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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Painted cells
Titibuuu   0
3 minutes ago
Source: Mock 3

A \( 2021 \times 2021 \) grid has \( k \) cells painted such that the following condition holds:
For every painted cell, at least one of its \emph{vertices} is also a vertex of \emph{another} painted cell.
Find the maximum possible value of \( k \).
0 replies
Titibuuu
3 minutes ago
0 replies
junior perpenicularity, 2 circles related
parmenides51   3
N 5 minutes ago by LeYohan
Source: Greece Junior Math Olympiad 2024 p2
Consider an acute triangle $ABC$ and it's circumcircle $\omega$. With center $A$, we construct a circle $\gamma$ that intersects arc $AB$ of circle $\omega$ , that doesn't contain $C$, at point $D$ and arc $AC$ , that doesn't contain $B$, at point $E$. Suppose that the intersection point $K$ of lines $BE$ and $CD$ lies on circle $\gamma$. Prove that line $AK$ is perpendicular on line $BC$.
3 replies
parmenides51
Mar 2, 2024
LeYohan
5 minutes ago
Sequence
Titibuuu   0
6 minutes ago
Source: Mock P2

Let \( a_1 = a \), and for all \( n \geq 1 \), define the sequence \( \{a_n\} \) by the recurrence
\[
a_{n+1} = a_n^2 + 1
\]Prove that there is no natural number \( n \) such that
\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]is a perfect square.
0 replies
Titibuuu
6 minutes ago
0 replies
Coolabra
Titibuuu   0
7 minutes ago
Source: Mock P1
Let \( a, b, c \) be distinct real numbers such that
\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]Find the maximum possible value of \( a \).
0 replies
Titibuuu
7 minutes ago
0 replies
Inspired by Bet667
sqing   5
N 9 minutes ago by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
5 replies
sqing
Thursday at 1:03 PM
sqing
9 minutes ago
A tangent problem
hn111009   0
19 minutes ago
Source: Own
Let quadrilateral $ABCD$ with $P$ be the intersection of $AC$ and $BD.$ Let $\odot(APD)$ meet again $\odot(BPC)$ at $Q.$ Called $M$ be the midpoint of $BD.$ Assume that $\angle{DPQ}=\angle{CPM}.$ Prove that $AB$ is the tangent of $\odot(APD)$ and $BC$ is the tangent of $\odot(AQB).$
0 replies
hn111009
19 minutes ago
0 replies
3-var inequality
sqing   4
N 21 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
4 replies
sqing
May 7, 2025
sqing
21 minutes ago
Inspired by Kosovo 2010
sqing   2
N 23 minutes ago by sqing
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
2 replies
sqing
Yesterday at 3:56 AM
sqing
23 minutes ago
Find all real numbers
sqing   6
N 24 minutes ago by sqing
Source: IMOC 2021 A1
Find all real numbers x that satisfies$$\sqrt{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}-\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}}+\sqrt{1-\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}}=x.$$2021 IMOC Problems
6 replies
sqing
Aug 11, 2021
sqing
24 minutes ago
I accidentally drew a 200-gon and ran out of time
justin1228   24
N 27 minutes ago by Ilikeminecraft
Source: USEMO P.5 2020
The sides of a convex $200$-gon $A_1 A_2 \dots A_{200}$ are colored red and blue in an alternating fashion.
Suppose the extensions of the red sides determine a regular $100$-gon, as do the extensions of the blue sides.

Prove that the $50$ diagonals $\overline{A_1A_{101}},\ \overline{A_3A_{103}},\ \dots,
\ \overline{A_{99}A_{199}}$ are concurrent.

Proposed by: Ankan Bhattacharya
24 replies
justin1228
Oct 25, 2020
Ilikeminecraft
27 minutes ago
IMO 2016 Problem 4
termas   54
N 35 minutes ago by OronSH
Source: IMO 2016 (day 2)
A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let $P(n)=n^2+n+1$. What is the least possible positive integer value of $b$ such that there exists a non-negative integer $a$ for which the set $$\{P(a+1),P(a+2),\ldots,P(a+b)\}$$is fragrant?
54 replies
termas
Jul 12, 2016
OronSH
35 minutes ago
Number theory sequences : sums divisible by n
Muradjl   43
N 41 minutes ago by OronSH
Source: IMO Shortlist 2017 N3
Determine all integers $ n\geq 2$ having the following property: for any integers $a_1,a_2,\ldots, a_n$ whose sum is not divisible by $n$, there exists an index $1 \leq i \leq n$ such that none of the numbers $$a_i,a_i+a_{i+1},\ldots,a_i+a_{i+1}+\ldots+a_{i+n-1}$$is divisible by $n$. Here, we let $a_i=a_{i-n}$ when $i >n$.

