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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
Past PUMaC results
mathkiddus   1
N 19 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
mathkiddus
2 hours ago
lpieleanu
19 minutes ago
study for AIME to qual for USJMO
Logicus14   0
30 minutes ago
hi

so this is my first year doing amc10 (9th grade) and I got a decent enough score (121.5) on amc10b and after some research I found out that I need to get AT LEAST 10 problems right to qualify for USJMO

did anyone get 10+ problems right? If you did how did you study? any general tips?


ty
0 replies
Logicus14
30 minutes ago
0 replies
Problem 2
evt917   47
N an hour ago by MC_ADe
Source: 2024 AMC 12B #2 / AMC 10B #2
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$
47 replies
evt917
Nov 13, 2024
MC_ADe
an hour ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N an hour ago by vsarg
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 $m+n$.
21 replies
megahertz13
Oct 13, 2024
vsarg
an hour ago
No more topics!
Who let MAA cook
HumanCalculator9   99
N Thursday at 5:51 PM by ArshaBajaj
Source: 2024 AMC 12B #10 / AMC 10B #15
A list of 9 real numbers consists of $1$, $2.2 $, $3.2 $, $5.2 $, $6.2 $, $7$, as well as $x, y,z$ with $x\leq y\leq z$. The range of the list is $7$, and the mean and median are both positive integers. How many ordered triples $(x,y,z)$ are possible?

$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) infinitely many}\qquad
$
99 replies
HumanCalculator9
Nov 13, 2024
ArshaBajaj
Thursday at 5:51 PM
Who let MAA cook
G H J
Source: 2024 AMC 12B #10 / AMC 10B #15
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mbappeisgoat
6 posts
#87
Y by
I got right lol
I got one solution with two of the same numbers and a different one (I forgot) but it said ORDERED triples so I thought it was 3.
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sdfgfjh
35 posts
#88
Y by
title is very based, this problem was very misplaced and of low quality (my favorite coping mechanism)
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pingpongmerrily
2424 posts
#89
Y by
sdfgfjh wrote:
title is very based, this problem was very misplaced and of low quality (my favorite coping mechanism)

honestly i hate this question it kinda ruined my score by 10 points bc i wasted too much time but the quality is fine
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stjwyl
643 posts
#90
Y by
pingpongmerrily wrote:
sdfgfjh wrote:
title is very based, this problem was very misplaced and of low quality (my favorite coping mechanism)

honestly i hate this question it kinda ruined my score by 10 points bc i wasted too much time but the quality is fine

Same
I was looking at #25 and got the system of equations but then time was up
If I had maybe 3 more minutes I would have gotten #25
and extra 6 pts
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mathfan2020
346 posts
#91
Y by
0,5,6.2-blocked
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Happy2Bee
6 posts
#92
Y by
Potato2017 wrote:
3.... worst problem ever even though i got it right

is 3 right ?
cause when i asked my sister who she told me infintely many
i also got that only
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asdf334
7547 posts
#93
Y by
cases on median easy money fr
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megahertz13
2915 posts
#94
Y by
HumanCalculator9 wrote:
A list of 9 real numbers consists of $1$, $2.2 $, $3.2 $, $5.2 $, $6.2 $, $7$, as well as $x, y,z$ with $x\leq y\leq z$. The range of the list is $7$, and the mean and median are both positive integers. How many ordered triples $(x,y,z)$ are possible?

$
\textbf{(A) }1 \qquad
\textbf{(B) }2 \qquad
\textbf{(C) }3 \qquad
\textbf{(D) }4 \qquad
\textbf{(E) infinitely many}\qquad
$

so sillyable :sob:
Z K Y
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Yiyj1
965 posts
#95
Y by
i got the x=0.1 z=7.1 case first somehow and then i got the x=0 and z=8 cases. a total of 3. spent like 72 hours trying to figure out a fourth case but failed.
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RollOver2020
342 posts
#96
Y by
Got 2 and found the last one after the test. This was a huge time sink and one of the dumbest questions the MAA has proposed
Z K Y
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FuturePanda
106 posts
#97
Y by
I don’t believe this was a bad question, and I don’t blame people for missing cases, but to be real, being organized during the test helps a LOT! Keeping track of the ordered pairs really helps you know every case, so you won’t get it wrong.
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Facejo
2779 posts
#98
Y by
OronSH wrote:
answer is C) 3 bruh i got trolled by the 0.1 7.1 case

I also got trolled by this :( highly unfortunate
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Demetri
800 posts
#99
Y by
Everyone else:
OMG WAS SO BASHY I SPENT 30 MINUTES ON PROBLEM AND GOT IT WRONG
OMG WAS SO BASHY I SPENT 30 MINUTES ON PROBLEM BUT SOMEHOW GOT IT RIGHT
Me:
Well i just skipped it :maybe:
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SpeedCuber7
1394 posts
#100
Y by
am i the only person in the world who put A
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ArshaBajaj
53 posts
#101
Y by
Missed one case after trying many cases
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N Quick Reply
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