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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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
Brilliant guessing game on triples
Assassino9931   6
N 17 minutes ago by sami1618
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
6 replies
Assassino9931
May 10, 2025
sami1618
17 minutes ago
Inequality
VicKmath7   18
N 20 minutes ago by lpieleanu
Source: Serbia JBMO TST 2020 P3
Given are real numbers $a_1, a_2,...,a_{101}$ from the interval $[-2,10]$ such that their sum is $0$. Prove that the sum of their squares is smaller than $2020$.
18 replies
VicKmath7
Sep 5, 2020
lpieleanu
20 minutes ago
CGMO6: Airline companies and cities
v_Enhance   17
N 26 minutes ago by eg4334
Source: 2012 China Girl's Mathematical Olympiad
There are $n$ cities, $2$ airline companies in a country. Between any two cities, there is exactly one $2$-way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of $n$.
17 replies
v_Enhance
Aug 13, 2012
eg4334
26 minutes ago
Every subset of size k has sum at most N/2
orl   52
N 27 minutes ago by lpieleanu
Source: USAMO 2006, Problem 2, proposed by Dick Gibbs
For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\tfrac{N}{2}.$
52 replies
+1 w
orl
Apr 20, 2006
lpieleanu
27 minutes ago
$$p_1 = 1, q_1 = 1, p_{n + 1} = 2q_n^2-p_n^2$$
sharifymatholympiad   4
N 34 minutes ago by Maths_Girl
Source: 2016 239 S4
The sequences of natural numbers $p_n$ and $q_n$ are given such that
$$p_1 = 1,\ q_1 = 1,\ p_{n + 1} = 2q_n^2-p_n^2,\  q_{n + 1} = 2q_n^2+p_n^2 $$Prove that $p_n$ and $q_m$ are coprime for any m and n.
4 replies
sharifymatholympiad
Oct 11, 2020
Maths_Girl
34 minutes ago
2024 IMO P1
EthanWYX2009   110
N 43 minutes ago by Just1
Source: 2024 IMO P1
Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$)

Proposed by Santiago Rodríguez, Colombia
110 replies
EthanWYX2009
Jul 16, 2024
Just1
43 minutes ago
Hard version of 2024 CSMC-3
EthanWYX2009   1
N 44 minutes ago by Phorphyrion
Source: 2023 December 谜之竞赛-6, also from an essay by Szemeredi
Determine the smallest positive real number \( \alpha \) such that there exists a constant \( c \) satisfying: For any prime \( p \) and any \( k \geq cp^\alpha \) distinct integers \( a_1, \cdots, a_k \) modulo \( p \), there exists a non-empty subset \( I \subseteq \{1, 2, \cdots, k\} \) such that
\[\sum_{i \in I} a_i \equiv 0 \pmod{p}.\]Proposed by Hanqing Huang, Peking University
1 reply
EthanWYX2009
Jul 14, 2025
Phorphyrion
44 minutes ago
Divisibility of a triple
goodar2006   55
N 44 minutes ago by Ilikeminecraft
Source: Iran TST 2013-First exam-2nd day-P5
Do there exist natural numbers $a, b$ and $c$ such that $a^2+b^2+c^2$ is divisible by $2013(ab+bc+ca)$?

Proposed by Mahan Malihi
55 replies
goodar2006
Apr 19, 2013
Ilikeminecraft
44 minutes ago
Two variables
Nguyenhuyen_AG   1
N an hour ago by RagvaloD
Let $x,\,y$ be non-negative real numbers. Prove that
\[4\left(\frac{1}{2x + y} + \frac{1}{2y + x}\right) \ge \frac{3}{x + y} + \frac{21(x + y)}{7x^2 + 22xy + 7y^2}.\]
1 reply
Nguyenhuyen_AG
Today at 2:26 PM
RagvaloD
an hour ago
3 var with parameter (on GMA 575)
mihaig   0
an hour ago
Source: VL
Find the least constant $K$ such that
$$a+b+c+K\cdot\frac{\left(a-c\right)^2}{a+c}\geq\sqrt{3\left(a^2+b^2+c^2\right)}$$for all $a\ge b\ge c\ge0,$ with $a>0.$
0 replies
mihaig
an hour ago
0 replies
Long game to go..
IndoMathXdZ   38
N an hour ago by cubres
Source: IMO SL 2018 C3
Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n + 1$ squares in a row, numbered $0$ to $n$ from left to right. Initially, $n$ stones are put into square $0$, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of these stones and moves it to the right by at most $k$ squares (the stone should say within the board). Sisyphus' aim is to move all $n$ stones to square $n$.
Prove that Sisyphus cannot reach the aim in less than
\[ \left \lceil \frac{n}{1} \right \rceil + \left \lceil \frac{n}{2} \right \rceil + \left \lceil \frac{n}{3} \right \rceil + \dots + \left \lceil \frac{n}{n} \right \rceil \]turns. (As usual, $\lceil x \rceil$ stands for the least integer not smaller than $x$. )
38 replies
IndoMathXdZ
Jul 17, 2019
cubres
an hour ago
complex numbers
EthanWYX2009   1
N an hour ago by alexheinis
Source: 2025 July 谜之竞赛 First Round-11
Let \( z_1, z_2\in\mathbb C \) satisfy \( |z_1 + 2z_2| \le 1 \) and \( |z_1^2+z_1z_2 + z_2^2| \le 1 \). Determine the maximum value of \( \max\{|z_1|, |z_2|\} \).

