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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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)!

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[*]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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0 replies
jlacosta
Yesterday at 11:16 PM
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
Geometry
VicKmath7   9
N a few seconds ago by tilya_TASh
Source: 8th European Mathematical Cup 2019 Junior Q3
Let $ABC$ be a triangle with circumcircle $\omega$. Let $l_B$ and $l_C$ be two lines through the points $B$ and $C$, respectively, such that $l_B  \parallel  l_C$. The second intersections of $l_B$ and $l_C$ with $\omega$ are $D$ and $E$, respectively. Assume that $D$ and $E$ are on the same side of $BC$ as $A$. Let $DA$ intersect $l_C$ at $F$ and let $EA$ intersect $l_B$ at $G$. If $O$, $O_1$ and $O_2$ are circumcenters of the triangles $ABC$, $ADG$ and $AEF$, respectively, and $P$ is the circumcenter of the triangle $OO_1O_2$, prove that $l_B  \parallel  OP \parallel l_C$.

Proposed by Stefan Lozanovski, Macedonia
9 replies
VicKmath7
Dec 26, 2019
tilya_TASh
a few seconds ago
2017 CNMO Grade 11 P5
minecraftfaq   3
N 3 minutes ago by CovertQED
Source: 2017 China Northern MO, Grade 11, Problem 5
Length of sides of regular hexagon $ABCDEF$ is $a$. Two moving points $M,N$ moves on sides $BC,DE$, satisfy that $\angle MAN=\frac{\pi}{3}$. Prove that $AM\cdot AN-BM\cdot DN$ is a definite value.
3 replies
minecraftfaq
Feb 24, 2020
CovertQED
3 minutes ago
Inspired by Nice inequality
sqing   0
11 minutes ago
Source: Own
Let $  a,b,c >0  $. Show that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq \frac{16}{k+3}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+k\right)$$Where $ k\geq 1.$
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq  \frac{b}{a}+\frac{c}{b}+\frac{a}{c}+13$$$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq \frac{16}{5}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+2\right)$$
0 replies
sqing
11 minutes ago
0 replies
Mmo 9-10 graders P5
Bet667   5
N 29 minutes ago by User141208
Let $a,b,c,d$ be real numbers less than 2.Then prove that $\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le4$
5 replies
+1 w
Bet667
Apr 3, 2025
User141208
29 minutes ago
3 var inequality
sqing   2
N 31 minutes ago by sqing
Source: Own
Let $ a,b,c>0 ,\frac{a}{b} +\frac{b}{c} +\frac{c}{a} \leq 2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). $ Prove that
$$a+b+c+2\geq abc$$Let $ a,b,c>0 , a^3+b^3+c^3\leq 2(ab+bc+ca). $ Prove that
$$a+b+c+2\geq abc$$
2 replies
sqing
Wednesday at 9:30 AM
sqing
31 minutes ago
Inspired by JK1603JK
sqing   4
N 34 minutes ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $  abc\neq 0$ and $ a+b+c=0.  $ Prove that
$$\left|\frac{a-b}{c}\right|+k\left|\frac{b-c}{a} \right|+k^2\left|\frac{c-a}{b} \right|\ge 3(k+1)$$Where $ k\geq 1.$
$$\left|\frac{a-b}{c}\right|+2\left|\frac{b-c}{a} \right|+4\left|\frac{c-a}{b} \right|\ge 9$$
4 replies
sqing
Yesterday at 9:44 AM
sqing
34 minutes ago
hard square root problem
kjhgyuio   1
N an hour ago by RagvaloD
........
1 reply
kjhgyuio
3 hours ago
RagvaloD
an hour ago
extremaly hard NT
gggzul   2
N an hour ago by thehound
Source: Cambodian IMO training camp
We will say that a set of $2025$ consecutive positive integers is cool if it contains exactly $13$ primes. Are there infinitely many cool sets?
2 replies
gggzul
2 hours ago
thehound
an hour ago
Tangent to two circles
Mamadi   0
2 hours ago
Source: Own
Two circles \( w_1 \) and \( w_2 \) intersect each other at \( M \) and \( N \). The common tangent to two circles nearer to \( M \) touch \( w_1 \) and \( w_2 \) at \( A \) and \( B \) respectively. Let \( C \) and \( D \) be the reflection of \( A \) and \( B \) respectively with respect to \( M \). The circumcircle of the triangle \( DCM \) intersect circles \( w_1 \) and \( w_2 \) respectively at points \( E \) and \( F \) (both distinct from \( M \)). Show that the line \( EF \) is the second tangent to \( w_1 \) and \( w_2 \).
0 replies
Mamadi
2 hours ago
0 replies
Number Theory problem
Mamadi   0
2 hours ago
Source: Own
Find all \( a, b \in \mathbb{N} \) such that \( a! + b \) and \( b! + a \) are both perfect squares.
0 replies
Mamadi
2 hours ago
0 replies
Two equal angles
jayme   0
2 hours ago
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
0 replies
jayme
2 hours ago
0 replies
Digits permutations all not divisible by 7
NicoN9   0
3 hours ago
Source: Japan Junior MO Preliminary 2020 P12
Find the number of possible quadruples $(a, b, c, d)$ with $1\le a<b<c<d\le 9$, satisfies the following property:

