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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.
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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]
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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
4th grader qual JMO
HCM2001   37
N 43 minutes ago by CatCatHead
i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier
37 replies
HCM2001
May 22, 2025
CatCatHead
43 minutes ago
Zsigmondy's theorem
V0305   3
N an hour ago by CatCatHead
Is Zsigmondy's theorem allowed on the IMO, and is it allowed on the AMC series of proof competitions (e.g. USAJMO, USA TSTST)?
3 replies
V0305
Yesterday at 6:22 PM
CatCatHead
an hour ago
interesting diophantiic fe in natural numbers
skellyrah   5
N an hour ago by skellyrah
Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \),
\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]
5 replies
skellyrah
Yesterday at 8:01 AM
skellyrah
an hour ago
Non-linear Recursive Sequence
amogususususus   3
N 2 hours ago by SunnyEvan
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
3 replies
amogususususus
Jan 24, 2025
SunnyEvan
2 hours ago
Inspired by 2025 Beijing
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
6 replies
sqing
Yesterday at 4:56 PM
sqing
2 hours ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   2
N 2 hours ago by sqing-inequality-BUST
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
sqing-inequality-BUST
2 hours ago
Nice "if and only if" function problem
ICE_CNME_4   14
N 2 hours ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
14 replies
ICE_CNME_4
Friday at 7:23 PM
wh0nix
2 hours ago
2-var inequality
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b> 0 , ab(a+b+1) =3.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{24}{(a+b)^2} \geq 8$$$$ \frac{a}{b^2}+\frac{b}{a^2}+\frac{49}{(a+  b)^2} \geq \frac{57}{4}$$Let $ a,b> 0 ,  (a+b)(ab+1) =4.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{40}{(a+b)^2} \geq 12$$$$\frac{a}{b^2}+\frac{b}{a^2}+\frac{76}{(a+ b)^2}  \geq 21$$
1 reply
sqing
3 hours ago
sqing
2 hours ago
Balkan Mathematical Olympiad
ABCD1728   1
N 3 hours ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
1 viewing
ABCD1728
Yesterday at 11:27 PM
ABCD1728
3 hours ago
area of quadrilateral
AlanLG   1
N 3 hours ago by Altronrren
Source: 3rd National Women´s Contest of Mexican Mathematics Olympiad 2024 , level 1+2 p5
Consider the acute-angled triangle \(ABC\). The segment \(BC\) measures 40 units. Let \(H\) be the orthocenter of triangle \(ABC\) and \(O\) its circumcenter. Let \(D\) be the foot of the altitude from \(A\) and \(E\) the foot of the altitude from \(B\). Additionally, point \(D\) divides the segment \(BC\) such that \(\frac{BD}{DC} = \frac{3}{5}\). If the perpendicular bisector of segment \(AC\) passes through point \(D\), calculate the area of quadrilateral \(DHEO\).
1 reply
AlanLG
Jun 14, 2024
Altronrren
3 hours ago
Inspired by 2025 SXTB
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b  $ be real number such that $ a^2+b^2=\frac12. $ Prove that
$$-\frac{\sqrt{9+6\sqrt 3}}{2}\leq(a+1)^2- (b-1)^2\leq\frac{\sqrt{9+6\sqrt 3}}{2}$$Let $ x $ be real number . Prove that
$$-\frac{2\sqrt 2}{3}\leq \frac{x}{x^2+1}+ \frac{ 2x}{x^2+4} \leq\frac{2\sqrt 2}{3}$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
IMO Shortlist 2014 G2
hajimbrak   14
N 3 hours ago by ezpotd
Let $ABC$ be a triangle. The points $K, L,$ and $M$ lie on the segments $BC, CA,$ and $AB,$ respectively, such that the lines $AK, BL,$ and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM, BMK,$ and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$.

Proposed by Estonia
14 replies
hajimbrak
Jul 11, 2015
ezpotd
3 hours ago
[TEST RELEASED] OMMC Year 5
DottedCaculator   132
N 5 hours ago by MathCosine
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
132 replies
DottedCaculator
Apr 26, 2025
MathCosine
5 hours ago
Base 2n of n^k
KevinYang2.71   50
N Today at 1:39 AM by ray66
Source: USAMO 2025/1, USAJMO 2025/2
Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n>N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$.
50 replies
KevinYang2.71
Mar 20, 2025
ray66
Today at 1:39 AM
Sums of pairs in a sequence
tenniskidperson3   56
N Apr 6, 2025 by Marcus_Zhang
Source: USAJMO 2010, Problem 2
Let $n > 1$ be an integer. Find, with proof, all sequences $x_1 , x_2 , \ldots , x_{n-1}$ of positive integers with the following three properties:
(a). $x_1 < x_2 < \cdots < x_{n-1}$ ;
(b). $x_i + x_{n-i} = 2n$ for all $i = 1, 2, \ldots , n - 1$;
(c). given any two indices $i$ and $j$ (not necessarily distinct) for which $x_i + x_j < 2n$, there is an index $k$ such that $x_i + x_j = x_k$.
56 replies
tenniskidperson3
Apr 29, 2010
Marcus_Zhang
Apr 6, 2025
Sums of pairs in a sequence
G H J
Source: USAJMO 2010, Problem 2
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kamatadu
480 posts
#44 • 1 Y
Y by HoripodoKrishno
I claim that the only sequence is $x_i=2i$. It is easy to see that such a sequence suffices the condition. Now we prove that this is the only one.

