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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
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in n^2-9 has 6 positive divisors than GCD (n-3, n+3)=1
parmenides51   9
N 5 minutes ago by megahertz13
Source: Greece JBMO TST 2016 p3
Positive integer $n$ is such that number $n^2-9$ has exactly $6$ positive divisors. Prove that GCD $(n-3, n+3)=1$
9 replies
parmenides51
Apr 29, 2019
megahertz13
5 minutes ago
J
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two subsets with no fewer than four common elements.
micliva   43
N 8 minutes ago by mudkip42
Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
43 replies
1 viewing
micliva
Apr 18, 2013
mudkip42
8 minutes ago
J
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Incenter and midpoint geom
sarjinius   95
N 8 minutes ago by Turkish_sniper
Source: 2024 IMO Problem 4
Let $ABC$ be a triangle with $AB < AC < BC$. Let the incenter and incircle of triangle $ABC$ be $I$ and $\omega$, respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ at $P \ne A$. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.
Prove that $\angle KIL + \angle YPX = 180^{\circ}$.

Proposed by Dominik Burek, Poland
95 replies
sarjinius
Jul 17, 2024
Turkish_sniper
8 minutes ago
J
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Your average linear recurrence relation question
NamelyOrange   1
N 10 minutes ago by NamelyOrange
Source: 2022 National Taiwan University STEM Development Program Admissions Test, P4
For positive integer $n$, define $a_n = (3+\sqrt{10})^n+(3-\sqrt{10})^n$.

(a) The constants $A$ and $B$ satisfy $a_{n+1} = Aa_n+Ba_{n-1}$ for all positive integer $n$. Find the value of these constants.
(b) Prove with induction that $a_n$ is an integer for all positive integer $n$.
(c) Prove that $n$ and $\lfloor (3+\sqrt{10})^n\rfloor$ have opposite parities for all integer $n$.
1 reply
NamelyOrange
Today at 4:01 AM
NamelyOrange
10 minutes ago
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No more topics!
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Easy combinatorics
Sadigly   5
N May 11, 2025 by GreekIdiot
Source: Azerbaijan Junior NMO 2018
First $20$ positive integers are written on a board. It is known that, after you erase a number from the board, there exists a number that is equal to the arithmetic mean of the rest of the numbers left on the board. Find all the numbers that could've been erased.
5 replies
Sadigly
May 11, 2025
GreekIdiot
May 11, 2025
J
Easy combinatorics
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Reveal topic tags
combinatorics
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Source: Azerbaijan Junior NMO 2018
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Sadigly
228 posts
Sadigly
#1 h  May 11, 2025, 9:23 PM
Y by
First $20$ positive integers are written on a board. It is known that, after you erase a number from the board, there exists a number that is equal to the arithmetic mean of the rest of the numbers left on the board. Find all the numbers that could've been erased.
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GreekIdiot
322 posts
GreekIdiot
#2 h  May 11, 2025, 9:28 PM
Y by
their sum before erasing equals $210$ and for the arithmetic mean to exist then if we let $x$ be the number erased we get that $19 \mid 210-x$
meaning $x \in \{1,20 \}$.
Z K Y
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Sadigly
228 posts
Sadigly
#3 h  May 11, 2025, 9:36 PM
Y by
GreekIdiot wrote:
their sum before erasing equals $210$ and for the arithmetic mean to exist then if we let $x$ be the number erased we get that $19 \mid 210-x$
meaning $x \in \{1,20 \}$.

I think you got the problem wrong. Let the erased number be $5$. Then, if we can choose a number satisfying the given condition (let this number be $n$), then we should have $\frac{210-5-n}{18}=n$
This post has been edited 1 time. Last edited by Sadigly, May 11, 2025, 9:36 PM
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GreekIdiot
322 posts
GreekIdiot
#4 h  May 11, 2025, 9:45 PM
Y by
From what I understand, the numbers from 1 to 20 are written, then we erase one of them and the arithmetic mean of the numbers must be one of the numbers left on the board. There are 19 numbers left, no?
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Sadigly
228 posts
Sadigly
#5 h  May 11, 2025, 10:03 PM
Y by
GreekIdiot wrote:
From what I understand, the numbers from 1 to 20 are written, then we erase one of them and the arithmetic mean of the numbers must be one of the numbers left on the board. There are 19 numbers left, no?

You deleted a number $a$, so 19 numbers are left

And you take a number $n$, and the arithmetic mean of the rest of the numbers (All 20 numbers except $a$ and $n$) is equal to $n$
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GreekIdiot
322 posts
GreekIdiot
#6 h  May 11, 2025, 10:08 PM
Y by
Oh wording was kinda weird...
Anyways then $210-x=19n$ meaning $19 \mid 210-x$ and finish as above
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