Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
Interesting inequalities
sqing   1
N 5 minutes ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
1 reply
1 viewing
sqing
an hour ago
sqing
5 minutes ago
functional equation with exponentials
produit   2
N 9 minutes ago by GreekIdiot
Find all solutions of the real valued functional equation:
f(\sqrt{x^2+y^2})=f(x)f(y).
Here we do not assume f is continuous
2 replies
produit
44 minutes ago
GreekIdiot
9 minutes ago
Interesting inequalities
sqing   7
N 29 minutes ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(bc+bd+cd)  \leq \frac{256}{81}$$$$ ab(a+2c+2d ) \leq \frac{256}{27}$$$$  ab(a+3c+3d )  \leq \frac{32}{3}$$$$ ab(c+d ) \leq \frac{64}{27}$$
7 replies
sqing
Yesterday at 1:25 PM
sqing
29 minutes ago
Geometry Finale: Incircles and concurrency
lminsl   174
N 31 minutes ago by LitleCabage0639
Source: IMO 2019 Problem 6
Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$.

Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.

Proposed by Anant Mudgal, India
174 replies
1 viewing
lminsl
Jul 17, 2019
LitleCabage0639
31 minutes ago
Self-evident inequality trick
Lukaluce   23
N 42 minutes ago by SunnyEvan
Source: 2025 Junior Macedonian Mathematical Olympiad P4
Let $x, y$, and $z$ be positive real numbers, such that $x^2 + y^2 + z^2 = 3$. Prove the inequality
\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]When does the equality hold?
23 replies
Lukaluce
May 18, 2025
SunnyEvan
42 minutes ago
Functional equations
mathematical-forest   0
an hour ago
Find all funtion $f:C\to C$, s.t.$\forall x \in C$
$$xf(x)=\overline{x} f(\overline{x})$$
0 replies
mathematical-forest
an hour ago
0 replies
cubefree divisibility
DottedCaculator   61
N an hour ago by sansgankrsngupta
Source: 2021 ISL N1
Find all positive integers $n\geq1$ such that there exists a pair $(a,b)$ of positive integers, such that $a^2+b+3$ is not divisible by the cube of any prime, and $$n=\frac{ab+3b+8}{a^2+b+3}.$$
61 replies
DottedCaculator
Jul 12, 2022
sansgankrsngupta
an hour ago
Cauchy and multiplicative function over a field extension
miiirz30   5
N 2 hours ago by AshAuktober
Source: 2025 Euler Olympiad, Round 2
Find all functions $f : \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}[\sqrt{2}]$ such that for all $x, y \in \mathbb{Q}[\sqrt{2}]$,
$$
f(xy) = f(x)f(y) \quad \text{and} \quad f(x + y) = f(x) + f(y),
$$where $\mathbb{Q}[\sqrt{2}] = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q} \}$.

Proposed by Stijn Cambie, Belgium
5 replies
miiirz30
3 hours ago
AshAuktober
2 hours ago
Find f
Redriver   6
N 2 hours ago by Unique_solver
Find all $: R \to R : \ \ f(x^2+f(y))=y+f^2(x)$
6 replies
Redriver
Jun 25, 2006
Unique_solver
2 hours ago
Circle
bilarev   2
N 2 hours ago by Blackbeam999
Let $k$ be a circle, $AB$ and $CD$ are parallel chords and $l$ is a
line from C, that intersects $AB$ in its middle point $L$ and $l\cap k=E$. $K$ is the middle of $DE$. Prove that $\angle AKE=\angle BKE.$
2 replies
bilarev
Oct 11, 2006
Blackbeam999
2 hours ago
Show that 2 triangles are bilogic
kosmonauten3114   0
2 hours ago
Source: My own
Given a scalene acute triangle $\triangle{ABC}$ with orthic triangle $\triangle{H_AH_BH_C}$, let $P_A$, $P_B$, $P_C$ be points such that $\triangle{AH_BH_C}\cup P_A \sim \triangle{H_ABH_C}\cup P_B \sim \triangle{H_AH_BC}\cup P_C$. Let $\ell_A$ be the trilinear polar of the polar conjugate of $P_A$ wrt $\triangle{ABC}$, and define $\ell_B$ and $\ell_C$ cyclically. Let $\triangle{A'B'C'}$ be the triangle bounded by $\ell_A$, $\ell_B$, $\ell_C$.
Show that $\triangle{ABC}$ and $\triangle{A'B'C'}$ are bilogic.
0 replies
kosmonauten3114
2 hours ago
0 replies
Interesting functions with iterations over integers
miiirz30   0
2 hours ago
Source: 2025 Euler Olympiad, Round 2
For any subset $S \subseteq \mathbb{Z}^+$, a function $f : S \to S$ is called interesting if the following two conditions hold:

1. There is no element $a \in S$ such that $f(a) = a$.
2. For every $a \in S$, we have $f^{f(a) + 1}(a) = a$ (where $f^{k}$ denotes the $k$-th iteration of $f$).

Prove that:
a) There exist infinitely many interesting functions $f : \mathbb{Z}^+ \to \mathbb{Z}^+$.

b) There exist infinitely many positive integers $n$ for which there is no interesting function
$$
f : \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}.
$$
Proposed by Giorgi Kekenadze, Georgia
0 replies
miiirz30
2 hours ago
0 replies
Moving stones on an infinite row
miiirz30   0
3 hours ago
Source: 2025 Euler Olympiad, Round 2
We are given an infinite row of cells extending infinitely in both directions. Some cells contain one or more stones. The total number of stones is finite. At each move, the player performs one of the following three operations:

1. Take three stones from some cell, and add one stone to the cells located one cell to the left and one cell to the right, each skipping one cell in between.

2. Take two stones from some cell, and add one stone to the cell one cell to the left, skipping one cell and one stone to the adjacent cell to the right.

3. Take one stone from each of two adjacent cells, and add one stone to the cell to the right of these two cells.

The process ends when no moves are possible. Prove that the process always terminates and the final distribution of stones does not depend on the choices of moves made by the player.

IMAGE

Proposed by Luka Tsulaia, Georgia
0 replies
miiirz30
3 hours ago
0 replies
Alice, Bob and 6 boxes
Anulick   1
N 3 hours ago by RPCX
Source: SMMC 2024, B1
Alice has six boxes labelled 1 through 6. She secretly chooses exactly two of the boxes and places a coin inside each. Bob is trying to guess which two boxes contain the coins. Each time Bob guesses, he does so by tapping exactly two of the boxes. Alice then responds by telling him the total number of coins inside the two boxes that he tapped. Bob successfully finds the two coins when Alice responds with the number 2.

What is the smallest positive integer $n$ such that Bob can always find the two coins in at most $n$ guesses?
1 reply
Anulick
Oct 12, 2024
RPCX
3 hours ago
Nonnegative integer sequence containing floor(k/2^m)?
polishedhardwoodtable   7
N Apr 19, 2025 by Maximilian113
Source: ELMO 2024/4
Let $n$ be a positive integer. Find the number of sequences $a_0,a_1,a_2,\dots,a_{2n}$ of integers in the range $[0,n]$ such that for all integers $0\leq k\leq n$ and all nonnegative integers $m$, there exists an integer $k\leq i\leq 2k$ such that $\lfloor k/2^m\rfloor=a_i.$

Andrew Carratu
7 replies
polishedhardwoodtable
Jun 21, 2024
Maximilian113
Apr 19, 2025
Nonnegative integer sequence containing floor(k/2^m)?
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO 2024/4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
polishedhardwoodtable
130 posts
#1 • 3 Y
Y by ehuseyinyigit, OronSH, ihatemath123
Let $n$ be a positive integer. Find the number of sequences $a_0,a_1,a_2,\dots,a_{2n}$ of integers in the range $[0,n]$ such that for all integers $0\leq k\leq n$ and all nonnegative integers $m$, there exists an integer $k\leq i\leq 2k$ such that $\lfloor k/2^m\rfloor=a_i.$

Andrew Carratu
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
YaoAOPS
1541 posts
#2 • 3 Y
Y by VicKmath7, shafikbara48593762, MS_asdfgzxcvb
We claim there are $2^n$ such sequences.
Define the $k$-interval as $\{a_k, a_{k+1}, \dots, 2k\}$. Then the condition requires that $k, \left\lfloor \frac{k}{2} \right\rfloor, \dots$ are in the $k$-interval. Call these the lamps for $k$.

