It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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.

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Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

Special AIME Problem Seminar B
Sat & Sun, Feb 1 - Feb 2 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

F=ma Problem Series
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0 replies
jlacosta
Feb 2, 2025
0 replies
six primes
IceWolf10   12
N a few seconds ago by chenghaohu
Source: BAMO 2022 D/2
Suppose that $p,p+d,p+2d,p+3d,p+4d$, and $p+5d$ are six prime numbers, where $p$ and $d$ are positive integers. Show that $d$ must be divisible by $2,3,$ and $5$.
12 replies
IceWolf10
Mar 3, 2022
chenghaohu
a few seconds ago
Inequality
Equinox8   5
N 14 minutes ago by sqing
Source: IrMO 2024 #8
Let $a,b,c$ be positive real numbers with $a \leq c$ and $b \leq c$. Prove that $$ (a +10b)(b +22c)(c +7a) \geq 2024 
 abc.$$
5 replies
Equinox8
6 hours ago
sqing
14 minutes ago
$a_{n+1}=\frac{1}{2\lfloor a_n \rfloor -a_n+1}$
Jjesus   2
N 15 minutes ago by mikestro
Source: Philippine Mathematical Olympiad 2024 P6
The sequence $\{a_n\}_{n\ge 1}$ of real numbers is defined as follows:
$$a_1=1, \quad \text{and}\quad a_{n+1}=\frac{1}{2\lfloor a_n \rfloor -a_n+1} \quad \text{for all} \quad n\ge 1$$Find $a_{2024}$.
2 replies
Jjesus
Feb 20, 2024
mikestro
15 minutes ago
adjacent striped balls
IceWolf10   7
N 18 minutes ago by chenghaohu
Source: BAMO-12 2022 P1
The game of pool includes $15$ balls that fit within a triangular rack as shown:
IMAGE
Seven of the balls are "striped" (not colored with a single color) and eight are "solid" (colored with a single color). Prove that no matter how the $15$ balls are arranged in the rack, there must always be a pair of striped balls adjacent to each other.
7 replies
IceWolf10
Mar 3, 2022
chenghaohu
18 minutes ago
No more topics!
The game cannot terminate
Fermat -Euler   4
N Jun 26, 2020 by TwilightZone
Source: IMO Shortlist 1994, C5
$ 1994$ girls are seated at a round table. Initially one girl holds $ n$ tokens. Each turn a girl who is holding more than one token passes one token to each of her neighbours.

a.) Show that if $ n < 1994$, the game must terminate.
b.) Show that if $ n = 1994$ it cannot terminate.
4 replies
Fermat -Euler
Oct 22, 2005
TwilightZone
Jun 26, 2020
The game cannot terminate
G H J
Source: IMO Shortlist 1994, C5
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Fermat -Euler
444 posts
#1 • 1 Y
Y by Adventure10
$ 1994$ girls are seated at a round table. Initially one girl holds $ n$ tokens. Each turn a girl who is holding more than one token passes one token to each of her neighbours.

a.) Show that if $ n < 1994$, the game must terminate.
b.) Show that if $ n = 1994$ it cannot terminate.
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orl
3647 posts
#2 • 2 Y
Y by Adventure10, Mango247
A harder variant of this problem can be discussed here.
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ophiophagous
79 posts
#3 • 1 Y
Y by Adventure10
Denote the girl originally holding all coins $A$. If $n<1994$, we just remove the girl opposite $A$ and consider the girls sitting in a row with A in the center. We use a string to represent the coins in each girl's hand. I'll just write out the numbers to the right of $A$.

0 ...
20...
02...
101...
201...
011...
111...
211...
021...
202...


Clearly, the string cycles between 20202020.... and 1111111... (in each cycle, another 20 and 11 appear)

Thus, if $n<1994$, we get 111..... ($A$ is left with no coins if $n$ is even, and one coin if $n$ is odd), so the girl opposite $A$ never gets a coin.

If n=1994, the 0's and 2's switch back and forth, i.e.

20202020.....
02020202.....

so the game does not end.
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Kayak
1298 posts
#4 • 2 Y
Y by RAMUGAUSS, Adventure10
@Above Sorry, your proof is probably very wrong - a person can have way more than $2$ cards.

A symmetric invariant for (a)

IMO Compendium assymetric invariant for (a)
This post has been edited 4 times. Last edited by Kayak, Sep 9, 2017, 11:25 AM
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TwilightZone
128 posts
#5
Y by
Solution to part b
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N Quick Reply
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