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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
k i A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
Posting Guidelines
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
How do I get a problem on the contest page?
How do I study for mathcounts?
Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
shortlisted problems being used in undergraduate competition
enter16180   0
6 minutes ago
Hello, I am posting here to let know ( clarified after a post in College Math forum) that Problem 10 of Open Mathematical Olympiad for University Students ( OMOUS-2025) held at Ashgabat, Turkmenistan on 13-18 April, 2025 is found to be A6 Shortlisted Problems IMO-2024.
Following is discussion on College Math Forum
https://artofproblemsolving.com/community/c7h3551018_omous2025_team_competition_p10


Image of problem from competition for reference below.
0 replies
enter16180
6 minutes ago
0 replies
Four variables
Nguyenhuyen_AG   0
7 minutes ago
Let $a,\,b,\,c,\,d$ non-negative real numbers. Prove that
\[\frac{abc}{(a+b+c)^3}+\frac{bcd}{(b+c+d)^3}+\frac{cda}{(c+d+a)^3}+\frac{dab}{(d+a+b)^3} \leqslant \frac{(a+b+c+d)^2}{27(a^2+b^2+c^2+d^2)}.\]
0 replies
Nguyenhuyen_AG
7 minutes ago
0 replies
Functional equation
shactal   1
N 21 minutes ago by Mathzeus1024
Source: Own
Hello, I found this functional equation that I can't solve, and I haven't got any hints. Could someone try and find the solution, it's actually quite difficult:
Find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that, for all $x, y \in \mathbb{R} $,
$$
f(x + f(y)) + f(y + f(x)) = f(x \, f(y) + y \, f(x)) + f(x + y)$$Thank you.
1 reply
shactal
Yesterday at 11:15 PM
Mathzeus1024
21 minutes ago
Probability Inequality
EthanWYX2009   0
33 minutes ago
Source: 2024 June 谜之竞赛-5
Determine the minimum real number \(\lambda\) such that for any $2024$ real numbers \(a_1, a_2, \cdots, a_{2024}\) satisfying
\[\sum_{i=1}^{2024} a_i = 0,\quad\sum_{i=1}^{2024} a_i^2 = 1,\]there exists a non-empty subset \(I\) of \(\{1, 2, \cdots, 2024\}\) for which
\[\sum_{i\in I} a_i \leq \lambda \cdot \min\{|I|, 2024 - |I|\}.\]Proposed by Tianqin Li, High School Affiliated to Renmin University of China
0 replies
EthanWYX2009
33 minutes ago
0 replies
9 MathCounts prep
ericheathclifffry   13
N Today at 3:02 AM by Aniharry
Redacted
13 replies
ericheathclifffry
Monday at 11:06 PM
Aniharry
Today at 3:02 AM
AMC 8 DHR
PhoenixMathClub   15
N Today at 1:50 AM by K1mchi_
Hey y'all so I just started preparing for the 2026 AMC 8 and I was wondering if you could give me some suggestions on what I should do so I can get DHR. I barely practice math competition and I just started to get serious this year so my scores might be bad so here it is:

5th Grade: 12
6th Grade: 16
6th Grade Summer/7th Grade: Mocking 20-24 on AMC 8 Tests

I saw many recommendations about buying AoPS books but not many about the number theory so I got the Introduction to Algebra, Introduction to Counting and Probability, and Introduction to Geometry Books. I was wondering what other suggestions that you would have for me to get better at AMC 8.
15 replies
PhoenixMathClub
Jul 10, 2025
K1mchi_
Today at 1:50 AM
9 Was this year's AMC 8 hard or easy
ChuMath   69
N Today at 1:30 AM by a.zvezda
Question: Was this year's AMC 8 harder or easier than last year's (or the historical average)

