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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
AMC12, AIME cutoff release date
sausagebun   3
N a minute ago by Challengees24
hi guys just wondering when the cutoffs will be released!
also, is a 84 on 12A a good shot at aime?
3 replies
sausagebun
33 minutes ago
Challengees24
a minute ago
List of Contests held on AoPS
DinoDragon186   4
N 3 minutes ago by DinoDragon186
I've just noticed that the ELMO is held on AoPS message board.
Are there any more such contests which are held ?
( Could I get a list ? )
4 replies
+1 w
DinoDragon186
31 minutes ago
DinoDragon186
3 minutes ago
9 How well do you think you will do on the AMC 8
TQ_Math   35
N 5 minutes ago by RocketScientist
I know it's November, but it's never not too late to prep, right?!

40 VOTES!!!
60 votess!
100 VOTES
120 votesdfgjkafjd
35 replies
TQ_Math
Nov 17, 2024
RocketScientist
5 minutes ago
9 I qualified to AIME what should I do now?
Bigtree   38
N 25 minutes ago by EaZ_Shadow
I got $112.5$ on AMC $10$ so I need $14$ on AIME right now I mock $5$
Any suggestions help (i’m in intermediate c&p, and precalc rn)
38 replies
Bigtree
Sunday at 9:58 PM
EaZ_Shadow
25 minutes ago
No more topics!
It's not like I share elements with you or anything, baka!
fidgetboss_4000   45
N Nov 19, 2024 by Mr.Sharkman
Source: AIME I #12
For any finite set $X$, let $|X|$ denote the number of elements in $X.$ Define $$S_n = \sum |A \cap B|,$$where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\{1, 2, 3, …, n\}$ with $|A| = |B|.$ For example, $S_2 = 4$ because the sum is taken over the pairs of subsets $$(A, B) \in \{ (\emptyset, \emptyset), (\{1\}, \{1\}), (\{1\}, \{2\}), (\{2\}, \{1\}), (\{2\}, \{2\}), (\{1, 2\}, \{1, 2\})\},$$giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4.$ Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by $1000.$
45 replies
fidgetboss_4000
Feb 9, 2022
Mr.Sharkman
Nov 19, 2024
It's not like I share elements with you or anything, baka!
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Source: AIME I #12
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Awesome_guy
862 posts
#36
Y by
Remarkable that SADGIME P6 is also linearity of expectation + enumerative combo

Edit: though the motivation here is less apparent
SADGIME P6 wrote:
Cherri walks from the bottom left square to the top right square of a $99 \times 99$ square grid, taking any one of the possible routes with equal probability. She leaves a trail of integers tracing her path. First, she places $1$ in her starting square. On her nth step, she travels one unit right or one unit up to reach a new square and places $n + 1$ in this new square for all integers $1 \leq n \leq 196$; provided that she ends on the top right square. Find the expected value of the sum of the integers Cherri places in the middle row of the grid.
This post has been edited 1 time. Last edited by Awesome_guy, Feb 13, 2022, 3:46 PM
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fidgetboss_4000
3423 posts
#37
Y by
SADGIME P6 was more slightly engineer's induction-able though
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Awesome3.14
1724 posts
#38 • 2 Y
Y by Mango247, Mango247
bruh such an easy problem, there were a lot of problems like this on intermediate C&P
couldn't get this during the test because I tried to bash 11 and 12 then added 1 instead of 3 on 13
This post has been edited 1 time. Last edited by Awesome3.14, Jun 12, 2022, 5:52 PM
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fidgetboss_4000
3423 posts
#39
Y by
idk though, for some reason it was statistically the second hardest problem on the AIME I
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channing421
1349 posts
#40 • 1 Y
Y by bryanguo
i think ppl just got scared by notation

because this was like rlly similar to 2020 AIME I #7
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samrocksnature
8783 posts
#41
Y by
Awesome3.14 wrote:
bruh such an easy problem, there were a lot of problems like this on intermediate C&P
couldn't get this during the test because I tried to bash 11 and 12 then added 1 instead of 3 on 13

ikr but I just skipped it ;-;
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brainfertilzer
1803 posts
#42
Y by
This reminds me of a problem in 112 combo, but I forgot which problem exactly