$24$ integers obtained by arranging four digits $a, b, c, d$ in some order, are all not divisible by $7$.
0 replies
NicoN9
3 hours ago
0 replies
8 times 8 grid and 64 coins
NicoN9   0
3 hours ago
Source: Japan Junior MO Preliminary 2020 P11
There are $8\times 8$ grid, and in each cell, there is a coin with one side white, and other side black. We start by all coin facing white. Alice and Bob executes the following operation:

First, Alice choose $8$ pairwise distinct cells, and turn over all of the coins in those cells. Next, Bob chooses one row or column, and turn over all of the coins in that row, or column.

Find the maximum possible positive integer $k$ with the following property:

No matter how Bob plays, Alice can always make $k$ coins facing black, after $2020$ turns.
0 replies
NicoN9
3 hours ago
0 replies
existence of a circle tangent to AB and AC
NicoN9   0
3 hours ago
Source: Japan Junior MO Preliminary 2020 P10
Let $ABC$ be a triangle with integer side lengths. Let $D, E$ be points on segment $BC$ such that $B, D, E,C$ are in this order, $BD=4$, and $EC=7$.
Suppose that there exists a circle which is tangent to sides $AB$ and $AC$, passes through $D, E$. Find the minimum of the perimeter of triangle $ABC$.
0 replies
NicoN9
3 hours ago
0 replies
IMO ShortList 2001, combinatorics problem 5
orl   12
N Apr 26, 2025 by Maximilian113
Source: IMO ShortList 2001, combinatorics problem 5
Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$, $0 \leq j \leq n$, $x_j$ equals the number of times $j$ appears in the sequence.
12 replies
orl
Sep 30, 2004
Maximilian113
Apr 26, 2025
IMO ShortList 2001, combinatorics problem 5
G H J
Source: IMO ShortList 2001, combinatorics problem 5
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orl
3647 posts
#1 • 3 Y
Y by Adventure10, Adventure10, Mango247
Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$, $0 \leq j \leq n$, $x_j$ equals the number of times $j$ appears in the sequence.
Attachments:
This post has been edited 1 time. Last edited by orl, Oct 25, 2004, 12:12 AM
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orl
3647 posts
#2 • 2 Y
Y by Adventure10, Mango247
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :)
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Brut3Forc3
1948 posts
#3 • 2 Y
Y by Adventure10, Mango247
Solution, from IMO Compendium
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riemannHypothesis2.71828
79 posts
#4 • 1 Y
Y by Adventure10
there must be something wrong with your solution, i haven't read it completely but i have found a counter-example.

Consider this for n = 6

3,2,1,1,0,0,0.

