Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 3:18 PM
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Yesterday at 3:18 PM
0 replies
mdk2013
Mar 30, 2025
pingpongmerrily
43 minutes ago
MOP Cutoffs Out?
Mathandski   29
N an hour ago by Mathandski
MAA has just emailed a press release announcing the formula they will be using this year to come up with the MOP cutoff that applies to you! Here's the process:

1. Multiply your age by $1434$, let $n$ be the result.

2. Calculate $\varphi(n)$, where $\varphi$ is the Euler's totient theorem, which calculates the number of integers less than $n$ relatively prime to $n$.

3. Multiply your result by $1434$ again because why not, let the result be $m$.

4. Define the Fibonacci sequence $F_0 = 1, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ for $n \ge 2$. Let $r$ be the remainder $F_m$ leaves when you divide it by $69$.

5. Let $x$ be your predicted USA(J)MO score.

6. You will be invited if your score is at least $\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$.

7. Note that there may be additional age restrictions for non-high schoolers.

See here for MAA's original news message.

.

.

.


Edit (4/2/2025): This was an April Fool's post.
Here's the punchline
29 replies
Mathandski
Tuesday at 11:02 PM
Mathandski
an hour ago
2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!
audio-on   19
N 2 hours ago by WhitePhoenix
Join the 2025 Math and AI 4 Girls Competition for a chance to win up to $1,000!

Hey Everyone, I'm pleased to announce the dates for the 2025 MA4G Competition are set!
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (@ 11:59pm PST).

Applicants will have one month to fill out an application with prizes for the top 50 contestants & cash prizes for the top 20 contestants (including $1,000 for the winner!). More details below!

Eligibility:
The competition is free to enter, and open to middle school female students living in the US (5th-8th grade).
Award recipients are selected based on their aptitude, activities and aspirations in STEM.

Event dates:
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (by 11:59pm PST)
Winners will be announced on June 28, 2025 during an online award ceremony.

Application requirements:
Complete a 12 question problem set on math and computer science/AI related topics
Write 2 short essays

Prizes:
1st place: $1,000 Cash prize
2nd place: $500 Cash prize
3rd place: $300 Cash prize
4th-10th: $100 Cash prize each
11th-20th: $50 Cash prize each
Top 50 contestants: Over $50 worth of gadgets and stationary


Many thanks to our current and past sponsors and partners: Hudson River Trading, MATHCOUNTS, Hewlett Packard Enterprise, Automation Anywhere, JP Morgan Chase, D.E. Shaw, and AI4ALL.

Math and AI 4 Girls is a nonprofit organization aiming to encourage young girls to develop an interest in math and AI by taking part in STEM competitions and activities at an early age. The organization will be hosting an inaugural Math and AI 4 Girls competition to identify talent and encourage long-term planning of academic and career goals in STEM.

Contact:
mathandAI4girls@yahoo.com

For more information on the competition:
https://www.mathandai4girls.org/math-and-ai-4-girls-competition

More information on how to register will be posted on the website. If you have any questions, please ask here!


19 replies
audio-on
Jan 26, 2025
WhitePhoenix
2 hours ago
Question about USAMO, self esteem, and college
xHypotenuse   18
N 3 hours ago by rayford
Hello everyone. I know this question may sound ridiculous/neagtive but I really want to know how the rest of the community thinks on this issue. Please excuse this yap session and feel free to ignore this post if it doesn't make sense, I don't think I really have a sane mind these days and something has gotten into my head.

I want your advice on what I should do in this situation. It has been my dream to make usamo since ~second semester of 9th grade and I started grinding from that time on. Last year, I qualified for the aime and got a 5. This year I really wanted to qualify for the olympiad and studied really hard. I spent my entire summer working on counting and probability, the subject I suck at the most. And yet, on amc 12, I fumbled hard. I usually mocked ~120-130s on amc 10s but on amc 12 this year, I got really mediocre scores ~100. So I had no chance of making usamo.

So during winter of 2024-2025 I kinda gave up on aime studying and I was like "hey, if I can't get into usamo, maybe ill qualify for usapho." Since I was pretty good at physics at that time. So I spended my winter hard grinding for f=ma and guess what? The test had stupid and ridiculous questions and I only got an 11. What really sucks is that even with the stupid amount of cheaters in f=ma, if I changed all of my "D" guesses to "C," then I would have qualified. Since I solved 10 actually and guessed the rest. Absolutely unfair that only 1 of my guesses were correct.

