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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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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
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   22
N 13 minutes ago by ijco
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 $m+n$.
22 replies
+1 w
megahertz13
Oct 13, 2024
ijco
13 minutes ago
9 AIME Qualification AMC 12A
studymoremath   1
N 14 minutes ago by studymoremath
Now with the AMC's over, I think everyone has more of an idea of cutoff ranges and difficulty levels this year. Are my chances high enough that I should start studying for AIME?
1 reply
studymoremath
Yesterday at 6:43 PM
studymoremath
14 minutes ago
Past PUMaC results
mathkiddus   1
N an hour ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
1 viewing
mathkiddus
2 hours ago
lpieleanu
an hour ago
study for AIME to qual for USJMO
Logicus14   0
an hour 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
an hour ago
0 replies
No more topics!
funny log fraction
OronSH   12
N Nov 13, 2024 by LostDreams
Source: 2024 AMC 12B #8
What value of $x$ satisfies \[\frac{\log_2x\cdot\log_3x}{\log_2x+\log_3x}=2?\]
$
\textbf{(A) }25\qquad
\textbf{(B) }32\qquad
\textbf{(C) }36\qquad
\textbf{(D) }42\qquad
\textbf{(E) }48\qquad
$
12 replies
OronSH
Nov 13, 2024
LostDreams
Nov 13, 2024
funny log fraction
G H J
Source: 2024 AMC 12B #8
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OronSH
1624 posts
#1 • 1 Y
Y by clarkculus
What value of $x$ satisfies \[\frac{\log_2x\cdot\log_3x}{\log_2x+\log_3x}=2?\]
$
\textbf{(A) }25\qquad
\textbf{(B) }32\qquad
\textbf{(C) }36\qquad
\textbf{(D) }42\qquad
\textbf{(E) }48\qquad
$
This post has been edited 1 time. Last edited by OronSH, Nov 13, 2024, 5:37 PM
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aopsonline2020888
64 posts
#2
Y by
C i think $       $
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yambe2002
1642 posts
#3
Y by
pretty sure 36 change bases and then just a bunch of calculations

very easy but long
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HonestCat
947 posts
#4
Y by
OronSH wrote:
What value of $x$ satisfies \[\frac{\log_2x\cdot\log_3x}{\log_2x+\log_3x}=2?\]$
\textbf{(A) }25\qquad
\textbf{(B) }32\qquad
\textbf{(C) }36\qquad
\textbf{(D) }42\qquad
\textbf{(E) }48\qquad
$

36
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Mathboy345
476 posts
#5
Y by
36 anyone confirm?
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fruitmonster97
2239 posts
#6 • 1 Y
Y by IbrahimNadeem
this was nice but trivial?

equal to $\tfrac{1}{\tfrac{1}{\log_2x}+\tfrac{1}{\log_3x}}=\tfrac{1}{\log_x6}=\log_6x,$ so $36,$ which is $\textbf{(C)}.$
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yambe2002
1642 posts
#7
Y by
fruitmonster97 wrote:
this was nice but trivial?

equal to $\tfrac{1}{\tfrac{1}{\log_2x}+\tfrac{1}{\log_3x}}=\tfrac{1}{\log_x6}=\log_6x,$ so $36,$ which is $\textbf{(C)}.$

nice solution i wasted too much time on this
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GAMER100
2088 posts
#8
Y by
What's wrong with this bogus solution?
yambe2002 wrote:
fruitmonster97 wrote:
this was nice but trivial?

equal to $\tfrac{1}{\tfrac{1}{\log_2x}+\tfrac{1}{\log_3x}}=\tfrac{1}{\log_x6}=\log_6x,$ so $36,$ which is $\textbf{(C)}.$

nice solution i wasted too much time on this

:wallbash:
This post has been edited 3 times. Last edited by GAMER100, Nov 13, 2024, 5:43 PM
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yambe2002
1642 posts
#9
Y by
what i did (don't do this)
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mathprodigy2011
58 posts
#10
Y by
oh for this one I wrote Log2(x)=a and log3(x) = b and then multiplying and subtracting gives ab-2a-2b=0. SFFT leads to (a-2)(b-2) = 36. so the answer is just 2^2 times 3^2 so 36
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pog
4905 posts
#11
Y by
Rewrite numerator as \[\dfrac{1}{\log_x 2 \cdot \log_x 3},\]and rewrite denominator as \[\dfrac{1}{\log_x 2} + \dfrac{1}{\log_x 3} = \dfrac{\log_x 2 + \log_x 3}{\log_x 2 \cdot \log_x 3}.\]This cancels, giving $\tfrac{1}{\log_x 2 + \log_x 3} = 2$. Hence $\log_x 6 = \tfrac12$, so $x = \boxed{\textbf{(C) }36}$.
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brainfertilzer
1802 posts
#12 • 1 Y
Y by LostDreams
change all the bases to $e$; you get $\ln x = \ln 36$ after some bash (doesn't matter what you change the base to ofc)
This post has been edited 2 times. Last edited by brainfertilzer, Nov 13, 2024, 7:28 PM
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LostDreams
125 posts
#13
Y by
Multiplying out gives

$\log_2x\cdot\log_3x=2\log_2x+2\log_3x$

By change of base

$\frac{\ln x}{\ln 2}\cdot\frac{\ln x}{\ln 3}=\frac{\ln x^2}{\ln 2}+\frac{\ln x^2}{\ln 3}$

Multiplying $\ln2\cdot\ln3$ and dividing $\ln x$ on both sides gives

$\ln x = 2(\ln 3 + \ln 2) = \ln 36$ so $x = \boxed{\textbf{(C) }36}$
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