This clearly works, and n is bigger than 3. So I think you MIGHT have a slight mistake somewhere.
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JBL
16123 posts
#5 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Probably you should reread the solution given -- $m$ and $n$ are different letters ;)
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pi37
2079 posts
#6 • 4 Y
Y by nmd27082001, Wizard_32, DCMaths, Adventure10
Note that
\[
x_0+x_1+x_2+\cdots = 0\cdot x_0 + 1\cdot x_1 + 2\cdot x_2 + \cdots
\]so
\[
x_0=x_2+2x_3+3x_4+\cdots
\]Now we do casework on $x_0$.

If $x_0=0$, then there is a $0$, so $x_0\ge 1$, a contradiction.
If $x_0=1$, then it's easy to see that the equation is satisfied only if $x_2=1$ and $x_3=x_4=\cdots = 0$. There must be at least one $2$, which means $x_1=2$. There is one $0$, so $n=3$, and $(1,2,1,0)$ is a valid sequence.
If $x_0=2$, then either $x_2=2$ and $x_3=x_4=\cdots = 0$, or $x_3=1$ and $x_2=x_4=x_5=\cdots = 0$. In the first case, since $x_3=x_4=\cdots = 0$ and there are two $2$s already, $x_1\ge 2$ is impossible. $x_1=0$ and $x_1=1$ yield the valid sequences $(2,0,2,0)$ and $(2,1,2,0,0)$. The second case is impossible
Finally, suppose $x_0=a$ with $a\ge 3$. This implies $x_a\ge 1$, and in fact $x_a=1$ since if $x_a \ge 2$, we get a contradiction in the equation. This implies the equation can only hold if $x_2=1$ and $x_3=\cdots = x_{a-1}=a_{a+1}=\cdots = 0$. Now since $x_2=1$, we need a $2$ somewhere, which can only come from $x_1=2$. Adjusting the length of the sequence with the correct number of $0$s yields the sequence $(a,2,1,0,0,\cdots 0,1,0,0,0)$ with $a$ total $0$s for each $a\ge 3$.
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Chandrachur
385 posts
#7 • 3 Y
Y by Wizard_32, Aryan-23, Adventure10
Answer

Solution

First Simple Key Observation See that \[ \sum_{i = 0}^n x_i = n+1\]This is because every number in the sequence is from {$0,1,..,n$} as each represent number of occurances of a particular $0<=j<=n$ in the sequence so each $x_i<=n+1$ and if one $x_i=n+1$ this can't hold(as there are only n remaining places to fill and $i$ can't be $n+1$). So every number in the sequence is either $0$,$1$ , ..., or $n$. And thus \[ \sum_{i = 0}^n x_i = n+1\]as $n+1$ is the total number of numbers in the sequence.
Second Simple Key Observation Let $x_0=k$. Observe that $x_0>0$ as if $x_0=0$ implies $0$ is not present in the sequence and simultaneously it is present once (a contradiction).
Main pivotal reasoning Let $x_0=k$ [$k>0$]. Now leaving $x_0$ and the $k$ other $x_i$ which have value $0$ there are $(n+1)-(k+1)=n-k$ other terms of the sequence whose sum is $n+1-k=(n-k)+1$.
Each of these takes values $>=1$(else that contradicts definition of $k$). After giving each of them a $1$ there is still a $1$ left which must go with one of them making that a $2$.(argument similar to proving Pigeon-Hole-Principle if you notice :yup: )
So exactly $n-k-1$ values among them are $1$ and only $1$ of them takes value $2$.($(n-k+1)=(n-k-1)*1+1*2$.

"So we see that there are always
i> $n-k-1$ entries of value $1$
ii> $1$ entry of value $2$
iii> $1$ entry of value $k$
iv> $k$ entries of value $0$". (Statement I)

Next we consider 3 cases :
Case 1 : $k>=3$
Case 2 : $k=2$
Case 3 : $k=1$

Case 1 : $k>=3$ The whole sequence is filled with $0,1,2$ and $k$. So for only these "$i$"s $x_i$ is non-zero. We know there is exactly one $2$, exactly one $k$. So $x_2=1$ and $x_k=1$ and $x_0=k>2$. We know there is a $2$ somewhere and thus $x_1$ must be $2$. This in turn suggests $n-k-1=2$ or , $k=n-3$. As we assumed $k>=3$ so here $n>=6$. Thus for every $n>=6$ there is exactly one sequence $(n-3,2,1,0,...,1,0,0,0)$.$\blacksquare$