69 replies
ChuMath
Jan 25, 2024
a.zvezda
Today at 1:30 AM
9 How many Math Olympiad concepts/theorems can you name?
a.zvezda   1
N Today at 1:15 AM by OWOW
I know these
1 reply
a.zvezda
Today at 1:09 AM
OWOW
Today at 1:15 AM
9 Worst math subject
a.zvezda   58
N Today at 1:13 AM by a.zvezda
Mine is geo and C&P because it's really annoying and last year, I got 16 on the AMC 8 when I sillied a few geo and C&P problems. :wallbash_red:
58 replies
a.zvezda
Jul 14, 2025
a.zvezda
Today at 1:13 AM
Random but useful theorems
booking   60
N Today at 1:12 AM by booking
There have been all these random but useful theorems
Please post any theorems you know, random or not, but please say whether they are random or not.
I'll start give an example:
Random
I am just looking for some theorems to study.
60 replies
booking
Jul 16, 2025
booking
Today at 1:12 AM
The 24 Game, but with a twist!
PikaPika999   416
N Today at 12:52 AM by Biglion
So many people know the 24 game, where you try to create the number 24 from using other numbers, but here's a twist:

You can only use the number 24 (up to 5 times) to try to make other numbers :)

the limit is 5 times because then people could just do $\frac{24}{24}+\frac{24}{24}+\frac{24}{24}+...$ and so on to create any number!

honestly, I feel like with only addition, subtraction, multiplication, and division, you can't get pretty far with this, so you can use any mathematical operations!

Banned functions
416 replies
PikaPika999
Jul 1, 2025
Biglion
Today at 12:52 AM
Troll Problem
giratina3   13
N Today at 12:14 AM by Ryanzzz
If $\frac{a}{a - 1} = \frac{b^2 + 2b - 1}{b^2 + 2b - 2}$, then what does $a$ equal in terms of $b$?

Hint 1
Hint 2
Hint 3
13 replies
giratina3
Jul 11, 2025
Ryanzzz
Today at 12:14 AM
counting problems
BlueAnglerfish42   8
N Today at 12:11 AM by Ryanzzz
1. How many ways are there for 5 skibidi toilets to swallow 30 different people if each toilet can swallow up to 5 people in one gulp? (A toilet can gulp more than once.)

2. Call a number "rizztastic" if it is divisible by more than 4 odd numbers. How many three-digit numbers are "rizztastic"?

3. 10 skibidi toilets don't have names, and 5 toilets have names. Each toilet name is three letters long. How many ways are there to give the rest of the toilets names?
8 replies
BlueAnglerfish42
Jul 17, 2025
Ryanzzz
Today at 12:11 AM
9 Easiest math competition
a.zvezda   51
N Today at 12:10 AM by Ryanzzz
MOEMS, Math Kangaroo, and Beestar are to me the easiest. :P
51 replies
a.zvezda
Jul 15, 2025
Ryanzzz
Today at 12:10 AM
Power sequence
TheUltimate123   8
N Jun 14, 2025 by ihategeo_1969
Source: ELMO Shortlist 2023 N2
Determine the greatest positive integer \(n\) for which there exists a sequence of distinct positive integers \(s_1\), \(s_2\), \(\ldots\), \(s_n\) satisfying \[s_1^{s_2}=s_2^{s_3}=\cdots=s_{n-1}^{s_n}.\]
Proposed by Holden Mui
8 replies
TheUltimate123
Jun 29, 2023
ihategeo_1969
Jun 14, 2025
Power sequence
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO Shortlist 2023 N2
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TheUltimate123
1753 posts
#1 • 5 Y
Y by ImSh95, p_ben, PikaPika999, ehuseyinyigit, cubres
Determine the greatest positive integer \(n\) for which there exists a sequence of distinct positive integers \(s_1\), \(s_2\), \(\ldots\), \(s_n\) satisfying \[s_1^{s_2}=s_2^{s_3}=\cdots=s_{n-1}^{s_n}.\]
Proposed by Holden Mui
Z K Y
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shinhue
158 posts
#2 • 3 Y
Y by ImSh95, PikaPika999, cubres
Very nice, any ideas, i think maximum value of $n$ would be $3$
Z K Y
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sman96
138 posts
#5 • 4 Y
Y by ImSh95, PikaPika999, ehuseyinyigit, cubres
shinhue wrote:
Very nice, any ideas, i think maximum value of $n$ would be $3$

