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Contests & Programs AMC and other contests, summer programs, etc.
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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   3
N 33 minutes ago by RainbowSquirrel53B
[center]IMAGE

Greetings, AoPS! IndyINTEGIRLS will be hosting a virtual math competition on May 25,
2024 from 12 PM to 3 PM EST.
Join other woman-identifying and/or non-binary "STEMinists" in solving problems, socializing, playing games, winning prizes, and more! If you are interested in competing, please register here![/center]

----------

[center]Important Information[/center]

Eligibility: This competition is open to all woman-identifying and non-binary students in middle and high school. Non-Indiana residents and international students are welcome as well!

Format: There will be a middle school and high school division. In each separate division, there will be an individual round and a team round, where students are grouped into teams of 3-4 and collaboratively solve a set of difficult problems. There will also be a buzzer/countdown/Kahoot-style round, where students from both divisions are grouped together to compete in a MATHCOUNTS-style countdown round! There will be prizes for the top competitors in each division.

Problem Difficulty: Our amazing team of problem writers is working hard to ensure that there will be problems for problem-solvers of all levels! The middle school problems will range from MATHCOUNTS school round to AMC 10 level, while the high school problems will be for more advanced problem-solvers. The team round problems will cover various difficulty levels and are meant to be more difficult, while the countdown/buzzer/Kahoot round questions will be similar to MATHCOUNTS state to MATHCOUNTS Nationals countdown round in difficulty.

Platform: This contest will be held virtually through Zoom. All competitors are required to have their cameras turned on at all times unless they have a reason for otherwise. Proctors and volunteers will be monitoring students at all times to prevent cheating and to create a fair environment for all students.

Prizes: At this moment, prizes are TBD, and more information will be provided and attached to this post as the competition date approaches. Rest assured, IndyINTEGIRLS has historically given out very generous cash prizes, and we intend on maintaining this generosity into our Spring Competition.

Contact & Connect With Us: Follow us on Instagram @indy.integirls, join our Discord, follow us on TikTok @indy.integirls, and email us at indy@integirls.org.

---------
[center]Help Us Out

Please help us in sharing the news of this competition! Our amazing team of officers has worked very hard to provide this educational opportunity to as many students as possible, and we would appreciate it if you could help us spread the word!
3 replies
Indy_Integirls
May 11, 2025
RainbowSquirrel53B
33 minutes ago
concyclic wanted, diameter related
parmenides51   3
N 43 minutes ago by Giant_PT
Source: China Northern MO 2023 p1 CNMO
As shown in the figure, $AB$ is the diameter of circle $\odot O$, and chords $AC$ and $BD$ intersect at point $E$, $EF\perp AB$ intersects at point $F$, and $FC$ intersects $BD$ at point $G$. Point $M$ lies on $AB$ such that $MD=MG$ . Prove that points $F$, $M$, $D$, $G$ lies on a circle.
IMAGE
3 replies
parmenides51
May 5, 2024
Giant_PT
43 minutes ago
Concurrency
Omid Hatami   14
N an hour ago by Ilikeminecraft
Source: Iran TST 2008
Suppose that $ I$ is incenter of triangle $ ABC$ and $ l'$ is a line tangent to the incircle. Let $ l$ be another line such that intersects $ AB,AC,BC$ respectively at $ C',B',A'$. We draw a tangent from $ A'$ to the incircle other than $ BC$, and this line intersects with $ l'$ at $ A_1$. $ B_1,C_1$ are similarly defined. Prove that $ AA_1,BB_1,CC_1$ are concurrent.
14 replies
Omid Hatami
May 20, 2008
Ilikeminecraft
an hour ago
<ACB=90^o if AD = BD , <ACD = 3 <BAC, AM=//MD, CM//AB,
parmenides51   2
N an hour ago by AylyGayypow009
Source: 2021 JBMO TST Bosnia and Herzegovina P3
In the convex quadrilateral $ABCD$, $AD = BD$ and $\angle ACD  = 3 \angle BAC$. Let $M$ be the midpoint of side $AD$. If the lines $CM$ and $AB$ are parallel, prove that the angle $\angle  ACB$ is right.
2 replies
parmenides51
Oct 7, 2022
AylyGayypow009
an hour ago
2v2 (bob lost the game)
GoodMorning   85
N an hour ago by maromex
Source: 2023 USAJMO Problem 5/USAMO Problem 4
A positive integer $a$ is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer $n$ on the board with $n+a$, and on Bob's turn he must replace some even integer $n$ on the board with $n/2$. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.

After analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of $a$ and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.
85 replies
GoodMorning
Mar 23, 2023
maromex
an hour ago
Good Permutations in Modulo n
swynca   9
N an hour ago by optimusprime154
Source: BMO 2025 P1
An integer $n > 1$ is called $\emph{good}$ if there exists a permutation $a_1, a_2, a_3, \dots, a_n$ of the numbers $1, 2, 3, \dots, n$, such that:
$(i)$ $a_i$ and $a_{i+1}$ have different parities for every $1 \leq i \leq n-1$;
$(ii)$ the sum $a_1 + a_2 + \cdots + a_k$ is a quadratic residue modulo $n$ for every $1 \leq k \leq n$.
Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.
9 replies
swynca
Apr 27, 2025
optimusprime154
an hour ago
Grid combo with tilings
a_507_bc   7
N an hour ago by john0512
Source: All-Russian MO 2023 Final stage 10.6
A square grid $100 \times 100$ is tiled in two ways - only with dominoes and only with squares $2 \times 2$. What is the least number of dominoes that are entirely inside some square $2 \times 2$?
7 replies
a_507_bc
Apr 23, 2023
john0512
an hour ago
sqrt(2)<=|1+z|+|1+z^2|<=4
SuiePaprude   3
N an hour ago by alpha31415
let z be a complex number with |z|=1 show that sqrt2 <=|1+z|+|1+z^2|<=4
3 replies
SuiePaprude
Jan 23, 2025
alpha31415
an hour ago
Simple but hard
Lukariman   5
N an hour ago by Giant_PT
Given triangle ABC. Outside the triangle, construct rectangles ACDE and BCFG with equal areas. Let M be the midpoint of DF. Prove that CM passes through the center of the circle circumscribing triangle ABC.
5 replies
Lukariman
Today at 2:47 AM
Giant_PT
an hour ago
inequalities
Ducksohappi   0
an hour ago
let a,b,c be non-negative numbers such that ab+bc+ca>0. Prove:
$ \sum_{cyc} \frac{b+c}{2a^2+bc}\ge \frac{6}{a+b+c}$
P/s: I have analysed:$ S_a=\frac{b^2+c^2+3bc-ab-ac}{(2b^2+ac)(2c^2+2ab)}$, similarly to $S_b, S_c$, by SOS
0 replies
Ducksohappi
an hour ago
0 replies
bulgarian concurrency, parallelograms and midpoints related
parmenides51   7
N an hour ago by Ilikeminecraft
Source: Bulgaria NMO 2015 p5
In a triangle $\triangle ABC$ points $L, P$ and $Q$ lie on the segments $AB, AC$ and $BC$, respectively, and are such that $PCQL$ is a parallelogram. The circle with center the midpoint $M$ of the segment $AB$ and radius $CM$ and the circle of diameter $CL$ intersect for the second time at the point $T$. Prove that the lines $AQ, BP$ and $LT$ intersect in a point.
7 replies
parmenides51
May 28, 2019
Ilikeminecraft
an hour ago
Interesting inequalities
sqing   3
N an hour ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ab+bc+ca- abc =3.$ Show that
$$a+k(b+c)\geq 2\sqrt{3 k}$$Where $ k\geq 1. $
Let $a,b,c \geq 0 $ and $2(ab+bc+ca)- abc =31.$ Show that
$$a+k(b+c)\geq \sqrt{62k}$$Where $ k\geq 1. $
3 replies
sqing
Today at 4:34 AM
sqing
an hour ago
Suggestions for preparing for AMC 12
peppermint_cat   3
N Today at 7:14 AM by Konigsberg
So, I have decided to attempt taking the AMC 12 this fall. I don't have any experience with math competitions, and I thought that here might be a good place to see if anyone who has taken the AMC 12 (or done any other math competitions) has any suggestions on what to expect, how to prepare, etc. Thank you!
3 replies
peppermint_cat
Today at 1:04 AM
Konigsberg
Today at 7:14 AM
Harmonic Mean
Happytycho   4
N Today at 4:42 AM by elizhang101412
Source: Problem #2 2016 AMC 12B
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?

