Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
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Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
Turkish JMO 2025?
bitrak   1
N 36 minutes ago by blug
Let p and q be prime numbers. Prove that if pq(p+ 1)(q + 1)+ 1 is a perfect square, then pq + 1 is also a perfect square.
1 reply
bitrak
Yesterday at 2:04 PM
blug
36 minutes ago
Combi Algorithm/PHP/..
CatalanThinker   0
42 minutes ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
5. [Czech and Slovak Republics 1997]
Each side and diagonal of a regular n-gon (n ≥ 3) is colored blue or green. A move consists of choosing a vertex and
switching the color of each segment incident to that vertex (from blue to green or vice versa). Prove that regardless of the initial coloring, it is possible to make the number of blue segments incident to each vertex even by following a sequence of moves. Also show that the final configuration obtained is uniquely determined by the initial coloring.
0 replies
CatalanThinker
42 minutes ago
0 replies
Combi Proof Math Algorithm
CatalanThinker   0
an hour ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
3. [Russia 1961]
Real numbers are written in an $m \times n$ table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
0 replies
CatalanThinker
an hour ago
0 replies
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   1
N an hour ago by Ritwin
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
1 reply
Primeniyazidayi
an hour ago
Ritwin
an hour ago
Do you need to attend mop
averageguy   5
N Yesterday at 5:55 PM by babyzombievillager
So I got accepted into a summer program and already paid the fee of around $5000 dollars. It's for 8 weeks (my entire summer) and it's in person. I have a few questions
1. If I was to make MOP this year am I forced to attend?
2.If I don't attend the program but still qualify can I still put on my college application that I qualified for MOP or can you only put MOP qualifier if you actually attend the program.
5 replies
averageguy
Mar 5, 2025
babyzombievillager
Yesterday at 5:55 PM
2014 amc 10 a problem 23
Rook567   3
N Yesterday at 4:13 PM by Rook567
Why do solutions assume 30 60 90 triangles?
If you assume 45 45 90 you get 5/6 as answer, don’t you?
3 replies
Rook567
Monday at 7:31 PM
Rook567
Yesterday at 4:13 PM
Inequality with a^2+b^2+c^2+abc=4
cn2_71828182846   72
N Yesterday at 4:12 PM by endless_abyss
Source: USAMO 2001 #3
Let $a, b, c \geq 0$ and satisfy \[ a^2+b^2+c^2 +abc = 4 . \] Show that \[ 0 \le ab + bc + ca - abc \leq 2. \]
72 replies
cn2_71828182846
Jun 27, 2004
endless_abyss
Yesterday at 4:12 PM
Special Points on $BC$
tenniskidperson3   40
N Yesterday at 3:43 PM by dipinsubedi
Source: 2013 USAMO Problem 6
Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]
40 replies
tenniskidperson3
May 1, 2013
dipinsubedi
Yesterday at 3:43 PM
Alcumus vs books
UnbeatableJJ   12
N Yesterday at 2:59 PM by pingpongmerrily
If I am aiming for AIME, then JMO afterwards, is Alcumus adequate, or I still need to do the problems on AoPS books?

I got AMC 23 this year, and never took amc 10 before. If I master the alcumus of intermediate algebra (making all of the bars blue). How likely I can qualify for AIME 2026?
12 replies
UnbeatableJJ
Apr 23, 2025
pingpongmerrily
Yesterday at 2:59 PM
Zsigmondy's theorem
V0305   17
N Yesterday at 7:22 AM by whwlqkd
Is Zsigmondy's theorem allowed on the IMO, and is it allowed on the AMC series of proof competitions (e.g. USAJMO, USA TSTST)?
17 replies
V0305
May 24, 2025
whwlqkd
Yesterday at 7:22 AM
USAJMO problem 3: Inequality
BOGTRO   106
N Monday at 11:12 PM by Learning11
Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
106 replies
BOGTRO
Apr 24, 2012
Learning11
Monday at 11:12 PM
what the yap
KevinYang2.71   31
N Monday at 3:26 PM by Blast_S1
Source: USAMO 2025/3
Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair $A,\,B$ of cities, they are connected with a road along the line segment $AB$ if and only if the following condition holds:
[center]For every city $C$ distinct from $A$ and $B$, there exists $R\in\mathcal{S}$ such[/center]
[center]that $\triangle PQR$ is directly similar to either $\triangle ABC$ or $\triangle BAC$.[/center]
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.