$s_1 = 6^6, s_2 = 6, s_3 = 36, s_4 = 18$ works
Z K Y
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predictableprimes
6 posts
#6 • 3 Y
Y by ImSh95, Sedro, cubres
Nice problem. The answer is $n=5$.
Claim: if $n>4$ then $s_{n-3}=2,s_{n-2}=16,s_{n-1}=4$.
Proof: suppose $s_1^{s_2}=s_2^{s_3}= ... =s_{n-1}^{s_n}$ and $n>4$ then $p|s_i$ $\implies$ $p|s_j$ for $i,j\in$ {$1,2,..n-1$}
also for any prime $p$ dividing $a_i$ we have $s_2V_p(s_1)=s_3V_p(s_2)= ..= s_{n-2}V_p(s_{n-1})=s_{n-1}V_p(s_n)$ $\implies$ $\frac{V_p(s_i)}{V_q(s_i)}$ is constant for all $i\in$ {$1,2,..n-1$} and any prime divisors $p $ and $q$ of $s_1$.
Since $\frac{V_p(s_i)}{V_q(s_i)}$ is constant , each $s_i$ must be a perfect power of some integer.
(This is true since exponent of primes of $s_i$ are in a fixed ratio)
Now let $s_i=a^{b_i}$ for all $i\in$ {$1,2,..n-1$}.
$\implies$$b_1a^{b_2}= .. =b_{n-4}a^{b_{n-3}}=b_{n-3}a^{b_{n-2}}=b_{n-2}a^{b_{n-1}}=b_{n-1}s_n$
For simplicity let $c_1=b_{n-4} , c_2=b_{n-3} , c_3=b_{n-2} , c_4=b_{n-1}$
We will prove that $c_{1}a^{c_2}=c_{2}a^{c_{3}}=c_{3}a^{c_{4}}=c_{4}s_n$ is not possible.
Case 1: $c_2>c_1$
$c_{1}a^{c_2}=c_{2}a^{c_{3}}$$\implies$ $a^{c_2}>a^{c_3}$$\implies$ $c_2>c_3$ also $c_{2}a^{c_{3}}=c_{3}a^{c_{4}}$$\implies$ $c_3>c_4$$\implies$$\frac{c_2}{c_3}$$=a^{c_4-c_3}$$\implies$$ c_3|c_2$
also $\frac{c_2}{c_1}$$=a^{c_2-c_3}$ since $c_3|c_2$ we have $c_3<$$\frac{c_2}{2}$$\implies$ $\frac{c_2}{c_1}$$ > $$a^{c_2/2}$ which is only possible if $(a,c_2,c_1)=(2,2,1),(2,3,1),(2,4,1)$.
Now for $(a,c_2,c_1)=(2,2,1),(2,3,1)$,$(c_1,c_2,c_3,c_4)=(1,2,1,2),(1,3,log_2(8/3),idk)$,Both of which are contradiction.
But $(a,c_2,c_1)=(2,4,1)$ gives the desired result.
Case 2:$c_2<c_1$
$c_{1}a^{c_2}=c_{2}a^{c_{3}}$$\implies$ $a^{c_2}<a^{c_3}$$\implies$ $c_2<c_3$ also $c_{2}a^{c_{3}}=c_{3}a^{c_{4}}$$\implies$ $c_3>c_4$
Now we prove that $c_4|c_3$ , if not then there exist a prime $P$ such that $V_P(c_4)>V_P(c_3)$ but $c_4s_n=c_3a^{c_4}$$\implies$$P$ divides a.
Let $V_P(a)=m , V_P(c_4)=l$ .Note that $c_3=c_2a^{c_3-c_4}$$\implies$$V_P(c_3)>(c_3-c_4)m$
and since $l>V_P(c_3)$ ,we have $V_P(c_3)|c_3,c_4$$\implies$ $V_P(c_3)>P^{V_P(c_3)}$$\frac{c_4-c_3}{P^{V_P(c_3)}}m$$\implies$$V_P(c_3)>P^{V_P(c_3)}$ which is a contradiction.
Hence $c_4|c_3$ now from $\frac{c_3}{c_2}$$=a^{c_3-c_4}$ we will get $\frac{c_3}{c_2}$$ > $$a^{c_3/2}$ which implies $(a,c_3,c_2)=(2,2,1),(2,3,1),(2,4,1)$
which would give contradiction as $(c_1,c_2,c_3,c_4)=(2,1,2,1),(4,1,3,log_2(8/3)),(2,1,4,4)$ for $(a,c_3,c_2)=(2,2,1),(2,3,1),(2,4,1)$ respectively ,all of which are contradiction.
Hence
$s_{n-3}=2,s_{n-2}=16,s_{n-1}=4$$\implies$$s_1^{s_2}=s_2^{s_3}= ... =s_{n-1}^{s_n}=2^{16}$
$\implies$$s_n=8,s_{n-4}=2^8$
Note that $s_{n-5}$ does not exist as if it did then $s_{n-5}^{2^8}=2^{16}$ which would be a contradiction.
Hence $n=5$ and the only solution is $(s_1,s_2,s_3,s_4,s_5)=(2^8,2,16,4,8)$.