And also since I didn't study for aime, I ended up being super rusty and so I only got a 7. Solved 9 tho. (I usually can consistently solve 10+ on aimes).

And now here's my senior year and ofc I want to apply to a prestigious college. But it feels stupid that I don't have any usamo or usapho titles like the people I know do. I think I will have good essays primarily due to a varied amount of life experiences but like, I don't feel like I will contribute much to the college without being some prestigious olympiad qualifier. So this led to me having a self esteem issue.

This also led me to the question: should I study one last year so that I can get into usamo in my senior year, or is there no point? Since like, colleges don't care about whatever the hell you do in your senior year, and also, it seems just 'weird' to be grinding math contests while the rest of the people from my school are playing around, etc. So this time around I've really been having an internal crisis between my self esteem (since getting into usamo will raise my self esteem a lot) and college/senior choices.

I know this may seem like a dumb question to some and you are free to completely ignore the post. That's fine. I just really want advice for what I should do in this situation and it would really help bring my life quality up

Thanks,
hypotenuse
18 replies
xHypotenuse
Today at 2:03 AM
rayford
3 hours ago
No more topics!
9c4 sperner's theorem
ccarolyn4   26
N Dec 3, 2024 by EaZ_Shadow
Source: 2024 AMC 10 P12
A group of $100$ students from different countries meet at a mathematics competition. Each student speaks the same number of languages, and, for every pair of students $A$ and $B$, student $A$ speaks some language that student $B$ does not speak, and student $B$ speaks some language that student $A$ does not speak. What is the least possible total number of languages spoken by all the students?

$
\textbf{(A) }9 \qquad
\textbf{(B) }10 \qquad
\textbf{(C) }12 \qquad
\textbf{(D) }51 \qquad
\textbf{(E) }100 \qquad
$
26 replies
ccarolyn4
Nov 13, 2024
EaZ_Shadow
Dec 3, 2024
9c4 sperner's theorem
G H J
Source: 2024 AMC 10 P12
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ccarolyn4
24 posts
#1 • 3 Y
Y by Soccerstar9, OronSH, akliu
A group of $100$ students from different countries meet at a mathematics competition. Each student speaks the same number of languages, and, for every pair of students $A$ and $B$, student $A$ speaks some language that student $B$ does not speak, and student $B$ speaks some language that student $A$ does not speak. What is the least possible total number of languages spoken by all the students?

$
\textbf{(A) }9 \qquad
\textbf{(B) }10 \qquad
\textbf{(C) }12 \qquad
\textbf{(D) }51 \qquad
\textbf{(E) }100 \qquad
$
This post has been edited 3 times. Last edited by ccarolyn4, Dec 23, 2024, 11:31 PM
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ccarolyn4
24 posts
#2 • 6 Y
Y by megarnie, solasky, Soccerstar9, lprado, Turtwig113, zhoujef000
confirmed 9 (A) from cm

my fav question from test <3
This post has been edited 2 times. Last edited by ccarolyn4, Nov 13, 2024, 5:21 PM
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gracemoon124
872 posts
#3
Y by
yeah what you do is $\tbinom nk$ where n is the number of languages in total & k is the number of languages each student speaks, we need $\tbinom nk \ge 100$, so we test out answer choices lol
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gladIasked
632 posts
#4 • 1 Y
Y by Andyluo
this is a really nice problem; unironically my favorite problem on the test
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paganiniana
211 posts
#5
Y by
9 choose 4 satisfies already, so A
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Squidget
433 posts
#6
Y by
I almost answered 100 but I came to my senses and I decided to leave unanswered. I only had 30 sec left anyway
This post has been edited 1 time. Last edited by Squidget, Nov 13, 2024, 5:25 PM
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zlrara01
335 posts
#7
Y by
this was a really cool problem, I just noticed it sounded like a mathcounts problem and did 9c4
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pingpongmerrily
3527 posts
#8
Y by
it was cool but then 9c4 seemed to easy so i quadruple checked it lol
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KevinYang2.71
410 posts
#9 • 5 Y
Y by solasky, LostDreams, OronSH, ranu540, Alex-131
this is why you read diestel
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MathRook7817
637 posts
#10
Y by
bro i skipped this it was easy
lol i sold
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aleyang
192 posts
#11
Y by
This was worded kinda weirdly but was just testing answer choices once you knew what it was saying
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Elephant200
1469 posts
#12
Y by
This is a nice problem, but of course I blanked on the test and I had no idea how to do it. Then 5 minutes after it ends I figure it out :wallbash_red:
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Nayantara5
64 posts
#13
Y by
Im really confused as to why it's not 100. Doesn't every student have to speak a language that 99 students don't speak?
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zlrara01
335 posts
#14
Y by
Nayantara5 wrote:
Im really confused as to why it's not 100. Doesn't every student have to speak a language that 99 students don't speak?

