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a My Retirement & New Leadership at AoPS
rrusczyk   888
N a minute ago by FlyingUFO11
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
888 replies
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rrusczyk
Yesterday at 6:37 PM
FlyingUFO11
a minute ago
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
Product of f(m) multiple of odd integers
buzzychaoz   23
N an hour ago by john0512
Source: China Team Selection Test 2016 Test 2 Day 2 Q4
Set positive integer $m=2^k\cdot t$, where $k$ is a non-negative integer, $t$ is an odd number, and let $f(m)=t^{1-k}$. Prove that for any positive integer $n$ and for any positive odd number $a\le n$, $\prod_{m=1}^n f(m)$ is a multiple of $a$.
23 replies
buzzychaoz
Mar 21, 2016
john0512
an hour ago
God do bosses have a hard job
AshAuktober   1
N an hour ago by tyrantfire4
Source: 05JPN5 (couldn't find on search function)
The boss has to assign ten job positions to ten candidates, considering two parameters: preference and ability. If candidate A prefers job $v$ to job $u$ and has a better ability in job $v$ than candidate B, but A is assigned job $u$ and B is assigned job $v$, then A will complain. Also, if it is possible to assign each job to a candidate with a higher ability, the director will complain. Show that the boss can assign the jobs so as to avoid any complaints.
1 reply
1 viewing
AshAuktober
an hour ago
tyrantfire4
an hour ago
Nordic 2025 P2
anirbanbz   5
N an hour ago by KAME06
Source: Nordic 2025
Let $p$ be a prime and suppose $2^{2p} \equiv 1 (\text{mod}$ $ 2p+1)$ is prime. Prove that $2p+1$ is prime$^{1}$

$^{1}$This is a special case of Pocklington's theorem. A proof of this special case is required.
5 replies
anirbanbz
5 hours ago
KAME06
an hour ago
Nice problem
hanzo.ei   6
N an hour ago by Ianis

Given two sequences $(a_n)$ and $(b_n)$ satisfying $(a_n + b_n)a_n \neq 0$ for all $n$, and both series
\[
\sum \frac{a_n}{b_n}, \quad \sum \frac{b_n}{a_n}
\]are convergent. Prove that the series
\[
\sum \frac{a_n}{a_n + b_n}
\]also converges.
6 replies
hanzo.ei
6 hours ago
Ianis
an hour ago
No more topics!
funny title
nguyenvana   1
N Mar 22, 2025 by pco
Source: no from book
Find all the functions f: R+ to R+ which satisfy the functional equation:
f(2f(x)+f(y))=2x+y (x,y R+)
1 reply
nguyenvana
Mar 22, 2025
pco
Mar 22, 2025
funny title
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Source: no from book
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nguyenvana
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Y by
Find all the functions f: R+ to R+ which satisfy the functional equation:
f(2f(x)+f(y))=2x+y (x,y R+)
This post has been edited 1 time. Last edited by nguyenvana, Mar 23, 2025, 8:53 AM
Reason: khác
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pco
23474 posts
#2 • 1 Y
Y by nguyenvana
nguyenvana wrote:
Find all the functions f: R+ to R+ which satisfy the functional equation:
f(2f(x)+f(y)+xy)=xy+2x+y (x,y R+)
Let $P(x,y)$ be assertion $f(2f(x)+f(y)+xy)=xy+2x+y$

$f(x)$ is obviously surjective
If $f(a)=f(b)$, comparaison of $P(a,b)$ with $P(b,a)$ implies $a=b$ and so $f(x)$ is injective.


$P(2,x)$ $\implies$ $f(2f(2)+f(x)+2x)=3x+4$
$P(x+1,1)$ $\implies$ $f(2f(x+1)+f(1)+x+1)=3x+4$
And so, since injective : $2f(x+1)=f(x)+x+2f(2)-f(1)-1$, and so $f(x+1)=\frac{f(x)+x}2+a$ for some constant $a$
So $f(x+2)=\frac{f(x+1)+x+1}2+a$ $=\frac{f(x)+3x}4+\frac{3a+1}2$

$P(x,2)$ $\implies$ $f(2f(x)+2x+f(2))=4x+2$
$P(1,2x)$ $\implies$ $f(f(2x)+2x+2f(1))=4x+2$

And so, since injective : $f(2x)+2x+2f(1)=2f(x)+2x+f(2)$ and so $f(2x)=2f(x)+b$ for some constant $b$

So $f(2(x+1))=2f(x+1)+b$ $=f(x)+x+2a+b$
But $f(2(x+1))=f(2x+2)=\frac{f(2x)+6x}4+\frac{3a+1}2$ $=\frac{2f(x)+b+6x}4+\frac{3a+1}2$
And so $f(x)+x+2a+b=\frac{2f(x)+b+6x}4+\frac{3a+1}2$
And so $f(x)=x+1-a-\frac{3b}2$ $\forall x>0$

Plugging $f(x)=x+u$ in original equation, we get $u=0$ and
$\boxed{f(x)=x\quad\forall x>0}$, which indeed fits
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