Note: $\triangle UVW$ is directly similar to $\triangle XYZ$ if there exists a sequence of rotations, translations, and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.
31 replies
KevinYang2.71
Mar 20, 2025
Blast_S1
Monday at 3:26 PM
Mustang Math Recruitment is Open!
MustangMathTournament   3
N Monday at 3:21 PM by Puzzlebooks206
The Interest Form for joining Mustang Math is open!

Hello all!

We're Mustang Math, and we are currently recruiting for the 2025-2026 year! If you are a high school or college student and are passionate about promoting an interest in competition math to younger students, you should strongly consider filling out the following form: https://link.mustangmath.com/join. Every member in MM truly has the potential to make a huge impact, no matter your experience!

About Mustang Math

Mustang Math is a nonprofit organization of high school and college volunteers that is dedicated to providing middle schoolers access to challenging, interesting, fun, and collaborative math competitions and resources. Having reached over 4000 U.S. competitors and 1150 international competitors in our first six years, we are excited to expand our team to offer our events to even more mathematically inclined students.

PROJECTS
We have worked on various math-related projects. Our annual team math competition, Mustang Math Tournament (MMT) recently ran. We hosted 8 in-person competitions based in Washington, NorCal, SoCal, Illinois, Georgia, Massachusetts, Nevada and New Jersey, as well as an online competition run nationally. In total, we had almost 900 competitors, and the students had glowing reviews of the event. MMT International will once again be running later in August, and with it, we anticipate our contest to reach over a thousand students.

In our classes, we teach students math in fun and engaging math lessons and help them discover the beauty of mathematics. Our aspiring tech team is working on a variety of unique projects like our website and custom test platform. We also have a newsletter, which, combined with our social media presence, helps to keep the mathematics community engaged with cool puzzles, tidbits, and information about the math world! Our design team ensures all our merch and material is aesthetically pleasing.

Some highlights of this past year include 1000+ students in our classes, AMC10 mock with 150+ participants, our monthly newsletter to a subscriber base of 6000+, creating 8 designs for 800 pieces of physical merchandise, as well as improving our custom website (mustangmath.com, 20k visits) and test-taking platform (comp.mt, 6500+ users).

Why Join Mustang Math?

As a non-profit organization on the rise, there are numerous opportunities for volunteers to share ideas and suggest projects that they are interested in. Through our organizational structure, members who are committed have the opportunity to become a part of the leadership team. Overall, working in the Mustang Math team is both a fun and fulfilling experience where volunteers are able to pursue their passion all while learning how to take initiative and work with peers. We welcome everyone interested in joining!

More Information

To learn more, visit https://link.mustangmath.com/RecruitmentInfo. If you have any questions or concerns, please email us at contact@mustangmath.com.

