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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.

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[*]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
Nice Collinearity
oVlad   10
N 15 minutes ago by Trenod
Source: KöMaL A. 831
In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear.

Proposed by Imolay András, Budapest
10 replies
+1 w
oVlad
Oct 11, 2022
Trenod
15 minutes ago
cute geo
Royal_mhyasd   1
N 23 minutes ago by Royal_mhyasd
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
1 reply
+1 w
Royal_mhyasd
3 hours ago
Royal_mhyasd
23 minutes ago
Hardest in ARO 2008
discredit   29
N 24 minutes ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
29 replies
discredit
Jun 11, 2008
JARP091
24 minutes ago
Macedonian Mathematical Olympiad 2019 problem 1
Lukaluce   5
N 28 minutes ago by AylyGayypow009
In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
5 replies
Lukaluce
Apr 20, 2019
AylyGayypow009
28 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   9
N 29 minutes ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
9 replies
OgnjenTesic
May 22, 2025
JARP091
29 minutes ago
DE is tangent to a fixed circle whose radius is half the radius of (O)
parmenides51   1
N 29 minutes ago by TigerOnion
Source: 2017 Saudi Arabia JBMO Training Tests 2
Let $ABC$ be a triangle inscribed in circle $(O)$ such that points $B, C$ are fixed, while $A$ moves on major arc $BC$ of $(O)$. The tangents through $B$ and $C$ to $(O)$ intersect at $P$. The circle with diameter $OP$ intersects $AC$ and $AB$ at $D$ and $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.
1 reply
parmenides51
May 28, 2020
TigerOnion
29 minutes ago
Orthocorrespondent of P on Euler line
Luis González   1
N 44 minutes ago by AuroralMoss
Let $O,G$ and $K$ be the circumcenter, centroid and symmedian point of $\triangle ABC,$ respectively. $P$ is an arbitrary point on Euler line $OG.$ Show that the orthocorrespondent of $P$ WRT $\triangle {ABC}$ falls on $GK.$
1 reply
Luis González
Feb 8, 2025
AuroralMoss
44 minutes ago
JBMO Shortlist 2023 N3
Orestis_Lignos   9
N an hour ago by Just1
Source: JBMO Shortlist 2023, N3
Let $A$ be a subset of $\{2,3, \ldots, 28 \}$ such that if $a \in A$, then the residue obtained when we divide $a^2$ by $29$ also belongs to $A$.

Find the minimum possible value of $|A|$.
9 replies
Orestis_Lignos
Jun 28, 2024
Just1
an hour ago
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   7
N an hour ago by sttsmet
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
7 replies
mathematics2003
Apr 14, 2021
sttsmet
an hour ago
a^2-bc square implies 2a+b+c composite
v_Enhance   41
N an hour ago by cursed_tangent1434
Source: ELMO 2009, Problem 1
Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Evan o'Dorney
41 replies
v_Enhance
Dec 31, 2012
cursed_tangent1434
an hour ago
IMO Shortlist 2008, Geometry problem 2
April   42
N an hour ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
an hour ago
3^n + 61 is a square
VideoCake   25
N an hour ago by endless_abyss
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
25 replies
VideoCake
Monday at 5:14 PM
endless_abyss
an hour ago
Problem 5
blug   3
N an hour ago by Jt.-.
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
3 replies
blug
May 19, 2025
Jt.-.
an hour ago
Inequality with xy+yz+zx=1
Kimchiks926   14
N 2 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
2 hours ago
A real sequence
Omid Hatami   14
N Mar 2, 2025 by HamstPan38825
Source: Iran National Olympiad (3rd Round) 2001
Does there exist a sequence $ \{b_{i}\}_{i=1}^\infty$ of positive real numbers such that for each natural $ m$: \[ b_{m}+b_{2m}+b_{3m}+\dots=\frac1m\]
14 replies
Omid Hatami
Jul 13, 2007
HamstPan38825
Mar 2, 2025
A real sequence
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran National Olympiad (3rd Round) 2001
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Omid Hatami
1275 posts
#1 • 6 Y
Y by Adventure10, Mango247, cubres, and 3 other users
Does there exist a sequence $ \{b_{i}\}_{i=1}^\infty$ of positive real numbers such that for each natural $ m$: \[ b_{m}+b_{2m}+b_{3m}+\dots=\frac1m\]
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perfect_radio
2607 posts
#2 • 3 Y
Y by Adventure10, Mango247, cubres
http://www.mathlinks.ro/viewtopic.php?search_id=2008794938&t=152341
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Vinxenz
56 posts
#3 • 2 Y
Y by Adventure10, cubres
Can we say this?

