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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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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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
Introducing a math summer program for middle school students
harry133   1
N 24 minutes ago by heheman
Introducing IITSP, an online math summer program designed for middle school students over summer.

The program is designed by Professor Shubhrangshu Dasgupta from the Department of Physics at the Indian Institute of Technology Ropar (IIT Ropar).

Please check out the webpage if you are interested in.

https://www.imc-impea.org/IMC/bbs/content.php?co_id=iitsp
1 reply
harry133
Today at 12:52 AM
heheman
24 minutes ago
Alcumus vs books
UnbeatableJJ   20
N 31 minutes ago by SirAppel
If I am aiming for AIME, then JMO afterwards, is Alcumus adequate, or I still need to do the problems on AoPS books?

I got AMC 23 this year, and never took amc 10 before. If I master the alcumus of intermediate algebra (making all of the bars blue). How likely I can qualify for AIME 2026?
20 replies
UnbeatableJJ
Apr 23, 2025
SirAppel
31 minutes ago
LTE or Binomial Theorem
P_Groudon   111
N 37 minutes ago by heheman
Source: 2020 AIME I #12
Let $n$ be the least positive integer for which $149^n - 2^n$ is divisible by $3^3 \cdot 5^5 \cdot 7^7$. Find the number of positive divisors of $n$.
111 replies
P_Groudon
Mar 12, 2020
heheman
37 minutes ago
Bring Back Downvotes
heheman   5
N 39 minutes ago by heheman
i would like to start a petition to bring back downvote, it you agree then write "bbd $    $" in threads
5 replies
heheman
2 hours ago
heheman
39 minutes ago
Brilliant Problem
M11100111001Y1R   4
N an hour ago by IAmTheHazard
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
4 replies
M11100111001Y1R
Yesterday at 7:28 AM
IAmTheHazard
an hour ago
Own made functional equation
Primeniyazidayi   1
N an hour ago by Primeniyazidayi
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
1 reply
Primeniyazidayi
May 26, 2025
Primeniyazidayi
an hour ago
not fun equation
DottedCaculator   13
N 2 hours ago by Adywastaken
Source: USA TST 2024/6
Find all functions $f\colon\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$,
\[f(xf(y))+f(y)=f(x+y)+f(xy).\]
Milan Haiman
13 replies
DottedCaculator
Jan 15, 2024
Adywastaken
2 hours ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   12
N 2 hours ago by atdaotlohbh
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
12 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
2 hours ago
Geometry with fix circle
falantrng   33
N 3 hours ago by zuat.e
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
33 replies
falantrng
Feb 25, 2018
zuat.e
3 hours ago
USAMO 2001 Problem 2
MithsApprentice   54
N 3 hours ago by lpieleanu
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
54 replies
MithsApprentice
Sep 30, 2005
lpieleanu
3 hours ago
German-Style System of Equations
Primeniyazidayi   1
N 3 hours ago by Primeniyazidayi
Source: German MO 2025 11/12 Day 1 P1
Solve the system of equations in $\mathbb{R}$

\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}
1 reply
Primeniyazidayi
3 hours ago
Primeniyazidayi
3 hours ago
gcd nt from switzerland
AshAuktober   5
N 3 hours ago by Siddharthmaybe
Source: Swiss 2025 Second Round
Let $a, b$ be positive integers. Prove that the expression
\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]is always a positive integer, and determine all possible values it can take.
5 replies
AshAuktober
4 hours ago
Siddharthmaybe
3 hours ago
Shortlist 2017/G1
fastlikearabbit   92
N 4 hours ago by Ilikeminecraft
Source: Shortlist 2017
Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.
92 replies
fastlikearabbit
Jul 10, 2018
Ilikeminecraft
4 hours ago
set construction nt
top1vien   2
N 4 hours ago by top1vien
Is there a set of 2025 positive integers $S$ that satisfies: for all different $a,b,c,d\in S$, we have $\gcd(ab+1000,cd+1000)=1$?
2 replies
top1vien
Yesterday at 10:04 AM
top1vien
4 hours ago
MAA finally wrote sum good number theory
IAmTheHazard   96
N May 22, 2025 by megahertz13
Source: 2021 AIME I P14
For any positive integer $a,$ $\sigma(a)$ denotes the sum of the positive integer divisors of $a.$ Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a.$ Find the sum of the prime factors in the prime factorization of $n.$
96 replies
IAmTheHazard
Mar 11, 2021
megahertz13
May 22, 2025
MAA finally wrote sum good number theory
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 AIME I P14
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thisismath1234
43 posts
#85
Y by
dolphinday wrote:
So, we can compute the least positive integer $n$ so that the expression is divisible by $43$ and $47$. By FLT, $a^{42} - 1 \equiv 0\pmod{43}$, for $a$ being relatively prime to $43$.
This is why $n$ has to be divisible by $43$, for the cases where $a$ is not relatively prime to $43$.

