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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
32 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
30 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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fidgetboss_4000
3423 posts
#1 • 4 Y
Y by centslordm, mathking999, megarnie, isache
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 $1000.$
This post has been edited 1 time. Last edited by fidgetboss_4000, Feb 9, 2022, 6:54 PM
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vsamc
3769 posts
#2 • 2 Y
Y by centslordm, programjames1
245, basically S_n = n(2n -2 choose n-1) from chairperson and vandermonde spam
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fidgetboss_4000
3423 posts
#3
Y by
proudest solve lesgo
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MortemEtInteritum
1330 posts
#4 • 2 Y
Y by J4wbr34k3r, rayfish
Consider how many times any given number $k$ is counted in the intersection of $A, B$, in the expression for $S_n$. If $A, B$ each contain $r$ numbers, then it is counted ${n-1\choose r-1}^2={n-1\choose r-1}{n-1\choose n-r}$ times, summed from $r=1$ to $r=n$. By Vandermonde's this sum becomes ${2n-2\choose n-1}$, so summing over all values of $k$, we get $S_n=n{2n-2\choose n-1}$. From here, a computation gives that the answer is $\frac{2022\cdot 2\cdot 4041}{2021^2}$, so taking mod 1000, $p+q=\boxed{245}$.
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peace09
5314 posts
#5
Y by
Uh I think I successfully engineered but panicked and forgot that Vandermonde's existed when I used it correctly for 2020 AIME I #7 :stretcher:
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fidgetboss_4000
3423 posts
#6
Y by
peace09 wrote:
Uh I think I successfully engineered

same here but took me the first 30 min of the test
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i3435
1346 posts
#7
Y by
We first find $S_n$ for any $n$.

Consider the expected value of $|A\cap B|$ when $|A|=|B|=k$. Of the $k$ elements of $A$, there is a $\frac{k}{n}$ chance any one of them are in $B$, so by Linearity of Expectation, this expected value is $\frac{k^2}{n}$. There are $\binom{n}{k}^2$ pairs $(A,B)$, so of all the pairs with $k$ elements each, the sum of $|A\cap B|$ is $\frac{k^2}{n}\cdot \binom{n}{k}^2=n\binom{n-1}{k-1}^2$. By Vandermonde's, summing this up for all $k$ gives $S_n=n\binom{2n-2}{n-1}$, as desired.

Now math gives $245$ as the answer.
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john0512
4146 posts
#8
Y by
engineer induction + polynomial interpolation
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asdf334
7547 posts
#9
Y by
spent thirty minutes bashing recursion and missed the two minute sol
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ike.chen
1162 posts
#10 • 1 Y
Y by rayfish
Once you realize how disgusting this problem seems, attacking from another perspective, i.e. Global, follows naturally :-D.
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CyclicISLscelesTrapezoid
355 posts
#11
Y by
I used expected value then Vandermonde's
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DottedCaculator
7224 posts
#12
Y by
engineer's induction go brrrr
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megarnie
5324 posts
#16
Y by
On the test, I thought in the wrong direction (or at least I couldn't find anything when thinking in that direction). Wasted over half an hour on this.

We'll find the number of times some integer $k$ from $1$ to $n$ is included in intersection.

We have that $k$ is included $\sum_{i=1}^n \binom{n-1}{i-1}^2=\sum_{i=1}^n \binom{n-1}{i-1}\binom{n-1}{n-i}=\binom{2n-2}{n-1}$ by Vandermonde's.

Summing over all $k$ gives $S_n=n\binom{2n-2}{n-1}$.

We have $\binom{4042}{2021}=\frac{4042!}{2021!2021!}$ and $\binom{4040}{2020}=\frac{4040!}{2020!2020!}$. Dividing gives $\frac{4041\cdot 4042}{2021^2}=\frac{4041\cdot 2}{2021}$.

So it's totally $\frac{2022\cdot 4041\cdot 2}{2021^2}$, so the answer is $22\cdot 41\cdot 2+441=1804+441=2\boxed{245}$.
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crazyeyemoody907
447 posts
#17
Y by
fidgetboss_4000 wrote:
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 $1000.$
Non-rigorous solution
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Reason: typo
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aryabhata000
245 posts
#18
Y by
Unusually low solve rate compared to 7 and 8, potentially due to placement after the fairly hard #11. But it’s basically easy double counting + vandermonde. Also, this was my first solve on the test before #1 weirdly.
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