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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 28 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
mathkiddus
2 hours ago
lpieleanu
28 minutes ago
study for AIME to qual for USJMO
Logicus14   0
39 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
39 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!
ewwwwwww
fruitmonster97   37
N Nov 14, 2024 by ryanjwang
Source: 2024 AMC 10B #4/2024 AMC 12B #4
Balls numbered $1,2,3,\ldots$ are deposited in $5$ bins, labeled $A,B,C,D,$ and $E$, using the following procedure. Ball $1$ is deposited in bin $A$, and balls $2$ and $3$ are deposted in $B$. The next three balls are deposited in bin $C$, the next $4$ in bin $D$, and so on, cycling back to bin $A$ after balls are deposited in bin $E$. (For example, $22,23,\ldots,28$ are despoited in bin $B$ at step 7 of this process.) In which bin is ball $2024$ deposited?

$\textbf{(A) }A\qquad\textbf{(B) }B\qquad\textbf{(C) }C\qquad\textbf{(D) }D\qquad\textbf{(E) }E$
37 replies
fruitmonster97
Nov 13, 2024
ryanjwang
Nov 14, 2024
ewwwwwww
G H J
Source: 2024 AMC 10B #4/2024 AMC 12B #4
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fruitmonster97
2239 posts
#1
Y by
Balls numbered $1,2,3,\ldots$ are deposited in $5$ bins, labeled $A,B,C,D,$ and $E$, using the following procedure. Ball $1$ is deposited in bin $A$, and balls $2$ and $3$ are deposted in $B$. The next three balls are deposited in bin $C$, the next $4$ in bin $D$, and so on, cycling back to bin $A$ after balls are deposited in bin $E$. (For example, $22,23,\ldots,28$ are despoited in bin $B$ at step 7 of this process.) In which bin is ball $2024$ deposited?

$\textbf{(A) }A\qquad\textbf{(B) }B\qquad\textbf{(C) }C\qquad\textbf{(D) }D\qquad\textbf{(E) }E$
This post has been edited 2 times. Last edited by jlacosta, Nov 13, 2024, 6:02 PM
Reason: Updating source field.
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pqr.
147 posts
#2
Y by
D $              $?
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fruitmonster97
2239 posts
#3
Y by
pqr. wrote:
D $              $?

Can confirm. Took a good 5-ish minutes :(
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vsamc
3769 posts
#4 • 1 Y
Y by GAMER100
63 * 64 / 2 < 2024 < 64 * 65/2 so d
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mathboy282
2966 posts
#5 • 1 Y
Y by GAMER100
63*64/2 = 2016 -> 63 == 3 mod 5, hence C is full, so the rest 8 balls must lie in D.
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Sedro
5739 posts
#6 • 1 Y
Y by IbrahimNadeem
D confirmed, spent 5 minutes finding the smallest triangular number less than $2024$ :|
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pingpongmerrily
2424 posts
#7
Y by
yeah i mighta misbubbled bc first i thought D then C but switched back to D lol
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EthanSpoon
635 posts
#8
Y by
It is D

I took 5 minutes on this
Skipped and moved on
and came back and took another 5 minutes
:|
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saturnrocket
1305 posts
#9
Y by
yeah it's D
This post has been edited 1 time. Last edited by saturnrocket, Nov 13, 2024, 5:47 PM
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elizhang101412
1060 posts
#10
Y by
i spent like 10 minutes
attempt 1: loops 12345 6 7 8 9 10, so nth repeat sum is 25n-10(n)/2
then I just dropped it here
attempt 2: loops in 15, 2024 mod 15 so E
attempt 3: 64*65/2>2024, 64 mod 5 gives D
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MathRook7817
234 posts
#11
Y by
D, let's go!!!
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exp-ipi-1
965 posts
#12
Y by
rlly hard for #4
This post has been edited 1 time. Last edited by exp-ipi-1, Nov 13, 2024, 6:04 PM
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LostDreams
125 posts
#13
Y by
Just sum to get as close to $2024$ and you see that $D$ has to be the next option since $1+...+63 = 2016$
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pingpongmerrily
2424 posts
#14
Y by
exp-ipi-1 wrote:
rlly hard for #4