https://link.mustangmath.com/join
3 replies
MustangMathTournament
May 24, 2025
Puzzlebooks206
Monday at 3:21 PM
[TEST RELEASED] OMMC Year 5
DottedCaculator   173
N Monday at 3:01 PM by drhong
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
173 replies
DottedCaculator
Apr 26, 2025
drhong
Monday at 3:01 PM
Distinct Integers with Divisibility Condition
tastymath75025   17
N May 9, 2025 by quantam13
Source: 2017 ELMO Shortlist N3
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu
17 replies
tastymath75025
Jul 3, 2017
quantam13
May 9, 2025
Distinct Integers with Divisibility Condition
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Source: 2017 ELMO Shortlist N3
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tastymath75025
3223 posts
#1 • 2 Y
Y by Adventure10, Mango247
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu
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nmd27082001
486 posts
#2 • 1 Y
Y by Adventure10
My solution:i will prove that there is no sequence sastisfied the problem
Let $p|C$ and $t=v_p(C)$ and $b_k=v_p(a_k)$
So we need $b_{k+1}.k<=tk+\sum_{i=1}$
Consider sequence $c_n=\sum_{k=1}^{n}\frac{1}{k}$
We can easily prove that $b_n<=[ta_n]+b_1$
But note that there exist m such that for all $n>m$,$n-b_1>[ta_n]$
Which implies there exist infinitive number i,j:$b_i=b_j$
Let S={(i,j),$b_i=b_j$}
Consider prime $q=!p$ of C
We found that in S there are infinitive (t,s):$v_p(a_t)=v_p(a_s)$ and $v_q(a_t)=v_q(a_s)$
Continue this with all prime divisor of C we are done
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ThE-dArK-lOrD
4071 posts
#3 • 2 Y
Y by Adventure10, Mango247
My lengthly solution
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jun 29, 2018, 8:03 PM
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Wizard_32
1566 posts
#4 • 2 Y
Y by centslordm, PRMOisTheHardestExam
Nice problem. I will give a sketch.
tastymath75025 wrote:
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.
The answer is no. Suppose not for a fixed $C.$ Consider any prime $p.$ Then the problem gives
$$k\nu_p (a_{k+1}) \le k\nu_p(C)+\nu_p(a_1)+\dots+\nu_p(a_k).$$Now we have the following key claim which can be proven by simple induction:
Claim: Let $\text{H}_n$ denote the $n$th harmonic number. Then
$$\nu_p( a_n)-\nu_p(a_1) \le \text{H}_n \nu_p(C).$$
The key hypothesis we need now is that $a_i$ are pairwise distinct. The claim gives $\nu_p(C)=0 \implies \nu_p(a_n) \le \nu_p(a_1).$ In particular $\nu_p(a_m)$ is eventually constant. So ignore primes for which $\nu_p(C)=0.$ This means we only have a finite set of prime factors to worry about for the sequence now.

The claim clearly gives $\nu_p(a_n/a_1) \le A \log n$ for some constant $A$ and all primes $p.$ Now it is not too hard to see that we can find arbitrarily large intervals over which $\left \lfloor A\log x \right \rfloor$ is constant. Since the set of prime factors of $\{a_k\}$ is finite now, hence by pigeonhole two terms $a_i,a_j$ will have the same $\nu_p$ for all primes $p,$ hence will be equal, a contradiction. $\square$

EDIT: Ok, I finally typed the elaborated version of the sketch: Full Proof
This post has been edited 3 times. Last edited by Wizard_32, Nov 6, 2020, 6:51 AM
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mathlogician
1051 posts
#5
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There is no such sequence for any $C$. For convenience, define $x_n = v_p(a_n)$ for every integer $n$. Furthermore, define $h_n = \frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}$. Finally, let $S = v_p(C)$. The pith of the problem lies in the following claim:

Claim: $x_n \leq x_1 + Sh_n$.

Proof: By strong induction. Remark that the condition tells us $$x_n \leq S + \frac{x_1+x_2+\dots+x_{n-1}}{n-1}.$$
This claim implies that for any prime $p$, there exists an arbitrary long sub-sequence of terms $a_i,a_{i+1},\dots,a_j$ such that $\nu_p(a_i) = \nu_p(a_{i+1}) = \dots = \nu_p(a_j)$ for sufficiently large $i$ and $j$. Note that for primes $p$ in $a_1$ but not $C$, $\nu_p(a_i) \leq \nu_p(a_1)$, so there are only a finite number of possible terms with equal $v_p$ for all primes $p \in C$. Therefore, for sufficiently large $i$ there will be two equal terms, the end.
This post has been edited 1 time. Last edited by mathlogician, Jan 25, 2021, 2:43 PM
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Sprites
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#6
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tastymath75025 wrote:
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu

Define $\mathbf{H}_n=1+\frac{1}{2}+.............+\frac{1}{n}$
The condition is equivalent to $k\nu_p(a_{k+1}) \le k \nu_p(C)+\sum_{j=1}^k \nu_p(a_j)$
Claim: $\nu_p(a_n)-\nu_p(a_1) \le\mathbf{H}_n \nu_p(C)$
Proof: Obvious by strong induction.
This implies that $\nu_p(a_j)$ is bounded and when $j$ is sufficiently large we will get $\nu_p(a_j)=\nu_p(a_{j+1}),\forall p \implies a_j=a_{j+1} $, contradiction.
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IvoBucata
46 posts
#7
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I'll prove that for no $C$ there exists such a sequence. I'll start with the following claim:

Claim: $v_p(a_{k+1})\leq (1+\frac12 + \cdots +\frac1k)v_p(C) + v_p(a_1)$ for every prime $p$.