No
Essentially, if you choose any two people out of the 100, then one person has to know a language that the other person does not know
This post has been edited 1 time. Last edited by zlrara01, Nov 13, 2024, 7:05 PM
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wuwang2002
1197 posts
#15
Y by
i think the problem statement could have been closer (makes me think it asks for the number of languages every single one of them speaks)
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orangebear
644 posts
#16
Y by
I am so dumb bruh, my brain just won’t work during the test.
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lpieleanu
2855 posts
#17 • 1 Y
Y by ranu540
Solution
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akliu
1764 posts
#18 • 1 Y
Y by Turtwig113
Oh man this was my favorite question on the test despite it being a combo question. This is what MAA needs to put on their tests.

Denote the set of total languages as $S$. Any two subsets $A$ and $B$ such that $A, B \in S$ and $|A| = |B|$ are guaranteed to satisfy the problem condition as long as $A \neq B$. Then, we just need to find the smallest value of $n$ such that for some $k$, $\tbinom{n}{k} \geq 100$. Manually checking the answer choices, $9$ works.
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saturnrocket
1306 posts
#19 • 1 Y
Y by Sabburi
9 took me 20 minutes to find out what it was saying lol
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CuriousMathBoy72
786 posts
#20
Y by
I somewhat grasped the idea of the problem first time (thoguht it was $\binom{n}2$ at first) then I read the problem and it was $\binom{n}{k}$ so I just made an entire pascals triangle
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RabtejKalra
55 posts
#21
Y by
Why did I say 12?
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Tetra_scheme
89 posts
#22
Y by
Another good one. Note that if they have the same number of languages, it is sufficient for each person to have a different total configuration of languages. This means n choose k is greater than 100 so 9 is the smallest.
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Facejo
2848 posts
#23
Y by
This was a really fun problem and probably my second favorite but it would have been even better if 9 wasn't the smallest option.
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jocaleby1
193 posts
#24
Y by
I feel like this problem could have been #5 - 10 if it weren't for the wording that made it a bit confusing. This problem felt really easy, so I kept second guessing myself. Overall, I liked this problem.

edit: 100th post :)
This post has been edited 1 time. Last edited by jocaleby1, Nov 16, 2024, 9:08 PM
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EaZ_Shadow
1154 posts
#25
Y by
Elephant200 wrote:
This is a nice problem, but of course I blanked on the test and I had no idea how to do it. Then 5 minutes after it ends I figure it out :wallbash_red:

I did it with one minute remaining lol
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youngjin
198 posts
#26
Y by
Im pretty sure the fact they all speak the same number of languages is irrelevant as $9$ is the smallest either way
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EaZ_Shadow
1154 posts
#27 • 1 Y
Y by WHO_LET_ME_COOK
ccarolyn4 wrote:
A group of $100$ students from different countries meet at a mathematics competition. Each student speaks the same number of languages, and, for every pair of students $A$ and $B$, student $A$ speaks some language that student $B$ does not speak, and student $B$ speaks some language that student $A$ does not speak. What is the least possible total number of languages spoken by all the students?

$
\textbf{(A) }9 \qquad
\textbf{(B) }10 \qquad
\textbf{(C) }12 \qquad
\textbf{(D) }51 \qquad
\textbf{(E) }100 \qquad
$

Misplace honestly shoulda been 16 or smth it tricked so many people
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N Quick Reply
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