$ \sum_{m=1}^{\infty}lim_{n\rightarrow{\infty}}(b_{m}+b_{2m}+...+b_{nm})=\sum_{m=1}^{\infty}{d(m)b_{m}}=\sum_{m=1}^{\infty}\frac{1}{m}$ (absurd)
where $ d(m)$ is the number of divisor of m
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Vinxenz
56 posts
#4 • 2 Y
Y by Adventure10, cubres
perfect_radio what do you think?It's possible.
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crude
41 posts
#5 • 5 Y
Y by narutomath96, Adventure10, Mango247, cubres, and 1 other user
I guess, I found something interesting on this problem.
The idea is the following:
Write the sum $ b_{m}+b_{2m}+b_{3m}+...=b_{m}+(b_{2m}+b_{4m}+...)+(b_{3m}+b_{6m}+...)+...-$
$ -(b_{6m}+b_{12m}+...)-(b_{8m}+b_{16m}+...)+...=b_{m}+\frac{1}{2m}+\frac{1}{3m}+\frac{1}{5m}+...-$
$ -\frac{1}{6m}-\frac{1}{10m}-\frac{1}{14m}...=\frac{1}{m}$, from this we can see that
$ b_{m}=\frac{1}{m}(1-\sum_{i\in P}{\frac{1}{i}}+\sum_{i,j\in P, i\neq j}{\frac{1}{ij}}-\sum_{i,j,k \in P, i\neq j\neq k, i\neq k}\frac{1}{ijk}+...)$, where $ P$ is the set of all prime numbers.
Clearly the sum in the brackets is independent of $ m$ and generally of anything. There are two options, either it does not exist or it is some constant let's say $ k$. In latter case we have that
$ k(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)=1$, which brings again to impossible result.
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toto1234567890
889 posts
#6 • 2 Y
Y by Adventure10, cubres
Yes. Exclusiom-Inclusion Principle kills it. :)
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SCP
1502 posts
#7 • 3 Y
Y by Adventure10, Mango247, cubres
Sure, the solution of post 5 is correct?
As there has been working with infinitely many infinite series, it seems there will be trouble with the order of executing that sum.
Also post 3 seems like not concluding anything as $d(m)$ isn't bounded.
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pad
1671 posts
#9 • 1 Y
Y by cubres
Quote:
Determine whether there exists a sequence $a_1,a_2,\ldots$ of nonnegative reals such that
\[ a_{n}+a_{2n}+a_{3n}+\cdots=\frac1n\]for every positive integer $n$.
No. Suppose there did. Let $S_k=\{ n\in \mathbb{N}:p_1\mid n, \text{ or } p_2\mid n, \ldots, \text{ or } p_k\mid n\}$ where we order the primes $\{p_1<p_2<\cdots\}$. The idea is to ``interlace'' the sequences for $n=p_1,\ldots,p_k$. Using CRT and Inclusion-Exclusion:
\[ \sum_{n\in S_k} a_n = \sum_{\substack{T\subset S_k \\ T\not=\emptyset}} (-1)^{|T|+1}\prod_{i\in T}\frac{1}{p_i} = 1-\prod_{i=1}^k\left(1-\frac{1}{p_i}\right) \]by expansion. Using the estimate $\log(1-\varepsilon)\le -0.01\varepsilon$ for $\varepsilon\in (0,1)$, we have
\[ \log \prod_{i=1}^k\left(1-\frac{1}{p_i}\right) = \sum_{i=1}^k \log\left(1-\frac{1}{p_i}\right) \le \sum_{i=1}^k -0.01\cdot \frac{1}{p_i}=-0.01P_k\]where $P_k:=\sum_{i=1}^k 1/p_i$. Therefore, since $\log$ is an increasing function, we have
\begin{align*} 
 \prod_{i=1}^k\left(1-\frac{1}{p_i}\right) \le e^{-0.01P_k} &\implies \sum_{n\in S_k}a_n \ge 1-\frac{1}{\left(e^{0.01} \right)^{P_k}}.
\end{align*}Hence
\begin{align*} 
 1-a_1=\sum_{i\ge 2}a_i \ge \sum_{n\in S_k} a_n \ge  1-\frac{1}{\left(e^{0.01} \right)^{P_k}}
 \implies a_1 \le \frac{1}{\left(e^{0.01} \right)^{P_k}}. 
\end{align*}But it is well-known that $P_k$ grows arbitrarily large varying $k$, so this forces $a_1=0$. Now, fix an $\ell$. Let $b_n=\ell a_{\ell n}$ for each $n$. Then $(b_n)$ is also a valid sequence, since
\[ b_n+b_{2n}+\cdots = \ell a_{\ell n}+\ell a_{2\ell n} + \cdots = \ell \cdot \frac{1}{\ell n} = \frac{1}{n}.\]Therefore, $b_1=0$, so $\ell a_\ell=0$, i.e. $a_\ell=0$. Therefore, the entire sequence is $0$, contradiction.