Proposed by Yiyan Lin
1 reply
EthanWYX2009
Today at 12:30 AM
alexheinis
an hour ago
Intersection on the circumcircle
jayme   2
N an hour ago by awesomeming327.
Dear Mathlinkers,

1. ABC a A-isoceles triangle
2. (O) the circumcircle
3. D a point on the segment AC
4. E the second point of intersection of (O) and BD
5. F the point of intersection of the tangent to (O) at E and AB
6. S le second point d'intersection de 0 avec (FC)
7. G, H the points of intersection of FD and BC, FC and GE.

Prove : EH and SD intersect on (O).

Sincerely
Jean-Louis

2 replies
jayme
Today at 2:23 PM
awesomeming327.
an hour ago
MKACRI UNSOLVED FE ON R+??!!
MR.1   3
N 2 hours ago by jasperE3
Source: own
find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$f(f(x))=x$ and
$f(x)+x$ is surjective on $ (0, \infty)$ for all $x, \in \mathbb{R}^{+}$
3 replies
MR.1
Jul 19, 2025
jasperE3
2 hours ago
combinatorics problem
henderson   6
N May 31, 2025 by aokmh3n2i2rt
$65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$
6 replies
henderson
Jan 23, 2016
aokmh3n2i2rt
May 31, 2025
combinatorics problem
G H J
G H BBookmark kLocked kLocked NReply
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henderson
312 posts
#1 • 1 Y
Y by Adventure10
$65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$
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jlammy
1099 posts
#2 • 2 Y
Y by Adventure10, Mango247
There are $\tbinom{65}{2}=2080$ possible sums, so we can find two such that $a+b\equiv c+d\pmod{2016}$.
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mjuk
196 posts
#3 • 1 Y
Y by Adventure10
henderson wrote:
$65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$

Do $a,b,c,d$ have to be different?
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henderson
312 posts
#4 • 2 Y
Y by Adventure10, Mango247
mjuk wrote:
henderson wrote:
$65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$

Do $a,b,c,d$ have to be different?

yes, they are distinct numbers.
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AopsUser101
1750 posts
#5 • 2 Y
Y by v4913, Adventure10
henderson wrote:
$65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$

SOURCE:
Azerbaijan Junior Mathematical Olympiad 2016

SOLUTION
Note that we only care about the numbers' remainders when divided by $2016$, so we can replace the $65$ natural numbers with numbers in the range $0$ to $2016$.
We can make the condition $a+b-c-d$ being divisible by $2016$ instead that $a+b-c-d = 0 \implies a+b=c+d$. Then, there are $2016 \cdot 2 = 4032$ possible sums (anything from $0$ to $4032$ inclusive). If two of the $65$ numbers have the same sum, then the problem is solved, so assume that no two of the $65$ numbers have the same sum. Then, there are $\binom{65}{2} = 4160$ possible sums. By the pigeonhole principle, we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016$.
This post has been edited 2 times. Last edited by AopsUser101, Oct 28, 2019, 4:00 AM
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wizixez
168 posts
#6
Y by
$\binom{65}{2} >2016$
This finishes the problem where $\binom{65}{2} $ equals to the number of all possible sums so we can always find such pairs
This post has been edited 1 time. Last edited by wizixez, Nov 15, 2024, 2:04 PM
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aokmh3n2i2rt
2 posts
#7
Y by
Let the $65$ natural numbers be $a_1,a_2, \dots , a_{65}$. For pairs $(a_i,a_j) ,i<j$, there are $\displaystyle \binom{65}{2}=2080$ possible pairs. By the Pigeonhole Principle, there exist
\[ \bigg\lfloor \dfrac{2080-1}{2016}\bigg\rfloor+1=2\]pairs with the same remainder when divided by 2016. That is,
\[ a_i+a_j \equiv a_m +a_k \pmod{2016}\]If $ \{a_i,a_j\} \ne \{a_k,a_m\}$ the proof is complete.
If $a_i=a_m$, then $ a_j \equiv a_k \pmod{2016}$. This implies $\{ a_j,a_k\} =\{0,2016\}$.
Among the $65$ numbers $a_1, \dots , a_{65}$, remove $2016$ (assume$ a_{65}=2016$).
The remaing $64$ numbers $a_1, \dots , a_{64}$ yield $\displaystyle \binom{64}{2}=2016$ pair
$\bullet$ If among these 2016 sums, two have the same remainder modulo, then:
\[ a_u+a_v \equiv a_r +a_t \pmod{2016}\]If $a_u=a_r$, then $a_v \equiv a_t \pmod{2016}$, which is impossible since $0 \le a_v, a_t <2015$ and all number are distinct.
Thus $ \{a_u,a_v\} \ne \{ a_r,a_t\}$, proving the claim.
$\bullet$ If all $2016$ sums have distinct remainders modulo $2016$, they cover all possible residues. Hence, there exists a pair $(a_p,a_q)$ such that:
\[ a_p+a_q \equiv 0 \pmod{2016}\]If $a_p=0$, then $a_q \equiv 0 \pmod{2016}$, which contradicts $ a_q \ne 0,2016$.
Therefore, $ \{a_p,a_q\} \ne 0$. Selecting the four distinct numbers $ a_p,a_q,0,2016$ gives:
\[ a_p +a_q \equiv 0 + 2016 \pmod{2016}\]
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