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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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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
something interesting...
SunnyEvan   0
an hour ago
Source: old result
Let $x$, $y$, $z$ be non-negative real numbers, no two of which are zero. Such that $ x+y+z=3.$ Prove that :
$$ \sum \frac{16(9-xyz)}{9(z+x)^2(y+3)^2} \geq \frac{2xyz}{\sum x^2} + \frac{\sum (z^2-x^2)(z^2-y^2)(z+x)(z+y)}{\sum(x+y)^3(y+z)^3} $$
0 replies
SunnyEvan
an hour ago
0 replies
Jbmo 2011 Problem 4
Eukleidis   13
N 2 hours ago by Adventure1000
Source: Jbmo 2011
Let $ABCD$ be a convex quadrilateral and points $E$ and $F$ on sides $AB,CD$ such that
\[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\]

If $S$ is the area of $AEFD$ show that ${S\leq\frac{AB\cdot CD+n(n-1)AD^2+n^2DA\cdot BC}{2n^2}}$
13 replies
Eukleidis
Jun 21, 2011
Adventure1000
2 hours ago
an algebra question
kjhgyuio   1
N 2 hours ago by aidan0626
.........
1 reply
kjhgyuio
2 hours ago
aidan0626
2 hours ago
we can find one pair of a boy and a girl
orl   16
N 2 hours ago by ezpotd
Source: Vietnam TST 2001 for the 42th IMO, problem 3
Some club has 42 members. It’s known that among 31 arbitrary club members, we can find one pair of a boy and a girl that they know each other. Show that from club members we can choose 12 pairs of knowing each other boys and girls.
16 replies
orl
Jun 26, 2005
ezpotd
2 hours ago
teleporting wizard starts on point (2017, 101), 4 moves
parmenides51   1
N 3 hours ago by jasperE3
Source: 2018 USAIMEO #2 p5 (Mock AIME -USAJMO) https://artofproblemsolving.com/community/c594864h1572209p9658908
A teleporting wizard starts on the point $(2017, 101)$ and can teleport to other Cartesian coordinates with only $1$ of $4$ moves: $(x, y) \to (x + y, y)$, $(x, y) \to (x - y, y)$ when $x > y$, $(x, y) \to (x, x + y)$, and $(x, y) \to (x, y - x)$ when $y > x$.

(a) Let $P(x)$ be any polynomial with positive integer coefficients that passes through $(0, 0)$. Show that for all such $P(x)$, there exists a unique point on the curve where the wizard can land on.

(b) For each $P(x)$, let $S$ be this unique point. Find the equation of the graph that contains all potential $S$.
1 reply
parmenides51
Nov 17, 2023
jasperE3
3 hours ago
IMO ShortList 2001, combinatorics problem 1
orl   33
N 3 hours ago by ihategeo_1969
Source: IMO ShortList 2001, combinatorics problem 1
Let $A = (a_1, a_2, \ldots, a_{2001})$ be a sequence of positive integers. Let $m$ be the number of 3-element subsequences $(a_i,a_j,a_k)$ with $1 \leq i < j < k \leq 2001$, such that $a_j = a_i + 1$ and $a_k = a_j + 1$. Considering all such sequences $A$, find the greatest value of $m$.
33 replies
orl
Sep 30, 2004
ihategeo_1969
3 hours ago
CooL geo
Pomegranat   0
3 hours ago
Source: Idk

In triangle \( ABC \), \( D \) is the midpoint of \( BC \). \( E \) is an arbitrary point on \( AC \). Let \( S \) be the intersection of \( AD \) and \( BE \). The line \( CS \) intersects with the circumcircle of \( ACD \), for the second time at \( K \). \( P \) is the circumcenter of triangle \( ABE \). Prove that \( PK \perp CK \).
0 replies
Pomegranat
3 hours ago
0 replies
Coefficient Problem
P162008   2
N 3 hours ago by cazanova19921
Consider the polynomial $g(x) = \prod_{i=1}^{7} \left(1 + x^{i!} + x^{2i!} + x^{3i!} + \cdots + x^{(i-1)i!} + x^{ii!}\right)$
Find the coefficient of $x^{2025}$ in the expansion of $g(x).$
2 replies
P162008
Yesterday at 12:16 PM
cazanova19921
3 hours ago
Shooting An Invisible Tank
Aryan27   0
3 hours ago
Source: 239 MO
An invisible tank is on a $100 \times 100 $ table. A cannon can fire at any $k$ cells of the board after that the tank will move to one of the adjacent cells (by side). Then the process is repeated. Find the smallest value of $k$ such that the cannon can definitely shoot the tank after some time.
0 replies
Aryan27
3 hours ago
0 replies
Showing Tangency
Itoz   1
N 4 hours ago by ja.
Source: Own
The circumcenter of $\triangle ABC$ is $O$. Line $AO$ meets line $BC$ at point $D$, and there is a point $E$ on $\odot(ABC)$ such that $AE \perp BC$. Line $DE$ intersects $\odot(ABC)$ at point $F$. The perpendicular bisector of line segment $BC$ intersects line $AB$ at point $K$, and line $AB$ intersects $\odot(CFK)$ at point $L$.

