ka March Highlights and 2025 AoPS Online Class Information
jlacosta0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.
Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!
Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.
Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form: Suggestion Form
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Please remember that if you find a bug outside of FTW! (after refreshing to make sure it’s not a glitch), make sure you’re following the How to write a bug report instructions and using the proper format
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2. Click on any thread.
3. Click quickly on a different thread. Expected behavior: To see the second thread. Frequency: Every time Operating system: Mac OS X Browser: Chrome and Firefox Additional information: Only happens in the Site Support forum. My tablet is school issued, but I have the problem at both school and home.
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Let be a positive integer. We say that a positive integer is -good if divides for some integer . We say a positive integer is -bad if it is not -good.
(a) Is it true that for every positive integer there exist consecutive -bad positive integers?
(b) Is it true that for every positive integer there exist consecutive -good positive integers?
Let be a set of at least three points of the plane in general position. Prove that there exists a non-intersecting polygon whose vertices are exactly the points of .
Let be real numbers such that Prove thatLet be real numbers such that Prove thatLet be real numbers such that Prove thatLet be real numbers such that Prove that
As outlined in this (locked) post, the forum/blog creation message is out of date. It has been over a year since that post, and it still has not changed. I don't know if this got lost somewhere in the "passing this along" chain or if it was determined to be irrelevant, but it seems like a relatively simple fix ;)
Pressing 'go down button' always creates a gray box on the last post
Craftybutterfly13
NMar 14, 2025
by Craftybutterfly
Summary of the problem: Pressing go down to last post button always creates a gray box overlapping last post
Page URL: any forum
Steps to reproduce:
1. Go to any topic in a forum
2. The gray box at the bottom overlaps part of the first post
Expected behavior: Should not show a gray box
Frequency: 100% of the time
Operating system(s): Linux HP EliteBook 835 G8 Notebook PC
Browser(s), including version: Chrome 133.0.6943.142 (Official Build) (64-bit) (cohort: Stable)
Additional information: It works on any other device, on my iPhone XR, a MacOS, and my iPad. Took the screenshot a month ago. The gray box still appears
k Somehow I broke the unique contest collection naming...
Equinox81
NMar 14, 2025
by jlacosta
Somehow I broke the unique contest collection naming scheme; for what it's worth, I also ran into an AJAX timeout error while trying to make this particular post collection for the first time, and pressed "create" twice.
Not sure if there's anything to be done here, but are there potentially unforeseen consequences here?
Summary of the problem: In the New Forums collection, many of the forums are duplicated, although some have slightly different information (likely due to the forum being edited by its admin(s)). Typically only one of the two has threads, and I believe this is the only way to access
Page URL: https://artofproblemsolving.com/community/c74_new_forums
Steps to reproduce:
1. Go to the link
2. You should see duplicates of some forums
Expected behavior: Each forum appears once
Frequency: Every time
Additional information: Typically only one of the two has threads, and from creating forums in the past I believe this is the only way to access the duplicate. Also not all of the forums are duplicated. Another reason to suspect these are duplicates are the forums with similar names have the same admin.
For the past while, the search function in private messages hasn’t been working. Whenever I search for anything, it says, “No topics here!” after trying to load for a while. I’ve tried different devices (laptop and ipad) and browsers (chrome on both devices, safari on ipad, and microsoft edge on laptop), and the results are the same. I’ve also had friends say the same happens for them.
I'm wondering how to look at my next classes. I'm not entirely sure if I'm registered for a class. I tried looking around on the My AoPS page, but didn't find anything that had to do with the course I wanted.
I attempted to abandon the Alcumus quest Hit The Gym: Accuracy. I clicked on it and selected abandon, but the popup disappeared and the quest was still there. There were no changes to the log, and my daily abandon was used, as when I clicked the quest again it said that all of my abandons were used.
Observe that , since . So, is a power of . Let . Note that both are prime, but one of , must be divisible by , so one of them must be . Of course, , so .
We have . Note that , so . means that is even, let . We have .
So, . Note for , for , by simply inducting. So, we have an obvious contradiction by size and we are reduced to a minimal case check which is left as an excercise to the reader.
The equation modulo gives implying is a power of . At least one of , and must be a multiple of , so it must be one of or . It thus follows that . Now Zsigmondy finishes as it implies . This gives us our only solution of .
Observe that the left hand side is divisible by and so for some Clearly, we have that is a prime. Hence, if then However, we already know that 3 divides one of It also can't divide as this would imply Hence, The rest is trivial by Zsigmondy
This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 7, 2025, 9:28 PM
Using modulo , we find , i.e., for a suitable . Suppose . As is a prime, we must have is a prime. At the same time, is also a prime, which is possible only when . Thus, and . Using modulo 3, we find that is even, set and factorize Clearly . So, . Since cannot be simultaneously divisible by 3, we must have and . The cases are easy, is the only solution. Let . Then, using modulo 8, we find is even. Moreover, using modulo 5, we find . So, where are both even, which doesn't have any solutions.
Since divides the RHS, it's clear that is a power of . Note that this means , so divides one of , from which we obviously have . By Zsigmondy (or just noting that is even and factoring) we can find that , so must hold, and it clearly works.
edit (even easier way to show ): If , then neither of are , so and , so divides the RHS, which is absurd as .
This post has been edited 2 times. Last edited by megarnie, Mar 7, 2025, 10:11 PM
Okay so inmediately gives that is a power of but also if then and thus which can't happen so , now by Zsigmondy we have for a contradiction so if then by checking so contradiction and if then and thus we are done .
Sketch of my in-contest solution:
1. Take mod to get for some
2. Take mod and by orders we get which means
3. Then by LTE we get or
4. Realize that since is a power of then one of is a multiple of which obviously implies
5. If then by mod we have which gives by orders.
I kinda forgot what I did afterwards. If I can get my submission/scratch work from my proctor I will complete the solution.
This post has been edited 4 times. Last edited by PEKKA, Mar 8, 2025, 7:25 PM
If I am not mistaken, heavy theorems can be very easily avoided?
The right-hand side is divisible by , so divides and hence . In particular, for some . But then and if , then is divisible by and hence not prime, contradiction. Hence .
The equation now reads . By mod we get that is even. With we factor and the greatest common divisor of the factors on the right is . Since and , we get and . Work only with the former -- if , then and no solution for . If , then by mod we get is even, so , factor and by mod , contradiction. Therefore, the only solution is .