Proof : It's not hard to show this by induction.

We can find arbitrarily large intervals over which $(1+\frac12 + \cdots +\frac1k)v_p(C) + v_p(a_1)$ is constant, but note that $a_{k+1}$ can take only values which are divisors of $a_1C^{X}$ where $X=\lceil (1+\frac12 + \cdots +\frac1k)v_p(C)\rceil $. Now at some point $d(a_1C^{X})$ is going to be smaller than the length of these intervals because $1+\frac12 + \cdots +\frac1k$ grows very slowly, so the $a_i$'s can't be distinct over these intervals.
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guptaamitu1
658 posts
#8
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Here's a different proof (we will bound the $v_p$'s differently without the harmonic number).
We will show for any $C$, such a sequence does not exist. Assume contrary. Fix $C$ and such a sequence. Call a prime nice if it divides some $a_i$. Note all nice primes must divide $C \cdot a_1$, in particular number of nice primes is finite. Fix a $c > 1$ such that $c > v_p(C)$ for all primes $p$. Fix any nice prime $p$. Let $b_i = v_p(a_i)$. Then we know that
$$k b_{k+1} \le kc + b_1 + b_2 + \cdots + b_k ~~ \forall ~ k \ge 2 \qquad \qquad (1)$$Intuitively, we will show that $b_i$'s grow slowly. Call an $i \ge 2$ a peak if $b_i > b_{i-1},\ldots,b_1$. Let $i_1+1 < i_2 + 1< i_3 + 1 < \cdots$ be all the peaks.


Claim 1: For all $n \ge 1$ we have
$$\frac{i_1 + i_2 + \cdots + i_n}{i_n} \le c ~ \iff ~ \frac{i_1 + \cdots + i_{n-1}}{i_n} \le c-1$$
Proof: Define $$f(m) = (b_m - b_1) + (b_m - 2) + \cdots + (b_m - b_{m-1})~ \forall ~ m \ge 2$$. Note $(1)$ is equivalent to
$$ \frac{f(k)}{k-1} \ge c  \qquad \qquad (2) $$Observe that for any $n \ge 1$ that
\begin{align*}
f(i_{n + 1} + 1) &= \sum_{j=1}^{i_{n+1}}\left( b_{i_{n+1} + 1} - b_j \right) = \sum_{j= i_n + 1}^{i_{n+1}} (b_{i_{n+1} + 1} - b_j) + \sum_{j=1}^{i_n} (b_{i_{n+1} + 1} - b_j) \\ &\ge  (i_{n+1} - i_n)(b_{i_{n+1} + 1} - b_{i_n + 1}) + \sum_{j=1}^{i_n} \bigg( (b_{i_{n+1} + 1} - b_{i_{n} + 1}) + (b_{i_n + 1} - b_j) \bigg) \\
&= i_{n+1}(b_{i_{n+1} + 1} - b_{i_n + 1}) + f(i_n + 1) \ge i_{n+1} + f(i_n + 1)
\end{align*}Now note $f(i_1 + 1) \ge i_1$. So it follows that
$$f(i_n + 1) \ge i_1 + i_2 + \cdots + i_n ~~ \forall ~ n \ge 1$$Plugging $k = i_n + 1$ in $(2)$ and using above implies our claim. $\square$


Claim 2: $\exists$ an $\alpha > 1$ such that $i_k \ge \alpha^{k-1} ~ \forall ~ k \ge c+2$.