Prove that $\odot(AFL)$ is tangent to $\odot (OBC)$.
1 reply
Itoz
Yesterday at 1:57 PM
ja.
4 hours ago
USAJMO problem 3: Inequality
BOGTRO   104
N 5 hours ago by justaguy_69
Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
104 replies
BOGTRO
Apr 24, 2012
justaguy_69
5 hours ago
2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!
audio-on   68
N 5 hours ago by RainbowSquirrel53B
Join the 2025 Math and AI 4 Girls Competition for a chance to win up to $1,000!

Hey Everyone, I'm pleased to announce the dates for the 2025 MA4G Competition are set!
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (@ 11:59pm PST).

Applicants will have one month to fill out an application with prizes for the top 50 contestants & cash prizes for the top 20 contestants (including $1,000 for the winner!). More details below!

Eligibility:
The competition is free to enter, and open to middle school female students living in the US (5th-8th grade).
Award recipients are selected based on their aptitude, activities and aspirations in STEM.

Event dates:
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (by 11:59pm PST)
Winners will be announced on June 28, 2025 during an online award ceremony.

Application requirements:
Complete a 12 question problem set on math and computer science/AI related topics
Write 2 short essays

Prizes:
1st place: $1,000 Cash prize
2nd place: $500 Cash prize
3rd place: $300 Cash prize
4th-10th: $100 Cash prize each
11th-20th: $50 Cash prize each
Top 50 contestants: Over $50 worth of gadgets and stationary


Many thanks to our current and past sponsors and partners: Hudson River Trading, MATHCOUNTS, Hewlett Packard Enterprise, Automation Anywhere, JP Morgan Chase, D.E. Shaw, and AI4ALL.

Math and AI 4 Girls is a nonprofit organization aiming to encourage young girls to develop an interest in math and AI by taking part in STEM competitions and activities at an early age. The organization will be hosting an inaugural Math and AI 4 Girls competition to identify talent and encourage long-term planning of academic and career goals in STEM.

Contact:
mathandAI4girls@yahoo.com

For more information on the competition:
https://www.mathandai4girls.org/math-and-ai-4-girls-competition

More information on how to register will be posted on the website. If you have any questions, please ask here!


68 replies
audio-on
Jan 26, 2025
RainbowSquirrel53B
5 hours ago
Question about AMC 10
MathNerdRabbit103   15
N 6 hours ago by GallopingUnicorn45
Hi,

Can anybody predict a good score that I can get on the AMC 10 this November by only being good at counting and probability, number theory, and algebra? I know some geometry because I took it in school though, but it isn’t competition math so it probably doesn’t count.

Thanks.
15 replies
MathNerdRabbit103
May 2, 2025
GallopingUnicorn45
6 hours ago
9 Did I make the right choice?
Martin2001   33
N Today at 2:11 AM by happypi31415
If you were in 8th grade, would you rather go to MOP or mc nats? I chose to study the former more and got in so was wondering if that was valid given that I'll never make mc nats.
33 replies
Martin2001
Apr 29, 2025
happypi31415
Today at 2:11 AM
Permutations Part 1: 2010 USAJMO #1
tenniskidperson3   69
N Apr 2, 2025 by akliu
A permutation of the set of positive integers $[n] = \{1, 2, . . . , n\}$ is a sequence $(a_1 , a_2 , \ldots, a_n ) $ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$. Let $P (n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1 \leq k \leq n$. Find with proof the smallest $n$ such that $P (n)$ is a multiple of $2010$.
69 replies
tenniskidperson3
Apr 29, 2010
akliu
Apr 2, 2025
Permutations Part 1: 2010 USAJMO #1
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ryanbear
1056 posts
#57
Y by
to form a permutation first arrange the perfect squares in the perfect square indexes --> this has $\lfloor \sqrt{n} \rfloor!$ ways
then arrange the perfect squares/2 in the perfect square/2 indexes --> this has $\lfloor \sqrt{n/2} \rfloor!$ ways
then arrange the perfect squares/3 in the perfect square/3 indexes --> this has $\lfloor \sqrt{n/3} \rfloor!$ ways
this repeats so $P(n)=(\lfloor \sqrt{n} \rfloor!)(\lfloor \sqrt{n/2} \rfloor!)..(\lfloor \sqrt{n/1434} \rfloor!)...$
To divide $2010$, it has to divide $67$, so $\lfloor \sqrt{n} \rfloor! \ge 67$ and $n \ge 67^2 = \boxed{4489}$
This post has been edited 1 time. Last edited by ryanbear, Aug 17, 2023, 2:51 PM
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joshualiu315
2534 posts
#58
Y by
realized i haven't made a post here

