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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
study for AIME to qual for USJMO
Logicus14   0
11 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
11 minutes ago
0 replies
Problem 2
evt917   47
N 33 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
1 viewing
evt917
Nov 13, 2024
MC_ADe
33 minutes ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N 34 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
megahertz13
Oct 13, 2024
vsarg
34 minutes ago
Stress during competitions
happypi31415   2
N an hour ago by GallopingUnicorn45
I noticed that during big tests, I think the percieved time pressure and high stakes mimic countdown rounds in mathcounts, so my brain defaults to a countdown round mentality. while this is fine in contests like mathcounts, where the first ~10 questions on sprint rounds are essentially countdown round questions and it gives me a chance to get into the flow of things, on tests like the AMC 10 this can really hurt -- notable examples of this include #2 on this year's 10A which was a system of equations. when i saw that the system didn't neatly add to get the desired answer, i panicked and spent ~3 minutes trying to find a neat way before ending up skipping it and coming back to it after problem 17. this is a really big time waste, and i probably spent ~5 minutes total working on it. however, if the problem was the only thing on the paper, I think i definitely would have solved it in <2.

(other examples could be #7 on this years 10B, which was on $7^{2024}+7^{2025}+7^{2026} \pmod{19}$. this one i panicked over because i misread $19$ as $17$ on this problem and didn't get an answer choice, so i again ended up skipping this one and spending way too long on it)

while these aren't necessarily huge time wastes, the psychological affect of skipping an early problem can really impair ur mental state while doing later problems as you won't be as confident

how do i counteract this, and get into the 'flow state' immediately? on mocks this doesn't happen, only in real contests.

also on the AIME this isn't a problem
2 replies
happypi31415
an hour ago
GallopingUnicorn45
an hour ago
No more topics!
PROBLEM 1
evt917   21
N Yesterday at 2:08 AM by physicskiddo
Source: 2024 AMC 12B #1 / 10B #1
In a long line of people, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?

$
\textbf{(A) }2021 \qquad
\textbf{(B) }2022 \qquad
\textbf{(C) }2023 \qquad
\textbf{(D) }2024 \qquad
\textbf{(E) }2025 \qquad
$
21 replies
evt917
Nov 13, 2024
physicskiddo
Yesterday at 2:08 AM
PROBLEM 1
G H J
Source: 2024 AMC 12B #1 / 10B #1
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evt917
1514 posts
#1 • 1 Y
Y by clarkculus
In a long line of people, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?

$
\textbf{(A) }2021 \qquad
\textbf{(B) }2022 \qquad
\textbf{(C) }2023 \qquad
\textbf{(D) }2024 \qquad
\textbf{(E) }2025 \qquad
$
This post has been edited 2 times. Last edited by jlacosta, Nov 13, 2024, 5:54 PM
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HonestCat
947 posts
#2
Y by
B 2022 I think
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Tem8
154 posts
#3
Y by
1013+1010-1=2022, so the answer is $\boxed{\textbf{B}}$
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HungryCalculator
500 posts
#4
Y by
B can confirm
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evt917
1514 posts
#5
Y by
sillying #1

ughhhhhhhhhhhhhhh AHHHHHHH

geez bro like how
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Squidget
372 posts
#6 • 1 Y
Y by ranu540
I thought the answer would be 2024 because they always do that, so I got nervous when I got 2022. But I was right (phew).

Just add them together and subtract 1
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alexanderhamilton124
242 posts
#7
Y by
rahhhhhhhhhh B very difficult for a #1
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TiguhBabeHwo
208 posts
#8 • 1 Y
Y by Sagnik123Biswas
ok but how did I get 25 and 24 and somehow add wrong on this
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osunm
355 posts
#9
Y by
welp i sillied that
i dont even remember how i got D during the contest but aghhhh
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Lhaj3
10 posts
#10
Y by
i almost sillied this one but changed my answer to B in the last 10 minutes
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MathRook7817
234 posts
#11
Y by
1012+1+1009 = 2022
lol i checked this about 3 times cuz the year is 2024 but the answer is 2022
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mathprodigy2011
58 posts
#12
Y by
evt917 wrote:
sillying #1

