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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 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
Hard T^T
Noname23   0
14 minutes ago
<problem>
0 replies
+1 w
Noname23
14 minutes ago
0 replies
i need help
MR.1   0
15 minutes ago
Source: help
can you guys tell me problems about fe in $R+$(i know $R$ well). i want to study so if you guys have some easy or normal problems please send me
0 replies
MR.1
15 minutes ago
0 replies
Flo0r functi0n
m4thbl3nd3r   3
N 33 minutes ago by m4thbl3nd3r
Find all positive integers such that $$n=\lfloor \sqrt{n}\rfloor^2+\lfloor \sqrt{n}\rfloor$$
3 replies
m4thbl3nd3r
an hour ago
m4thbl3nd3r
33 minutes ago
Changeable polynomials, can they ever become equal?
mshtand1   4
N 42 minutes ago by CHESSR1DER
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.5
Initially, two constant polynomials are written on the board: \(0\) and \(1\). At each step, it is allowed to add \(1\) to one of the polynomials and to multiply another one by the polynomial \(45x + 2025\). Can the polynomials become equal at some point?

Proposed by Oleksii Masalitin
4 replies
mshtand1
Yesterday at 12:47 AM
CHESSR1DER
42 minutes ago
Guessing polynomial by its maximum values on segments
NO_SQUARES   3
N 43 minutes ago by pco
Source: Kvant 2025 no. 1 M2828 and The XIX Southern Mathematical Tournament
Maxim has guessed a polynomial $f(x)$ of degree $n$. Sasha wants to guess it (knowing $n$). During a turn, Sasha can name a certain segment $[a;b]$ and Maxim will give in response the maximum value of $f(x)$ on the segment $[a;b]$. Will Sasha be able to guess $f(x)$ in a finite number of steps?
M. Didin
3 replies
NO_SQUARES
Yesterday at 3:21 PM
pco
43 minutes ago
easy number theory
MuradSafarli   1
N an hour ago by Tuvshuu
\[
v_p(n!) \leq \frac{n}{p - 1}
\]
1 reply
MuradSafarli
2 hours ago
Tuvshuu
an hour ago
Inspired by Kazakhstan 2017
sqing   1
N an hour ago by sqing
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a+b+c=2. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 1$$Let $a,b,c\ge \frac{1}{3}$ and $a+b+c=1. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 9$$
1 reply
1 viewing
sqing
3 hours ago
sqing
an hour ago
Collinear geometry problem with incircle
ilovemath0402   1
N an hour ago by deraxenrovalo
Given acute $\triangle ABC$ not isosceles, the incircle $(I)$. $D,E,F$ is the intersection of $(I)$ with $BC,CA,AB$. $P$ is the projection of $D$ onto $EF$. $DP$ cut $(I)$ at the second point $K$. $L$ is the projection of $A$ onto $IK$. $(LEC), (LFB)$ cut $(I)$ at the second point $M,N$ respectively. Prove $M,N,P$ are collinear
1 reply
ilovemath0402
Jul 22, 2023
deraxenrovalo
an hour ago
Wait wasn&#039;t it the reciprocal in the paper?
Supercali   6
N 2 hours ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
6 replies
Supercali
Jul 9, 2023
kes0716
2 hours ago
About old Inequality
perfect_square   0
3 hours ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
perfect_square
3 hours ago
0 replies
inquality
karasuno   1
N 3 hours ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
karasuno
4 hours ago
sqing
3 hours ago
Number Theory
karasuno   0
4 hours ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
4 hours ago
0 replies
Minimum number of values in the union of sets
bnumbertheory   5
N 4 hours ago by Rohit-2006
Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3
For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
5 replies
bnumbertheory
Oct 14, 2023
Rohit-2006
4 hours ago
two triangles have equal circumradii
littletush   5
N 5 hours ago by Taha.kh
Source: Italy TST 2009 p5
Two circles $O_1$ and $O_2$ intersect at $M,N$. The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$. Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.
5 replies
littletush
Mar 10, 2012
Taha.kh
5 hours ago
K-pop sequences
L567   8
N Dec 21, 2024 by alexanderhamilton124
Source: India EGMO TST 2023/5
Let $k$ be a positive integer. A sequence of integers $a_1, a_2, \cdots$ is called $k$-pop if the following holds: for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, \cdots , a_{n+k} \}$. Determine, as a function of $k$, how many $k$-pop sequences there are.

Proposed by Sutanay Bhattacharya
8 replies
L567
Dec 10, 2022
alexanderhamilton124
Dec 21, 2024
K-pop sequences
G H J
G H BBookmark kLocked kLocked NReply
Source: India EGMO TST 2023/5
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L567
1184 posts
#1
Y by
Let $k$ be a positive integer. A sequence of integers $a_1, a_2, \cdots$ is called $k$-pop if the following holds: for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, \cdots , a_{n+k} \}$. Determine, as a function of $k$, how many $k$-pop sequences there are.