Click to reveal hidden text
This post has been edited 1 time. Last edited by joshualiu315, Oct 1, 2023, 6:14 AM
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peace09
5419 posts
#59
Y by
From OTIS
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blackbluecar
303 posts
#60
Y by
(sketch) Let $S \subseteq\mathbb{N}$ be the set of squarefree numbers. Moreover, for any $s \in S$ let $f_s(n)$ be the number of positive integers $k \leq n$ where $sk$ is a square number. We explicitly give the formula \[ P(n) = \prod_{s \in S} f_s(n)! \]Note that if $2010$ divides $P(n)$ then $67$ divides $ \prod_{s \in S} f_s(n)!$. Note that $s \in S$ and $s>1$ then $f_s(n)<f_1(n)$, so if $67$ divides $ \prod_{s \in S} f_s(n)! $ then $67$ divides $f_1(n)!$. So, $f_1(n) \geq 67 \implies n \geq 67^2$. Note that $2010$ divides $P(67^2)$ so we are done
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EpicBird08
1751 posts
#61
Y by
Yay, a combo I can actually solve!

We claim that the answer is $67^2 = 4489.$ We will find an explicit formula for $P(n)$ to show this.

Suppose we have an integer $1 \le k \le n.$ Write $k = xk_0^2,$ where $x$ is as small as possible (note that $x$ is squarefree). Thus we can actually categorize the different $1 \le k \le n$: those that are of the form $x^2,$ those of the form $2x^2,$ those of the form $3x^2,$ and so on, for every squarefree integer. If $m_k$ is the $k$th squarefree integer, then call each category $d_i$ the set of integers of the form $m_i x^2.$ Clearly each number in each category must be paired with itself in our final result. In particular, there are $\left\lfloor \sqrt{\frac{n}{m_k}} \right\rfloor$ numbers in the $d_k.$ Therefore,
\[
P(n) = \prod_{i=1}^{\infty} \left\lfloor \sqrt{\frac{n}{m_i}} \right\rfloor !.
\]Note that $2010$ is divisble by the prime number $67,$ so one of the terms has to be $67!$ if $n$ is minimized. It clearly must be $\lfloor \sqrt{n} \rfloor !$ since any other term would result in a larger value of $n$ (and $\lfloor \sqrt{n} \rfloor !$ would also contain the factor of $67$ already). The smallest value of $n$ that satisfies this is $n = 67^2.$ This works since $2010 \mid 67!.$
This post has been edited 3 times. Last edited by EpicBird08, Dec 17, 2023, 9:34 PM
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shendrew7
794 posts
#63
Y by
Let $S$ denote the set of squarefree integers. Then $P(n)$ can be expressed in the form
\[\prod_{k \in S} \left \lfloor \sqrt{\frac nk} \right \rfloor !.\]
We have $2010 = 2 \cdot 3 \cdot 5 \cdot 67$, so we need a factor of 67 in this product. The first time this occurs is when $n = \boxed{67^2}$ and $k=1$, so nothing less works. Clearly, we also have $2 \cdot 3 \cdot 5 \mid 67!$, so this value of $n$ is indeed valid. $\blacksquare$
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gracemoon124
872 posts
#64
Y by
Note that any perfect square may be paired up with any other perfect square— so that gives a “sub-permutation” of $\lfloor\sqrt n\rfloor !$ ways so far (as there are $\lfloor \sqrt n\rfloor$ perfect squares in our allowed range).

If it’s $2$ times a perfect square, or $2k^2$ for some $k$, it may only be paired up with others of the same form. This adds another factor of $\lfloor\sqrt{\tfrac n2}\rfloor!$, using similar logic as above.

