How many geometric progressions?

First let us consider geometric progressions with common ratio $r \in \mathbb{Z^+}$ (positive integers).

If $r=1$ , then the sequence will not be increasing.

If $r=2$ , then the first term $a_1$ must be less than $\frac{1000}{2^5}$ so that the final term (and therefore all terms) are less than $1000$ . $\frac{1000}{2^5}=\frac{125}{4}=31.25$ , so $a_1$ can be any positive integer from $1$ to $31$ . So far we have $31$ distinct progressions defined by $r=2$ and $a_1 \in \{1,2,3,...,31\}$ .

If $r=3$ , then $a_1$ must be less than $\frac{1000}{3^5}=\frac{1000}{243} \approx 4.1$ , so $a_1$ can be any positive integer from $1$ to $4$ . We now have $31+4=35$ progressions.

If $r=4$ , then $a_1$ must be less than $\frac{1000}{4^5}=\frac{1000}{1024}<1$ , but $a_1$ must be a positive integer, so it can't be less than $1$ , so there exist no progressions for $r \geq 4$ because one of the terms would end up being greater than $1000$ .

If we let $r \in \mathbb{Q}$ (rational numbers), then we must be careful that all of the terms in the progression are positive integers. If $r$ is of the form $\frac{x}{y}$ , then $a_1$ must be of the form $k \times y^5, k \in \mathbb{Z^+}$ , so that the $y$ term in the denominator of $r$ will cancel for all six terms, leaving an integer. In addition, $a_1 \times r^5=k \times y^5 \times (\frac{x}{y})^5=k \times x^5$ must still be less than $1000$ . For this reason, $x$ cannot be greater than or equal to $4$ , so $x$ can only be $2$ or $3$ . And because $a_1$ is of the form $k \times y^5$ , which must also be less than $1000$ , $y$ can only be $1$ , $2$ or $3$ .

The case $y=1$ is the positive integers covered above. For the remaining four possible values of of $r=\frac{x}{y} \{\frac{2}{2}, \frac{2}{3}, \frac{3}{2}, \frac{3}{3}\}, \frac{2}{2}=\frac{3}{3}=1$ and $\frac{2}{3}$ cause a non-increasing sequence, so the only other possible rational value of $r$ is $\frac{3}{2}$ .

$a_1 = k \times y^5=k \times 2^5 =32k$ , must be less than $\frac{1000}{r^5}=\frac{1000}{(3/2)^5}=\frac{32 \times 1000}{243}$ , so $k$ must be less than $\frac{\frac{32 \times 1000}{243}}{32}=\frac{1000}{243} \approx 4.1$ . So $k$ can be any positive integer from $1$ to $4$ . This gives us four progressions defined by $r=\frac{3}{2}$ and $a_1 \in \{2^5, 2 \times 2^5, 3 \times 2^5,4 \times 2^5\}$ . Adding these $4$ to the $35$ accounted for before, we get $39$ progressions.

We have now exhausted all possible values of $r$ , so we have $39$ progressions.

For this problem, we can represent the geometric sequence as:

$ra^{5}$ , $ra^{4}b$ , $ra^{3}b^{2}$ , $ra^{2}b^{3}$ , $rab^{4}$ , $rb^{5}$

Since $4^{5} = 1024$ , we know that $b < 4$ .

We also know that $a < b$ because the sequence is strictly increasing.

For $a = 1$ and $b =2$ , there are $\frac {1000} {2^5}$ values of $r$ ( $r$ ranges from $1$ to $31$ ).

For $a = 1$ and $b = 3$ , there are $\frac {1000} {3^5}$ values of $r$ ( $r$ ranges from $1$ to $4$ ).

For $a = 2$ and $b = 3$ , there are $\frac {1000} {3^5}$ values of $r$ ( $r$ ranges from $1$ to $4$ ).

The sum of all of these is $39$ .

Consider the progression starting with the term $a$, with a multiplicative factor of $r$. As the progression has 6 terms, the last term is $a*r^5$.

As all the terms must be integers, $a$ must be an integer. $r$ can be also an integer or a rational number.

Case 1: $r$ is integer.

Since the last term is $a*r^5$, and $a$ is an integer, $r^5$ must be less than 1000. $r$ can be 2 ($2^5$=32) or 3 ($3^5$=243), but cannot be 4 ($4^5$=1024). $r$ cannot be 1 because the progression is strictly increasing.

If $r$=2, $a$ can be any positive integer below 32, since $2^5$ 31=992 and $2^5$ 32=1024. Then, there are 31 progressions with $r$=2.

If $r$=3, $a$ must be less than 5. Then, there are 4 progressions with $r$=3.

Case 2: $r$ is rational.

Suppose that $r=\frac{p}{q}$, with gcd(p,q)=1. Then, $a r^5 = \frac{a p^5}{q^5}$, and therefore $a=k q^5$. Otherwise, the last term would not be integer. This means that the last term is $k p^5$, with $p$ integer, which reduces to the previous case.