Let $k\in \{1, 2, \dots, n\}$ be an integer, and consider how many times $k$ contributes to $S_n$. This happens when $k\in A$ and $k\in B$. It remains to add some number of elements of $\{1, 2, \dots, n\}\backslash \{k\}$ to $A$ and an equal number of elements to $B$. There are $$\binom{n-1}{0}^2 + \binom{n-1}{1}^2 + \dots + \binom{n-1}{n-1}^2 = \binom{2n-2}{n-1}$$ways to do this. Therefore, each $k\in \{1, 2, \dots, n \}$ contributes $\binom{2n-2}{n-1}$ to the total sum. Hence, $S_n = n\binom{2n-2}{n-1}$. The rest is computation:
\[ \frac{S_{2022}}{S_{2021}} = \frac{2022\binom{4042}{2021}}{2021\binom{4040}{2020}} = \frac{2022}{2021}\cdot \frac{4042!}{2021!^2}\cdot\frac{2020!^2}{4040!} = \frac{4044\cdot 4041}{2021^2}.\]We get $p + q = 4044\cdot 4041 + 2021^2\equiv 44\cdot 41 + 21^2\equiv \boxed{245}\pmod{1000}$.
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daijobu
505 posts
#43
Y by
Video Solution
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peace09
5314 posts
#44 • 1 Y
Y by gracemoon124
Quote:
Consider how many times any given number $k$ is counted in the intersection of $A, B$
how did i not see this

... well, at least the following motivates a solution for one who sees two messy ways but not the clean third to set up a sum
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awesomeparrot
154 posts
#45
Y by
Wow clean math work I wish :P
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john0512
4146 posts
#46 • 1 Y
Y by fidgetboss_4000
let's actually not engineer induct

Consider all pairs of $m$ elements subsets. In each pair, the expected value of the intersection by LOE is $$n\cdot (m/n)^2=\frac{m^2}{n}.$$Since there are $\binom{m}{n}^2$ such pairs, we have $$S_n=\sum_{m=0}^n \binom{m}{n}^2 \cdot \frac{m^2}{n}=1/n \sum_{m=0}^n \binom{m}{n}^2 \cdot m^2=1/n (n^3 C_{n-1})=n\binom{2n-2}{n-1}.$$Simply plugging in gives the answer of 245.
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cxrupptedpat
223 posts
#47 • 1 Y
Y by bjump
John post on your youtube channel pls :)))
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ryanbear
1043 posts
#48
Y by
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eg4334
413 posts
#49
Y by
First we find a general form for $S_n$. As we typically do in these problems, count each individual element seperately based on how many times it appears in $S_n$. In it clear that every element is symmetric, so just consider $1$. It is counted only once for the one element susbets. It is counted $(n-1)^2$ times for the two element subsets because for $A$ you choose one more element out of the remainding $n-1$ and the same for $B$ (remember, ordered and not nessacerily distinct). Similarly for three elements it is counted $\left( \binom{n-1}{2} \right)^2$, and on. Summing over all elements, $$S_n = n \left(  \left( \binom{n-1}{0} \right)^2 + \left( \binom{n-1}{1} \right)^2 + \left( \binom{n-1}{2} \right)^2 + \dots + \left( \binom{n-1}{n-1} \right) ^2 \right) = n \cdot \binom{2n-2}{n-1}$$So $$\frac{S_{2022}}{S_{2021}} = \frac{2022 \cdot \frac{4042!}{2021! \cdot 2021!}}{2021 \cdot \frac{4040!}{2020! \cdot 2020!}} = \frac{2 \cdot 2022 \cdot 4041}{2021 \cdot 2021}$$And the answer is $$2 \cdot 2022 \cdot 4041 + 2021 \cdot 2021 \equiv 804+441 \equiv \boxed{245} \pmod{1000}$$
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Mr.Sharkman
389 posts
#50
Y by
We use indicator functions. We have
$$S_{n} = \sum_{i=1}^{n} \sum_{|A| = |B| = i} |A \cap B| = \sum_{i=1}^{n} \sum_{|A|=|B| = i} \sum_{j = 1}^{n} 1_{j \in A}1_{j \in B} = \sum_{i=1}^{n} \left(\sum_{|A|=i}^{n}\sum_{j=1}^{n} 1_{j \in A}\right)^{2}.$$Now,
$$\sum_{|A|=i}^{n}\sum_{j=1}^{n} 1_{j \in A} = \sum_{j=1}^{n} \sum_{j \in A, |A| = i} 1 = \sum_{j=1}^{n}{n-1 \choose i-1} = n {n-1 \choose i-1},$$so we get
$$S_{n} = \sum_{i=1}^{n} n {n-1 \choose i-1}^{2} = n {2n-2 \choose n-1}.$$When we divide these, we get
$$\frac{S_{2022}}{S_{2021}} = \frac{2022 \cdot 8082}{2021^{2}},$$and since the numerator and denominator here are relatively prime, we get $2022 \cdot 8082+2021^{2} \equiv \boxed{245} \pmod 1000$
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