Beware, Just Integers!

Note that we want maximum terms. Suppose the common ratio is $2$ . Then the GP with maximum term is the one whose first term is smallest i.e $100$ . So the GP is $100,200,400,800$ . But this is not the GP with maximum terms.Since the terms are distinct the common ratio is not equal to $1$ . So we conclude that if $r$ is common ratio , to get maximum terms in GP , optimally , the value of $r$ must be bound : $1 < r < 2$ .

So let $r=\dfrac{3}{2}$ . For the terms of GP to be an integer and to have maximum terms , the first term must optimally be as small as possible and can be represented as $2^n$ where $n$ must be maximum as possible. And we have $2^7 = 128$ . Thus the GP required is $128,192,288,432,648,972$ with $\boxed{6}$ terms.

Lets's try to assume that there is a GP with at least 7 terms. This means that we have a starting number $a$ and a ratio $r$ such that

$a, ar, ar^2, \ldots , ar^6 \in \left\{100, 101, \ldots , 1000 \right\} \subset \mathbb N \setminus \left\{ 0 \right\}$

Now, since all the terms must be integers this means that $a \in \mathbb N$ and $ar \in \mathbb N$ . This implies that $r \in \mathbb Q$ in fact if $r$ is irrational, and since $a$ is natural, we would have $ar$ irrational. Morever $r$ is positive otherwise $ar$ would be not positive as it must be since it belongs to $\left\{ 100, \ldots, 1000 \right\}$ . So there are two positive integers $m,d \in \mathbb N$ such that

$r = \dfrac{m}{d}$

Note that $m \not = d$ otherwise $r=1$ and all the ters of the GP would be the same (they must be distinct).

Since $ar^6 = a \dfrac{m^6}{d^6}\in \ \mathbb N$ this implies $d^6 \vert a$ . How many $d$ are such that $d^6$ divides the number $a$ belonging to $\left\{ 100, \ldots, 1000 \right\}$ ? We have few choices for $d$ .

First choiche $d=1$ . That implies $a$ can be whatever integer number from $100$ to $1000$ . But we must also have $am^6 \in \left\{ 100, \ldots, 1000 \right\}$ and must be $m \geq 2$ , and $a \geq 100$ . Now $am^6 \geq 100 \cdot 2^6 = 6400$ and it is not in $\left\{ 100, \ldots, 1000 \right\}$ . This means $d$ cannot be $1$ .

Second choiche $d=2$ . That means $64$ divides $a$ . So that $a$ can be one of these values $128 = 64 \cdot 2, 192 = 64 \cdot 3, \ldots , 960 = 64 \cdot 15$ so that we have $a = 64 \cdot x$ with $x \in \left\{ 2,3,, \ldots , 15 \right\}$ .What about $m$ ? $m$ could be $1$ but this would led to $a r^6 = a \dfrac{1}{64} = x \in \left\{ 100, \ldots, 1000 \right\}$ while $x \leq 15$ . So $m \not = 1$ . $m \geq 3$ but this would let to $ar^6 = xm^6 \geq 2\cdot 729 = 1458 \in \left\{ 100, \ldots, 1000 \right\}$ that is not true. This means no vlues for $m$ can suit, and definitely $d$ cannot be 2.

Third choice $d = 3$ . This means $729$ divides $a$ that leaves only one possibility $a = 729$ . The fact that $ar^6 \in \left\{ 100, \ldots, 1000 \right\}$ transaltes in $m^6 \in \left\{ 100, \ldots, 1000 \right\}$ but the only integer whose sixth power belongs in that range is $3$ and $m$ must be distinct from $d=3$ . So there are no choices for $m$ and thid means $d$ cannot be $3$ .

Fourth choice $d \geq 4$ . This leads immediately to a contraddiction because it would imply that $d^6 \geq 4096$ is a divisor of $a \leq 1000$ . So it can't be $d \geq 4$ .

Every choiche for $d$ takes to a contraddiction, so we proved that there are no GP with at least 7 terms in $\left\{ 100, \ldots, 1000 \right\}$ .

Otherwise it is easy to find a GP with 6 terms (the same argument from the previous proof can be used). Namely the one we get with $a = 128$ and $r = \dfrac{3}{2}$

$128, \ 192, \ 288, \ 432, \ 648, \ 972.$

So we can conclude that the maximum number of terms the GP can have is 6.

Result of search:

1
2
3
4
5
6

128
192
288
432
648
972

Only one set is available with a = 128 and r = 1.5 for maximum of 6 terms. There are six sets for 5 terms and zero set for 7 terms. Search fineness was 0.0001 to 3 for r, with a = 100 to 999. Finest set was found for 5 terms of a = 256 and r = 1.25 which is {256, 320, 400, 500, 625}.

Answer: $\boxed{6}$