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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
40 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
40 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!
Sakupen circles
EpicBird08   28
N Oct 29, 2024 by JH_K2IMO
Source: 2023 AMC 10A Problem 15
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
28 replies
EpicBird08
Nov 9, 2023
JH_K2IMO
Oct 29, 2024
Sakupen circles
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 AMC 10A Problem 15
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EpicBird08
1679 posts
#1 • 1 Y
Y by centslordm
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
Attachments:
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ryanbear
1043 posts
#2
Y by
its E because the shaded region equals to $\pi(1+2+...+n) > 2023\pi$ and then use the sum of arithmetic sequence formula to get that $(n)(n+1) > 4046$ then plug in the answer choices
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wuwang2002
1127 posts
#3
Y by
i got E (64)
This post has been edited 1 time. Last edited by wuwang2002, Nov 9, 2023, 5:04 PM
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ostriches88
1441 posts
#4
Y by
can confirm E
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MrThinker
616 posts
#5
Y by
64 really ez
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OlympusHero
17012 posts
#6
Y by
This was E, pretty easy problem to silly though if you miswrite the summation
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gicyuraok2
1059 posts
#7
Y by
It is definitely E bad solution

sakupen circles :skull:
This post has been edited 1 time. Last edited by gicyuraok2, Nov 9, 2023, 7:41 PM
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HonestCat
947 posts
#8
Y by
I got 46 :(
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Alex-131
5144 posts
#9
Y by
gicyuraok2 wrote:
It is definitely E bad solution

I got the correct inequality and though there is no way n = 48 or answer choices works and skipped. I forgot to read all the circles
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peace09
5313 posts
#10 • 12 Y
Y by brainfertilzer, ryanbear, ihatemath123, mahaler, gracemoon124, Zhaom, Significant, EpicBird08, mathmax12, akliu, centslordm, Alex-131
Click to reveal hidden text

Y'all really need to take the WMC.
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think4l
342 posts
#11
Y by
Solution: https://primesets.blogspot.com/2023/11/2023-amc-10a-15.html
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MC413551
2225 posts
#12
Y by
2^2-1^2+4^3-3^2+...
Factorize to 1+2+3+4+...
n(n+1)/2>2023
n=64
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joshualiu315
2431 posts
#14
Y by
peace09 wrote:
Click to reveal hidden text

Y'all really need to take the WMC.

lmao real. i managed to silly that wmc problem sadly

The formula is simply $1+2+\dots+n$ for even $n$. Thus, $n=\boxed{\textbf{(E) } 64}$.
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Shreyasharma
589 posts
#15
Y by
This was eh.
Note that we basically have $\pi \cdot \left( (2^2 - 1^2) + (4^2 - 3^2) + (6^2 - 5^2) + \dots \right)$. The terms in the parentheses form a arithmetic series of the form $\{3, 7, 11, \dots\}$. Thus we have the total area after $2k$ circles is,
\begin{align*}
\pi \cdot \frac{(4k+2) \cdot k}{2} = \pi \cdot (2k+1) \cdot k
\end{align*}Now noting that $k \approx 32$ is when the sum above exceeds $2023 \pi$, we find $2k=64$.
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ryanbear
1043 posts
#17
Y by
https://www.quora.com/How-do-you-find-the-sum-of-series-100%C2%B2-99%C2%B2-98%C2%B2-97%C2%B2-96%C2%B2-95%C2%B2-2%C2%B2-1%C2%B2
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HighWater
379 posts
#18
Y by
I got (E) 64

Also why Sakupen Circles
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I-_-I
2047 posts
#19 • 6 Y
Y by ihatemath123, centslordm, pog, akliu, mathmax12, rayford
5 seconds
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MathKid16
70 posts
#20 • 1 Y
Y by thegamer1049
no way gometry dash reference??? AMC10 was sponsoredby robtop??
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xHypotenuse
548 posts
#21
Y by
Is that a geometry dash reference????????


oops sniped
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gracemoon124
871 posts
#22 • 1 Y
Y by peace09
edit problem statement to be the below, thanks


An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
[asy]
size(6cm);
pen greywhat;
greywhat = RGB(105,105,105);
filldraw(circle((8, 0), 8), greywhat);
filldraw(circle((7, 0), 7), white);
filldraw(circle((6, 0), 6), greywhat);
filldraw(circle((5, 0), 5), white);
filldraw(circle((4, 0), 4), greywhat);
filldraw(circle((3, 0), 3), white);
filldraw(circle((2, 0), 2), greywhat);
filldraw(circle((1, 0), 1), white);
[/asy]
$\textbf{(A) }46\qquad\textbf{(B) }48\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }64$
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akliu
1705 posts
#23
Y by
Somehow managed to not silly the WMC question, but sillied and got computationally blocked on this, tilting my entire testing rhythm.

There's always going to be that one question on the test though unless you're @epicbird08 :o
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StevePlayer
7 posts
#24
Y by
E, i swear i saw this problem a few years ago
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FireK47
140 posts
#25
Y by
How I did it
A lot of text but in my mind it went quickly
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programmeruser
2454 posts
#26
Y by
dᅟamn i thought this was some weird circle geo, should've taken more than a 5 second read on this
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ryanbear
1043 posts
#27
Y by
E, i swear i saw this problem a few years ago
2003 AIME I P2
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vincentwant
1135 posts
#28
Y by
I went from P15 to P1 on 10A so I started with this one lol

nice title
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paixiao
1704 posts
#29
Y by
I-_-I wrote:
5 seconds

May I ask how I got that?
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bachisnotcool7
391 posts
#30
Y by
Work through a few small cases and see that the area increases at a rate increasing by 4, so you can create an equation.
$\frac{n^2+n}{2}=A$, where $A$ is area in $\pi$ and $n$ is the number of circles. We can clearly see $n=63$ is just low and $n=64$ is at least $2023\pi$ and $64$ is even so it works $(E)$
This post has been edited 1 time. Last edited by bachisnotcool7, Dec 30, 2023, 12:36 PM
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JH_K2IMO
46 posts
#32
Y by
The shaded areas can be described as follows: starting from the rightmost shaded region, the area for the first region is calculated as 4π−1π=3π, for the second region as 16π−9π=7π, and for the third region as 36π−25π=11π. Observing this pattern, it can be deduced that the area of the n-th shaded region is given by the formula 4n−1.

Thus, the total shaded area can be expressed as:(4×1−1)+(4×2−1)+(4×3−1)+…+(4×n−1)=4{n(n+1)/2}-n
To determine the minimum n such that the total shaded area is at least 2023π, the following inequality must hold: 2n^2+n≥2023.
Solving this quadratic inequality will yield the smallest integer value for n. After performing the necessary calculations, it is found that the minimum
n satisfying this condition is 32. Therefore, the total number of circles is given by 2n, resulting in 32×2=64.

Thus, the answer is 64.
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