Proposed by Warut Suksompong, Thailand
43 replies
Muradjl
Jul 10, 2018
OronSH
41 minutes ago
Rootiful sets
InternetPerson10   37
N an hour ago by OronSH
Source: IMO 2019 SL N3
We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_0, a_1, \cdots, a_n \in S$, all integer roots of the polynomial $a_0+a_1x+\cdots+a_nx^n$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^a - 2^b$ for positive integers $a$ and $b$.
37 replies
InternetPerson10
Sep 22, 2020
OronSH
an hour ago
IMO 2020 Problem 5
rcorreaa   70
N an hour ago by OronSH
Source: IMO 2020 Problem 5
A deck of $n > 1$ cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.
For which $n$ does it follow that the numbers on the cards are all equal?

Proposed by Oleg Košik, Estonia
70 replies
rcorreaa
Sep 22, 2020
OronSH
an hour ago
Easy IMO 2023 NT
799786   133
N Apr 24, 2025 by Maximilian113
Source: IMO 2023 P1
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
133 replies
799786
Jul 8, 2023
Maximilian113
Apr 24, 2025
Easy IMO 2023 NT
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2023 P1
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799786
1052 posts
#1 • 7 Y
Y by mmkkll, feranjos, CrazyFok, Littlelame, emmelin, KhaiMathAddict, cubres
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
This post has been edited 7 times. Last edited by v_Enhance, Sep 18, 2023, 12:33 AM
Reason: minor latex pet peeve
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qwedsazxc
167 posts
#2 • 9 Y
Y by srijan30pq, math90, akliu, Mehrshad, combo_nt_lover, Math.1234, KhaiMathAddict, cubres, Want-to-study-in-NTU-MATH
Assume $p<q$ be the smallest two prime divisors of $n$. Then $d_{k-1}=\frac{n}{p}$.
Assume $d_m=\frac{n}{q}$ for some $m$, and $d_{m+1}=\frac{n}{p^{c+1}}$ and $d_{m+2}=\frac{n}{p^c}$ for some nonnegative integer $c$.
Then, since $d_m\mid d_{m+1}+d_{m+2}$, $p^{c+1}\mid q+pq$ and $p|q$ which is a contradiction.
Therefore $n$ does not have two distinct prime factors; $n$=$p^t$ for some prime $p$ and a positive integer $t\neq1$. It's easy to show that this suffices.
This post has been edited 4 times. Last edited by qwedsazxc, Jul 8, 2023, 1:46 PM
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Tintarn
9042 posts
#3 • 13 Y
Y by PNT, math90, IAmTheHazard, Assassino9931, combo_nt_lover, Joider, Math.1234, Jia_Le_Kong, dhfurir, Sedro, Acorn-SJ, aidan0626, H_Taken
Solution
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Seungjun_Lee
526 posts
#5 • 2 Y
Y by dgkim, Kingsbane2139
From $d_{k-2} \mid d_{k-1} + d_{k}$ one can get that $d_{k-2} \mid d_{k-1}$ since $d_{k-2} \mid n = d_k$
Then $d_2 = \frac{n}{d_{k-1}} \mid \frac{n}{d_{k-2}} = d_3$ so $d_2 \mid d_3$. Here, since $d_2 \mid d_3 + d_4$, we get that $d_2 \mid d_4$.
Again, we can easily show $d_{k-3} | d_{k-1}$ in the same way, which leads to $d_{k-3} \mid d_{k-2}$

By induction, $d_1 \mid d_2 \mid \cdots \mid d_k$

If $p \mid n$ and $q \mid n$ for prime $p > q$
$\frac{n}{p} \mid \frac{n}{q}$ so $q \mid p$ and this is contradiction

Hence, $n = p^{k-1}$
This indeed fits the condition
This post has been edited 3 times. Last edited by Seungjun_Lee, Jun 24, 2024, 1:57 AM
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ihatemath123
3446 posts
#6 • 4 Y
Y by Inconsistent, centslordm, channing421, Zhaom
The answer is prime powers, which obviously work.

Note that the three largest divisors of $n$ are either $\left \{ \tfrac{n}{q}, \tfrac{n}{p}, n \right \}$ for distinct prime divisors $p$ and $q$ of $n$, or $\left \{ \tfrac{n}{p^2}, \tfrac{n}{p}, n \right \}$ for some prime divisor $p$ of $n$.

In the former case we have a contradiction, since
\[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]obviously not an integer.

So, the three largest divisors of $n$ are $\frac{n}{p^2}, \frac{n}{p}$ and $n$.

With similar reasoning, now consider the fourth largest divisor. It is either $\tfrac{n}{q}$ or $\tfrac{n}{p^3}$; the former option fails, since
\[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]obviously not an integer either.

