Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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
Sunday, Jun 29 - Oct 26
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
Cyclic Quads and Parallel Lines
gracemoon124   17
N a few seconds ago by Siddharthmaybe
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
17 replies
gracemoon124
Aug 16, 2023
Siddharthmaybe
a few seconds ago
Problem 1 (First Day)
Valentin Vornicu   137
N 3 minutes ago by Siddharthmaybe
1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.
137 replies
Valentin Vornicu
Jul 12, 2004
Siddharthmaybe
3 minutes ago
Concentric Circles
MithsApprentice   62
N 3 minutes ago by Siddharthmaybe
Source: USAMO 1998
Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
62 replies
MithsApprentice
Oct 9, 2005
Siddharthmaybe
3 minutes ago
four points lie on a circle
pohoatza   77
N 8 minutes ago by Siddharthmaybe
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
77 replies
pohoatza
Jun 28, 2007
Siddharthmaybe
8 minutes ago
No more topics!
Very easy number theory
darij grinberg   102
N Apr 29, 2025 by ND_
Source: IMO Shortlist 2000, N1, 6th Kolmogorov Cup, 1-8 December 2002, 1st round, 1st league,
Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have \[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]
102 replies
darij grinberg
Aug 6, 2004
ND_
Apr 29, 2025
Very easy number theory
G H J
Source: IMO Shortlist 2000, N1, 6th Kolmogorov Cup, 1-8 December 2002, 1st round, 1st league,
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#1 • 12 Y
Y by Adventure10, asdf334, donotoven, mathematicsy, TFIRSTMGMEDALIST, megarnie, adityaguharoy, ImSh95, Rounak_iitr, Mango247, MS_asdfgzxcvb, and 1 other user
Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have \[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]
This post has been edited 1 time. Last edited by djmathman, Oct 3, 2016, 3:32 AM
Reason: adjusted formatting
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
grobber
7849 posts
#2 • 6 Y
Y by Adventure10, Jan-Huang, Dzadzo, ImSh95, Mango247, kiyoras_2001
Indeed, it's quite easy: It says that the inverse of any $a$ in $\mathbb Z_n$ is actually $a$. This means that in the group $U(\mathbb Z_n)$ each element has order $2$. It's easy to see that unless $n$ is $1,2,3,4$, we will be able to find a number which is coprime with $n$, $>1$ and $<\sqrt n$. That's because if $n$ is prime we take $2$ and if $n$ isn't prime we take $p-1$, where $p$ is the smallest prime factor of $n$.

Edit:

Looks like I keep missing cases :): if $2|n$ then this might not work; Anyway, we know that $\phi(n)$ must be a power of $2$, which tells us that $n$ has form $2^tp_1p_2\ldots p_k$, with $p_i$ odd primes of the form $2^a+1$. Morevoer, we know that there are no numbers $a$ s.t. $(a,n)=1$ and $1<a\le\sqrt n$. I'll see what I can do.
This post has been edited 4 times. Last edited by grobber, Aug 6, 2004, 1:56 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
harazi
5526 posts
#3 • 4 Y
Y by Adventure10, Adventure10, ImSh95, Mango247
I may be wrong, but I think that grobber is wrong. I don't this are the answers. And I wouldnt call it exactly very easy number theory. As far as I remember this problem was given in a TST in Hungary, it also appeared on IMO Shortlist and finally a more general case was given in our TST this year (and contrary to most peoples expectations, it didnt turn out to be easy). It isnt difficult, but not very easy.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
grobber
7849 posts
#4 • 3 Y
Y by Adventure10, ImSh95, Mango247
Do you by any chance know what the answer is?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
harazi
5526 posts
#5 • 4 Y
Y by Illuzion, Adventure10, ImSh95, Mango247
Answer: 1,2,3,4,6,8,12,24.
This post has been edited 1 time. Last edited by darij grinberg, Aug 2, 2011, 1:23 PM
Reason: post corrected as explained by its author
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#6 • 3 Y
Y by Adventure10, ImSh95, Mango247
Huh, sorry, I must have been wrong...

darij
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
grobber
7849 posts
#7 • 3 Y
Y by Adventure10, ImSh95, Mango247
I'm terribly sorry I rushed into it like that. Ok, so after the above observations, the proof isn't hard to come by: if $n\ge 49$ then we take $a=7$. $7$ doesn't divide $n$ because it's not a Fermat prime, and $1<7\le\sqrt n$. On the other hand, if $n\ge 25$, then $30=2\cdot 3\cdot 5|n$. This means that the only number $>25$ that we have to check is $30$, and it doesn't work because we take $a=7$ again.

The solutions can now be found manually.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pbornsztein
3004 posts
#8 • 3 Y
Y by Adventure10, ImSh95, Mango247
I think Harazi means that one :
http://www.kalva.demon.co.uk/short/soln/sh00n1.html

Pierre.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
harazi
5526 posts
#9 • 4 Y
Y by Adventure10, ImSh95, Mango247, and 1 other user
Yes, I rushed a little bit, but it's practically the same problem. Anyway, what's with these complicated solutions? The solution given by Sasha to my problem for the TST is just perfect: smart and very short.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Peter
3615 posts
#10 • 3 Y
Y by Adventure10, ImSh95, Mango247
ab+1 divides n isn't the same as ab = 1 mod n... :D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
harazi
5526 posts
#11 • 3 Y
Y by Adventure10, ImSh95, Mango247
Really? Well, Peter, when I said they are practically the same I wasn't wrong. Think a little bit: they both ask the numbers n for which any relatively prime a with n has order 2 in Z_n? Isn't this the same?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Peter
3615 posts
#12 • 3 Y
Y by Adventure10, ImSh95, Mango247
harazi wrote:
they both ask the numbers n for which any relatively prime a with n has order 2 in Z_n? Isn't this the same?

Sorry hazari, I just learnt the euler phi function yesterday, orders are planned for in a couple of days, currently I got even no clue what they are :D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
isotomion
28 posts
#13 • 3 Y
Y by Adventure10, ImSh95, Mango247
Peter VDD wrote:
ab+1 divides n isn't the same as ab = 1 mod n... :D

He replaced b by -b. Then, of course, the condition "ab + 1 is divisible by n" becomes "ab = 1 mod n", and the condition "a + b is divisible by n" becomes "a = b mod n".

Darij
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Peter
3615 posts
#14 • 3 Y
Y by Adventure10, ImSh95, Mango247
Yes, sorry

didn't think of that since both problems state natural numbers :blush:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#15 • 3 Y
Y by Adventure10, ImSh95, Mango247
Sorry, of course I meant: replace b by any number equivalent to -b modulo n. This number needs not to be negative.

Darij
Z K Y
G
H
=
a