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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
A property of divisors
rightways   8
N 14 minutes ago by de-Kirschbaum
Source: Kazakhstan NMO 2016, P1
Prove that one can arrange all positive divisors of any given positive integer around a circle so that for any two neighboring numbers one is divisible by another.
8 replies
rightways
Mar 17, 2016
de-Kirschbaum
14 minutes ago
Nice problem
hanzo.ei   1
N 40 minutes ago by Mathzeus1024
Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that
\[
f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1,
\quad \forall x, y \in \mathbb{R}.
\]
1 reply
+1 w
hanzo.ei
2 hours ago
Mathzeus1024
40 minutes ago
ortho conf DEF, radius MD, intersect ME,MF, collinear H,K,L
star-1ord   0
an hour ago
Source: Estonia Final Round 2025 12-3
Let $ABC$ be an acute-angled triangle with $|AB|<|AC|$. The altitudes $AD,BE$ and $CF$ intersect at $H$. Let $M$ be the midpoint of $BC$. Point $K$ is chosen on the extension of $EM$ beyond $M$ and point $L$ is chosen on the segment $FM$ such that $|MK|=|ML|=|MD|$. Prove that points $K, L$ and $H$ are collinear.

a little harder version
0 replies
star-1ord
an hour ago
0 replies
Funny system of equations in three variables
Tintarn   10
N 2 hours ago by Marcus_Zhang
Source: Baltic Way 2020, Problem 5
Find all real numbers $x,y,z$ so that
\begin{align*}
    x^2 y + y^2 z + z^2 &= 0 \\
    z^3 + z^2 y + z y^3 + x^2 y &= \frac{1}{4}(x^4 + y^4).
\end{align*}
10 replies
Tintarn
Nov 14, 2020
Marcus_Zhang
2 hours ago
a^{2m}+a^{n}+1 is perfect square
kmh1   1
N 2 hours ago by kmh1
Source: own
Find all positive integer triplets $(a,m,n)$ such that $2m>n$ and $a^{2m}+a^{n}+1$ is a perfect square.
1 reply
kmh1
Mar 20, 2025
kmh1
2 hours ago
Interesting problem
deraxenrovalo   0
2 hours ago
Given $\triangle$$ABC$ with circumcenter $O$$.\;$Let $P$ be an arbitrary point on $(BOC)$ such that $P$ is outside $(ABC)$$.\;$Let $Q$ be an arbitrary point on $(ABC)$$.\;$$AB$ cuts $(ACP)$ again at $E$ and $AC$ cuts $(ABP)$ again at $F$$.\;$The intersection of $BF$ and $CE$ is $R$$.\;$Let $X$ and $Y$ be the intersection of $EF$ with $(PQC)$ and $(PQR)$ respectively such that $X$, $Y$, $P$ are pairwise distinct.
Show that : $(APX)$, $(BPY)$, $(QPE)$ are coaxial circles

hint
0 replies
deraxenrovalo
2 hours ago
0 replies
Vieta Jumping Unsolved(Reposted)
Eagle116   0
2 hours ago
Source: MONT, Vieta Jumping part
The question is:
Let $x_1$, $x_2$, $\dots$, $x_n$ be $n$ integers. If $k>n$ is an integer, prove that the only solution to
$$x_1^2 + x_2^2 + \dots + x_n^2 = kx_1x_2\dots x_n $$is is $x_1 = x_2 = \dots = x_n = 0$.
0 replies
Eagle116
2 hours ago
0 replies
Geometry with parallel lines.
falantrng   32
N 2 hours ago by endless_abyss
Source: RMM 2018,D1 P1
Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .
32 replies
falantrng
Feb 24, 2018
endless_abyss
2 hours ago
sum divides n-th moment
navi_09220114   1
N 2 hours ago by ja.
Source: Own. Malaysian IMO TST 2025 P9
Given four distinct positive integers $a<b<c<d$ such that $\gcd(a,b,c,d)=1$, find the maximum possible number of integers $1\le n\le 2025$ such that $$a+b+c+d\mid a^n+b^n+c^n+d^n$$
Proposed by Ivan Chan Kai Chin
1 reply
navi_09220114
Yesterday at 1:07 PM
ja.
2 hours ago
Find all functions
Jackson0423   0
3 hours ago
Find all functions F:R->R such that
1/(F(F(x))-F(x))=F(x)
I know x+1/x works..
0 replies
Jackson0423
3 hours ago
0 replies
2x+1 is a perfect square but the following x+1 integers are not.
Sumgato   7
N 3 hours ago by Davut1102
Source: Spain Mathematical Olympiad 2018 P1
Find all positive integers $x$ such that $2x+1$ is a perfect square but none of the integers $2x+2, 2x+3, \ldots, 3x+2$ are perfect squares.
7 replies
Sumgato
Mar 17, 2018
Davut1102
3 hours ago
Prove that P1(x), P2(x) ,... Pn(x) = k has no root
truongphatt2668   2
N 3 hours ago by truongphatt2668
Let $n \in \mathbb{N}^*$ and $P_1(x),P_2(x), \ldots P_n(x) \in \mathbb{Z}[x]$ such that $\mathrm{deg} P_i = 2, \forall i = \overline{1,n}$. Prove that exists many $k \in \mathbb{N}$ such that every equation: $P_i(x) = k, \forall i = \overline{1,n}$ has no real roots
2 replies
truongphatt2668
Today at 2:26 AM
truongphatt2668
3 hours ago
Geo: incircle, escircle, isotomic conjugate
XAN4   1
N 3 hours ago by deraxenrovalo
Source: Own
For $\triangle{ABC}$, Its incircle $\odot I$ and $A-$escircle $\odot I_A$ are tangent to $BC$ at $D$ and $E$ respectively. $AI$ intersects line $BC$ at $J$. Line $AD$ intersects $\odot I$ at $F$, and line $AE$ intersects $\odot I_A$ at $G$. Line $FG$ intersects $BC$ at $H$. Prove that $BJ=CH$.
1 reply
XAN4
Mar 19, 2025
deraxenrovalo
3 hours ago
special sets
ChubbyTomato426   0
3 hours ago
Let $n$ be a positive integer. A subset $\{a, b, c, d\} \subseteq \{1, 2, . . . , 4n\}$ with four distinct elements is special if there exists a rearrangement $(x, y, z, w)$ of $(a, b, c, d)$ such that $xy -zw = 1$. Prove that the set $\{1, 2, . . . , 4n \}$ cannot be partitioned into $n$ special disjoint sets.
0 replies
ChubbyTomato426
3 hours ago
0 replies
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
587 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
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#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
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#11 • 1 Y
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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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