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
Yesterday at 11:16 PM
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
Yesterday at 11:16 PM
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
positive integers forming a perfect square
cielblue   0
42 minutes ago
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
0 replies
cielblue
42 minutes ago
0 replies
Infinite product problem
ReticulatedPython   6
N an hour ago by ReticulatedPython
Compute $$\prod_{n=1}^{\infty}3^{\frac{1}{2^{n-1}}}+1$$
hint

The solution to this problem is pretty short once you find out the trick. :D
6 replies
ReticulatedPython
2 hours ago
ReticulatedPython
an hour ago
Function equation
LeDuonggg   6
N an hour ago by MathLuis
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
6 replies
LeDuonggg
Yesterday at 2:59 PM
MathLuis
an hour ago
A nice and easy gem off of StackExchange
NamelyOrange   0
an hour ago
Source: https://math.stackexchange.com/questions/3818796/
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.
0 replies
NamelyOrange
an hour ago
0 replies
Math and AI 4 Girls
mkwhe   37
N an hour ago by deeptisidana
Hey everyone!

The 2025 MA4G competition is now open!

Apply Here: https://xmathandai4girls.submittable.com/submit


Visit https://www.mathandai4girls.org/ to get started!

Feel free to PM or email mathandai4girls@yahoo.com if you have any questions!
37 replies
mkwhe
Apr 5, 2025
deeptisidana
an hour ago
at everystep a, b, c are replaced by a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)
NJAX   8
N an hour ago by Assassino9931
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem 8
Three positive integers are written on the board. In every minute, instead of the numbers $a, b, c$, Elbek writes $a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)$ . Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
Note. $\gcd(x,y)$ - Greatest common divisor of numbers $x$ and $y$

Proposed by Sergey Berlov, Russia
8 replies
NJAX
May 31, 2024
Assassino9931
an hour ago
What's the chance that two AoPS accounts generate with the same icon?
Math-lover1   0
2 hours ago
So I've been wondering how many possible "icons" can be generated when you first create an account. By "icon" I mean the stack of cubes as the first profile picture before changing it.

I don't know a lot about how AoPS icons generate, so I have a few questions:
- Do the colors on AoPS icons generate through a preset of colors or the RGB (red, green, blue in hexadecimal form) scale? If it generates through the RGB scale, then there may be greater than $256^3 = 16777216$ different icons.
- Do the arrangements of the stacks of blocks in the icon change with each account? If so, I think we can calculate this through considering each stack of blocks independently.
0 replies
Math-lover1
2 hours ago
0 replies
9 Have you participated in the MATHCOUNTS competition?
aadimathgenius9   29
N 3 hours ago by hashbrown2009
Have you participated in the MATHCOUNTS competition before?
29 replies
aadimathgenius9
Jan 1, 2025
hashbrown2009
3 hours ago
9 What is the most important topic in maths competition?
AVIKRIS   67
N Today at 2:31 PM by Craftybutterfly
I think arithmetic is the most the most important topic in math competitions.
67 replies
AVIKRIS
Apr 19, 2025
Craftybutterfly
Today at 2:31 PM
9 AMC 8 Scores
ChromeRaptor777   125
N Today at 2:12 PM by Soupboy0
As far as I'm certain, I think all AMC8 scores are already out. Vote above.
125 replies
ChromeRaptor777
Apr 1, 2022
Soupboy0
Today at 2:12 PM
Facts About 2025!
Existing_Human1   257
N Today at 1:49 PM by Charizard_637
Hello AOPS,

As we enter the New Year, the most exciting part is figuring out the mathematical connections to the number we have now temporally entered

Here are some facts about 2025:
$$2025 = 45^2 = (20+25)(20+25)$$$$2025 = 1^3 + 2^3 +3^3 + 4^3 +5^3 +6^3 + 7^3 +8^3 +9^3 = (1+2+3+4+5+6+7+8+9)^2 = {10 \choose 2}^2$$
If anyone has any more facts about 2025, enlighted the world with a new appreciation for the year


(I got some of the facts from this video)
257 replies
Existing_Human1
Jan 1, 2025
Charizard_637
Today at 1:49 PM
Berkeley mini Math Tournament Online is June 7
BerkeleyMathTournament   7
N Today at 1:39 PM by Inaaya
Berkeley mini Math Tournament is a math competition hosted for middle school students once a year. Students compete in multiple rounds: individual round, team round, puzzle round, and relay round.