Initially I did not checked the case for $(a,c_2,c_1)=(2,4,1)$ and fake solved that $n=4$ :blush:
This post has been edited 2 times. Last edited by predictableprimes, Jun 29, 2023, 7:50 PM
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DottedCaculator
7418 posts
#7 • 3 Y
Y by ImSh95, predictableprimes, cubres
Let $s_i=c^{a_i}$ where $c\neq1$ for $i\leq n-1$. Then, $a_1c^{a_2}=a_2c^{a_3}=\ldots=a_{n-2}c^{a_{n-1}}$, so $a_i=xc^{b_i}$ for $i\leq n-2$. Therefore, we get $b_1+xc^{b_2}=b_2+xc^{b_3}=\ldots=b_{n-3}+xc^{b_{n-2}}=N$. Let $j$ be the largest number such that $N\geq xc^j$. Since the numbers are distinct, we have $b_1\neq b_2\neq\ldots\neq b_{n-2}$, so at least $n-5$ of the $b_i$ for $i\geq2$ satisfy $N-xc^{b_i}=b_{i-1}\geq xc^{j-1}(c-1)$ so $xc^{n-6+xc^{j-1}(c-1)}\leq N<xc^{j+1}$. Solving gives $n-6+xc^{j-1}(c-1)<j+1$. Since $c>1$, we get
$$j+1>n-6+xc^{j-1}(c-1)\geq n-6+2^{j-1},$$or $$n<j+1-2^{j-1}+6\leq7.$$Therefore, $n\leq6$.