Claim: We induct on the following claims:
  1. Each $k$-interval consists of all integers $\{0, 1, \dots, k\}$.
  2. $\{a_{2k-1}, a_{2k}\} = \{a_{k-1}, k\}$
  3. $a_{k-1}$ is a lamp for $k$.
Proof. We prove this inductively. The base case of $k = 0, 1$ works.
Now, we show that $a_{k-1}$ is a lamp for $a_k$. Suppose that $k$ is even. Then we get that $\{a_{k-1}, a_{k}\} = \{a_{\frac{k}{2}-1}, \frac{k}{2}\}$, and note that $a_{\frac{k}{2}-1}$ is a lamp for $\frac{k}{2}$ which is a lamp for $\frac{k}{2}$ which implies the result.
Likewise, if $k$ is odd, we consider $\{a_{k-1}, a_{k-2}\} = \{\frac{k-1}{2}, a_{\frac{k-1}{2}-1}\}$ which are both lamps for $k$. As such, since the $a_{k-1}, \dots, a_{2k-2}$ contains $a_k$ exactly once, we get that $a_k, \dots, a_{2k-2}$ doesn't contain $a_k$. It also can't contain $k$. Since $a_k$ and $k$ are lamps for $k$, $a_k, \dots, a_{2k}$ must contain them, which implies that $a_{2k-1, a_{2k}} = \{a_{k-1}, k\}$. Then we get that $\{a_k, \dots, a_{2k}\} = \{a_{k-1}, \dots, a_{2k-2}, k\} = \{0, \dots, k\}$ for the third claim. $\blacksquare$
Notably, we also have that if $\{a_{2k-1}, a_{2k}\} = \{a_{k-1}, k\}$ and $a_0, \dots, a_{2k-2}$ is a valid sequence, then $a_0, \dots, a_{2k}$ is a valid sequence.
This constraint gives us $2^n$ total options for building up $a_0, \dots, a_{2k}$ by first choosing $\{a_1, a_2\}$ and so forth.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Usernumbersomething
62 posts
#3
Y by
See https://artofproblemsolving.com/community/c6h44479p281572 for something similar. I feel like this problem is like a more accessible version of the 2005 IMO problem.
This post has been edited 3 times. Last edited by Usernumbersomething, Jun 23, 2024, 4:03 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GrantStar
821 posts
#4 • 1 Y
Y by Rajukian
Answer: $2^n$. The key claim is the following:

Claim: $a_k, \dots, a_{2k}$ is a permutation of $(0,1, \dots, k)$ for all $k\leq n$.
Proof. We strong induct on $n$. The base cases of $k=1$ and $k=2$ can be checked.
Now, we know that $a_{k-1}, \dots, a_{2k-2}$ is a permutation of $(0,\dots, k-1)$. Thus $a_k, \dots, a_{2k-2}$ are not $k=\lfloor k/2^0\rfloor$, so $a_{2k-1}$ or $a_{2k}$ is $k$. It suffices to show that $a_{k-1}$ appears in $a_k, \dots, a_{2k}$ or $a_{k-1}=a_{2k-1},a_{2k}$. By our inductive hypothesis and the reasoning above, $\{a_{2l-1}, a_{2l}\} = \{a_{l-1},l\}$ for $l<k$.

Let $j=\lfloor k/2\rfloor$. Thus $\{a_{2j-1},a_{2j}\}=\{a_{j-1},j\}$. Also, $k-1=2j-1$ or $2j$ by the floor definition of $k$. Thus $a_{k-1}=a_{j-1}$ or $k$. If $a_{k-1}=j$, then since $\lfloor k/2\rfloor = a_j$ must appear in $a_k, \dots, a_{2k}$ we conclude. Else, $a_{k-1}=a_{j-1}$. We can repeat the same argument to find $a_{j-1}=\lfloor j/2\rfloor$ or $a_{\lfloor j/2\rfloor -1}$. Repeat this inductively to get $a_{k-1}=\lfloor k/2^m\rfloor$ for some $m$. Thus $a_{k-1}$ is in $a_k, \dots, a_{2k}$ and the claim is proven. $\blacksquare$