Case 2 : $k=2$ As $k=2$ put that in the Statement I we get that there will be exactly $2$ "$2$"s in the sequence, exactly $n-2-1=n-3$ "$1$"s and $2$ zeroes. Again as the sequence is filled with only $0$,$1$ and $2$ only $x_0,x_1$ and $x_2$ can be non-zero and $x_0=x_2=2$. What remains is only $x_1$. $x_1$ can't be $2$ (then $x_2=3$ which is not). So $x_1=0$ or $x_1=1$.
If $x_1=0$ $n-3=0$ or, $n=3$ and the sequence is $(2,0,2,0)$. $\blacksquare$
If $x_1=1$ $n=4$ and the sequence is $(2,1,2,0,0)$. $\blacksquare$

Case 3 : $k=1$ Again put $k=1$ in Statement I and we get that there are exactly $1$ "$2$", exactly $1$ "$0$" and exactly $n-1-1+1=n-1$ "$1$"s in the sequence. As there is one $2$ somewhere in the sequence and except $x_0,x_1$ and $x_2$ all other $x_i=0$ and $x_0=x_2=1<2$ so $x_1$ must be that $2$. This in turn gives $n=3$ and our solution is $(1,2,1,0)$. $\blacksquare$ $\blacksquare$ $\blacksquare$ :play_ball:
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yayups
1614 posts
#8 • 1 Y
Y by Adventure10
We claim the sequences that work are $(1,2,1,0)$, $(2,0,2,0)$, $(2,2,0,1,0)$, and $(n-3,2,1,0,\ldots,0,1,0,0,0)$ for any $n\ge 6$. It is easy to see that these work, so we now show that they are the only ones.

Note that $0\le x_k\le n+1$ as $x_k$ is the size of some subset of $\{0,1,\ldots,n\}$. Furthermore, if $x_k=n+1$, then all the $x_i$s are $k$, so the sequence is $(n+1,\ldots,n+1)$, which doesn't work. Therefore, $0\le x_k\le n$ for all $k$. We have the following key claim.

Claim: We have \[n+1=x_0+\cdots+x_n=x_1+2x_2+\cdots+(n-1)x_{n-1}+nx_n.\]
Proof: Note that $x_0+\cdots+x_n$ is the sum of the number of times $k$ appears in the sequence over all $k$. Since each term of the sequence is between $0$ and $n$, this means that $x_0+\cdots+x_n$ is counting the number of terms in the sequence, which is $n+1$, so \[x_0+\cdots+x_n=n+1.\]Furthermore, $x_0+\cdots+x_n$ is the sum of the terms of the sequence, which is \[\sum_{k=0}^n k\cdot(\text{number of times }k\text{ appears})=\sum_{k=0}^n kx_k,\]which proves the claim. $\blacksquare$

In particular, we have \[\boxed{x_0=x_2+2x_3+\cdots+(n-1)x_n}.\]We now split into cases based on the value of $x_0$.

Case 1: Suppose $x_0=0$. However this can't happen, as the sequence then has at least one $0$, so $x_0\ge 1$.

Case 2: Suppose $x_0=1$. Thus, \[1=x_2+2x_3+\cdots+(n-1)x_n,\]so $x_2=1$ and $x_3=\cdots=x_n=0$. But we have exactly one $0$, so we must have $n\le 3$. Now, \[x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(1+1)=n-1,\]so the sequence is $(1,n-1,1,0)$ or $(1,n-1,1)$. The first gives to $(1,2,1,0)$ and the second gives $(1,1,1)$. We see that $(1,1,1)$ doesn't work, so the only solution from this case is $\boxed{(1,2,1,0)}$.