fax even if you get 63 triangular i almost thought that would go in box C
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Tem8
154 posts
#15
Y by
I made a pyramid of distributions of balls lol.
$A$: 1
$B$: $\binom{2}{2}+1$, $\binom{3}{2}$
$C$: $\binom{3}{2}+1$, $\binom{3}{2}+2$, $\binom{4}{2}$
$D$: $\binom{4}{2}+1$, $\binom{4}{2}+2$, $\binom{4}{2}+3$, $\binom{5}{2}$
$E$: $\binom{5}{2}+1$, $\binom{5}{2}+2$, $\binom{5}{2}+3$, $\binom{5}{2}+4$, $\binom{6}{2}$
$A$: $\binom{6}{2}+1$, $\binom{6}{2}+2$, $\binom{6}{2}+3$, $\binom{6}{2}+4$, $\binom{6}{2}+5$, $\binom{7}{2}$.
$\cdots$
Because ball $2024$ lies between $\binom{64}{2}+1=2016$ and $\binom{64}{2}$, it follows that ball $2024$ was deposited to bin number $64 \equiv 4 \pmod{5}$, or $\boxed{\textbf{(D) } D}$.
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mathprodigy2011
58 posts
#16
Y by
this question wasn't too bad. because 63*32=2016 is like common knowledge so it would have to be64. 64 = 4 mod 5 so D
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brainfertilzer
1802 posts
#17
Y by
Hard; equivalent to find minimal $n$ such that $n(n+1) > 2\cdot 2024$, which took like 6-7 minutes to do.
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mathprodigy2011
58 posts
#18
Y by
brainfertilzer wrote:
Hard; equivalent to find minimal $n$ such that $n(n+1) > 2\cdot 2024$, which took like 6-7 minutes to do.

well if you approximate then n^2 is around 4000 and so n is between 60 and 70. try 65 and it almost works. or just remember that 63*64 = 4032(2016 = 63*32)
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plang2008
271 posts
#19
Y by
brainfertilzer wrote:
Hard; equivalent to find minimal $n$ such that $n(n+1) > 2\cdot 2024$, which took like 6-7 minutes to do.

trivial by 2023 AMC once again
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bot1132
80 posts
#20
Y by
sqrt2 * 45 :)
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eg4334
391 posts
#21
Y by
$\frac{64 \cdot 65}{2} > 2024$ but $\frac{63 \cdot 64}{2} < 2024$, so it happens on the $64$th pass. $64 \equiv 4 \pmod{5}$, so $D$.
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pog
4905 posts
#22
Y by
Recall that $2016$ is the $63$rd triangular number, so our $64$th bin is the answer. Hence $\boxed{\textbf{(D) }D}$.
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yambe2002
1642 posts
#23
Y by
do people actually know this stuff because i dont
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williamxiao
2457 posts
#24
Y by
If you have over 60 triangular numbers memorized you’re a freak

Just estimate based on the square root of 4048
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pingpongmerrily
2424 posts
#25
Y by
yambe2002 wrote:
do people actually know this stuff because i dont

no just estimate
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greenAB08
7 posts
#26
Y by
My prep day of the test saved me a lot of time. Always good to know the triangular numbers that exceed this year
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wuwang2002
1127 posts
#27
Y by
horrible problem.
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giratina3
243 posts
#28
Y by
This problem was complete bash… not well written…
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yambe2002
1642 posts
#29
Y by
pingpongmerrily wrote:
yambe2002 wrote:
do people actually know this stuff because i dont

no just estimate

fair enough, i took a way longer route
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akliu
1705 posts
#30
Y by
Some notes on this question I guess? I found the thing relatively quickly because we want $n$ such that $n(n+1) > 4048$. $4048$ is pretty close to $4225$ which is a perfect square of $65$, so $n$ is probably around $63$ to $65$. Indeed, $n=64$ here.
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bot1132
80 posts
#31
Y by
fruitmonster97 wrote:
Balls numbered $1,2,3,\ldots$ are deposited in $5$ bins, labeled $A,B,C,D,$ and $E$, using the following procedure. Ball $1$ is deposited in bin $A$, and balls $2$ and $3$ are deposted in $B$. The next three balls are deposited in bin $C$, the next $4$ in bin $D$, and so on, cycling back to bin $A$ after balls are deposited in bin $E$. (For example, $22,23,\ldots,28$ are despoited in bin $B$ at step 7 of this process.) In which bin is ball $2024$ deposited?

$\textbf{(A) }A\qquad\textbf{(B) }B\qquad\textbf{(C) }C\qquad\textbf{(D) }D\qquad\textbf{(E) }E$

D
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abbominable_sn0wman
34 posts
#32
Y by
honestly this wasn't too bad it was just a bit early (im also a combo main--)
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amaops1123
1644 posts
#33
Y by
Why is this P4
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pingpongmerrily
2424 posts
#34 • 1 Y
Y by kosarsi
its not hard, its just very bashy and the computations get ugly so its easy to silly
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pog
4905 posts
#35
Y by
williamxiao wrote:
If you have over 60 triangular numbers memorized you’re a freak

Just estimate based on the square root of 4048

I only know 2016
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jb2015007
591 posts
#36
Y by
pqr. wrote:
D $              $?

how is this d bruh idk I left it blank
This post has been edited 1 time. Last edited by jb2015007, Nov 14, 2024, 3:00 PM
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MrMustache
2450 posts
#37 • 1 Y
Y by kosarsi
amaops1123 wrote:
Why is this P4

So that people who don't find non-bushy ways to solve the problem spend room achy time and do worse.
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ryanjwang
10 posts
#38
Y by
pingpongmerrily wrote:
yeah i mighta misbubbled bc first i thought D then C but switched back to D lol
same :sob:
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