BmMT 2025 Online will be held on June 7th, and registration is OPEN! Registration is $8 per student. Our website https://berkeley.mt/events/bmmt-2025-online/ has more details about the event, past tests to practice with, and frequently asked questions. We look forward to building community and inspiring students as they explore the world of math!

3 out of 4 of the rounds are completed with a team, so it’s a great opportunity for students to work together. Beyond getting more comfortable with math and becoming better problem solvers, our team is preparing some fun post-competition activities!

Registration is open to students in grades 8 or below. You do not have to be local to the Bay Area or California to register for BmMT Online. Students may register as a team of 1, but it is beneficial to compete on a team of at least 3 due to our scoring guideline and for the experience.

We hope you consider attending, or if you are a parent or teacher, that you encourage your students to think about attending BmMT. Thank you, and once again find more details/register at our website,https://berkeley.mt.
7 replies
BerkeleyMathTournament
Yesterday at 7:37 AM
Inaaya
Today at 1:39 PM
1234th Post!
PikaPika999   259
N Today at 12:14 PM by PikaPika999
I hit my 1234th post! (I think I missed it, I'm kinda late, :oops_sign:)

But here's a puzzle for you all! Try to create the numbers 1 through 25 using the numbers 1, 2, 3, and 4! You are only allowed to use addition, subtraction, multiplication, division, and parenthesis. If you're post #1, try to make 1. If you're post #2, try to make 2. If you're post #3, try to make 3, and so on. If you're a post after 25, then I guess you can try to make numbers greater than 25 but you can use factorials, square roots, and that stuff. Have fun!

1: $(4-3)\cdot(2-1)$
259 replies
PikaPika999
Apr 21, 2025
PikaPika999
Today at 12:14 PM
9 What competitions do you do
VivaanKam   5
N Today at 4:45 AM by valisaxieamc

I know I missed a lot of other competitions so if you didi one of the just choose "Other".
5 replies
VivaanKam
Apr 30, 2025
valisaxieamc
Today at 4:45 AM
Number theory or function ?
matematikator   15
N Apr 23, 2025 by YaoAOPS
Source: IMO ShortList 2004, algebra problem 3
Does there exist a function $s\colon \mathbb{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer.

Proposed by Dan Brown, Canada
15 replies
matematikator
Mar 18, 2005
YaoAOPS
Apr 23, 2025
Number theory or function ?
G H J
Source: IMO ShortList 2004, algebra problem 3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
matematikator
110 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Does there exist a function $s\colon \mathbb{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer.

Proposed by Dan Brown, Canada
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 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Where did you find this problem? It is realy interesting.

I believe the answer to be 'yes', and I think I have a way of constructing the function, but it's only verified empirically, and I haven't fixed my ideas well enough to provide a full proof. Here's what I thought we could do:

It suffices to determine the function on the positive rationals, of course, so let's restrict our attention to those. Construct the Stern-Brocot tree step by step. At each step, to each leaf you attach a leaf to the left and a leaf to the right. Label the leaf to the left with $-1$ and the leaf to the right with $1$. It seems to work.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Myth
4464 posts
#3 • 2 Y
Y by Adventure10, Mango247
We have $s(x)=s(1+x)=-s(1/x)$. I see a direct way to use continuous fractions here, since having these rules we can obtain $s(x)$ from $s(1)$. :?
Did I say something stupid?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
matematikator
110 posts
#4 • 2 Y
Y by Adventure10, Mango247
Solution: Let x=a/b be a positive rational, where a,b are coprime positive integers. Consider the squence of consecutive remainders given by euclidean algorithm for the ordered pairs (a,b). if u mod v denotes the least non negative remainder of u modulo v, this sequence can be written as
r_0=a, r_1=b, r_2=a mod b, ... , r_(i+1)=r_(i-1) mod r_i, ... , r_n=1, r_(n+1)=0 (1)