Case 3: Suppose $x_0=2$. Thus, we have \[2=x_2+2x_3+3x_4+\cdots+(n-1)x_n,\]so $x_4=\cdots=x_n=0$. We have two subcases.
  • Case 3.1: Suppose $x_2=2$ and $x_3=0$. Then, $x_3=\cdots=x_n=0$, so $n\le 4$ as $x_0=2$. Now, \[x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(2+2)=n-3,\]so the sequence is $\boxed{(2,0,2,0)}$ or $\boxed{(2,1,2,0,0)}$.
  • Case 3.2: Suppose $x_2=0$ and $x_3=1$. Then, $x_2=x_4=\cdots=x_n=0$, so $n\le 4$ as $x_0=2$. Now, \[x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(2+1)=n-2.\]Since $x_3=1$, we have $x_1\ge 1$, so $n=3,4$. These give $(2,1,0,1)$ and $(2,2,0,1,0)$. Neither of these work, so no solutions here.

Case 4: Suppose $x_0\ge 3$. Let $a=x_0$. Then, we have $x_a\ge 1$, and we have \[a=x_2+2x_3+\cdots+(n-1)x_n\ge (a-1)x_a.\]If $x_a\ge 2$, then we have $a\ge 2(a-1)$, or $a\le 2$, which isn't the case. Therefore, $x_a=1$, so \[\sum_{\substack{k=2\\ k\ne a}}^n (k-1)x_k=1.\]This means $x_2=1$ and $x_k=0$ for $k\in\{3,\ldots,n\}\setminus\{a\}$. Now, $a\ge 3$, so the sequence has exactly two $1$s except for possibly $x_1$. This means that $x_1=2$. But we have \[x_0+\cdots+x_n=n+1,\]so \[a+2+1+1=n+1,\]so $a=n-3$. This gives the sequence $\boxed{(n-3,2,1,0,\ldots,0,1,0,0,0)}$ for any $n\ge 6$ (as $a=n-3\ge 3$).

This completes the proof.
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william122
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#9 • 2 Y
Y by Mango247, Mango247
We claim that the only sequences are $(1,2,1,0)$, $(2,0,2,0)$, $(2,1,2,0,0)$, and $(n-3,2,1,0,\ldots,0,1,0,0,0)$ for all $n\ge 6$.

First, note that $x_0+\ldots+x_n=n+1$. As there are $x_0$ $0$s, there are $n-x_0$ positive numbers with positive index. However, they sum to $n+1-x_0$, so they must be comprised of all 1s and a single 2. This means that the only positive numbers in our sequence have indices a subset of $\{0,1,2,x_0\}$. Now, we break into cases based on the value of $x_0$.

If $x_0=1$, note that we only have one $2$, so $x_2=1$. Now, we must have $x_1=2$. No other numbers can be positive, so this only yields $(1,2,1,0)$.

If $x_0=2$, our sequence now has two $2$s, so $x_2=2$. Now, we can have $x_1=0,1$, leading to the solutions $(2,0,2,0)$ and $(2,1,2,0,0)$.

Finally, if $x_0\ge 3$, we must have $x_2=1$ and $x_{x_0}=1$, so $x_1=2$. This yields the aforementioned infinite class of solutions.
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math_pi_rate
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#10
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Nice problem! Here's my solution: The only possible sequences are $$(2,1,2,0,0),(1,2,1,0),(2,0,2,0) \text{ and } (n-3,2,1,\underbrace{0,0, \dots ,0}_{(n-6) \text{ times}},1,0,0,0) \quad \forall n>5$$
Notice that the sequence consists of only $0,1, \dots ,n$. Consider the sets $A$ and $B$ given by $$A=\{i \text{ such that }\exists k \text{ with } x_k=i\} \text{ and } B=\{i \text{ such that } x_i \neq 0\}$$Note that the given condition implies that $A=B$. Now, we have $x_0+x_1+ \dots +x_n=n+1$ as the LHS simply represents the total number of times each number $0,1, \dots ,n$ is present in the sequence. Also, $x_1+2x_2+\dots +nx_n$ is the sum of the elements of the sequence. Thus, we get $$x_0+x_1+ \dots+x_n=x_1+2x_2+ \dots +nx_n=n+1$$$$\Rightarrow x_0=x_2+2x_3+ \dots +(n-1)x_n$$If there exist $3$ numbers more than $1$ in set $B$, then let $i>j>k \geq 2$ be the three maximal integers in $B$. This gives $$x_0 \geq (i-1)x_i+(j-1)x_j+(k-1)x_k \geq i+j+k-3 \geq i+3+2-3=i+2$$But then, since $x_0$ lies in set $A$, so $x_0 \in B$ as well, which contradicts the fact that $i$ is the maximal element in $B$. So there are at most $2$ elements in $B$ which are more than $1$. If there are exactly $2$ such elements, then let $i>j \geq 2$ be these numbers. Then $$x_0=(i-1)x_i+(j-1)x_j \geq i+j-2 \geq i$$As shown above, $x_0 \in B$, which means that we must have $x_0 \leq i$ as well. Thus, equality must hold everywhere, which gives $x_0=i,$ $x_i=x_j=1$ and $j=2$. Then there are $2$ ones in the sequence, and so $x_1=2$. Since $x_0+x_1+x_i+x_j=n+1$, so this gives $x_0=n-3$, and hence $i=n-3$. As $i>2$, so we must also have $n-3>2$, i.e. $n>5$. This gives the infinite class of sequences mentioned in the beginning.