Firstly, for some $n$, we prove that $x_i=i\cdot x_1$. We proceed using induction. The base case $x_1=1\cdot x_1=x_1$ is clearly true. Now assume the induction hypothesis for some $k-1<n-1$, and we prove it for $k$.

Consider the following sequence.\[x_1+x_{n-k},x_2+x_{n-k},\ldots,x_{k-1}+x_{n-k}.\]Note that this sequence has $k-1$ terms in it. Firstly, we have that $x_k+x_{n-k}=2n$ and the fact that $x_i<x_k$ for all $1\le i\le k-1$. So this gives us that $x_i+x_{n-k}<x_k+x_{n-k}=2n$ which means that we will get some $x_j$ for each $i$ such that $x_j=x_i+x_{n-k}$. Now clearly from the increasing condition of the sequence, we get that $j\ge n-k+1$. So each $x_j$ belongs to the following sequence.\[x_{n-(k-1)},x_{n-(k)},\ldots,x_{n-1}.\]So note that this sequence also has $k-1$ terms. So this forces that the equality occurs in each of the values of $x_j$ from both the sequences. So this gives that $x_1+x_{n-k}=x_{n-(k-1)}$, which further gives that $x_1+(2n - x_{n-(n-k)})=2n - x_{n-((n-k+1)}\implies x_1 - x_k = -x_{k-1}\implies x_1+x_{k-1}=x_k$. Now using our induction hypothesis, we get that $x_{k-1}=(k-1)x_1$ which finally gives $x_k=k\cdot x_1$. Now then, we finally get that $x_{n-1}=(n-1)x_1$. Putting this into $x_1+x_{n-1}=2n$, we get that $x_1=2$ and we are done.
Z K Y
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trk08
614 posts
#45
Y by
We claim the only possible sequence is if $x_i=2i$. This sequence clearly works, and we now prove it is the only one.//

First of all, consider $x_1+x_{n-2}$. As $x_1>0$, it has to be more than $x_{n-2}$, which implies that it must be $x_{n-1}$. As each of the following sums are in decreasing order, from top to bottom:
\[x_1+x_{n-2}\]\[x_1+x_{n-3}\]\[\dots\]\[x_1+x_1,\]and there are $n-2$ of them, we can say that:
\[x_1+x_{n-2}=x_{n-1}\]\[x_1+x_{n-3}=x_{n-2}\]\[\dots\]\[x_1+x_1=x_2.\]

Therefore, it is easy to see that $x_{i}=ix_1$. Therefore, $nx_1=2n$, or $x_1=2$, as desired.
Z K Y
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Shreyasharma
683 posts
#46
Y by
Note that we are given,
  • $x_1 < x_2 < \dots < x_{n-1}$
  • $x_1 + x_{n-1} = x_2 + x_{n-2} = x_3 + x_{n-3} = \dots = 2n$
From the final condition we find $x_1 + x_2$, $x_1 + x_3$, $\dots$, $x_1+x_{n-2}$ all belong to $(x_i)$. Clearly $x_1 + x_{n-2} = x_{n-1}$ or that $$x_{n-2} = x_{n-1} - x_1$$Then $x_1 + x_{n-3} = x_{n-2}$, which in turn gives, $$x_{n-3} = x_{n-1} - 2x_1$$and so on. Inducting downwards we find $$x_{n-1-k} = x_{n-1} - kx_1$$However then $x_1 = x_{n-1} - (n-2)x_1$ which gives $nx_1 = 2n$, or $x_1 = 2$. Now from our induction we find the sequence of $(x_i)$ are simply the even integers from $2$ to $2n - 2$.
Z K Y
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joshualiu315
2534 posts
#47
Y by
Consider the sequence $x_1+x_1, x_1+x_2, \dots, x_1+x_{n-1} = 2n$. Each of the $n-2$ terms before $2n$ are obviously less than $2n$, and they must be part of the sequence $x_1, x_2, \dots, x_{n-1}$. This means we must necessarily have