To finish, repeatedly applying the claim gives $a_0=0$ and $i \in \{a_{2i-1}, a_{2i}\}$. Now, I claim the answer is $2^n$. This is from choosing which of $a_{2i-1}, a_{2i}$ is $i$. It suffices to show each choice gives a unique sequence. This is true by applying the claim on $1$ through $n$ to get the other number in each pair.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
v_Enhance
6877 posts
#5 • 2 Y
Y by Amkan2022, Rajukian
Solution from Twitch Solves ISL:

The answer is $2^n$. We note $a_0=0$ and ignore it going forward, focusing only on $a_i$ for $i \ge 1$.
In what follows, for each positive integer $t$ we let \[ D(t) \coloneqq \left\{ t, \left\lfloor t/2 \right\rfloor, \left\lfloor t/4 \right\rfloor, \dots \right\}. \]For example, $D(13) = \{13, 6, 3, 1, 0\}$. Then the problem condition is equivalent to saying that every element of $D(k)$ appears in $\{a_k, \dots, a_{2k}\}$.
We prove the following structure claim about all the valid sequences.
Claim: In any valid sequence, for each $0 \le k \le n$,
  • $a_{2k-1}$ and $a_{2k}$ are elements of $D(k)$; and
  • $a_k$, \dots, $a_{2k}$ consist of all the numbers from $0$ to $k$ each exactly once.
Proof. We proceed by induction; suppose we know it's true for $k$ and want it true for $k+1$. By induction hypothesis:
  • $\{a_k, \dots, a_{2k}\}$ contains each of $0$ to $k$ exactly once;
  • $a_k$ is an element of $D(\left\lceil k/2 \right\rceil)$;
  • We also know $\{a_{k+1}, \dots, a_{2k+2}\}$ contains all elements of $D(k+1)$ by problem condition.
However, note that \[ D\left( \left\lceil k/2 \right\rceil \right) \subseteq D(k+1) \]so that means either \[ (a_{2k+1} = a_k \text{ and } a_{2k+2} = k+1) \quad\text{OR}\quad (a_{2k+1} = k+1 \text{ and } a_{2k+2} = a_k). \]$\blacksquare$
We return to the problem of counting the sequences. It suffices to show that if $(a_0, \dots, a_{2n})$ is a valid sequence, there are exactly two choices of ordered pairs $(x,y) \in \{0, \dots, n+1\}$ such that $(a_0, \dots, a_{2n}, x, y)$ is a valid sequence. However, the structure claim above implies that $\{x,y\} = \{a_n, n+1\}$, so there are at most two choices. Moreover, both of them work by the structure claim again (because $k=n=1$ is the only new assertion when augmenting the sequence, and it holds also by the structure claim). This completes the proof.

Remark: Here are some examples to follow along with. When $n=4$ the $16$ possible values of $(a_4, a_5, a_6, a_7, a_8)$ are \[ \begin{array}{cc} (2,1,3,4,0) & (2,0,3,4,1) \\ (2,1,3,0,4) & (2,0,3,1,4) \\ (0,1,3,4,2) & (1,0,3,4,2) \\ (0,1,3,4,2) & (1,0,3,4,2) \\ (0,1,3,2,4) & (1,0,3,2,4) \\ (2,3,1,4,0) & (2,3,0,4,1) \\ (2,3,1,0,4) & (2,3,0,1,4) \\ (0,3,1,4,2) & (1,3,0,4,2) \\ (0,3,1,2,4) & (1,3,0,2,4) \\ \end{array} \]Now the point is that when moving to $n=5$, the element $a_4 \in \{0,1,2\} = D(2) \subseteq D(5)$ is chopped-off, and $a_9$ and $a_{10}$ must be $5$ and the chopped-off element in some order. So each of these sequences extends in exactly two ways, as claimed.
This post has been edited 1 time. Last edited by v_Enhance, Oct 26, 2024, 1:20 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
CANBANKAN
1301 posts
#6 • 1 Y
Y by GeoKing
The answer is $2^n$.