Now $B$ must have atleast one element more than $1$, since otherwise $x_0=0$ (which is not possible as this means that $0 \in A$ but $0 \not \in B$). So suppose $B$ has exactly $1$ element more than $1$, and let this element be $i$. Then $x_0=(i-1)x_i$. If $i>3$, then $x_0 \in B$ gives that some multiple of $i-1$ also lies in $B$, which is contrary to the fact that $i$ is the only element more than $1$ in set $B$. So we have $i=2$, and $x_0=x_2$. Then we get $2x_0+x_1=n+1$. But $x_3,x_4, \dots ,x_n$ are all zeros, and so $x_0 \geq n-2$. Also, $x_2 \leq 2$ since it represents the number of times $2$ appears in the sequence, and we cannot have $x_0=x_1=x_2=2$. Then, as $x_0=x_2$, so we get $2 \leq x_0 \leq n-2$, which gives $n \leq 4$. A simple case bash then yield the remaining answers (in particular, remember that $1 \leq x_0 \leq 2$). $\blacksquare$
This post has been edited 2 times. Last edited by math_pi_rate, May 9, 2020, 9:06 AM
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awesomeming327.
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#11 • 1 Y
Y by Mango247
Let $S$ be the sum $x_0+x_1+\dots+x_n.$ While $S$ is that, $S$ is also zero times the number of times zero appears in the sequence, plus one times the number of times one appears in the sequence, and so on adding to $x_1+2x_2+3x_3+\dots+9x_9.$ Where $x_i=0$ if $x_i$ does not ever happen in the sequence.

$~$
Note that $x_{10}=x_{11}=\dots=0$ because numbers $\ge 10$ aren't digits. We have $9\ge x_0=x_2+2x_3+\dots+8x_9.$ When $x_0=0$ this is impossible. When $x_0=1$, then $x_2=1$ and that's it for index greater than one. Thus, $(1,2,1,0)$

$~$
Note that $x_0=2$ implies that $(x_2,x_3)=(2,0)$ or $(0,1).$ This gives $(2,0,2,0)$ or $(2,1,2,0,0)$ and none for $(0,1)$ because there's a two.

$~$
For $x_0\ge 3$, note that $(x_0-1)x_{x_0}\ge x_0-1$ so there is one other two and that's it. This gives $(x_0,2,1,0,\dots,0,1,0,0,0).$
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megarnie
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#12
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The only solutions are $\boxed{(1,2,1,0), (2,0,2,0), (2,1,2,0,0), (x-3,2,1,0,0,\ldots, 0,1,0,0,0)}$ for any $x\ge 6$ (where the entries in $\ldots$ are zero). We can check that these all work.

Notice that \[x_0 + x_1 + \cdots + x_n = 0\cdot x_0 + 1\cdot x_1 + \cdots + n \cdot x_n,\]so
\[x_0 = x_2 + 2x_3 + 3x_4 + \cdots + (n-1)x_n\]
Clearly $x_0>0$.