We can continue this to get that $P(n)$ is
\[\prod_{1\le \ell\le n}\left\lfloor\sqrt{\frac{n}{\ell}}\right\rfloor!\qquad \text{for all squarefree $\ell$.}\]Since $2010=2\cdot 3\cdot 5\cdot 67$, $67$ must be a factor of one of the $\lfloor\sqrt{\tfrac{n}{\ell}}\rfloor!$s. If we want to minimize $n$, we can take $67$ to be a factor of $\lfloor \sqrt{n}\rfloor!$, and then $n=67^2$. It’s easy to check that $n=67^2$ works, so that is our answer. $\square$
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peppapig_
281 posts
#65
Y by
I claim that the answer is $67^2$, or $4489$.

Definitions.
Define $S_{(n,b)}$ for some positive integer $n$ and positive integer $b$ not divisible by the square of any prime to be the set of all integers $0<k\leq n$ that can be expressed as $a^b$ for some positive integer $a$. In other words, $k\in S_{(n,b)}$ if and only if $\sqrt k=a\sqrt b$ for some $a$ in the set of natural numbers. Additionally, define a "good" permutation $P(n)$ to be a permutation such that $ka_k$ is a perfect square for all integers $k$ such that $1\leq k\leq n$.

--

Now I claim that $4489$ works. Let the number of "good" permutations be $m$. Note that for any $k\in S_{(n,b)}$ for any $(n,b)$, the mapping of $k$ after the permutation, or $k$, must also be in $S_{(n,b)}$ in order for $ka_k$ to be a perfect square. Additionally, if $a_k\in S_{(n,b)}$, then $ka_k$ is indeed a perfect square. Therefore;
\[m=\Pi_{b\leq n} |S_{(n,b)}|!,\]for all $b$ not divisible by the square of any prime. Note that since $S_{(4489,1)}=67$, we have that $67!\mid m$. Since $2010\mid 67!$, we must have that $2010\mid m$, as desired.

Finally, note that since $67\mid 2010$, we must have that $67\mid|S_{(n,b)}|!$ for some $(n,b)$. Since $67$ is prime, this implies that
\[|S_{(n,b)}|\geq 67 \iff n\geq 67^2b \iff n\geq 4489,\]since $b\geq1$. This is what we wished to prove, finishing the problem.
This post has been edited 1 time. Last edited by peppapig_, Feb 24, 2024, 3:26 AM
Reason: Organization
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Markas
105 posts
#66
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Let the numbers from 1 to n be divided into sets: ${1, 4, \cdots, a^2}$; ${2, 8, \cdots, 2a_1^2}$; ${3, 12, \cdots, 3a_2^2}$, and we continue in the same manner.

If we order these sets, we get a permutation.
Clearly, there are $\lfloor \sqrt{n} \rfloor! \cdot \lfloor \frac{\sqrt{n}}{2} \rfloor! \cdots$ permutations $\Rightarrow$ ${P(n)= \prod_{k=1}^{n}\lfloor\sqrt{\frac{n}{k}}}\rfloor!$

We have 2010 = 2.3.5.67 $\Rightarrow$ since we search the minimum of n it occurs when $67 \mid \lfloor \sqrt{n} \rfloor!$ $\Rightarrow$ $\lfloor \sqrt{n} \rfloor \geq 67$ $\Rightarrow$ $\sqrt{n} \geq 67$ $\Rightarrow$ $n \geq 67^2 = 4489$ $\Rightarrow$ n = 4489.
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blueprimes
351 posts
#67
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For a squarefree positive integer $s$, define $G_s$ as the set of positive integers at most $n$ of the form $sk^2$ where $k$ is an integer.

$\textbf{Lemma 1.}$ Every integer in $\{1, 2, \dots, n \}$ belongs to some $G_s$.
Proof. Simply divide any positive integer by its largest square factor, and let the result be $s$. Then that integer belongs to $G_s$ by definition.

$\textbf{Lemma 2.}$ All $G_s$ are disjoint.
Proof. For the sake of contradiction assume that some positive integer belongs in both $G_{s_A}$ and $G_{s_B}$ where $s_A$ and $s_B$ are distinct squarefree integers. Then for some integers $u$, $v$ we have $s_A u^2 = s_B v^2$ implying $s_A s_B$ is a square. By a simple $\nu_p$ argument it is easy to see that the latter can only occur when $s_A = s_B$, a contradiction.

These two lemmas readily imply that the disjoint union of all $G_s$ is $\{1, 2, \dots, n \}$. Now the problem falls apart due to the following claim:

$\textbf{Claim 1.}$ If $a$, $b$ are positive integers, then $ab$ is a square if and only if some $s$ exists where $a, b \in G_s$.
Proof. The "if" part is obvious. For the "only if" part, for the sake of contradiction let $a = s_A u^2$ and $b = s_B v^2$ where $s_A$ and $s_B$ are distinct squarefree integers. Then $ab = s_A s_B u^2v^2$ is a square, which implies $s_A s_B$ is a square. Again, this can only happen when $s_A = s_B$, a contradiction.

