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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
1 viewing
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
USAMO MERCH
elasticwealth   4
N 10 minutes ago by hashbrown2009
Jane street sent:
- A T-shirt
- Deck of cards
- Portable charger (battery)
- 12 oz mug
- Jane Street Hat
4 replies
1 viewing
elasticwealth
4 hours ago
hashbrown2009
10 minutes ago
9 ARML Location
deduck   11
N 21 minutes ago by dragoon
UNR -> Nevada
St Anselm -> New Hampshire
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
11 replies
deduck
2 hours ago
dragoon
21 minutes ago
Problem 1 — Symmetric Squares, Symmetric Products
RockmanEX3   8
N 31 minutes ago by Baimukh
Source: 46th Austrian Mathematical Olympiad National Competition Part 1 Problem 1
Let $a$, $b$, $c$, $d$ be positive numbers. Prove that

$$(a^2 + b^2 + c^2 + d^2)^2 \ge (a+b)(b+c)(c+d)(d+a)$$
When does equality hold?

(Georg Anegg)
8 replies
RockmanEX3
Jul 14, 2018
Baimukh
31 minutes ago
Math solution
Techno0-8   0
39 minutes ago
Solution
0 replies
Techno0-8
39 minutes ago
0 replies
D1027 : Super Schoof
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $p>11$ a prime number with $a=\text{card}\{(x,y) \in \mathbb Z/ p \mathbb Z: y^2=x^3+1\}$ and $b=\dfrac 1 {((p-1)/2)! \times ((p-1)/3)! \times ((p-1)/6)!} \mod p$ when $p \mod 3=1$.



Is it true that if $p \mod 3=1$ then $a \in \{b,p-b, \min\{b,p-b\}+p\}$ else $A=p$.
0 replies
Dattier
an hour ago
0 replies
Interesting inequality
sealight2107   0
an hour ago
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
0 replies
sealight2107
an hour ago
0 replies
Bosnia and Herzegovina 2022 IMO TST P1
Steve12345   3
N 2 hours ago by waterbottle432
Let $ABC$ be a triangle such that $AB=AC$ and $\angle BAC$ is obtuse. Point $O$ is the circumcenter of triangle $ABC$, and $M$ is the reflection of $A$ in $BC$. Let $D$ be an arbitrary point on line $BC$, such that $B$ is in between $D$ and $C$. Line $DM$ cuts the circumcircle of $ABC$ in $E,F$. Circumcircles of triangles $ADE$ and $ADF$ cut $BC$ in $P,Q$ respectively. Prove that $DA$ is tangent to the circumcircle of triangle $OPQ$.
3 replies
Steve12345
May 22, 2022
waterbottle432
2 hours ago
Functional equation of nonzero reals
proglote   6
N 2 hours ago by pco
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
6 replies
proglote
Oct 24, 2013
pco
2 hours ago
k Funcional equation problem
khongphaiwminh   1
N 2 hours ago by jasperE3
Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$$f(x+f(y))=f(x+y)+f(y)$$for any $x, y \in \mathbb R^+$. Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$.
1 reply
khongphaiwminh
2 hours ago
jasperE3
2 hours ago
AO and KI meet on $\Gamma$
Kayak   29
N 2 hours ago by Mathandski
Source: Indian TST 3 P2
Let $ABC$ be an acute-angled scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Suppose $AB < AC$. Let $H$ be the orthocenter and $I$ be the incenter of triangle $ABC$. Let $F$ be the midpoint of the arc $BC$ of the circumcircle of triangle $BHC$, containing $H$.

Let $X$ be a point on the arc $AB$ of $\Gamma$ not containing $C$, such that $\angle AXH = \angle AFH$. Let $K$ be the circumcenter of triangle $XIA$. Prove that the lines $AO$ and $KI$ meet on $\Gamma$.

Proposed by Anant Mudgal
29 replies
Kayak
Jul 17, 2019
Mathandski
2 hours ago
4 Variables Cyclic Ineq
nataliaonline75   1
N 2 hours ago by NO_SQUARES
Prove that for every $x,y,z,w$ non-negative real numbers, then we have:
$\frac{x-y}{xy+2y+1}+\frac{y-z}{yz+2z+1} + \frac{z-w}{zw+2w+1} + \frac{w-x}{wx+2x+1} \geq 0$
1 reply
nataliaonline75
2 hours ago
NO_SQUARES
2 hours ago
IMO Genre Predictions
ohiorizzler1434   56
N 2 hours ago by Theoryman007
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
56 replies
ohiorizzler1434
May 3, 2025
Theoryman007
2 hours ago
I'm trying to find a good math comp...
ysn613   7
N 2 hours ago by Gavin_Deng
Okay, so I'm in sixth grade. I have been doing AMC 8 since fourth grade, but not anything else. I was wondering what other "good" math competitions there are that I am the right age for.

I'm also looking for prep tips for math competitions, because when I (mock)ace 2000-2010 AMC 8 and then get a 19 on the real thing when I was definitely able to solve everything, I feel like what I'm doing isn't really working. Anyone got any ideas? Thanks!
7 replies
ysn613
Apr 30, 2025
Gavin_Deng
2 hours ago
USAMO Medals
YauYauFilter   24
N 2 hours ago by deduck
YauYauFilter
Apr 24, 2025
deduck
2 hours ago
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
G H J
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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
795 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
Y by
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
Y by
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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