Case 1: $x_0 = 1$.
Then we must obviously have $x_2 = 1$ and $x_3 = x_4 \cdots = x_n = 0$. Since there is only one zero in the sequence, we have $n=3$. Since $x_0 = x_2 = 1$, we have $x_1  = 2$. This gives $(1,2,1,0)$.

Case 2: $x_0 = 2$.
Then we must have $x_4 = x_5 = \cdots = x_n = 0$ and either $x_3 = 1, x_2 = 0$ or $x_2 = 2, x_3 = 0$.

If $x_3 = 1, x_2 = 0$, then we must have $x_i = 3$ for some $i$, which means $x_1 = 3$, but this is obviously impossible.

Thus, $x_2 = 2, x_3 = x_4 = \cdots = x_n = 0$. Notice that $x_i\ne 1\forall i\ne 1$, so $x_1\in \{0,1\}$. We must have exactly two zeroes, so $x_1 = 0$ implies $n=3$ and $x_1 = 1$ implies $n=4$. This gives $(2,0,2,0)$ and $(2,1,2,0,0)$.

Case 3: $x_0 > 2$.
Let $x_0 = k$. Since $k\le n$, we have $(k-1)x_k \le x_0 = k$, so $x_k$ is at most $1$. Since $x_0 = k$, we have $x_k\ge 1$, so $x_k = 1$. Now, $(k-1)x_k = k-1$, so \[\sum_{i=2, i\ne k}^n (i-1)x_i  = 1,\]so we must have $x_2 = 1$ and $x_3 = \cdots = x_{k-1} = x_{k+1} = \cdots = x_n = 0$. Then $x_1\ne 0$, so there are exactly $n-3$ zeroes in the sequence, which means $x_0 = n-3$ and $n>5$. Now $x_2 = x_k = 1$, so $x_1 = 2$ must hold. This gives $(n-3,2,1,0,0,\ldots, 0, 1,0,0,0)$, as desired.
This post has been edited 2 times. Last edited by megarnie, Dec 21, 2022, 2:55 PM
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Maximilian113
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First note that they should all be nonnegative integers. Observe that $$0x_0+1x_1+2x_2+\cdots + nx_n = x_0+x_1+x_2 + \cdots + x_n = n+1.$$The first equation holds because the LHS counts the sum of the numbers in the sequence, while the second equation holds because the LHS counts the number of numbers in the sequence.

Let $x_0=k.$ Clearly $k > 0.$ Then $$x_1+x_2+\cdots + x_n = (n-k)+1,$$but the LHS has $n-k$ positive terms. Thus $n-k-1$ of them are $1$s, there is a $2,$ and the rest are $0$s. Therefore, at most $x_1, x_2, x_0, x_k$ can be positive, so clearly $n-k+1 \leq 4 \implies n-k \leq 3.$ Meanwhile since we have at least $(n-k-1)$ $1$s it follows that $1 \leq n-k.$

If $n-k=1,$ we have $x_0=n-1,$ no $1$s, and $1$ $2$, with everything else being $0.$ Thus $x_2, x_{n-1}$ are both positive, so they must equal. This yields $(2, 0, 2, 0).$

If $n-k=2,$ we have $x_0=n-2,$ so there is one $1, 2$ in the sequence while the rest is zero. This makes both $x_1, x_2$ positive and casework on which one is $2$ yields $(1, 2, 1, 0), (2, 1, 2, 0, 0).$

If $n-k=3,$ we have $x_0=k=n-3,$ so there are $(n-3)$ $0$s, $(2)$ $1$s, $(1)$ $2,$ and $(1)$ $n-3.$ Thus $x_{n-3}, x_1, x_2$ are all positive. Casework on the size of $n$ yields $(k, 2, 1, \{0, \cdots, 0 \}, 1, 0, 0, 0)$ where the set in the middle has $k-3$ $0$s. Also note that $k \geq 3.$

Thus the solutions are $(2, 0, 2, 0), (1, 2, 1, 0), (2, 1, 2, 0,0), (k, 2, 1, \{0, \cdots, 0 \}, 1, 0, 0, 0)$ where $k \geq 3.$
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