Now it is obvious that $P(n) = \prod_{s \text{ squarefree}} |G_s|!$. The largest of the $G_s$ is $G_1$, and since the largest prime factor of $2010$ is $67$ we must have $|G_1| \ge 67 \implies n \ge 67^2$. Now $n = 67^2$ works since $2010 \mid 67!$. The final answer is $67^2 = 4489$.
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WheatNeat
224 posts
#68
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Just wondering, did you have to find the general formula for the solution, or could you write the solution without it?
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eg4334
637 posts
#69
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The answer is $\boxed{67^2}$, a number far too large to compute in the timespan of the USAJMO. The key is that all squares must be permuted among each other giving us $\lfloor \sqrt{\frac{n}{k}} \rfloor !$ and generalizing this obvious statement. We then split the numbers into groups based on their largest squarefree factor $k$. In the previous case, $k=1$. Its not hard to see that all groups must be permuted within each other, giving us the answer of $$P(n) = \prod_{k \in \text{squarefree integers}} \lfloor \sqrt{\frac{n}{k}} \rfloor !$$We need a factor of $67$ in this and the conclusion immediately follows. No number smaller than $67^2$ will produce a factor of $67$ because the number being factorialied is smaller than that.

@below yes
This post has been edited 1 time. Last edited by eg4334, Jan 13, 2025, 4:51 AM
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Maximilian113
575 posts
#70
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@above ru doing entry combo from otis lol

I wrote this solution before but never had a chance to post it:

Notice that for $k$ being a perfect square, since there are $\lfloor \sqrt{n} \rfloor$ perfect squares less than or equal to $n,$ and each of these squares match with such a $k,$ there are $\lfloor \sqrt{n} \rfloor!$ ways to order the perfect squares. Similarly, in general we apply this same logic to squarefree integers $k,$ (since if it wasn't squarefree it would have been counted when considering some squarefree number), we get that $$P(n) = \prod_{k \text{ squarefree}} \left(\left\lfloor \sqrt{\frac{n}{k}} \right\rfloor! \right).$$Clearly for $P(n)$ to be a multiple of $2010,$ it must be a multiple of $67,$ so one of the terms in the product is a multiple of $67.$ Thus to minimize $n,$ it must be $\lfloor \sqrt{n} \rfloor!$ so $n \geq 67^2.$ We can then see that clearly $n=67^2 = \boxed{4489}$ works, so this is our answer.
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NAMO29
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#74
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Claim 1. The squares can be permuted amongst themselves.
If $a^2$ goes to the position of $b^2$, $a^2b^2=(ab)^2$

Claim 2. Numbers of the form $ab^2$ can be permuted amongst themselves.(where a doesn't change for making permutations possible and a is a non-square number)
$ab^2*ac^2=(abc)^2$

Let the numbers proposed in Claim 2 be termed as $A_a$(where $A_2$ are the numbers with a=2)

Claim 3. Number of squares in [n]$ \geq$ Number of $A_a$ s in [n] for a particular a.

Sequence of squares
$t_1=1$
$t_2=4$
$t_3=9$
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Sequence of $A_a$ s
$t_1=a$
$t_2=4a$
$t_3=9a$
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Clearly, density of squares$ \geq$ density of A_a s.

Claim 4. Due to permutation,
P(n)=(Number of squares)!(Number of $A_2$)!(Number of $A_3$)!...

Since, 2010=2*3*5*67,
67|P(n)

If 67|k!, 2,3, and 5 also divide k!

As density of squares is more, we take #squares as 67.
Hence, smallest possible value of $n=67^2=\boxed{4489}$
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akliu
1800 posts
#75
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Note that for all squarefree numbers $s$, any permutation of the numbers $(s, 4s, 9s, \dots)$ will still result in a valid sequence, and this applies for all squarefree numbers. We arrive at this conclusion from noting that all squares are permutable with each other, and then all values $ab^2$, and so on. Therefore, our answer is $(\lfloor \sqrt{\frac{n}{s_1}} \rfloor)! (\lfloor \sqrt{\frac{n}{s_2}} \rfloor)! \dots$ where $s_i$ is the $i$-th squarefree integer and $s_1 = 1$. This value will first be divisible by $67$ when $n = 67^2$, and we can check that it is indeed also a multiple of $2010$.
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