\begin{align*}
    x_1+x_1 &= x_2, \\
    x_1+x_2 &= x_3, \\
    \dots &\phantom{=} \\
    x_1+x_{n-2} &= x_{n-1}.
\end{align*}
Hence, $x_k = kx_1$, and plugging this in gives

\[x_1+(n-1)x_1=2n \implies x_1=2.\]
Thus, $x_n = \boxed{2n}$, which is easily checked to be true.
Z K Y
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dolphinday
1328 posts
#48
Y by
We find that
\[x_1 < x_1 + x_1 < x_1 + x_2 \dots x_1 + x_{n-2}\]are all in $x_i$, due to $(c)$, and since there are $n-1$ terms, this sequence must be equivalent to
\[x_1, x_2, \dots, x_{n-1}\]So, we can rewrite each term as $ix_1$.
$\newline$
\[x_1 + x_{n-1} = 2n\]can be rewritten as
\[x_1 + (n-1)x_1 = 2n \implies nx_1 = 2n \implies x_1 = 2\]Due to the fact that each term is equal to $ix_1$, this sequence is just $x_i = 2i$.
This post has been edited 1 time. Last edited by dolphinday, Jan 6, 2024, 11:58 AM
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shendrew7
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#49
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Due to the third condition, we know the sequence $x_2, x_3, \ldots, x_{n-1}, 2n$ must correspond with the terms of
\[x_1+x_1 < x_1+x_2 < \ldots < x_1+x_{n-2} < x_1+x_{n-1} = 2n.\]
Thus we know $x_k = k \cdot x_1$ for $1 \leq k \leq n-1$, from which the second condition tells us our only solution is $\boxed{x_i = 2i}$. $\blacksquare$
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peppapig_
280 posts
#50
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I goofed last time;

I claim that the only possible sequence is $2$, $4$, $6$, $\dots$, $2n-2$. Note that by problem conditions, we have that
\[x_1+x_{n-1}=2n,\]but we also have that
\[x_1+x_{n-2}=x_j,\]for some integer $j$. However, since $x_1<x_2<\dots<x_{n-1}$, we must have that $x_{n-2}=x_{n-1}-x_1$. Similarly, we get that
\[x_k=x_1+x_{n-3}<x_1+x_{n-2}=x_{n-1},\]meaning that $k=n-2$. Continuing this, we get that our sequence must be $x_1$, $2x_1$, $\dots$, $(n-1)x_1$, and since $x_1+x_{n-1}=2n$, we also get that $x_1$ must be equal to $2$. Therefore the only possible sequence is $2$, $4$, $6$, $\dots$, $2n-2$, which indeed works, finishing the problem.
This post has been edited 1 time. Last edited by peppapig_, Feb 24, 2024, 3:33 AM
Reason: Missed a $
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two_steps
102 posts
#51
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is this correct? why is my solution different from everyone else lol

For any two terms $x_i < x_j$, notice that the following terms are also in the sequence
\[x_i, x_j \rightarrow 2n-x_j \rightarrow 2n-x_j+x_i \rightarrow x_j-x_i\]This is essentially a stronger version of the converse of condition 3. From here, note that we may prove that each element is a multiple of $x_1$ by induction. The only possible sequences are now $1,2,\dots, n-1$ and $2,4,\dots,2(n-1)$. The first sequence doesn't work, while the second one does.
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ItsBesi
147 posts
#52
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My solution is different from others so did I fakesolve?

We separate the problem into $2$ cases which both are analogous.

Case 1. $n-\text{is even}$
Click to reveal hidden text

Case 2. $n-\text{is odd}$
Click to reveal hidden text

Since in both cases we have: $\{x_1,x_{k_1}, \dots , x_{k_{n-2}} \}=\{x_1,x_2, \dots, x_{n-1} \}$ we get:

$x_1=x_1 , x_2=x_{k_1} , x_3=x_{k-2} ,  \dots , x_{n-1}=x_{k_{n-2}}$
So
$x_1+x_1=x_{k_1}=x_2 \implies x_1+x_1=x_2 \implies x_2=2 \cdot x_1$
$x_1+x_2=x_{k_2}=x_3 \implies x_1+x_2=x_3 \implies x_3=x_1+x_2=x_1 + 2 \cdot x_1=3 \cdot x_1 \implies x_3=3 \cdot x_1$
$x_2+x_2=x_{k_3}=x_4 \implies x_2+x_2=x_4 \implies x_4=2 \cdot x_2=2 \cdot 2 \cdot x_1= 4 \cdot x_1 \implies x_4=4 \cdot x_1$
$x_2+x_3=x_{k_4}=x_5 \implies x_2+x_3=x_5 \implies x_5=x_2+x_3=2 \cdot x_1 + 3 \cdot x_1 =5 \cdot x_1 \implies x_5= 5 \cdot x_1$
$x_3+x_3=x_{k_5}=x_6 \implies x_3+x_3=x_6\implies x_6=x_3+x_3=3 \cdot x_1 + 3 \cdot x_1=6 \cdot x_1 \implies x_6=6 \cdot x_1$
$\dots$
$x_{\frac{n}{2}-1}+x_{\frac{n}{2}-1}=x_{k_{n-3}}=x_{n-2} \implies x_{\frac{n}{2}-1}+x_{\frac{n}{2}-1}=x_{n-2} \implies x_{n-2}= x_{\frac{n}{2}-1}+x_{\frac{n}{2}-1}=2 \cdot  x_{\frac{n}{2}-1} = 2 \cdot (\frac{n}{2}-1) \cdot x_1=(n-2) \cdot x_1 \implies x_{n-2}=(n-2) \cdot x_1$

Now again by $(b) \implies$

$x_2+x_{n-2}=2n \implies 2 \cdot x_1 + (n-2) \cdot x_1 =2n \implies n \cdot x_1=2n \implies x_1=2
So x_1=2, x_2=2 \cdot x_1=2 \cdot 2=4 , x_3=3 \cdot x_1= 3 \cdot 2=6 , \dots x_{n-1}=(n-1) \cdot x_1=(n-1) \cdot 2=2n-2$

Hence $(x_1,x_2, \dots , x_{n-1}=(2,4, \dots ,2n-2)$
This post has been edited 1 time. Last edited by ItsBesi, Sep 26, 2024, 1:12 PM
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Maximilian113
575 posts
#53
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Observe that property (b) is equivalent to $$i+j < n \iff x_i+x_j < 2n.$$Therefore, $$x_1 < x_1+x_1 < x_1+x_2 < \cdots < x_1+x_{n-2} < 2n.$$However, each of these sums are some $x_k$ but since $$x_2<x_3<\cdots < x_{n-1} < 2n$$it follows that $x_1+x_k=x_{k+1}$ for each $k=1, 2, \dots, n-2.$ Thus $x_2=2x_1, x_3=3x_1,$ etc. This yields $2n=x_1+x_{n-1}=x_1+(n-1)x_1 \implies x_1=2,$ and it follows that the sequence is $$2, 4, 6, \cdots, 2n-2.$$It is easy to show that this works.
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eg4334
636 posts
#54
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Goofy problem.

The conditions imply that the sums, in increasing order, $\{ x_1+x_1, x_1+x_2, \dots x_1+x_{n-1} \}$ must biject to $x_2, x_3, \dots x_{n-1}$. Therefore, $x_1+x_1=x_2$, $x_1+x_2=x_3$, and likewise. Then its not hard to see that $x_i = i x_1$ and from the second condition we obtain $x_1=2$. Therefore the only sequence is the increasing one of evens.
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Ihatecombin
66 posts
#55
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Notice that \(a_1,a_1 + a_1, a_1 + a_2, \dots. a_1 +a_{n-2}\) are all different,
by an easy induction we obtain the only sequence that works is \(a_i = 2i\).
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bjump
1033 posts
#56 • 2 Y
Y by KenWuMath, imagien_bad
Solution from over a year ago ...

The answer is $x_{k}= 2n$
Note that
$$x_1 + x_{n-1}=2n$$$$x_{1}+ x_{n-2} = x_{n-1}$$$$\vdots$$$$x_{1}+x_{1}=x_{2}$$I claim that $x_{n} = nx_1$ we will prove this via induction with a trivial base case of $n=1$. Assume that this is true for $1 \le n \le k$ then $x_{k+1} = x_{k}+ x_1=kx_1+x_1= (k+1)x_1$. Now by condition (b) $x_i+x_{n-i} = 2n \iff (n-i+i)x_1=2n \iff x_1=2$. Therefore $x_k = k x_1 = 2k$.
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de-Kirschbaum
202 posts
#57
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Note that $x_{n-2}<x_1+x_{n-2}<x_2+x_{n-2}=2n$ so by property 3 we must have $x_{n-2}+x_1=x_{n-1}$. By a similar reasoning this works for all $x_i$ so we have that $x_1, x_2,\ldots, x_{n-1}$ is an arithmetic sequence with common difference $x_1$, and by property 1 we must have $nx_1=2n \implies x_1=2$ so the only sequence that works is $x_i=2i$.
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Marcus_Zhang
980 posts
#58
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