Clearly $a_0=0$, and $\{a_1,a_2\} = \{0,1\}$.

The key structural claim is that $\{a_{2k-1},a_{2k}\} = \{a_{k-1},k\}$, call $P(k)$. It is clearly true for $k=1$.

Note that one can show $P(1),P(2),\cdots,P(k)$ imply $(a_k,\cdots,a_{2k})$ is a permutation of $\{0,\cdots,k\}$ via induction. Furthermore, $P(1),\cdots,P(k-1)$ and $\{a_k,\cdots,a_{2k}\}$ being permutation of $\{0,\cdots,k\}$ imply that $\{a_{2k-1},a_{2k}\} = \{a_{k-1},k\}$, because I remove $a_{k-1}$, add $a_{2k},a_{2k-1}$, and end up just adding $k$ to the set.

I will use $P(1),\cdots,P(k-1)$ to show $\{a_k,\cdots,a_{2k}\}\supset\{0,\cdots,k\}$, which suffices. Let $x \in \{0,\cdots,k\}$. If $x=k$ we are done, since the problem condition tells us that $$\{a_k,\cdots,a_{2k}\} \supset \left\{x,\lfloor \frac x2\rfloor, \cdots, \lfloor \frac{x}{2^m}\rfloor, \cdots, 1,0\right\}. (*)$$
Henceforth assume $x<k$. Then by $P(x)$, we have $x \subset \{a_{2x+1},a_{2x+2}\}$.

By $P(2x+1),P(2x+2)$, if $2x+2 < k$ we have $\{a_{2x+1},a_{2x+2} \} \subset \{a_{4x+3},\cdots,a_{4x+6}\}$.

(If $2x+1\ge k$ then we also have $\{a_{2x+1},a_{2x+2}\}\subset \{a_k,\cdots,a_{2k}\}$ since $x<k$. If $2x+2=k$ then we have $\{a_{2x+1},a_{2x+2} \} = \{a_k, a_{2k-1},a_{2k}\}$ by $P(k-1)$)

We iterate the inductive step. Set a counter $e=2$. At each $e$ we have a set $\{a_{2^ex + (2^e-1)}, \cdots, a_{2^ex + 2^{e+1}-2}\}$ containing $x$, and we want to show it is contained in $\{a_k,\cdots,a_{2k}\}$. If $2^ex + 2^{e+1}-2 < k$ then by inductive hypothesis on $2^ex+(2^e-1)$ to $2^ex + (2^{e+1}-2)$ we have

$$ \{a_{2^ex + (2^e-1)}, \cdots, a_{2^ex + 2^{e+1}-2}\} \subset \{  a_{2(2^ex + (2^e-1))+1}, a_{2(2^ex + (2^e-1))+2}, \cdots, a_{2(2^ex + (2^{e+1}-2))+1}, a_{2(2^ex + (2^{e+1}-2))+2} \} = \{a_{2^{e+1}x + (2^{e+1}-1)}, \cdots, a_{2^{e+1}x + 2^{e+2}-2}\} $$
Thus we set $e \leftarrow e+1$. Otherwise, $2^ex + 2^{e+1}-2\ge k$, so $ 2^ex + (2^e-1) > \frac 12 (2^ex + 2^{e+1}-2) \ge \frac k2$. For everything from $2^ex + 2^e-1$ to $k-1$ (possibly there is nothing there) I apply inductive hypothesis, so this ends up giving us a subset of $\{a_k,\cdots,a_{2k}\}$. This proves $P(k)$, as desired.

Since the only restrictions we have are $\{a_{2k-1},a_{2k}\} = \{a_{k-1},k\}$, given any $a_0,\cdots,a_{2k-2}$ there are 2 choices for $(a_{2k-1},a_{2k})$. We make $n$ such choices so the answer is $2^n$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1727 posts
#7
Y by
Let the $k$-set be the set of all nonnegative integers that can be the result of
\[\left\lfloor\frac{k}{2^m}\right\rfloor\]for some nonnegative integer $m$. For example, the $5$-set is $\{5,2,1,0\}$.