the index n=n(x) of the least nonzero remainder r_n=1 is uniquily determined by x, so
t(x)=(-1)^n(x) is a well-defined function from the positive rationals into {-1,1}. Now define s: Q-->{-1,1} by

s(x)=t(x) for x in Q, x>0 or s(x)=-t(-x) for x in Q, x<0 or s(0)=1

We prove that s has the desired properties. Let x and y be distinct rational numbers.
* If x+y=0, let x>0 and y<0 (x,y are nonzero). Then , by definition, s(x)=t(x),
s(y)=-t(-y)=-t(x), hence s(x)s(y)=-1.
**If xy=1 then x and y are of the same sign and neither equals 1 or -1. Suppose first
that x=a/b >0, y=b/a >0, with a,b coprime positive integers, one may also assume
that a>b. Euclidean algorithm for (a,b) starts a,b, (a mod b), ... .On the other hand,
the algorithm for pair (b,a) gives b,a,b,(a mod b)m,... .
Because a>b implies r_2=(b mod a)=b. Each term in (1) depends only on the previous
two, so the sequence for x has length by 1 less than the sequence for y, that is,
n(y)=n(x)+1. Hence t(y)=-t(x), and so s(y)=-s(x), as needed. Now the case of x,y
follows from the definition...
**If x+y=1 then ....

by the same logic it'll be true...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Myth
4464 posts
#5 • 2 Y
Y by Adventure10, Mango247
We see that $t(x)$ is a number of subfractions in the continuous fraction representing $x$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
matematikator
110 posts
#6 • 2 Y
Y by Adventure10, Mango247
absolutely...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Raúl
69 posts
#7 • 2 Y
Y by Adventure10, Mango247
I knew it's imoral relive topic but I have a question: we can prove that $ s(1) = 0$ if we use $ s(x) = - s(\frac {1}{x})\Leftrightarrow s(1) = - s(1)\Leftrightarrow s(1) = 0$ contradiction.

I dont really understand :maybe:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bearkiller
37 posts
#8 • 2 Y
Y by Adventure10, Mango247
x and y are distinct rational numbers if you use s(x) =-s(1/x) for x=1 then we have x=1=1/1=1/x=y.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wolstenholme
543 posts
#9 • 7 Y
Y by bcp123, jt314, JasperL, meowme, Adventure10, Mango247, Sedro
Every positive rational number has a unique continued fraction of the form $ a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\ddots\frac{1}{a_{m - 1} + \frac{1}{a_m}}}}} $ for some $ m \in \mathbb{N} $ where $ a_i \in \mathbb{N} $ for all $ i $. From now on this will be denoted as $ [a_0; a_1, a_2, \dots, a_m] $.

For every $ a \in \mathbb{Q}^{+} $, let $ s(a) = (-1)^m $. Then let $ s(0) = -1 $ and for all $ b \in \mathbb{Q} \setminus \{0\} $ let $ s(-b) = -s(b) $. I claim that this $ s $ satisfies the desired properties. To prove this, consider two rational numbers $ x < y $.

$ 1) $ If $ x + y = 0 $ then by definition $ s(x)s(y) = -1 $ as desired.

$ 2) $ If $ x + y = 1 $ and $ x = 0 $ then by definition $ s(x)s(y) = -1 $ as desired since $ s(1) = 1 $.

$ 3) $ If $ x + y = 1 $ and $ x < 0 $ then to prove that $ s(x)s(y) = -1 $ it suffices to show that $ s(-x) = s(y) $. Let $ -x = [a_0; a_1, a_2, \dots, a_m] $. Then $ y = [a_0 + 1; a_1, a_2, \dots, a_m] $ and so $ s(-x) = s(y) = (-1)^m $ as desired.

$ 4) $ If $ x + y = 1 $ and $ x > 0 $ we can let $ y = [0; 1, a_2, \dots, a_m] $ so that $ x = [0; a_2 + 1, a_3, \dots, a_m] $ so $ s(y) = (-1)^m $ and $ s(x) = (-1)^{m - 1} $ which implies that $ s(x)s(y) = -1 $ as desired.