$\textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 45 \qquad
\textbf{(C)}\ 504 \qquad
\textbf{(D)}\ 1008 \qquad
\textbf{(E)}\ 2015 $
4 replies
Happytycho
Feb 21, 2016
elizhang101412
Today at 4:42 AM
Stressed spelled backwards
centslordm   25
N Apr 20, 2025 by NicoN9
Source: AIME 2025 #3
The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a singlescoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let $N$ be the number of different assignments of flavors to players that meet these conditions. Find the remainder when $N$ is divided by $1000.$
25 replies
centslordm
Feb 7, 2025
NicoN9
Apr 20, 2025
Stressed spelled backwards
G H J
G H BBookmark kLocked kLocked NReply
Source: AIME 2025 #3
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centslordm
4786 posts
#1 • 6 Y
Y by Sedro, Unicode_Master03B8, clarkculus, AmethystC, megarnie, Mogmog8
The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a singlescoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let $N$ be the number of different assignments of flavors to players that meet these conditions. Find the remainder when $N$ is divided by $1000.$
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Mathandski
759 posts
#2 • 3 Y
Y by IbrahimNadeem, AmethystC, redbean
Can anyone confirm $2\boxed{016}$
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agong
3 posts
#3
Y by
yes 016 ^
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anduran
481 posts
#4
Y by
I got $16$ as well.
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AlexWin0806
50 posts
#5
Y by
i also got 2016 from casework on the amount of chocolate, vanilla, and strawberry ice creams: (6, 2, 1), (5, 3, 1), (4, 3, 2)
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EaZ_Shadow
1273 posts
#6 • 2 Y
Y by AmethystC, Ad112358
Mathandski wrote:
Can anyone confirm $2\boxed{016}$

yes I got 2016
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Challengees24
1098 posts
#7 • 3 Y
Y by studymoremath, forestcaller2010, ilikemath247365
can anyone else agree easiest problem on test???
Z K Y
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plang2008
337 posts
#8 • 1 Y
Y by scannose
knew the combinations instantly from killer sudoku
Z K Y
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bot1132
145 posts
#9
Y by
i counted 1,4,4 case :(
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MathPerson12321
3786 posts
#10
Y by
Solution

How did I silly in real contest like bro?????
dont even remember what i did :skull:
i swear to god if i added wrong
This post has been edited 1 time. Last edited by MathPerson12321, Feb 7, 2025, 4:44 PM
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xHypotenuse
783 posts
#11
Y by
How tf did I silly this in contest

edit: it's because I did 9 * 8 * 7 and added 1 to each of c, v, s by default whyyy
This post has been edited 1 time. Last edited by xHypotenuse, Feb 7, 2025, 4:52 PM
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megahertz13
3183 posts
#12
Y by
Let $x,y,z$ denote the number of people who chose each of the three flavors, where $x+y+z=9$ and $x>y>z>0$. By inspection, we have three cases for $(x,y,z):(2,3,4),(1,2,6),(1,3,5)$. These give $$\frac{9!}{2!3!4!}+\frac{9!}{1!2!6!}+\frac{9!}{1!3!5!}=1260+252+504=2\boxed{016}.$$
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junlongsun
70 posts
#13
Y by
$c+v+s=9$, $c>v>s>0$
First do casework to find all possible cases of $(c,v,s)$