Proposed by Sutanay Bhattacharya
This post has been edited 1 time. Last edited by L567, Dec 10, 2022, 7:00 PM
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L567
1184 posts
#2 • 1 Y
Y by VicKmath7
First, note that the sequence is non-decreasing. Also, $a_{i+1} - a_i$ must be either $0$ or $1$, and it depends on whether $a_{i+1+k} - a_{i+k} = 0$ or $1$.

Define $b_i = a_{i+1} - a_i$, then the condition means that $b_{i+k} = b_i$, and so each residue class mod $k$ has a constant value of $b_i$. Since this is the only constraint, there are $2^k$ ways to choose this.

If the sequence of $b_i$ is determined, $a_1$ is also uniquely determined (it is the number of $1$'s among the first $k$ $b_i$ + 1), so there are overall exactly $2^k$ sequences. $\blacksquare$
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gghx
1069 posts
#4
Y by
Rewrite as $f$. Obviously $f(n)\le f(n+1)\le f(n)+1$. Note that if $f(n+1)=f(n)$ then $f(n+1+k)=f(n+k)$, while if $f(n+1)=f(n)+1$ then $f(n+1+k)=f(n+k)$. Hence the function is determined by $f(1),f(2),\cdots f(k+1)$ alone. Conversely, if there are $f(1)$ distinct elements amongst them, and $f(i)\le f(i+1)\le f(i)+1$ for all $1\le i\le k$, then we can construct such a function. Suppose $f(1)=t$, then we need $t$ "+1"s between adjacent numbers in $f(1)$ to $f(k+1)$. There are $\binom{k}{t}$ ways to do this. Summing over all $t$, this gives $2^k$.
This post has been edited 1 time. Last edited by gghx, Mar 5, 2023, 1:57 PM
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kamatadu
466 posts
#5 • 2 Y
Y by HoripodoKrishno, whencence
Here is a solution written with every possible motivation I had. :P Also, I just totally loved this problem! :D

The answer if $f(k)=2^k$.
Note that $a_n\le a_{n+1}\le a_n+1$ as we are just adding one more term into the previous set which is either a new term, or one of the pre-existing ones.

Now note that $a_i$ is the number of distinct elements in\[a_1,a_2,\ldots,a_{k+i}\]and that $a_{i+1}$ is the number of distinct elements in\[a_1,a_2,\ldots,a_{k+i},a_{k+i+1}.\]So if $a_{i+1}=a_i$, then both the sequences have the same number of distinct elements, which means $a_{k+i+1}\in\left\{a_1,a_2,\ldots,a_{k+i}\right\}$. Now using the fact that $a_{k+i+1}in{a_{k+i},a_{k+i}+1}$ we get that this forces $a_{k+i+1}=a_{k+i}$. Similarly, if $a_{i+1}=a_i+1$, then we can get $a_{k+i+1}=a_{k+i}+1$.

Now let $+$ denote the operation $a_i\rightarrow a_{i+1}$ when $a_{i+1}=a_i+1$ and $\square$ denote the operation $a_i\rightarrow a_{i+1}$ when $a_{i+1}=a_i$. Due to what we derived above, we get that the operation of $a_{k+i}\rightarrow a_{k+i+1}$ is same as that of the operation $a_i\rightarrow a_{i+1}$. This gives us that if we determine the order of operations on the sequence $\left\{a_1,a_2,\ldots, a_{k+1}\right\}$, then we automatically get the order of operations on the entire sequence which is described as below.
\begin{align*}
    a_1&\rightarrow a_2\rightarrow\cdots\rightarrow a_{k+1}\\
    a_{k+1}&\rightarrow a_{k+2}\rightarrow\cdots\rightarrow a_{2k+1}
.\end{align*}
So we just need to determine the sequence $\left\{a_1,a_2,\ldots, a_{k+1}\right\}$. Let there be $m$ many $+$ operations, and thus $k-m$ many $\square$ operations in the sequence of first $k+1$ numbers; where $0\le m\le k$. So we get that there are $m+1$ distinct elements in the sequence. But we also had that this value equals $a_1$, and so $a_1=m+1$. Thus after picking the ordering of the $+$ and the $\square$ operations, we get the sequence from $a_1$ to $a_{k+1}$, which on extending further gives us our entire sequence.