$ 5) $ If $ xy = 1 $, we can assume WLOG that $ 0 < x < 1 $. Then letting $ x = [0; a_1, a_2, \dots, a_m] $ we have that $ y = [a_1; a_2, a_3, \dots, a_m] $ so $ s(x) = (-1)^m $ and $ s(y) = (-1)^{m - 1} $ which implies that $ s(x)s(y) = -1 $ as desired.

This implies that my claim is true and so we are done. Note that this also implies that Grobber's idea about the Stern-Brocot tree works as well.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dgrozev
2463 posts
#10 • 1 Y
Y by Adventure10
Note also that this function $s$ is unique up to multiplication by $-1$. Indeed let $x\in\mathbb{Q}^{+}$ and $x=[a_0;a_1,\ldots,a_m]$. Using $s(x+1)=s(x)$ and $s(x)=-s(1/x)$ it's easily obtained $s(x)=(-1)^m s(1)$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_pi_rate
1218 posts
#11 • 3 Y
Y by AlastorMoody, amar_04, AmirKhusrau
Fake alg, pure NT :furious:. Anyway, here's my solution: For two integers $0<p<q$, let $f(p,q)$ denote the number of steps taken in the Euclidean Algorithm to find $\gcd(p,q)$. In other words, if we let $b_0=q,$ $b_1=p$ and write the system of equations $$b_r=a_{r+1}b_{r+1}+b_{r+2} \text{ such that } 0<b_{r+1}<b_r \quad \forall r \in \{0,1, \dots ,n\} \text{ and } b_{n+2}=0$$then $f(p,q)=n,$ where $n \geq 0$. Then we claim that the following function works-
The ProGawd Function wrote:
  1. $s(n)=1$ for $n \in \mathbb{N},$ and $s(-m)=-1$ for $m \in \mathbb{N} \cup \{0\}.$
    $\text{ }$
  2. $s \left(\frac{-1}{2} \right)=1.$
    $\text{ }$
  3. $s(1+x)=s(x)$ for $x \neq \frac{-1}{2},0.$
    $\text{ }$
  4. For integers $p,q$ with $0<p<q$, $$s \left (\frac{p}{q} \right)=(-1)^{f(p,q)+1}$$
It's easy to see that this uniquely defines the image of all rationals. Now we show that this indeed satisfies all the given properties.
  • First we prove $s(x)s(-x)=-1$ when $x \neq 0$. For $x \in \mathbb{N}$ and $x=\pm \frac{1}{2}$, this easily follows from the definition, so we can ignore these cases. WLOG assume $x>0$ and suppose $x \in (n-1,n)$ for some $n \in \mathbb{N}$ with $2x \neq 1$. Then from property 3 of our function, we have $s(x)=s(x-(n-1))$ and $s(-x)=s((n-1)-x)$. Since $0<x-(n-1)<1,$ so it suffices to show that $s(x) \cdot s(-x)=-1$ for $x \in (0,1)$.

