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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 AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year 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!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section 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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0 replies
jlacosta
Nov 1, 2024
0 replies
Circles Tangent to a Points: Find Area of Region
hesa57   45
N 2 minutes ago by xTimmyG
Source: 2015 AIME 2 Problem 15
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

IMAGE
45 replies
hesa57
Mar 26, 2015
xTimmyG
2 minutes ago
Counterexample
avn   36
N 2 minutes ago by mhgelgi
Source: 2019 AMC 10B #2 / 12B #2
Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?

$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$
36 replies
avn
Feb 14, 2019
mhgelgi
2 minutes ago
9 How well do you think you will do on the AMC 8
TQ_Math   33
N 3 minutes ago by elizhang101412
I know it's November, but it's never not too late to prep, right?!

40 VOTES!!!
60 votess!
100 VOTES
120 votesdfgjkafjd
33 replies
TQ_Math
Nov 17, 2024
elizhang101412
3 minutes ago
10B Score Thread
BS2012   111
N 3 minutes ago by eugenewang1
$\begin{tabular}{c|c|c|c|c}Username & Grade & 10B \\ \hline
BS2012 & 9 & 144  \\
\end{tabular}$
EDIT: I found out i didn't silly #19, so i got 144
111 replies
BS2012
Nov 13, 2024
eugenewang1
3 minutes ago
No more topics!
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   37
N Yesterday at 3:43 AM by Skillfulcotton
Orz to the Legends (Music, AMC version - MegaMath Channel original)

Get pumped up for the AMCs!

[youtube]https://www.youtube.com/watch?v=KmiNI00uo-s&ab_channel=MegaMathChannel[/youtube]

Join our discord! https://discord.gg/hh7vntTb2E

Math Problem (AMC 10/12): Let $M$ be the smallest positive integer satisfying the property that $M^6$ is a multiple of both $2024^2$ and $2025^3$. How many positive divisors does $M$ have?

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
37 replies
megahertz13
Oct 13, 2024
Skillfulcotton
Yesterday at 3:43 AM
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
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megahertz13
2957 posts
#1 • 7 Y
Y by Shan3t, Marcus_Zhang, kilobyte144, alexanderhamilton124, alphanova, evt917, orangebear
Orz to the Legends (Music, AMC version - MegaMath Channel original)

Get pumped up for the AMCs!



Join our discord! https://discord.gg/hh7vntTb2E

Math Problem (AMC 10/12): Let $M$ be the smallest positive integer satisfying the property that $M^6$ is a multiple of both $2024^2$ and $2025^3$. How many positive divisors does $M$ have?

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
This post has been edited 9 times. Last edited by megahertz13, Nov 16, 2024, 2:53 AM
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RedFireTruck
4057 posts
#2 • 2 Y
Y by Amkan2022, bomberdoodles
2024 is like 2^3*11*23 and 2025=3^4*5^2 so M has to be like 2*3^2*5*11*23 so the ans is like 48 me thinks

im a bit rusty at this math thing tho so I might be trippin
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DylanPShen
709 posts
#3
Y by
It says the video is restricted
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kilobyte144
622 posts
#4
Y by
DylanPShen wrote:
It says the video is restricted

It works for me. Does it work if you go directly to the link? https://www.youtube.com/watch?v=KmiNI00uo-s
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megahertz13
2957 posts
#5
Y by
RedFireTruck wrote:
2024 is like 2^3*11*23 and 2025=3^4*5^2 so M has to be like 2*3^2*5*11*23 so the ans is like 48 me thinks

im a bit rusty at this math thing tho so I might be trippin

This is what I got, nice!
DylanPShen wrote:
It says the video is restricted

Maybe sign out of your account so that there's no age-restriction issues
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megahertz13
2957 posts
#6
Y by
Here's another problem

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
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Aaronjudgeisgoat
631 posts
#7
Y by
megahertz13 wrote:
Here's another problem

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

hola
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megahertz13
2957 posts
#8
Y by
Aaronjudgeisgoat wrote:
megahertz13 wrote:
Here's another problem

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

hola

Did you find the nice solution or did you alg bash it?
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megahertz13
2957 posts
#11
Y by
get hyped for the AMC B!!
This post has been edited 1 time. Last edited by megahertz13, Nov 8, 2024, 12:40 AM
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mathnerd_101
1444 posts
#12
Y by
Beat is gas, lyrics are... more... questionable...? sorry bro
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HighWater
379 posts
#13
Y by
Hopefully I can get a decent score on 10B, I failed 10A with at most a 123...
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Existing_Human1
123 posts
#14
Y by
Why is this so catchy?
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Existing_Human1
123 posts
#15
Y by
Existing_Human1 wrote:
Why is this so catchy? (In a good way)
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megahertz13
2957 posts
#17
Y by
HighWater wrote:
Hopefully I can get a decent score on 10B, I failed 10A with at most a 123...

