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AMC and other contests, summer programs, etc.
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Contests & Programs 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
Logical guessing game!
Mathdreams   22
N an hour ago by JH_K2IMO
Source: 2021 Fall AMC10B P10
Fourty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
22 replies
Mathdreams
Nov 17, 2021
JH_K2IMO
an hour ago
Possibility of USAMO?
MathXplorer10   4
N an hour ago by MathXplorer10
Hi guys!


I got a 118.5 on the 12B test this year. I am wondering if it is possible to make USAMO (what do you think the cutoffs would be this year?)

For some background, I got 121.5/127.5 on the 10s last year, and got a 7 on AIME with no extra prep. Is it possible to go from a 7 to a 10 (or whatever I need to get on AIME)?

Thank you!
4 replies
MathXplorer10
4 hours ago
MathXplorer10
an hour ago
10a vs 10b
golden_star_123   111
N an hour ago by happyfish0922
Post the difference between your 10a and 10b score!
111 replies
golden_star_123
Wednesday at 6:24 PM
happyfish0922
an hour ago
What do next?
FuturePanda   2
N an hour ago by Tem8
Hi everyone,

I think I got an 81 and 102 for 12A and 10B, sillying way too much on both. I read all of the AOPS books, and I know most of the theorems for the AMC’s I just don’t know which ones to apply to solve the problems. Additionally, I suck at trig, complex, and logarithms. What should I do to improve?

For example, should I be grinding past AIME’s?
I plan on reading most of the Awesomemath books for L3
2 replies
FuturePanda
3 hours ago
Tem8
an hour ago
No more topics!
ewwwwwww
fruitmonster97   37
N Yesterday at 6:00 AM 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
Wednesday at 5:34 PM
ryanjwang
Yesterday at 6:00 AM
ewwwwwww
G H J
Source: 2024 AMC 10B #4/2024 AMC 12B #4
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fruitmonster97
2234 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, Wednesday at 6:02 PM
Reason: Updating source field.
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pqr.
147 posts
#2
Y by
D $              $?
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fruitmonster97
2234 posts
#3
Y by
pqr. wrote:
D $              $?

Can confirm. Took a good 5-ish minutes :(
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vsamc
3768 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
2394 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, Wednesday at 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
233 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, Wednesday at 6:04 PM
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LostDreams
124 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
2394 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
153 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
57 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
57 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
4903 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
2455 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
2394 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
1126 posts
#27
Y by
horrible problem.
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giratina3
242 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
33 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
1645 posts
#33
Y by
Why is this P4
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pingpongmerrily
2394 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
4903 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
574 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, Yesterday at 3:00 PM
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MrMustache
2449 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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