Remarks
This post has been edited 2 times. Last edited by pad, Dec 8, 2020, 9:35 PM
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TheUltimate123
1740 posts
#10 • 1 Y
Y by cubres
The answer is no. Assume for contradiction such a sequence exists, and let \(P_n\) denote the first \(n\) primes. By the Principle of Inclusion-Exclusion, \[\frac1n-a_n=\lim_{k\to\infty}\sum_{T\subset P_k}\frac{(-1)^{|T|}}{n\prod_{p\in T}p}=\frac1n\lim_{k\to\infty}\sum_{T\subset P_k}\frac{(-1)^{|T|}}{\prod_{p\in T}p}=\frac{1-a_1}n.\]Solving, \(a_n=\tfrac1na_1\) for all \(n\), so \[1=\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty\frac{a_1}k=a_1\sum_{k=1}^\infty\frac1k.\]Since \(\sum_{k=1}^\infty a_k\) diverges, \(a_1=0\), which means that \(a_n=0\) for all \(n\). This yields the desired contradiction.
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laikhanhhoang_3011
637 posts
#11 • 1 Y
Y by cubres
i think the idea of this problem related to Euler formula in Zeta Function
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TheHazard
93 posts
#12 • 1 Y
Y by cubres
Indeed.

Note that $a_1 + a_2 + \dots = 1$ and that $a_n \le \frac{1}{n}$. However, we also have that
\[
    \sum_{k}^{N} \mu(k) (a_k + a_{2k} + \dots) = \sum_{k}^N \frac{\mu(k)}{k}
\]eventually contains all of $a_1$ through $a_N$ and thus has a limit of $1$. This contradicts the fact that $\lim_{s \to 1^+} \frac{1}{\zeta(s)} = \lim_{s \to 1^+} \sum_k \frac{\mu(k)}{k}$ approaches $0$ which implies that this has a limit of $0$.
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emi3.141592
71 posts
#13 • 1 Y
Y by cubres
Let $D_i$ be the set of all positive integers with exactly $i$ prime factors, not necessarily distinct. Observe that
$$
a_n = 1 - \sum_{k=2}^{\infty} a_{kn} = 1 - \sum_{k=1}^{\infty} (-1)^{k+1} \left( \sum_{x \in D_k} \left( \sum_{i=1}^{\infty} a_{xkn} \right) \right) = \frac{a_1}{n}.
$$$$
=\frac{1}{n}\left(1 - \sum_{k=1}^{\infty} (-1)^{k+1} \left( \sum_{x \in D_k} \left( \sum_{i=1}^{\infty} a_{xk} \right) \right)\right) = \frac{a_1}{n}.
$$But this implies that
$$
a_1 + a_2 + a_3 + \dots = a_1 + \frac{a_1}{2} + \frac{a_1}{3} + \dots \neq 1
$$which is a contradiction.
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MathLuis
1556 posts
#14 • 1 Y
Y by cubres
Let $S_k$ denote the set of the first $k$ primes, now from PIE just notice that:
\[ \frac{1}{n}-b_n=\lim_{\ell \to \infty} \sum_{S \subset S_{\ell}} \frac{(-1)^{|S|+1}}{n \cdot \prod_{p \in S} p}=\frac{1}{n} \cdot \lim_{\ell \to \infty} \sum_{S \subset S_{\ell}} \frac{(-1)^{|S|+1}}{\prod_{p \in S} p}=\frac{1-b_1}{n} \implies b_n=\frac{b_1}{n} \]But now we have $b_1 \left(1+\frac{1}{2}+\cdots \right)=1$ which by divergence of harmonic series gives $b_1=0$ but then $b_i=0$ for all positive integer $i$ which clearly fails, so no such sequence exists.
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Ritwin
158 posts
#15
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No, there does not. We can prove by PIE that for any $n = gk$, \[ \sum_{\gcd(m,n)=g} a_m = \frac{\varphi(n/g)}{n} = \frac1g \cdot \frac{\varphi(k)}{k}. \]On the other hand, $\tfrac{\varphi(k)}{k}$ can be arbitrarily small, so the left sum must be $0$. Since $g$ was arbitrary, every $a_i = 0$. This is obviously a contradiction.

Proof (of $\varphi(k)/k$ arbitrarily small). Enumerate the primes $p_1 < p_2 < \cdots$ and take $k = p_1 p_2 \cdots p_N$. We have \[ \frac{\varphi(k)}{k} = \prod_{i=1}^N \left(1 - \frac{1}{p_i}\right) \leq \exp \left(-\sum_{i=1}^N \frac{1}{p_i}\right) \]since $e^{-t} \geq 1-t$. Since $\textstyle{\sum} \tfrac{1}{p_i}$ diverges, the right side is arbitrarily small, meaning $\varphi(k)/k$ gets arbitrarily small. $\square$
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HamstPan38825
8868 posts
#16
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This problem is ridiculously cute. The answer is no.

Let $k$ be any positive integer, and let $S_k$ denote the set of all positive integers relatively prime to the first $k$ primes $p_1, p_2, \dots, p_k$. By principle of inclusion-exclusion, \[\lim_{k \to \infty} \sum_{d \in S_k} a_d = \lim_{k \to \infty} \frac{\varphi(p_1p_2 \dots p_k)}{p_1 p_2 \dots p_k} = \lim_{k \to \infty} \prod_{i=1}^k \left(1-\frac 1{p_i}\right).\]This converges to zero as its reciprocal is lower bounded by \[\lim_{k \to \infty}\prod_{i=1}^k \left(1+\frac 1{p_i - 1}\right) > \lim_{k \to \infty}\prod_{i=1}^k \left(1+\frac 1{p_i}\right) = \sum_{i=1}^\infty \frac 1i = \infty.\]But $a_1$ always falls in the given set, so $a_1 = 0$. Similarly, by scaling the sequence, we get $a_n = 0$ for every positive integer $n$.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 2, 2025, 5:21 AM
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