    Take $x=\frac{p}{q}$ for $0<p<q$, and first suppose $p< \frac{q}{2}$. We have $$s \left( \frac{-p}{q} \right)=s \left(1-\frac{p}{q} \right)=(-1)^{f(q-p,q)+1}$$But, as $q-p>\frac{q}{2}>p$, so the first step of our Euclidean Algorithm while finding $\gcd(q-p,q)$ will be $q=(q-p)+p$. So $f(q-p,q)=1+f(p,q-p)$. Also, using $q>2p$, we get $f(p,q-p)=f(p,q)$. Thus, we have $$s \left( \frac{-p}{q} \right)=(-1)^{f(q-p,q)+1}=(-1)^{1+f(p,q)+1}=-s \left( \frac{p}{q} \right)$$Thus, for $x \in \left(0, \frac{1}{2} \right),$ we have $s(x)s(-x)=-1$. Finally, if $x \in \left(\frac{1}{2},1 \right),$ then $s(x)=s(x-1)$ and $s(-x)=s(1-x)$. Since $0<1-x<\frac{1}{2}$, then the previous result gives $s(1-x)s(-(1-x))=-1$. Combined with the above equalities, we get $s(x)s(-x)=-1$. Thus, the result is true for all $x \in \mathbb{Q}$.
    $\text{ }$
  • Now we show that $s(x)s(1-x)=-1$ for $x \neq \frac{1}{2}$. For $x=1$, the result is true by definition. And for $x \neq 1,\frac{1}{2}$, by the above result, we have $s(1-x)=-s(x-1)=-s((x-1)+1),$ which directly gives the desired property.
    $\text{ }$
  • Finally we show that $s(x)s \left(\frac{1}{x} \right)=-1$ when $x \neq 0,\pm 1$. Since $s(-x)=-s(x)$ (proved before), so it suffices to show the result for positive $x$. Also, WLOG we can take $0<x<1$, and write $x=\frac{p}{q}$ with $p<q$ and $p,q \in \mathbb{N}$. Then $$s \left(\frac{1}{x} \right)=s \left( \frac{q}{p} \right)=s \left( \frac{q}{p} -1\right)= \dots =s \left( \frac{q \pmod{p}}{p} \right)=(-1)^{f(q \pmod{p},p)+1}$$But, while finding $\gcd(q\pmod{p},p)$ we simply start from the second step in the Euclidean Algorithm to find $\gcd(p,q)$. So we have $f(q \pmod{p},p)+1=f(p,q)$. This gives us $$s \left(\frac{1}{x} \right)=(-1)^{f(q \pmod{p},p)+1}=(-1)^{f(p,q)}=-s \left(\frac{p}{q} \right)=-s(x)$$Thus, we have $s(x)s(-x)=-1$ as desired. $\blacksquare$
This post has been edited 1 time. Last edited by math_pi_rate, Mar 29, 2020, 8:57 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blacksheep2003
1081 posts
#12
Y by
Solution
This post has been edited 1 time. Last edited by blacksheep2003, Oct 13, 2020, 10:29 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
yayups
1614 posts
#13 • 5 Y
Y by Mathematicsislovely, TechnoLenzer, Mango247, Mango247, richrow12
The answer is yes, and we construct such a function. First, note that the only condition on $s(0)$ is that $s(0)=-s(1)$, which we can manually enforce at the end. Thus, we assume that the inputs to $s$ are nonzero rationals.

Write $x=a/b$ where $\gcd(a,b)=1$ and $b>0$. We set \[s(x) = \begin{cases}1 &\text{if } (a^{-1}\mod b)\le b/2 \\ -1 &\text{if } (a^{-1}\mod b)>b/2.\end{cases}\]From this definition, it is clear that $s(x)= - s(-x)$ and $s(x)=-s(1-x)$, so it suffices to show that $s(x)=-s(1/x)$. Since we know $s$ is odd, it suffices in fact to show $s(x)=-s(1/x)$ for $x>0$.

Lemma: If $a,b\ge 1$ are relatively prime positive integers, then \[a\cdot(a^{-1}\mod b) + b\cdot(b^{-1}\mod a) = ab+1.\]
Proof: Indeed, letting $T$ be the left side of the above equation, we see that $T\equiv 1\pmod{a}$ and $T\equiv 1\pmod{b}$, so $T\equiv 1\pmod{ab}$ since $\gcd(a,b)=1$.

Furthermore, we have \[a+b\le T\le a(b-1)+b(a-1)<2ab,\]so we must in fact have $T=ab+1$, as desired. $\blacksquare$

The above lemma implies that $(a^{-1}\mod b)\le b/2$ if and only if $(b^{-1}\mod a)>a/2$, which implies that $s(a/b) = -s(b/a)$.

Remark: The condition is basically the same as ISL 2017 N8, which motivates the classification of $s$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1711 posts
#14
Y by
We define $n(x)$ on the positive rational numbers to be the number of terms in the continued fraction of $x$. Then let $s(x)=-1^{n(x)}$ and $s(-x)=-s(x)$. Then, clearly $s(x)s(-x)=-1$. If $xy=1$ and $x>1$ then $y = 0+\tfrac{1}{x}$ and therefore $n(y)=1+n(x)$, so $s(x)s(y)=-1$.