$c=6$
$v+s=3 \rightarrow v=2, s=1$
$(6,2,1)$

$c=5$
$v+s=4\rightarrow v=3, s=1$
$(5,3,1)$

$c=4$
$v+s=5\rightarrow v=3, s=2$
$(4,3,2)$

If $c$ was any smaller there would be no cases:
Now we just add up the amount of ways to order the cases $(6,2,1)$, $(5,3,1)$, and $(4,3,2)$;

$$(6,2,1) \rightarrow {\binom{9}{6}}{\binom{3}{2}} = 252$$
$$(5,3,1) \rightarrow {\binom{9}{5}}{\binom{4}{3}} = 504$$
$$(4,3,2) \rightarrow {\binom{9}{4}}{\binom{5}{3}} = 1260$$
$$1260+504+252=2016$$
$$(2016 \equiv 16)\mod 1000$$
$$\fbox{016}$$
This post has been edited 1 time. Last edited by junlongsun, Feb 7, 2025, 9:26 PM
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Elephant200
1472 posts
#14
Y by
I almost forgot at least one player chose each flavor--glad I didn't
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junlongsun
70 posts
#15
Y by
On the test i included (7,1,1) as a case
:(
Rip
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apple143
62 posts
#16
Y by
Challengees24 wrote:
can anyone else agree easiest problem on test???

idk abt easiest but it was relatively easy yes.
i also got 16.
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RedFireTruck
4223 posts
#17
Y by
We decide to use PIE because RedFireTruck isn't very clever (I unironically did this on the test).

We will first calculate the number of assignments where each flavor is liked by at least one person, and no two flavors are liked by the same number of people.

After this, we can simply divide by $6$ for symmetry reasons.

The total number of flavor assignments is $3^9$.

The number of flavor assignments where flavor $x$ isn't assigned to anybody is $2^9$.

The number of flavor assignments where flavors $x$ and $y$ aren't assigned to anybody is $1$.

By PIE, this means that the total number of flavor assignments where each flavor is liked by at least $1$ person is $3^9-3\cdot 2^9+3\cdot 1$.

Now, we must subtract the number of these assignments who have $2$ flavors with the same number of people liking each of them.

This number is $3(\frac{9!}{4!4!1!}+\frac{9!}{2!2!5!}+\frac{9!}{1!1!7!})+\frac{9!}{3!3!3!}$.

Therefore, our answer is

\begin{align*}
\frac{3^9 - 3 \cdot 2^9 + 3 - 3(\frac{9!}{4!4!1!} + \frac{9!}{2!2!5!} + \frac{9!}{1!1!7!}) - \frac{9!}{3!3!3!}}{6} &= \frac{3(3^8+1)-3\cdot 2^9-3(630+756+72)-1680}{6} \\
&= \frac{6562}{2}-256-\frac{1458}{2}-280 \\
&= 3281-256-729-280 \\
&= 2\boxed{016}.
\end{align*}
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ilikemath247365
256 posts
#18
Y by
MathPerson12321 wrote:
Solution

How did I silly in real contest like bro?????
dont even remember what i did :skull:
i swear to god if i added wrong

This is exactly what I did! Somehow I managed not to silly this problem, considering I always silly combination problems.
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Alex-131
5391 posts
#19
Y by
RedFireTruck wrote:
We decide to use PIE because RedFireTruck isn't very clever (I unironically did this on the test).

We will first calculate the number of assignments where each flavor is liked by at least one person, and no two flavors are liked by the same number of people.

After this, we can simply divide by $6$ for symmetry reasons.

The total number of flavor assignments is $3^9$.