Let $f(n)$ be the desired answer. We will show that $f(n)=2^n$. First, note that $f(0)=1$ is clear because the $0$-set, $\{0\}$, is a subset of $\{a_0\}$, implying that $a_0=0$. $f(1)=2$ because the $1$-set, $\{1,0\}$, is a subset of $\{a_1,a_2\}$ on top of the fact that $a_0=0$, so both $(0,0,1)$ and $(0,1,0)$ work.

We now work recursively, showing that $f(n)=2f(n-1)$ for all $n\ge 2$. Drop the condition that the sequence's terms are in $[0,n]$. We will inductively show that this is forced anyway.

It suffices to show that there are exactly $2$ ways to select $a_{n-1}$ and $a_n$, because $\{a_{2n-1},a_{2n}\}$ is determined, for any selection of $a_1$, $a_2$, $\dots$, $a_{2n-2}$ that satisfies the conditions for $0\le k\le n-1$.

Claim 1:
  • For any sequence that is a solution to the $n-1$ problem, $a_1$, $a_2$, $\dots$, $a_{2n-2}$, we have that the values $a_{n},a_{n+1},\dots,a_{2n-2}$ contain all but two values from the $n$-set.
  • $a_{2n-1}$ and $a_{2n}$ are both in the $n$-set.
  • $a_{2n-1}$ and $a_{2n}$ are different from everything in $a_{n},a_{n+1},\dots,a_{2n-2}$.

We proceed with strong induction, where the base case has already been proved above. Clearly, since $n$ is not in any $k$-set for $k<n$, $a_{n},a_{n+1},\dots,a_{2n-2}$ does not contain $n$. Since $a_{n-1}$ is in the $\lfloor n/2\rfloor$-set, it is in the $n$-set. By the Inductive Hypothesis, $a_{n},a_{n+1},\dots,a_{2n-2}$ are all different from $a_{n-1}$. Thus, the first part of the claim is proved. The next two parts follow naturally because $a_{2n-1},a_{2n}$ must be the two remaining values.
We are done.
This post has been edited 1 time. Last edited by awesomeming327., Jan 27, 2025, 4:23 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#8
Y by
For some fixed $k$ let the set $f(x)$ be the range of values $\left \lfloor \frac{k}{2^m} \right \rfloor$ can attain as $m$ varies. A key observation is that $f(\lfloor x/2 \rfloor) = f(x) /\{x\}.$ Note that $a_0=0.$ We prove the following proposition by induction on $k$:

For all $k \geq 1,$ $\{a_k, a_{k+1}, \cdots, a_{2k}\}$ is a permutation of $\{0, 1, 2, \cdots, k\},$ and $a_{2k}, a_{2k-1} \in f(k).$

The base case $k=1$ is trivial. Now assume that the proposition holds for some $k.$ Then for $k+1,$ observe that as $\{a_k, a_{k+1}, \cdots, a_{2k}\}$ is a permutation of $\{0, 1, 2, \cdots, k\}$ one of $a_{2k+1}, a_{2k+2}$ equals $k+1 \in f(k+1).$ Meanwhile $$a_{k} \in f(\lceil k/2 \rceil) = f(\lfloor (k+1)/2 \rfloor) = f(k+1)/\{k+1\}.$$so as $a_{k+1}, a_{k+2}, \cdots, a_{2k}$ are distinct from $a_k$ we have that one of $$a_{2k+1}, a_{2k+2}=a_k \in f(k+1).$$Thus as $a_k \neq k+1$ it follows that $$\{a_{2k+1}, a_{2k+2}\} = \{k+1, a_k\} \subseteq f(k+1).$$We can also see from here that $$\{a_{k+1}, a_{k+2}, \cdots, a_{2k+2}\}$$is a permutation of $\{0, 1, 2, \cdots, k+1\}.$ So the induction is complete.

Now from above we can induct to show that the answer is $\boxed{2^n},$ essentially for every next $\{a_{2k+1}, a_{2k+2}\} = \{a_k, k+1\}$ there are two possible ways to assign them.
Z K Y
N Quick Reply
G
H
=
a