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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 7 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
+1 w
mathkiddus
an hour ago
lpieleanu
7 minutes ago
study for AIME to qual for USJMO
Logicus14   0
18 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
1 viewing
Logicus14
18 minutes ago
0 replies
Problem 2
evt917   47
N 40 minutes 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
40 minutes ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N 41 minutes 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
1 viewing
megahertz13
Oct 13, 2024
vsarg
41 minutes ago
No more topics!
9c4 pigeon hole
ccarolyn4   21
N Today at 9:24 AM by Tetra_scheme
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
$
21 replies
+1 w
ccarolyn4
Nov 13, 2024
Tetra_scheme
Today at 9:24 AM
9c4 pigeon hole
G H J
Source: 2024 AMC 10 P12
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ccarolyn4
22 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 2 times. Last edited by ccarolyn4, Nov 13, 2024, 5:18 PM
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ccarolyn4
22 posts
#2 • 5 Y
Y by megarnie, solasky, Soccerstar9, lprado, Turtwig113
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
871 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
592 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
201 posts
#5
Y by
9 choose 4 satisfies already, so A
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Squidget
372 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
312 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
2424 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
348 posts
#9 • 4 Y
Y by solasky, LostDreams, OronSH, ranu540
this is why you read diestel
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MathRook7817
234 posts
#10
Y by
bro i skipped this it was easy
lol i sold
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aleyang
178 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
1451 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
312 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
1127 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
357 posts
#16
Y by
I am so dumb bruh, my brain just won’t work during the test.
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lpieleanu
2248 posts
#17 • 1 Y
Y by ranu540
Solution
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akliu
1705 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
1305 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
772 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
12 posts
#21
Y by
Why did I say 12?
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Tetra_scheme
21 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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