If equality holds, then $c=2$ and $x=1$, so $a_i$ is a power of $2$. Therefore, $s_i$ must be a power of $2$ for all $i\leq n$, which gives a contradiction after the same steps. Therefore, $\boxed{n=5}$. A construction for $n=5$ is given by $256^2=2^{16}=16^4=4^8=65536$.
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IAmTheHazard
5007 posts
#8 • 4 Y
Y by GeoKing, ImSh95, centslordm, cubres
The answer is $5$; construction is $256^2=2^{16}=16^4=4^8$.
Observe that for any $i<n$ and prime $p$, we have $\tfrac{\nu_p(s_i)}{\nu_p(s_1)}=\tfrac{s_{i+1}}{s_{2}}$, so for any $i$, $s_i$ is some rational power of $s_1$. This implies that there exists some natural number $P$ such that $s_i=P^{a_i}$ for all $i$, where $a_i$ is a positive integer. The equation then becomes
$$a_1P^{a_2}=a_2P^{a_3}=\cdots=a_{n-1}P^{a_n}.$$But then for any $i<j<n$, the ratio $\tfrac{a_i}{a_j}$ is a power of $P$, hence we can let $a_i=cP^{b_i}$ for all $i<n$, where $c$ is a (fixed) natural number. The equation then becomes
$$b_1+cP^{b_2}=b_2+cP^{b_3}=\cdots=b_{n-2}+cP^{b_{n-1}}=b_{n-1}+a_n.$$
The key claim here is that for $2 \leq i \leq n-1$ we must have $b_i \leq 2$. Indeed, by taking $i$ such that $b_i$ is maximal, we find that
$$b_{i-1}+cP^{b_i}=b_i+cP^{b_{i+1}} \implies b_{i-1}=b_i-c(P^{b_i}-P^{b_{i+1}})\leq b_i-cP^{b_i-1}\leq b_i-2^{b_i-1},$$but if $b_i \geq 3$ then this final value is negative. Since we need $a_i$ to be a positive integer, we have $cP^{b_{i-1}} \geq 1$, so $cP^{b_i-1}=cP^{b_{i-1}}\cdot P^{b_i-b_{i-1}-1}\geq P^{b_i-b_{i-1}-1}$, so $b_{i-1} \leq b_i-P^{b_i-b_{i-1}-1} \implies P^{b_i+(-b_{i-1})-1} \leq b_i+(-b_{i-1})$, but this can clearly be checked to be impossible (for $b_i \geq 3$).

This finishes the problem, since the $b_i$ must be distinct, so between $2$ and $n-1$ inclusive there are at most $3$ of them. This misses $b_1$ and also $a_n$ (since $b_n$ doesn't exist), so $n=5$ is maximal. $\blacksquare$
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The_Turtle
254 posts
#9 • 2 Y
Y by GeoKing, cubres
My problem!
Original problem statement wrote:
Determine the longest possible length of a sequence of distinct positive integers $s_1$, $s_2$, $\ldots$, $s_n$ for which
\[s_1^{s_2} = s_2^{s_3} = \ldots = \left(s_{n-1}\right)^{s_n}.\]