Proof: We will only use each $i_n \ge 1$ and Claim 1 to prove this. Observe $i_1 + \cdots + i_{c+1} > c+1$, since all of them cannot be $(1)$ (by Claim 1 for $n=c+1$). Now choose a very small $\alpha > 1$ such that:
\begin{align*}
i_1 + i_2 + \cdots + i_{c+1} \ge 1 + \alpha + \cdots + \alpha^c \\
\frac{1}{\alpha} + \frac{1}{\alpha^2} + \cdots + \frac{1}{\alpha^c} \ge c-1
\end{align*}We show this $\alpha$ works. Suppose for some $t \ge c+1$ it holds that $i_1 + \cdots + i_t \ge 1 + \alpha + \cdots + \alpha^{t-1}$. We will show $i_{t+1} \ge \alpha^t$ (note this would imply our claim by induction). $k=t+1$ in $(3)$ gives
$$i_t \ge \frac{i_1 + \cdots + i_t}{c-1} \ge \frac{1 + \alpha + \cdots + \alpha^{t-1}}{c-1} = \frac{\alpha^t - 1}{(c-1)(\alpha - 1)}  $$So it suffices to show
\begin{align*}
 \frac{\alpha^t - 1}{(c-1)(\alpha - 1)} \ge \alpha^t \iff (c-1)(\alpha-1) \le \frac{\alpha^t - 1}{\alpha^t} = 1 - \frac{1}{\alpha^t} \iff (c-1)(\alpha -1) + 1 - \frac{1}{\alpha^t} \le 0
\end{align*}Indeed,
\begin{align*}
(c-1) (\alpha -1) + 1 - \frac{1}{\alpha^t} &= (\alpha -1) \left( (c-1) - \left( \frac{1}{\alpha} + \frac{1}{\alpha^2} + \cdots + \frac{1}{\alpha^{t-1}} \right) \right) \\ &\le (\alpha -1) \left(c-1 -\left( \frac{1}{\alpha} + \frac{1}{\alpha^2} + \cdots + \frac{1}{\alpha^c} \right) \right)  \le 0
\end{align*}This proves our claim. $\square$


Claim 3 (Key Result): For large $t$, all of $b_1,b_2,\ldots,b_{\alpha^t}$ are at most $c(t+1)$.

Proof: Observe that $b_{i_{n+1} + 1} - b_{i_n + 1} \le c ~ \forall ~ n \ge 1$. As $b_{i_1 + 1} = b_2 \le b_1 + c$, so it follows $$b_{i_m + 1} \le b_1 + cm ~~ \forall ~ m \ge 1$$Now $i_{t+1} + 1 \ge \alpha^t+1$, thus all of $b_1,b_2,\ldots,b_{\alpha^t-1}$ are $\le b_{i_t + 1} \le b_1 + tc$, and $b_1 + tc \le t(c+1)$ for large $t$. This proves our claim. $\square$


Now let $p_1,p_2,\ldots,p_k$ be all the nice primes, and $\alpha_1,\alpha_2,\ldots,
\alpha_k > 1$ be any numbers for which Claim 3 is true. Let $\alpha = \min(\alpha_1,\alpha_2,\ldots,\alpha_k)$. For a large $t$, look at the numbers
$$ a_1,a_2,\ldots,a_{\alpha^t} $$By Claim 3 we know that for any $p_i$ adic valuation of any of these numbers is $\le c(t+1)$. It follows number of distinct numbers between them is at most
$$ \bigg(c(t+1) + 1 \bigg)^k$$As all $a_i$'s are distinct, so this forces
$$ \bigg( c(t+1) + 1 \bigg)^k \ge \alpha^t \qquad \text{for all large } t $$But this is a contradiction as exponential functions grow faster than polynomial functions. This completes the proof. $\blacksquare$


Motivation: The problem isn't even true if $a_i$'s are not given to be distinct, so we somehow had to prove that. So we basically had to prove the $v_p$'s don't grow fast. Now $(1)$ was only interesting for peaks. So it was natural to consider peaks. Now after getting Claim 1, I was sure we only have to use Claim 1 and ignore all other conditions on peaks, which along with $b_{i_{n+1} + 1} - b_{i_n + 1} \le c$ would give us the sequence $\{b_k\}_{k \ge 1}$ doesn't grow fast. So we only had to prove Claim 2. Basically, we conjecture $i_k \ge \alpha^{k-1}$ for all $k$ (for some $\alpha > 1$). We then find the sufficient conditions on $\alpha$ for which we can prove this by induction. The two equations in proof of Claim 2 were precisely those. Now we had some problems like it might happen $i_1 = i_2 = 1$, but that was easy to fix by showing $i_k \ge \alpha^{k-1}$ for large $k$.
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IAmTheHazard
5003 posts
#9 • 1 Y
Y by centslordm
The answer is no such $C$. Let $p$ be some prime, and let $\nu_p(C)=c$, $x_i=\nu_p(a_i)$ for $i \geq 1$. Viewing the divisibility condition in $p$-adic terms only, it is equivalent to
$$kx_{k+1} \leq kc+x_1+\cdots+x_k.$$Let $(H_n)$ denote the sequence of harmonic numbers. The crux of the problem is the following:

Claim: $x_n-x_1 \leq cH_{n-1}$.
Proof: Shifting $(x_i)$ doesn't modify the truth of the condition, so WLOG let $x_1=0$. We now use strong induction:
\begin{align*}
(n-1)c+x_1+\cdots+x_{n-1}&\leq(n-1)c+c\left(\left(\frac{1}{1}\right)+\left(\frac{1}{1}+\frac{1}{2}\right)+\cdots+\left(\frac{1}{1}+\cdots+\frac{1}{n-2}\right)\right)\\
&=c\left(1+(n-2)+\frac{n-2}{1}+\frac{n-3}{2}+\cdots+\frac{1}{n-2}\right)\\
&=c\left(\frac{n-1}{1}+\frac{n-1}{2}+\cdots+\frac{n-1}{n-2}+\frac{n-1}{n-1}\right)\\
&=c(n-1)H_{n-1},
\end{align*}so $(n-1)x_n \leq c(n-1)H_{n-1} \implies x_n \leq cH_{n-1}$ as desired.

The claim implies that $\nu_p(a_n)$ is zero if $p \nmid a_1C$, $O(1)$ if $p \nmid C$ but $p \mid a_1$, and $O(\log n)$ otherwise. Suppose there are $a$ distinct primes dividing $C$ and $b$ distinct primes dividing $a_1$ but not $C$. Then there are $O(1^b(\log n)^a)\sim O((\log n)^a)$ choices for the value of $a_n$we have $O(1)$ options for $\nu_p(a_n)$ if $p \mid a_1$ and $p \nmid C$, and $O(\log n)$ options for $\nu_p(a_n)$ if $p \mid C$. But $n$ dominates $O((\log n)^a)$, so by Pigeonhole for sufficiently large $n$ there must exist two non-distinct elements of the sequence: contradiction. $\blacksquare$
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PianoPlayer111
40 posts
#10 • 2 Y
Y by Mango247, Mango247
The anwser is, that there is no sequnce, which satisfies the problem.
Proof:
Now choose any $a_{k+1} < a_{k+2}$.
First of all we see, that $a_{k+1}^k \mid C^k a_1 a_2 ... a_k$ $\Rightarrow$ $k \cdot v_p(a_{k+1}) \le k \cdot v_p(C) + v_p(a_1) + v_p(a_2) + ... +v_p(a_k)$ $\Rightarrow$ $k \cdot (v_p(a_{k+1} - v_p(C)) \le v_p(a_1) + v_p(a_2) + ... +v_p(a_k)$ $(1)$. Similarly and because of $a_{k+1} < a_{k+2}$, we get $(k+1) \cdot (v_p(a_{k+2} - v_p(C)) \le v_p(a_1) + v_p(a_2) + ... + v_p(a_{k+1})$ $(2)$, for every prime dividing both sides of the equations. Now we calculate $(2) - (1)$, which is equivalent after some boring basics in arithmetic to $(k+1) \cdot [v_p(a_{k+2}) - v_p(a_{k+1})] \le v_p(C)$ $\Rightarrow$ $v_p((\frac{a_{k+2}}{a_{k+1}})^{k+1}) \le v_p(C)$ $\Rightarrow$ $(\frac{a_{k+2}}{a_{k+1}})^{k+1} \mid C$ $\Rightarrow$ $(\frac{a_{k+2}}{a_{k+1}})^{k+1} \le C$. But this inequality is not true for a large $k$, hence we get a contradiction and we are done :ninja:
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aaabc123mathematics
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#11
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PianoPlayer111 wrote:
{a_{k+1}})^{k+1} \le C$. But this inequality is not true for a large $k$, hence we get a contradiction and we are done  :ninja:[/quote] 
Why ?If {a_{k+1}})^{k+1} \le C$. isn't very big ,how can explain the numer is larger than C?
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oty
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#12
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Nice problem
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DongerLi
22 posts
#13
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How do the $\nu_p$'s grow? Define $S$ as the set of prime divisors of $a_1C$. Note that the given sequence has no prime factors outside of $S$. Let prime $p \in S$ be arbitrary. Denote $b_n = \nu_p(a_n)$ for all positive integers $n$. The given condition rewrites as:
\[kb_{k + 1} \leq k\nu_p(C) + b_1 + b_2 + \cdots + b_k.\]Define $s_n = \frac{b_1 + b_2 + \cdots + b_n}{n}$ as the average of the first $n$ elements of our new sequence. Adding $k(b_k + \cdots + b_1)$ to both sides of this inequality and dividing through by $k(k + 1)$ lends us a useful inequality.
\[\frac{b_{k+1}+\cdots + b_1}{k+1} \leq \frac{1}{k+1} \nu_p(C) + \frac{b_k + \cdots + b_1}{k}.\]Applying this inequality repeatedly gives us the following inequality.
\[s_k \leq \left(\frac{1}{k} + \cdots + \frac{1}{2}\right)\nu_p(C) + s_1.\]Plugging back into the original inequality and bounding the harmonic series gives us:
\[b_{k+1} \leq \nu_p(C) + s_k \leq \left(\frac{1}{k} + \cdots + \frac{1}{1}\right)\nu_p(C) + b_1 \leq \ln(k+1) + b_1.\]Something is seriously wrong, and we are very happy about it. Since the above is true for all primes $p$, there are $O((\ln k)^{|S|})$ different possibilities for the values of $a_1, \dots, a_k$. Since
\[O((\ln k)^{|S|}) < k\]as $k$ approaches infinity, there must exist two terms in the sequence that are equal.
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thdnder
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#15
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Since $a_{k + 1}^k \mid C^k a_{1}a_{2}\dots a_{k}$ and $C$ is constant, so the set of primes dividing an element in $(a_{n})_{n \ge 1}$is finite. Let $p$ be a arbitrary prime dividing one of $a_{1}, a_{2}, \dots$. Then the divisibility condition becomes $k\nu_{p}(a_{k + 1}) \le k\nu_{p}(C) + \nu_{p}(a_{1}) + \nu_{p}(a_{2}) + \dots + \nu_{p}(a_{k})$. Now consider the following claim:

Claim:
For $n \ge 2$, we have $\nu_{p}(a_{n}) \le \nu_{p}(a_{1}) + \nu_{p}(C)(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n - 1})$.

Proof:

Apply strong induction on $n$. Base case is clear. For the induction step, $\nu_{p}(a_{n + 1}) \le \nu_{p}(C) + \frac{1}{n}(\nu_{p}(a_{1}) + \nu_{p}(a_{2}) + \dots + \nu_{p}(a_{n})) \le \nu_{p}(C) + \nu_{p}(a_{1}) + \frac{1}{n}(\nu_{p}(C)((\frac{1}{1}) + (\frac{1}{1} + \frac{1}{2}) + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3}) + \dots + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}))) = \nu_{p}(a_{1}) + \nu_{p}(C)(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n})$. $\blacksquare$

Since the set of primes dividing an element in $(a_{n})_{n \ge 1}$is finite, let $a$ be a number of elements in the set of primes dividing an element in $(a_{n})_{n \ge 1}$. Then taking large $N$, we see that there are most $(O(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{N})^k) = O((\ln N)^k) < N$ distinct values in $a_{1}, a_{2}, \dots, a_{N}$, a contradiction. Therefore there are no such $C$. $\blacksquare$
This post has been edited 1 time. Last edited by thdnder, Sep 29, 2023, 2:24 PM
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Mathandski
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#16
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Subjective Rating (MOHs) $       $
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cursed_tangent1434
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#17
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The first observation is that the set of all primes dividing some term of $(a_i)$ is finite as any prime dividing $a_i$ for $i >1$ must divide $Ca_1$ by induction due to the given divisibility condition. Now, we prove our key bound.

