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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
New Math Textbook
AkshajK   0
3 hours ago
Hey! Wanted to share a sneak peak of the new MathDash Book - hopefully it can help with your math training!

Included in this sneak peak are chapters on Primes and Divisors, Power of a Point, Ptolemy's Theorem, Linearity of Expectation, and Generating Functions, along with over 100 practice problems in total - in a unique timed contest format (different from any other book!).

Would love to hear feedback - this will help us add more content and improve it!

https://mathdash.com/book
(currently only works on Web)
0 replies
AkshajK
3 hours ago
0 replies
Logical guessing game!
Mathdreams   22
N 4 hours 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
4 hours ago
Possibility of USAMO?
MathXplorer10   4
N 4 hours 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
Today at 5:06 AM
MathXplorer10
4 hours ago
10a vs 10b
golden_star_123   111
N 4 hours ago by happyfish0922
Post the difference between your 10a and 10b score!
111 replies
golden_star_123
Wednesday at 6:24 PM
happyfish0922
4 hours ago
No more topics!
generic combo #22
zhoujef000   22
N Today at 3:30 AM by SomeRandomGuyOnline
Source: 2024 AMC 10B #22/2024 AMC 12B #16
A group of $16$ people will be partitioned into $4$ indistinguishable $4$-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^r M,$ where $r$ and $M$ are positive integers and $M$ is not divisible by $3.$ What is $r?$

$\textbf{(A) }5 \qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
22 replies
zhoujef000
Wednesday at 5:34 PM
SomeRandomGuyOnline
Today at 3:30 AM
generic combo #22
G H J
Source: 2024 AMC 10B #22/2024 AMC 12B #16
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zhoujef000
215 posts
#1 • 2 Y
Y by clarkculus, tanxin002
A group of $16$ people will be partitioned into $4$ indistinguishable $4$-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^r M,$ where $r$ and $M$ are positive integers and $M$ is not divisible by $3.$ What is $r?$

$\textbf{(A) }5 \qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
This post has been edited 1 time. Last edited by zhoujef000, Wednesday at 7:07 PM
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maxamc
318 posts
#2
Y by
16 select 4 * 12 select 4 * 8 select 4 * 4 select 4 /4! *(4*3)^4 is our answer, has 5 factors of 3
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nsking_1209
81 posts
#3
Y by
Some combo and legendres theorem gives A
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mathboy282
2966 posts
#4
Y by
16 choose 4,4,4,4, divide over 4! to acount for indistinguishable committees, chairperson and secretary -> (4*3)^5

in numerator: floor(16/3)+floor(16/9)=5+1=6, + 4 factors of 3 from 4*3 ^ 4
ind emoniator: 5 factors of 3, 1 each from 4!

6+4-5=5
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Sedro
5739 posts
#5
Y by
Phew thought I sillied, answer is A. Also #16 on the 12B.
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Michw08
48 posts
#7
Y by
I thought you had to find M I'm done :(
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mathprodigy2011
57 posts
#8 • 2 Y
Y by mathnerd_101, songyanxin
this was so easy, misplaced fs
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aleyang
171 posts
#9
Y by
I learned from my mistakes from that disjoint problem on A. This time divided by 4! and got it right :D
This post has been edited 2 times. Last edited by aleyang, Wednesday at 6:09 PM
Reason: from 10a
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LostDreams
124 posts
#10
Y by
bruh what am i doing i literally did this problem and fricking forgot to bubble in actually brain dead
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EaZ_Shadow
279 posts
#11
Y by
bro just use generating functions on $(a+b+c+d)^16$ and find the coefficient of the $a^4b^4c^4d^4$ term and then multiply by6 4 times, then divide by 4! to get 5 as the answer since there is 5 factors of 3 left
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lele0305
22 posts
#12
Y by
oops me when i do $\text{floor}\Big(\frac{5}{3}\Big)=2$ :noo:
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Andrew2019
2216 posts
#13
Y by
$\frac{1}{4!}\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4}(3 \cdot 2)^4$ gives 5, i forgot to divide by $4!$ on the test
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andrewcheng
482 posts
#14 • 1 Y
Y by songyanxin
16C4*12C4*8C4*4C4(4C2)^2
r/4!emoving all non 3 factors gives us
15*12*9*6*3/24*24*24*24*24 *3^4
gives us 3*3^4 after removing all powers of 2 which gives us 5
we don't make the 10A mistake again
This post has been edited 1 time. Last edited by andrewcheng, Wednesday at 6:29 PM
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pingpongmerrily
2393 posts
#15
Y by
Andrew2019 wrote:
forgot to divide by $4!$ on the test
new version of "i forgot to divide by $3!$ on 10a :skull:
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eg4334
391 posts
#16 • 1 Y
Y by Sedro
$$v_3\left( \frac{\binom{16}{4} \cdot \binom{12}{4} \cdot \binom{8}{4} \cdot \binom{4}{4} }{4!}  \cdot (4 \cdot 3)^4\right) = \boxed{5}$$
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Seaspace1250
20 posts
#17
Y by
i could solve this problem but not #5 lol misplaced?
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ryanbear
1043 posts
#18 • 3 Y
Y by Sedro, pog, aidan0626
I forgot to divided by $4!$ but I accidentally wrote $12^3$ instead of $12^4$ so I got it right
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EaZ_Shadow
279 posts
#19
Y by
ryanbear wrote:
I forgot to divided by $4!$ but I accidentally wrote $12^3$ instead of $12^4$ so I got it right

Lolllllll :rotfl: no way bro
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MathPerson12321
3160 posts
#20
Y by
eg4334 wrote:
$$v_3\left( \frac{\binom{16}{4} \cdot \binom{12}{4} \cdot \binom{8}{4} \cdot \binom{4}{4} }{4!}  \cdot (4 \cdot 3)^4\right) = \boxed{5}$$

Exactly what I did
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Cheerfulfrog
898 posts
#21
Y by
Seaspace1250 wrote:
i could solve this problem but not #5 lol misplaced?

5 was time consuming
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mathprodigy2011
57 posts
#22
Y by
bro how was this #22 lol, I took 12b and it was #16. the question was easier than so many.
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megahertz13
2913 posts
#23
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SomeRandomGuyOnline
33 posts
#24
Y by
Does this solution work: arrange the people in a line so that the first 4 are in the first group, next 4 in second group and etc. so there are 16! ways to do this. Then divide by 4! since there are that many ways to arrange the groups, and divide 4!^4 since in each group we can arrange the order of the people 4! ways. so (16!)/4!^5 which contains a 3^1. Then since we need to pick a senator and chair guy it's 4*3 for each group. so 3^1*3^4 which is 3^5 . So (A) 5? (unfortunately I didn't come up with this in the competition...)
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