}$ where $n > 1, \ n^2 \mid k, \ m > n, \ \gcd(m, n) = 1$ is given by $f(n) = \sum\lfloor \frac{100}{m^2} \rfloor $. So the number of GPs with non-integer common ratio $\sum_{n= 2}^9f(n)$. Thus the total number of GP's in question is $53 + \sum_{n= 2}^9f(n) = 105.$

My questions:

*

*When $n \in \mathbb Z_+$, we have the following
$k, 2k, 4k \iff 25, 50, 100$
$k, 3k, 9k \iff 11, 33, 99$
$k, 4k, 16k \iff 6, 24, 96$
$\ldots$
$k, 10k, 100k \iff 1, 10, 100$
That is, nine triples in geometric order. I am confused as to what exactly $\sum_{n = 2}^{10}\lfloor \frac{100}{n^2} \rfloor$ counts.


*When the common ratio is rational, we have $km^2 \le 100n^2$. Why, then, is the number of $k$ is given by $\lfloor \frac{100}{m^2} \rfloor$ instead of, say, $\lfloor \frac{100n^2}{m^2} \rfloor$?

Your first patterN, with a common factor of $2$, can start with any number from $1$ through $25$.  $\lfloor \frac {100}{n^2} \rfloor$ is the number of progressions with common factor $n$.  The third number is $n^2$ times the first, so the progression must start with a number small enough that the third term is no greater than $100$.  That is all numbers from $1$ through $\lfloor \frac {100}{n^2} \rfloor$, where the brackets represent the floor function, so $\lfloor \frac {100}{9} \rfloor=11$
For the second, $k$ must be a multiple of $n^2$, so you should not have $n^2$ in the numerator.