Suppose $x+y=1$. WLOG, $x > \tfrac{1}{2}$. If $x < 1$ then $x=[0;1,a,\dots]$ and $y=[0;1+a,\dots]$ where the $\dots$ is the same. If $x>1$ then $x=(-y)+1$ so they have the same number of terms in the continued fraction, thus $s(x)=s(-y)=-s(y)$ as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
asdf334
7585 posts
#15 • 1 Y
Y by megarnie
The answer is yes. We construct $s$ directly.
First assume $s(0)=1$. For any $x>0$ we have
\[s(x)=-s(-x)=s(x+1)\]and we also have $s(1)=-1$. Hence for any $n\in \mathbb{N}$ we have $s(n)=-1$.
For each $q\ge 2$ (starting from the smallest), we now choose $s\left(\frac{p}{q}\right)$ for $p$ starting from $1$ and relatively prime to $q$ according to the following rules:
  • If $p<q$ then $s\left(\frac{p}{q}\right)=-s\left(\frac{q}{p}\right)$.
  • If $p>q$ then $s\left(\frac{p}{q}\right)=s\left(\frac{p-q}{q}\right)$.
Evidently all outputs for nonnegative inputs have been chosen with no problems.
Now we prove that for distinct positive rationals $x$ and $y$ the conditions hold. Importantly our selection of outputs satisfies $s(x)=s(x+1)$ as well. (The only restriction on $s(0)$ is $s(0)=-s(1)$ which clearly holds.)
  • If $xy=1$ then WLOG $x=\frac{p}{q}$ with $p<q$. From the first rule in the previous section we are fine.
  • Notice that $x+y=0$ cannot hold. If $x+y=1$ then write $x=\frac{a}{a+b}$ and $y=\frac{b}{a+b}$. Then
    \[s(x)=-s(1/x)=-s(1/x-1)=-s(b/a)\]\[s(y)=-s(1/y)=-s(1/y-1)=-s(a/b)\]thus $s(x)s(y)=-1$.

Now we write $s(-x)=-s(x)$ for positive $x$.
  • If $x$ and $y$ are distinct, negative and satisfy $xy=1$ we are okay: $s(-x)s(-y)=-1\implies s(x)s(y)=-1$.
  • If $x$ and $y$ are distinct and satisfy $x+y=0$ we are okay by definition.
  • If $x+y=1$ and WLOG $x$ is negative while $y$ is positive, then
    \[s(y)=s(y-1)=-s(1-y)=-s(x)\]and we are okay.

All cases are exhausted and the construction works. $\blacksquare$
This post has been edited 1 time. Last edited by asdf334, Jun 2, 2024, 5:01 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
YaoAOPS
1535 posts
#16 • 1 Y
Y by NicoN9
This problem gets no big booms.


we claim the answer is yes. We first define a function $f: \mathbb{Q}^+ \times \mathbb{Q}^+ \to \{+1, -1\}$ for coprime nonzero inputs as follows: let $f(1, 1) = +1, f(1, a) = -1, f(a, 1) = +1$ for all $a \ne 1$. Then we define recursively
\[
	f(a+b,b) = f(a,a+b) = f(a,b).
\]This is well defined as $f$ is defined on input by Euclidean algorithm, and $f(a,b)$ is defined uniquely by whichever of $a, b$ is smaller.

Now note that $f(a, b) = -f(b, a)$ follows since $f(1, a) = -f(a, 1)$ recursively with the Euclidean algorithm. Furthermore, for $a < b$, $f(b-a, b) = f(b-a, a) = f(b,a) = -f(b,a) = -f(a,b)$ $(\heartsuit)$.

We now extend $f$ to negative coprime inputs by defining $f(-a, b) = f(a, -b) = -f(a, b)$. By $(\heartsuit)$ and by the definition, it follows that for $b \ne 1$
\[
	f(a, b) = f(a+b, b), f(b, a) = f(b, a+b)
\]holds for all $a$. We finally define $s\left(\frac{a}{b}\right) = f(a, b)$ for coprime $a, b$. Then $f(a,b) = -f(b,a)$ holds over the extension for the same reasons, $f(a,b) = -f(-a,b)$ holds by definition, and $f(a,b) = -f(b-a,b)$ holds since
\[
	f(b-a,b) = f(-a,b) = -f(a,b)
\]so we are done.
Z K Y
N Quick Reply
G
H
=
a