We repeat this reasoning to deduce that $n$ is just $p^{k-1}$.
This post has been edited 4 times. Last edited by ihatemath123, Jul 18, 2024, 11:33 PM
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GrantStar
821 posts
#7 • 4 Y
Y by centslordm, jeff10, OronSH, MulhamAgam
HUH how is this IMO level?
sol
This post has been edited 1 time. Last edited by GrantStar, Jul 8, 2023, 5:27 AM
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bobthegod78
2982 posts
#8 • 1 Y
Y by centslordm
The answer is $p^e$ for all primes $p$ and $e>1$, which obviously works. Let $p$ be the smallest prime factor of $n$, we claim that $p$ is the only prime factor of $n$. We use induction downward to prove $d_i = \frac{n}{p^{k-i}}$. The base case is obvious. For the inductive step, if we instead had a different prime $q$, then
\[
\frac{n}{q} \mid \frac{n}{p^{k-i}} + \frac{n}{p^{k-i-1}},
\]but this is obviously false. We are done.
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LoloChen
479 posts
#9 • 1 Y
Y by GeoKing
This NT seems familiar. Maybe it's an old problem?
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carefully
240 posts
#10
Y by
LoloChen wrote:
This NT seems familiar. Maybe it's an old problem?
The only related problem I can think of is IMO 2002 P4, but it's not even close.
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Seungjun_Lee
526 posts
#11
Y by
LoloChen wrote:
This NT seems familiar. Maybe it's an old problem?

Maybe JBMO 2002 is similar with this problem
but this is even easier than that
https://artofproblemsolving.com/community/c6h58610p358021
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Seungjun_Lee
526 posts
#12 • 1 Y
Y by ihatemath123
ihatemath123 wrote:
The answer is prime powers, which obviously work.

Note that the three largest divisors of $n$ are either $\left \{ \frac{n}{q}, \frac{n}{p}, n \right \}$ for distinct prime divisors $p$ and $q$ of $n$, or $\left \{ \frac{n}{p^2}, \frac{n}{p}, n \right \}$ for some prime divisor $p$ of $n$.

In the former case we have a contradiction, since
\[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]obviously not an integer.

So, the three largest divisors of $n$ are $\frac{n}{p^2}, \frac{n}{p}$ and $n$.

With similar reasoning, now consider the fourth largest divisor. It is either $\frac{n}{q}$ or $\frac{n}{p^3}$; the former option fails, since
\[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]obviously not an integer either.

We repeat this reasoning to deduce that $n$ is just $p^k$.

I think you have a very small mistake
isn't it $p^{k-1}$??
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Anzoteh
126 posts
#13
Y by
I'll only show the more technical part below.

Partial solution
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beansenthusiast505
26 posts
#14 • 1 Y
Y by Tellocan
For any $n$, the 3 largest divisors are $n,\frac{n}{p},\frac{n}{q}$ for primes $p<q$ or $n,\frac{n}{p},\frac{n}{p^2}$.

However, first case doesn't work because $\frac{\frac{n}{p}+n}{\frac{n}{q}}=\frac{q}{p}+q$ which is not an integer.

Repeat for further cases. All should not work for $q$ as you will get $\frac{q(p+1)}{p^k}$ which are not integers. Thus only $p^{k-1}$ work


well there goes 10M N prediction more like 0M now
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Supertinito
45 posts
#15 • 38 Y
Y by Tintarn, JingheZhang, khan.academy, steppewolf, pieater314159, math90, oolite, Filipjack, eibc, TSandino, Creeper1612, ike.chen, TheStrayCat, Aryan-23, Sagnik123Biswas, leonhard8128, akliu, rg_ryse, spiritshine1234, LolitaLaLolita, Packito, GreatKillaOE, ihatemath123, EpicBird08, Madyyy, Alex-131, jestrada, Reakniseb, starchan, Sedro, antimonio, ATGY, khina, Marcus_Zhang, MELSSATIMOV40, aidan0626, NicoN9, SteppenWolfMath
This problem was proposed by me, Santiago Rodriguez from Colombia. I hope that you enjoy it. :)
This post has been edited 2 times. Last edited by Supertinito, Jul 16, 2023, 7:45 PM
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yofro
3151 posts
#16
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Far too easy for the IMO. It's just prime powers, which trivially work. Suppose $n$ isn't a prime power. Clearly the list of divisors goes $\left(1,\cdots,\frac{n}{q}, \frac{n}{p^k},\cdots,\frac{n}{p},n\right)$ for some primes $p,q$ and some $k\in\mathbb{N}$. Thus $\frac{n}{q}\mid \frac{n}{p^k}+\frac{n}{p^{k-1}}$ and hence
$$\frac{q(p+1)}{p^{k-1}}\in\mathbb{Z}$$Which is impossible unless $k=1$, and for $k=1$ we need $\frac{n+n/p}{n/q}$ to be an integer, which is impossible.
This post has been edited 1 time. Last edited by yofro, Jul 8, 2023, 6:15 AM
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