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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 6 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
6 minutes ago
study for AIME to qual for USJMO
Logicus14   0
17 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
17 minutes ago
0 replies
Problem 2
evt917   47
N 39 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
39 minutes ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N 40 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
40 minutes ago
No more topics!
Problem 5
evt917   19
N Nov 13, 2024 by MathPerson12321
Source: 2024 AMC 12B #5 / AMC 10B #5
In the following expression, Melanie changed some of the plus signs to minus signs: $$ 1 + 3+5+7+\cdots+97+99$$When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?

$
\textbf{(A) }14 \qquad
\textbf{(B) }15 \qquad
\textbf{(C) }16 \qquad
\textbf{(D) }17 \qquad
\textbf{(E) }18 \qquad
$
19 replies
evt917
Nov 13, 2024
MathPerson12321
Nov 13, 2024
Problem 5
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 12B #5 / AMC 10B #5
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evt917
1514 posts
#1
Y by
In the following expression, Melanie changed some of the plus signs to minus signs: $$ 1 + 3+5+7+\cdots+97+99$$When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?

$
\textbf{(A) }14 \qquad
\textbf{(B) }15 \qquad
\textbf{(C) }16 \qquad
\textbf{(D) }17 \qquad
\textbf{(E) }18 \qquad
$
This post has been edited 3 times. Last edited by jlacosta, Nov 13, 2024, 6:00 PM
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christmasbaby
76 posts
#2
Y by
15, I think?
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yambe2002
1642 posts
#3
Y by
15 please
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Sedro
5739 posts
#4
Y by
I got 15 too.
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elizhang101412
1060 posts
#5
Y by
its b def
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fruitmonster97
2239 posts
#6
Y by
Yep, note that $35^2<1250<36^2$ so everything after $2(35)-1$ should be negated $\implies15,\textbf{(B)}$
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MathRook7817
234 posts
#7
Y by
yeah 15, just bash it out
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Tem8
154 posts
#8
Y by
Note that to change the least amount of plus signs, Melanie should change signs for $99$, $97$, $95$ and so on until the expression becomes negative. Assume that Melanie added the terms until $1+3+5+\cdots+(2n-1)$ and after that she subtracted $-[(2n+1)+\cdots+99]$.

\begin{align*}
1+3+5+\cdots+(2n-1)&=n^2\\
(2n+1)+\cdots+(2(49)+1)&=\frac{(100+2n)(50-n)}{2}=(50+n)(50-n)\\
1+3+5+\cdots+(2n-1)&-[(2n+1)+\cdots+99] \\
&=n^2-2500+n^2\\
&=2n^2-2500 < 0\\
n^2 &< 1250\\
n &\le 35.
\end{align*}The least $n$ is $35$, thus, Melanie changed signs for numbers from $99$ to $(2n+1)=71$, which is exactly for $\boxed{\textbf{(B) } 15}$ numbers.
This post has been edited 2 times. Last edited by Tem8, Nov 14, 2024, 4:20 AM
Reason: Typo
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alexanderhamilton124
242 posts
#9
Y by
Dayum this was pretty hard for a 5. Got B.
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LostDreams
125 posts
#10
Y by
yea it was 15 i just went from right to left changing the signs based on answer choices from least to greatest
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Elephant200
1451 posts
#11 • 1 Y
Y by Deinoncyus_Ban
I spent so so much time on this. I think this problem is almost the sole reason I ran out of time on this test...
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eg4334
391 posts
#12
Y by
Very computationally intensive for a p5??

We want $99+97+\dots + n > 1250$ for some odd $n$. $n=71$ yields $1275$, and it is apparently that anything higher does not work. $\frac{99-71}{2} + 1 = \boxed{15}$
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meduh6849
313 posts
#13
Y by
eg4334 wrote:
Very computationally intensive for a p5??

We want $99+97+\dots + n > 1250$ for some odd $n$. $n=71$ yields $1275$, and it is apparently that anything higher does not work. $\frac{99-71}{2} + 1 = \boxed{15}$


Check post #8, still a terrible bash for p5 though
This post has been edited 1 time. Last edited by meduh6849, Nov 13, 2024, 7:36 PM
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pog
4905 posts
#14
Y by
It's not *that* bashy just list out $99+97+95+93+91+89+87+85+83+81+79+77+75+73$ and the average here is $86$ so the sum is $86 \cdot 14 = 1204$. Then $1204+71=1275$ works so we're done.
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joshualiu315
2431 posts
#15 • 3 Y
Y by pog, ranu540, Tem8
huh didn't find this bashy


The total sum is $50^2 = 2500$. Thus, we wish to find the largest value of $x$ such that $1+3+\dots+(2x-1) = x^2 < 1250$; it is easy to see that $x = 35$ is our desired value, so the answer is $\boxed{15}$.
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Shreyasharma
589 posts
#16
Y by
I only didn't silly this because I failed computation the first time around.
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Jndd
1404 posts
#17
Y by
you just need 1 + 3 + ... + 2n - 1 < 1250, so n^2 < 1250. n = 35 works fine, so 2*35 + 1 and onwards must be negative. this is (99-71)/2 + 1 = 15.
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RedFireTruck
4057 posts
#18
Y by
aight say the first n (1, 3, ..., 2n-1) stay positive and the rest turn negative

the sum is then n^2-(50-n)(2n+1+99)/2=n^2-(50-n)(n+50)=2n^2-2500

we want largest n s.t. 2n^2<2500 so n^2<1250 so 35^2=1225 by that one trick so answer is 50-35=$\boxed{15}$
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RabtejKalra
12 posts
#19
Y by
I hated this question so much. Ended up just leaving it unanswered.
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MathPerson12321
3168 posts
#20
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fruitmonster97 wrote:
Yep, note that $35^2<1250<36^2$ so everything after $2(35)-1$ should be negated $\implies15,\textbf{(B)}$

Yep
I just assumed we subtract everything from 99-81 and bashed
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