:sob: good luck!
Existing_Human1 wrote:
Why is this so catchy?

maybe because it helps decrease stress
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vsarg
191 posts
#18
Y by
How AI generated is the song? Ik the voice and thumbnail but did u make the lyrics or smth
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megahertz13
2957 posts
#24
Y by
vsarg wrote:
How AI generated is the song? Ik the voice and thumbnail but did u make the lyrics or smth

AI created a draft of the lyrics, but I edited them.
This post has been edited 1 time. Last edited by megahertz13, Nov 11, 2024, 2:31 PM
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Squidget
380 posts
#26
Y by
Haha I thought that was actually you(megahertz) singing. I don’t know how i didn’t realize that it has to be AI.
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vsarg
191 posts
#28 • 2 Y
Y by kilobyte144, giratina3
Squidget wrote:
Haha I thought that was actually you(megahertz) singing. I don’t know how i didn’t realize that it has to be AI.

Im pretty sure megahertz is a girl though. (Not trying to be offensive or discriminatory or anti-pride or suggestive or opinionated or hating or mean or assuming or...)
This post has been edited 1 time. Last edited by vsarg, Nov 16, 2024, 2:24 AM
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FuturePanda
143 posts
#29
Y by
he's not.
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vsarg
191 posts
#31
Y by
Oh ok! $$ $$
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ijco
20 posts
#33
Y by
FuturePanda wrote:
he's not.

i think you are confused, megahertz isnt a girl
This post has been edited 1 time. Last edited by ijco, Nov 17, 2024, 1:51 AM
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vincentwant
1158 posts
#36
Y by
as someone who knows megahertz personally he is not a girl
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wikjay
222 posts
#37
Y by
vincentwant wrote:
as someone who knows megahertz personally he is not a girl

I went to mp with her ik her personally as well
This post has been edited 1 time. Last edited by wikjay, Nov 16, 2024, 5:10 PM
Reason: her
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Happycat2
668 posts
#38
Y by
SO IS MEGAHERTZ A GIRL OR A BOY OMG
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amaops1123
1678 posts
#39
Y by
AMCS MAKE SCHOOL MATH LOOK LIKE A JOKE :FIRE: :FIRE: :SPEAKING_HEAD:
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Mathdreams
1383 posts
#40
Y by
amaops1123 wrote:
AMCS MAKE SCHOOL MATH LOOK LIKE A JOKE :FIRE: :FIRE: :SPEAKING_HEAD:

This line is currently in question :/
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alphanova
10 posts
#41 • 1 Y
Y by kilobyte144
Happycat2 wrote:
SO IS MEGAHERTZ A GIRL OR A BOY OMG

He's a boy. See here and scroll down.
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vsarg
191 posts
#43
Y by
Wow... This turned into a gender fight. He's a boy yall I just though he was a girl because of his voice... So he's a boy and no more comments about this
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NoSignOfTheta
1590 posts
#47
Y by
vincentwant wrote:
as someone who knows megahertz personally he is not a girl
wikjay wrote:
vincentwant wrote:
as someone who knows megahertz personally he is not a girl

I went to mp with her ik her personally as well

POV catfish
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FuturePanda
143 posts
#50
Y by
I know him too he is a boy!
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mathMagicOPS
809 posts
#51
Y by
I can confirm that megahertz is definitely nota girl, we teamed for a competition.
This post has been edited 2 times. Last edited by mathMagicOPS, Nov 18, 2024, 3:44 AM
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Embershed97
691 posts
#52
Y by
lol cant imagine megahertz being a girl tho
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vsarg
191 posts
#54
Y by
? bro someone lock this bruh
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Amkan2022
1922 posts
#55
Y by
Embershed97 wrote:
lol cant imagine megahertz being a girl tho

is this genderism?
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EaZ_Shadow
381 posts
#56
Y by
Why are y’all discussing about gender rn bro that’s not important nor is it relevant someone should lock this thread up or delete this topic :/
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vsarg
191 posts
#58
Y by
Im rlly sorry I started this.
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Skillfulcotton
150 posts
#59
Y by
megahertz13 wrote:
Math Problem (AMC 10/12): Let $M$ be the smallest positive integer satisfying the property that $M^6$ is a multiple of both $2024^2$ and $2025^3$. How many positive divisors does $M$ have?
Here's the solution to #1.
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Skillfulcotton
150 posts
#60
Y by
megahertz13 wrote:
Here's another problem

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

I'm not sure how to solve this but is this a finite series? Click to reveal hidden text
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