This line is just wrong. We care about divisibility by of n by 43 when p is 1 mod 43, not when p (or a) is 0 mod 43. Did you actually do the problem or just read the solutions above and post something that went along the same lines
This post has been edited 1 time. Last edited by thisismath1234, Oct 31, 2023, 2:16 AM
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dolphinday
1329 posts
#86
Y by
thisismath1234 wrote:
dolphinday wrote:
So, we can compute the least positive integer $n$ so that the expression is divisible by $43$ and $47$. By FLT, $a^{42} - 1 \equiv 0\pmod{43}$, for $a$ being relatively prime to $43$.
This is why $n$ has to be divisible by $43$, for the cases where $a$ is not relatively prime to $43$.

This line is just wrong. We care about divisibility by of n by 43 when p is 1 mod 43, not when p (or a) is 0 mod 43. Did you actually do the problem or just read the solutions above and post something that went along the same lines

Thank you for catching that mistake but it's quite rude of you to assume that, when I genuinely misunderstood something.
I will make sure to correct that, but all it seems you've been doing is criticizing other people, calling out the problem for being too easy and just generally being rude. Can you avoid that please
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thisismath1234
43 posts
#87
Y by
I'm sorry if my words seemed harsh; I didn't mean to be unkind and will be more mindful in the future. I shouldn't have assumed that you did not do the problem.
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dolphinday
1329 posts
#88
Y by
thisismath1234 wrote:
I'm sorry if my words seemed harsh; I didn't mean to be unkind and will be more mindful in the future. I shouldn't have assumed that you did not do the problem.

It's ok, no worries and also thanks for catching my mistake. Probably if you didn't, I wouldn't have ever noticed :)
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OlympusHero
17020 posts
#89
Y by
mira74 wrote:
wait I think i see a couple posts saying this so imma just point out that you dont need $\varphi(2021)$ to divide $n$. just having $\varphi(2021)/2$ divide it is enough.

In general, for $a^k \equiv 1 \pmod{n}$ for all $a$ relatively prime to $n$, we don't need $\varphi(n) \mid k$. The actual number is given by the Carmichael Function, which is sorta the lcm of the totients of the prime powers, but is weird when there's a power of $2$ dividing $n$.

I was just solving this problem and had a question about it regarding this: I had the exact solution as @vsamc in #4, but this is technically wrong since it would be the LCM of $2021$ and HALF the totient of $2021$, not the totient of $2021$. I was wondering how, in general, you would know whether it is half the totient or something else instead of just the totient itself? I looked and didn't find any general formula for the Carmichael function, so I was curious about this. For this problem, it wouldn't make a difference to the answer, but it might for some other problem so I wanted to know. Thanks in advance!
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OlympusHero
17020 posts
#90
Y by
Sorry to double post but I'm pretty sure I figured it out - 2021 is 43 * 47, so you split it up into mod 43 and 47, which are both prime, so their totients are 42 and 46. Thus the Carmichael function would give lcm(42,46) which is indeed half of the totient of 2021.
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v_Enhance
6878 posts
#91 • 1 Y
Y by sanaops9
Sorry to double post but I'm pretty sure I figured it out - 2021 is 43 * 47, so you split it up into mod 43 and 47, which are both prime, so their totients are 42 and 46. Thus the Carmichael function would give lcm(42,46) which is indeed half of the totient of 2021.

Yes, that's exactly right.