Solution
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MathLuis
1597 posts
#10 • 1 Y
Y by cubres
For some prime $p$ basically we have $s_2 \nu_p(s_1)=s_3 \nu_p(s_2) = \cdots =s_n \nu_p(s_{n-1})$, also notice if a prime $p$ divides some $s_i$ here for $i$ between $1,n-1$, then it divides all as well and combined we have that for any two primes $p \ne q$ that $\frac{\nu_p(a_i)}{\nu_q(a_i)}$ is constant for all $1 \le i \le n-1$ (as long as $q$ divides one of the terms at least ofc), then from fixing the pick of $q$ and moving $p$ over all primes dividing the terms $s_1, \cdots, s_{n-1}$ we have that each of them is the power of a same positive integer $P$, so now let $s_i=P^{a_i}$ for some positive integer $a_i$ for all $1 \le i \le n-1$, then we have that $a_1 \cdot P^{a_2}= \cdots =a_{n-2} \cdot P^{a_{n-1}}=a_{n-1}s_n$, also notice that for $n=5$ we have the construct $256^2=2^{16}=16^4=4^8$, so now FTSOC suppose we can have $n \ge 6$.
The idea is that we can check the $a_2 \cdot P^{a_3}=a_3 \cdot P^{a_4}=a_4 \cdot P^{a_5}$ then $P^{a_4}=\frac{a_2}{a_3} \cdot P^{a_3}$ so $\frac{a_3}{a_2}$ is $P^k$ for some $k \in \mathbb Z$ that is nonzero, and thus $a_3=a_2 \cdot P^k$ but also $a_4=a_3-k=a_2 \cdot P^k-k$ and thus we have that $P^{a_5}=\frac{P^{a_2 \cdot P^k}}{P^k-\frac{k}{a_2}}$ and so the denominator also has to be a rational power of $P$, now if $k>0$ then $P^k-k \le P^k-\frac{k}{a_2} \le P^{k-1}$ which gives that $k \ge P^{k-1}(P-1)$ and if $P \ge 3$ then you will just never reach it, otherwise you must have $P=2$ and $k=1,2$ and thus equality always holds regardless which means that $a_2=1$ so if $k=2$ then $a_3=4$ and thus $a_4=2$ but then remember $2a_1=16$ so $a_1=8$ and then $2^{a_5}=8$ so $a_5=3$ but then we have $a_5 \mid 16$ which can't happen so contradiction, and if $k=1$ then $a_3=2$ but also $a_1 \cdot 2=4$ so $a_1=2$ which can't happen.
Therefore $k>0$ can't happen so if $k<0$ then we have $1-\frac{k}{a_3}$ to be a power of $P$ as well which is $1-\frac{k \cdot P^{-k}}{a_2}$ being an integer power of $P$. So if $-k \ge a_2$ then its a power of $P$ greater than $P^{-k}$ but as a result $P^{-k} \mid a_2$ so $-k \ge P^{-k}$ and this is just never true so we must have $a_2 \ge 1-k$, in fact notice we just also have $a_2=a_3 \cdot P^{-k} \ge P^{-k}$ but it happens now that $a_1=a_2 \cdot P^{a_2(P^k-1)}=a_3 \cdot P^{a_3(1-P^{-k})-k} \ge 1$ which means that $a_3 \ge P^{a_3(P^{-k}-1)+k} \ge 2^{a_3(2^{-k}-1)+k}$ so if $a_3=1$ we have a trivial contradiction, if $a_3=2$ then we have $1 \ge 2^{1-k}-2+k$ so $3-k \ge 2^{1-k}$ which can only happen when $k=-1$ so $a_2=4$ but as equality holds we have $a_1=1$ and then $a_4=3$ but that would mean $a_4 \mid 16$ which is false, so if $a_3 \ge 3$ then think that $a_3(2^{-k}-1)+k \ge a_3-1$ and thus $a_3 \ge 2^{a_3-1}$ which is only true when $a_3=1,2$ so contradiction!.
Therefore either way $n \ge 6$ is not feasable, thus $n=5$ is maximun thus we are done :cool:.
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ihategeo_1969
284 posts
#11 • 1 Y
Y by cubres
See that we have $s_i=a^{x_i}$ where $a$ is not a perfect square and hence we are solving \[x_1a^{x_2}=\dots=x_{n-1}a^{x_n}\]Now $x_i/x_j$ is a perfect square for $1 \le i,j \le n-1$ and hence write $x_i=Ca^{y_i}$ for $1 \le i \le n-1$ and finally we are solving \[y_1+Ca^{y_2}=\dots=y_{n-2}+Ca^{y_{n-1}}\]where $y_i \ge 0$ are pairwise distinct. Assume $a \ge 2$ otherwise $n=2$ and assume from now $n$ is as large as it can be.

Let $y_k=\max\{y_2,\dots,y_{n-1}\}$ and hence let $i$ be in this range with $i \ne 2,k$ and so \[a^{y_k-1} \le Ca^{y_k-1}(a-1) \le Ca^{y_k}-Ca^{y_i}=y_{i-1}-y_{k-1} \le y_k \implies y_k \le 2\]Now as $y_2$, $\dots$, $y_{n-1}$ are all pairwise distinct and in range $[0,2]$ we have $n-2 \le 3$ so $n \le 5$.

The construction is $256^2=2^{16}=16^4=4^8$ and hence $\boxed{n=5}$ is the answer.
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