ughhhhhhhhhhhhhhh AHHHHHHH

geez bro like how

dang, its about mindset on testing day
Z K Y
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andrewcheng
482 posts
#13
Y by
confirm B
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eg4334
391 posts
#14
Y by
$1012$ to the left of the person and $1009$ to the right, ergo $1012+1009+1 = \boxed{2022}$
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joshualiu315
2431 posts
#15
Y by
There are $1012$ people to the left of the person and $1009$ to the right of the person, so the answer is $1012+1009+1 = \boxed{2022}$.
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Mr.Sharkman
386 posts
#16
Y by
mathprodigy2011 wrote:
evt917 wrote:
sillying #1

ughhhhhhhhhhhhhhh AHHHHHHH

geez bro like how

dang, its about mindset on testing day

Bruh this happened to me on HMMT and it cost me top 10 :wallbash_red: (On general)
This post has been edited 1 time. Last edited by Mr.Sharkman, Nov 13, 2024, 8:48 PM
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Mr.Sharkman
386 posts
#17
Y by
Add the numbers. Then, notice that the person that is in this position is counted twice, giving $1013+1010-1 = 2022.$
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RedFireTruck
4057 posts
#18
Y by
aint no way we got PIE on a p1 before gta 6

1013+1010-1=2022
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aaja3427
1918 posts
#19
Y by
I just tested for a small number of people (2nd from the left and 2nd from the right in a line of 3 ppl, for example) to check that you add the two numbers together and subtract 1
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SpeedCuber7
1394 posts
#20
Y by
TiguhBabeHwo wrote:
ok but how did I get 25 and 24 and somehow add wrong on this

final five was free

i get it (although i skipped 25 because i don't know geo)
This post has been edited 1 time. Last edited by SpeedCuber7, Thursday at 5:41 PM
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player01
1695 posts
#21 • 1 Y
Y by physicskiddo
A very timesink problem.

Because this is an early counting problem, we will resort to one of the most basic counting techniques: counting.

WLOG let the person of interest be named Bob. Firstly, we are given that Bob is the 1013th from the left. Thus, we can symbolize this by drawing 1013 evenly spaced dots, marking the rightmost one red (which will represent Bob):

[asy]
size(10cm);
for (int i=0;i<1012;++i) {
    dot((i,0));
}
dot((1012, 0), red);

draw((1012,-200)--(1012,200), red);
[/asy]

Note about diagrams

I have drawn a red line segment through the rightmost dot so that its position is more easily verifiable. Now, we are also given that Bob is the 1010th from the right. Thus, we can draw 1010 dots, such that Bob is the leftmost dot:

[asy]
size(10cm);
for (int i=0;i<1009;++i) {
    dot((i,0));
}
dot((-1, 0), red);

draw((-1,-200)--(-1,200), red);
[/asy]

Again, I have drawn a red line through the leftmost dot in order to clarify the position of Bob. Now, we realize that we can match the two diagrams at the red line, because that is where Bob stands in both:

[asy]
size(10cm);

for (int i=0;i<1012;++i) {
    dot((i,100));
}
dot((1012, 100), red);

draw((1012,-200)--(1012,300), red);

for (int i=0;i<1009;++i) {
    dot((i + 1013,0));
}
dot((-1 + 1013, 0), red);

draw((-1 + 1013,-200)--(-1 + 1013,200), red);
[/asy]

Now, we can "merge" the two parts together:

[asy]
size(10cm);

for (int i=0;i<1012;++i) {
    dot((i,0));
}
dot((1012, 0), red);

draw((1012,-200)--(1012,200), red);

for (int i=0;i<1009;++i) {
    dot((i + 1013,0));
}
dot((-1 + 1013, 0), red);

draw((-1 + 1013,-200)--(-1 + 1013,200), red);
[/asy]

We have a single line of dots, which means that we can finally count them up to find the total number of people in the line. Note

After a little counting, we arrive at the final answer of $\textbf{(B) }2022$. $\blacksquare$

Additionally, we can conclude that it will take Bob a very long time to get out of line.
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physicskiddo
812 posts
#22 • 1 Y
Y by player01
@above is the solution i used in the test, very optimal
Z K Y
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