The general formula for the Carmichael function is stated at https://en.wikipedia.org/wiki/Carmichael_function#Recurrence_for_%CE%BB(n) (although it's called a "recurrence" in Wikipedia right now, I think that's a bit misleading). As you've already figured out, it's just the LCM of Carmichael function on each individual prime (power).
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de-Kirschbaum
202 posts
#92
Y by
If $2021|\sigma(a^n)-1$ for all integers $a$, then we can consider prime integers first. If $a=p$, then we want $\sigma(p^n)-1 \equiv 0 \mod{43}$ and $\sigma(p^n)-1 \equiv 0 \mod{47}$. First consider the 43 case. Writing out the LHS we have that $\frac{p^{n+1}-1}{p-1}-1 \equiv 0 \mod{43} \implies \frac{p^{n+1}-1}{p-1} \equiv 1 \mod{43}$. Now there are two things we must consider. First, if $p-1, 43$ are coprime, then division is defined in mod 43 and we can just multiply it out. In that case, $p^{n+1}-1 \equiv p-1 \mod{43} \implies p(p^n-1) \equiv 0 \mod{43}$. Of course, if $p=43$ this is true, but we need to ensure this is true for all $p \neq 43$. In order to do that, we must have $p^n-1 \equiv 0 \mod{43}$. FLT guarantees this is true when $42 | n$, and we will take it because there probably exists some prime that has that as the smallest period.

If $p-1, 43$ aren't coprime, then their gcd is 43 since 43 is a prime. That means we can write $p-1=43k \implies p=43k+1$. Thus consider the original expression of the sigma function $\sigma(p^n)=1+p+...+p^n \equiv 1+1+...+1 \equiv n+1 \mod{43}$. We want $n+1-1 \equiv n \equiv 0 \mod{43}$, so $43|n$. By similar analysis we can get $46|n, 47|n$. So we know that the least $n$ right now is $lcm(42,43,46,47)$.

Now consider any composite number $a=p_1^{e_1}p_2^{e_2}...p_m^{e_m}$. We have that $\sigma(a^n)-1 \equiv \frac{p_1^{ne_1+1}-1}{p_1-1}...\frac{p_m^{ne_m+1}-1}{p_m-1} -1 \mod{2021}$. Note that $n$ actually ensures each part of this multiplication to be $1 \mod{2021}$, as $n$ guarantees $\sigma(p^n) \equiv 1 \mod{2021}$ for any p. Thus, we just have $1-1 \equiv 0 \mod{2021}$ and we do not need to modify $n$ further.
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gladIasked
648 posts
#93
Y by
fairly straightforward

We want $\sigma(a^n) \equiv 1\pmod{2021}$. Because of the multiplicativity of $\sigma$, we only need check $a = p^k$, where $p$ is a prime. Note that $$\sigma((p^k)^n) = 1 + p + p^2 + \cdots + p^{kn}$$. Our original modular congruence becomes \begin{align*}1 + p + p^2 + \cdots + p^{kn}&\equiv 1\pmod{2021}\\ \iff p + p^2 + \cdots p^{kn} &\equiv 0\pmod{2021}\\ \iff 1 + p + \cdots + p^{kn-1}&\equiv 0\pmod{2021} \\ \iff \frac{p^{kn}-1}{p-1}&\equiv 0\pmod{2021}\end{align*}This implies that $p^{kn} \equiv 1\pmod{2021}$. Breaking this up with CRT, we have $p^{kn} \equiv 1\pmod{43}$ and $p^{kn}\equiv 1\pmod{47}$, from which we deduce (via FLT) that $42\mid n$ and $46\mid n$. However, this fails when either $v_{43}(p^{kn} - 1) = v_{43}(p-1)$ or $v_{47}(p^{kn} - 1) = v_{47}(p-1)$. Using LTE, we see that $$v_{43}(p^{kn} - 1) = v_{43}(p-1) + v_{43}(n)$$. We need $v_{43}(n) > 0$, so $43\mid n$. We can similarly deduce that $47\mid n$. Thus, our answer will just be $n = \text{lcm}(42, 43, 46, 47)$, which gives us the final answer of $\boxed{125}$.
This post has been edited 2 times. Last edited by gladIasked, Jan 14, 2024, 5:07 PM
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Cusofay
85 posts
#94
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We know that $\sigma (a^n)-1=\prod_{1\leq i\leq k}(1+p_i+p_i^2+\dots+p_i^{ne_i})-1$. Thus, we contend just need to find the smallest $n$ for which $\frac{p(p^n-1)}{p-1} \equiv 0 \pmod{2021}$. If $43,47\mid p-1$ then using LTE we find that $2021\mid$. Otherwise we use euler's totient theorem and if $p=47,43$ then $ord_{43},ord_{47}\mid \Phi(2021)$. Hence $n=1952286$