The number of flavor assignments where flavor $x$ isn't assigned to anybody is $2^9$.

The number of flavor assignments where flavors $x$ and $y$ aren't assigned to anybody is $1$.

By PIE, this means that the total number of flavor assignments where each flavor is liked by at least $1$ person is $3^9-3\cdot 2^9+3\cdot 1$.

Now, we must subtract the number of these assignments who have $2$ flavors with the same number of people liking each of them.

This number is $3(\frac{9!}{4!4!1!}+\frac{9!}{2!2!5!}+\frac{9!}{1!1!7!})+\frac{9!}{3!3!3!}$.

Therefore, our answer is

\begin{align*}
\frac{3^9 - 3 \cdot 2^9 + 3 - 3(\frac{9!}{4!4!1!} + \frac{9!}{2!2!5!} + \frac{9!}{1!1!7!}) - \frac{9!}{3!3!3!}}{6} &= \frac{3(3^8+1)-3\cdot 2^9-3(630+756+72)-1680}{6} \\
&= \frac{6562}{2}-256-\frac{1458}{2}-280 \\
&= 3281-256-729-280 \\
&= 2\boxed{016}.
\end{align*}

nou RFT is very clever
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Unicode_Master03B8
263 posts
#20 • 1 Y
Y by ilikemath247365
This would have been nice to have in 2016.
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sansgankrsngupta
143 posts
#21
Y by
OG! I don't understand what the question wants to say, First you say that each person takes one scoop of exactly one of the three flavours. Next line you say some player chose all the 3 flavours?
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Rice_Farmer
921 posts
#22
Y by
Bro i just realized i did the PIE wrong i did $3^9-6*4-3$ instead of $3^9-3*2^9+3*1$ :wallbash_red:
This post has been edited 1 time. Last edited by Rice_Farmer, Feb 9, 2025, 3:29 PM
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Andyluo
977 posts
#23
Y by
Rice_Farmer wrote:
Bro i just realized i did the PIE wrong i did $3^9-6*4-3$ instead of $3^9-3*2^9+3*1$ :wallbash_red:

wait how do you use pie
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sadas123
1306 posts
#24
Y by
Easiest question on the test should be on the AMC 10 problem 9 or 10Solution

Sorry my latex is bad but I was to lazy to put the choosing latex.
This post has been edited 3 times. Last edited by sadas123, Feb 9, 2025, 4:22 PM
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jasperE3
11346 posts
#25
Y by
centslordm wrote:
The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a singlescoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let $N$ be the number of different assignments of flavors to players that meet these conditions. Find the remainder when $N$ is divided by $1000.$

Let $c$ be the number of players who chose chocolate, $v$ vanilla, $s$ strawberry. We do casework on $(c,v,s)$, given that $c,v,s\in\mathbb N$ and $c+v+s=9$ and $c>v>s$:

Case 1: $(c,v,s)=(4,3,2)$
There are $\binom92$ ways to choose which members of the team had strawberry ice cream, then $\binom73$ ways to choose which of the remaining members had vanilla ice cream, then the rest of the members must have had chocolate. Hence $\binom92\binom73$ assignments for this case.
Case 2: $(c,v,s)=(5,3,1)$
Similarly, $\binom91\binom83$ assignments.
Case 3: $(c,v,s)=(6,2,1)$
Similarly, $\binom91\binom82$ assignments.

Our answer is $\binom92\binom73+\binom91\binom83+\binom91\binom82=2016\Rightarrow\boxed{016}$.
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NicoN9
156 posts
#26
Y by
The idea is just to brute force. Let the number of members choose chocolate, vanilla, strawberry be $a$, $b$, $c$, respectively, then we have $a+b+c=9$, and $a>b>c$. we have\[
(a, b, c)=(6, 2, 1), (5, 3, 1), (4, 3, 2).
\]and from a little easy calculation, we conclude that $16$ is the answer.
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