Now to finish, we have can select a value for $m$ from $0\le m\le k$, and then, we will have $\binom{k}{m}$ ways to choose the positions of the $+$, which sets the position for the rest of the $\square$ operations too. So we get that our final answer is $f(k)=\displaystyle\sum_{m=0}^k \binom{k}{m}=2^k$ and we are done. :stretcher:
This post has been edited 1 time. Last edited by kamatadu, Jul 22, 2023, 1:39 PM
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HoRI_DA_GRe8
584 posts
#7
Y by
Solution
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rama1728
800 posts
#8
Y by
Wow! Neat bijection. I guess Im getting good C practice before RMO :P

The answer is $2^k$. First note that $a_i$ is a non-decreasing sequence due to it's definition. So, $a_{i+1}-a_i=0,1$ if $a_{k+i+1}-a_{k+i}=0,1$ and so if we define $d_i=a_{i+1}-a_i$, then $d_i$ is periodic with period $k$. Now what matters is the choice of $d_1,\hdots , d_k$, which can be chosen in $2^k$ ways, so we are done.

Remark. Many times, what is important with such problems which create strong 'dependencies' is to utilize those 'dependencies' and identify uniquely determined objects. The problem then would be very easy.
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Pyramix
419 posts
#9
Y by
We claim that there are exactly $2^k$ such sequences for any $k$.

Claim: For any positive integer $n$, we have $a_{n+k+1}-a_{n+k}=a_{n+1}-a_n$.
Proof. By definition, we have $0\leq a_{n+1}-a_n\leq 1$. Note that if $a_{n+k+1}=a_{n+k}$ then $a_{n+1}=a_n$, and if $a_{n+k+1}\ne a_{n+k}$, then their difference is 1, but then $a_{n+1}-a_n=1$. So, $a_{n+k+1}-a_{n+k}=a_{n+1}-a_n$. $\blacksquare$
A simple induction gives $a_{n+tk+1}-a_{n+tk}=a_{n+1}-a_n$ for any $t$. So, we really just need to ensure that $a_1$ is the number of distinct elements in the set $\{a_1,a_2,\ldots,a_{1+k}\}$. In this set, call a number $1\leq i\leq k$ good if $a_{i+1}=a_i+1$ and bad otherwise. We have that $a_1=1+\#(\text{good numbers between }1\text{ and }k\text{ inclusive})$.
This can be done very easily, let there be $g$ good numbers from $1,2,\ldots,k$. Then, $a_1=1+g$ and choose $g$ good numbers in $\binom{n}{g}$ ways. Hence,
\[\#(k-\text{pop sequences})=\sum_{g=0}^{k}\binom kg=2^k,\]as claimed.
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AshAuktober
905 posts
#10
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Note that $a_{i+1} - a_i \in \{0, 1 \} \forall i$.
Further, any $k$-pop sequence must satisfy $a_{i+1} = a_i \iff a_{i+k+1} = a_{i+k}$. And conversely, all sequences satisfying this and the fact that $a_1 = |\{a_1, \dots, a_{k+1}\}|$ work.
So now we choose $a_{i+1} - a_i$ for $i = 1, 2, \dots, k$ out of $\{0, 1\}$ to get a working $k$-pop sequence. Since there are $2^k$ ways to do this, the final answer is in fact $\boxed{2^k}$.

EDIT: WAIT POST #700 LESGOOOOOOOO
This post has been edited 1 time. Last edited by AshAuktober, Dec 17, 2024, 3:11 PM
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alexanderhamilton124
374 posts
#11 • 1 Y
Y by L13832
This is a really nice problem :)

Claim 1: $a_i \leq a_{i + 1} \leq a_i + 1$
Observe that $a_i$ is the number of distinct elements in $\{a_1, \dots, a_{k + i}\}$, while $a_{i + 1}$ is the number of distinct elements in $\{a_1, \dots, a_{k + i}, a_{k + i + 1}\}$. Note that $a_{k + i + 1}$ can either add one distinct element or the number of distinct elements remain the same, so our claim is proved.

Claim 2: The sequence is dependent on the selection of $\{a_1, \dots, a_{k + 1}\}$.
Again, $a_i$ is the number of distinct elements in $\{a_1, \dots, a_{k + i}\}$, while $a_{i + 1}$ is the number of distinct elements in $\{a_1, \dots, a_{k + i}, a_{k + i + 1}$. If $a_{i + 1} = a_i$, then $a_{k + i + 1}$ must not add any distinct element, or $a_{k + i + 1} = a_{k + i}$. If $a_{i + 1} = a_i + 1$, then $a_{k + i + 1}$ must add a distinct element, or $a_{k + i + 1} = a_{k + i} + 1$.

This proves our claim. To finish, note that for each $a_j$ where $j \leq k$, we can either choose $a_{j + 1} = a_j$ or $a_{j + 1} = a_j + 1$, and $a_1$ is merely the number of pairs $(a_i, a_{i + 1}) + 1$, where $a_{i + 1} = a_i + 1$. So there will be $2^k$ selections (dependent on $a_1$) for $(a_1, a_2, \dots, a_{k + 1}$, which means there are $2^k$ sequences.
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