$p$ cannot be 2, because $r>1$ (the progression is increasing).

If $p=3$, $q$ can be only 2. Then $a=k 2^5$, and the last term is $k 3^5$. As in the previous case, $k$ must be less than 5, so there are only 4 progressions.

To sum up, if the first term is $a$ each term is multiplied by $r$, the possibilities are:

• $r=2, a= 1..31$ (31 cases)
• $r=3, a= 1..4$ (4 cases)
• $r=3/2, a= 32, 64, 96, 128$ (4 cases)

In total, 31+4+4=39 possibilities.

Since it is strictly increasing progression, then the last term ar^5 must be less than 1000. The common ratio, r must be either integers or fractions.

Case 1: r is integer Smallest r would be 2, and since 4^5 = 1024 > 1000, so r can only be 2 and 3. For r=2, since ar^5 = 243a < 1000 \Rightarrow 1 \leq a \leq 31. Similarly for r=3, 1 \leq a \leq 4.

Case 2: r is fraction Let r= \frac {m}{n}. The last term become \frac {am^5}{n^5}. But since all terms are integers and \gcd(m,n)=1, a must be divisible by n^5, or a=kn^5. The last term now becomes km^5. As the progression is increasing, \frac {m}{n} > 1 \Rightarrow m > n. And n \neq 1 (so that r is not an integer) and km^5 < 1000, therefore n=2 and m=3 and 1 \leq k \leq 4. Total number of such progressions = 31+4+4 = 39

Suppose the first term of the progression is $a$ and the ratio is $r$ . Then $a$ is a positive integer and $r$ is a rational number with $r>1$ . (Note that $r$ does not have to be an integer.)

Suppose $r=\frac{n}{m},$ where $n$ and $m$ have no common factors. The last term of the progression, $ar^5=\frac{an^5}{m^5}$ must be an integer. So $m^5$ divides $a,$ thus $a\geq m^5.$ Therefore, $ar^5\geq \frac{m^5n^5}{m^5}=n^5.$ Because $ar^5<1000,$ this implies $n^5<1000.$ Note that $4^5=1024,$ so $n$ can only be $2$ or $3$ .

If $n=2,$ then $m=1$ (because $r>1$ ). In this case the progression consists of numbers $a,2a,...,32a.$ This works if and only if $1\leq a$ and $32a<1000$ , which is equivalent to $1\leq a \leq 31.$

If $n=3,$ then $m=1$ or $m=2.$ If $m=1,$ then the progression consists of numbers $a,3a,...,243a.$ This works if and only if $1\leq a$ and $243a<1000$ , which is equivalent to $1\leq a \leq 4.$

If $m=2,$ then, from the argument above, $2^5$ must divide $a$ , so $a=32b$ for some integer $b$ . In this case, the progression consists of the numbers $32b, 48b, 72b, 108b, 162b, 243b$ . This works if and only if $1\leq b$ and $243b<1000$ , which is equivalent to $1\leq b \leq 4.$

Putting this all together, we have $31+4+4=39$ geometric progressions.

Since 4^5 > 1000, we can say that common ratio has to be less than 4

If ratio is 2, then terms are a, 2a, 4a, 8a, 16a, 32a 32a < 1000 => a < 32 So, 31 different GP's

If ratio is 3, then terms are a, 3a, 9a, 27a, 81a, 243a 243a < 1000 a < 5 So, 5 different GP's

Now, ratio can also be 3/2, terms will be a, 3a/2, 9a/4, 27a/8, 81a/16, 243a/81 => a has to be a multiple of 81, say 81k => 243k < 1000 => k < 5 So, 4 more GP's

Now, if ratio is p/q, then q has to be 2 or 3 (as q^5 for integers more than 3 is more than 1000) In case of 2, p can be 3 or 4 (both are already considered) and no other number s possible (as then p^5 will exceed 1000) In case of 3, p has to be more than 3 (but for p > 3, p^5 will exceed 1000)

So, only 31 + 4 + 4 = 39 series are possible

Even the answer is wrong: 35, not 39. The fractional ratios are missing, the whole argument is very non-rigorous.

We have to find how many couples of integers $(a;q)$ are there such as $a \times q^5 \leq 1000$ In fact, $f = e \times q = d \times q^2 = c \times q^3 = b \times q^4 = a \times q^5$ , and f must be less or equal to 1000. We see that must be $q^5 \leq \frac {1000}{a}$ , so q can be 2 or 3 for $1 \leq a \leq 4$ , and can be only 2 for $5 < a \leq 31$ . If $a > 31$ then $q = 1$ , but it can't be, because the progression must be strictly increasing. So, the number of couples $(a;q)$ is $4 \times 2 + 31 = 39$