Claim : For all primes $p \mid Ca_1$ and positive integers $k$,
\[\nu_p(a_k) \le \nu_p(a_1)+\nu_p(C)\left(\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k-1}\right)\]
Proof : This is a simple calculation. Note that the condition implies that for all $k \ge 1$,
\[(k-1)\nu_p(a_k)\le (k-1)\nu_p(C) + \nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-1})\]Now replacing $\nu_p(a_2),\dots , \nu_p(a_{k-1})$ inductively we obtain,
\begin{align*}
        \nu_p(a_k) & \le \nu_p(C) + \frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-1})}{k-1}\\
        & \le \nu_p(C) + \frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-2})}{k-1} + \frac{\nu_p(C)+\frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-2})}{k-2}}{k-1}\\
        &= \nu_p(C) + \frac{\nu_p(C)}{k-1} + \frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(k-2)}{k-2}\\
        & \vdots\\
        & \le \nu_p(C)+\frac{\nu_p(C)}{k-1} + \frac{\nu_p(C)}{k-2}+\dots + \frac{\nu_p(C)}{2} + \frac{\nu_p(a_1)}{2} +\frac{\nu_p(a_1)}{2}\\
        &= \nu_p(a_1) + \nu_p(C) \left (\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k-1}\right)
    \end{align*}which proves the claim.

Note that we may not have $a_k < k$ for all sufficiently large $k$. This is because if $a_k <k$ for all $k \ge N$ then for all $k \ge M=\max(a_1,a_2,\dots , a_{N-1})$, $a_k <M$ but this is a set of $M$ distinct positive integers bounded above by $M-1$ which is a clear contradiction.

However, revisiting our claim, this means for all $k$ the number of distinct positive integers that are possible for $a_1,a_2,\dots , a_k$ is
\[ \prod_{i=1}^r \left(\left \lfloor \nu_{p_i}(a_1) + \nu_{p_i}(C)H_{k-1}\right \rfloor +1\right) = O\left(\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k}\right)^{r} = O(\ln k)^r < k\]for all sufficiently large $k$ where $p_1,p_2,\dots , p_r$ is the set of all primes dividing $Ca_1$. But this is a contradiction to what we noted above so the desired is impossible for all $C>1$ and we are done.
This post has been edited 1 time. Last edited by cursed_tangent1434, Apr 22, 2025, 9:12 AM
Reason: minor details
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ihategeo_1969
242 posts
#18
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We will prove $\boxed{\text{no such sequence exists}}$.

See that $\text{rad}(a_1a_2 \dots) \mid \text{rad}(a_1C)$; hence we can write each $a_i$ as $p_1^{e_1} \dots p_t^{e_t}$ for fixed primes $p_1$, $\dots$, $p_t$ ($t$ is finite).

Claim: $\nu_p(a_n) \le O( \log n)$ for each $p \in \{p_1,\dots,p_t\}$.
Proof: See that \[\nu_p(a_n) \le \nu_p(C)+\frac{\nu_p(a_1)+\dots+\nu_p(a_{n-1})}{n-1}\]By simple induction we have that \begin{align*}
\nu_p(a_n) & \le \nu_p(a_1)+\nu_p(C) \left(1+\dots+\frac{1}{n-1} \right) 
 \le \nu_p(a_1)+ \nu_p(C) \left(1+\int_1^{n-1} \frac{dx}{x} \right) 
 = \nu_p(a_1)+\nu_p(C) \left(1+\log(n-1) \right)=O(\log n) \end{align*}As required. $\square$

Now assign a sequence $(\nu_{p_1}(a_n), \dots, \nu_{p_t}(a_n))$ to each $a_n$. Now fix some large $N$ and look at the number of distinct sequences among $a_1$, $\dots$, $a_N$ which is \[O(\log N)^t \ll N \]And hence by PHP, two the sequences must coincide and hence their corresponding numbers are same which is a contradiction.
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quantam13
115 posts
#19
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The main thing is to bound $\nu_p$ for a prime $p$. The condition gives $$k\nu_p (a_{k+1}) \le k\nu_p(C)+\nu_p(a_1)+\dots+\nu_p(a_k)$$Now the key claim is that $$\nu_p( a_n)-\nu_p(a_1) \le \text{H}_n \nu_p(C)$$for all integers $n$ where $\text{H}_n$ is the $n$th harmonic number, which can be proven with simple strong induction. Now this $\nu_p$ can be combined with some analytic estimates to get that the $\nu_p$ of $a_i$ becomes eventually constant for any prime $p\mid C$ and bounded for any prime $p\nmid C$ and these with PHP and polylogarithmic bounds, we get that $a_i$ is not an injective sequence, a contradiction
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