$$\mathbb{Q.E.D.}$$
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L13832
268 posts
#95 • 2 Y
Y by S_14159, CRT_07
\begin{align*}
    &\text{We need}\; \sigma(a^n)-1 \equiv 0 \pmod {2021}.
    \;\text{Checking for values:} \\
    &[b]a=1:[/b] \sigma(1^n)-1=0 \;\text{so true for all n.}\\
    &[b]a=p[/b], \text{(p prime)}: \;\sigma(p^n)-1=\frac{p(p^n-1)}{p-1} \equiv 0 \pmod{2021}.\\
    &\text{Case I: p-1 has no 43s or 47s in its prime factorisation then it is easy to see that}\\&\; p(p^n-1)\equiv0 \pmod {2021} \; \text{we have to show when} \; p^n \equiv 1 \pmod{43} \; \text{and} \\ &\; p^n \equiv 1 \pmod{47},\; \text{n=42c and n=46d.} \;\text{So,} \;\text{n}=\text{LCM}(42,46)=42 \cdot 23.\\
    &\text{Case II: p-1 has 43 or 47 in prime factorisation}, \\& \text{First we look if 43 is there}\; p-1=43^ke \\
    &\Rightarrow p^n-1=43^{k+1}f\Rightarrow \boxed{p-1 \equiv 43^ke\pmod{43^{k+1}}} \\& \Rightarrow p^n \equiv (1+43^ke)^n \equiv n43^ke+1 \equiv 1\pmod{43^{k+1}} \\& \Rightarrow n43^ke \equiv 0 \pmod {43^{k+1}} \Rightarrow 43 \vert n.
   \\& \text{Similarly we try for when 47 is there in prime factorisation of $p-1$ we get}\;  47\vert n\\
   & \Rightarrow n=\operatorname{lcm}(42,43,46,47)\\
   &[b]a is composite[/b]\\
   &\text{Let}\; a=\prod_{i=1}^{u}p_i^{e_i} \;\text{Note that} \; \sigma(a) \text{is a multiplicative} \\
&\text{function:} \;\sigma(a)=\sigma\left(\prod_{i=1}^{u}p_i^{e_i}\right)=\prod_{i=1}^{u}\sigma\left(p_i^{e_i}\right), \; \text{as this product of} \text{is over all divisors of a.}\\
&\text{At}\;  n=\operatorname{lcm}(42,43,46,47),\; \text{we have}\; 
\sigma(a^n)-1=(\prod_{i=1}^{u}\sigma(p_i^{e_in}))-1 \equiv(\prod_{i=1}^{u}1)-1 \equiv 0\pmod{2021}.\\
&\text{The problems asks for the sum of the factors of n.}\\&n=\operatorname{lcm}(42,43,46,47)= 2\cdot3\cdot7\cdot23\cdot43\cdot47.\\
&\text{Therefore,}\; n=2+3+7+23+43+47=\boxed{\textbf{125}}
\end{align*}
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blueprimes
362 posts
#96
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Recall that $2021 = 43 \cdot 47$. We claim the minimal $n$ is $\text{lcm}(42, 43, 46, 47) = 2 \cdot 3 \cdot 7 \cdot 23 \cdot 43 \cdot 47$, yielding $\boxed{125}$ as the requested answer.

Consider an arbitrary prime $\gcd(p - 1, 2021) = 1$, allowing $a = p$ we want
\[1 + p + p^2 + \dots + p^n = \dfrac{p^{n + 1} - 1}{p - 1} \equiv 1 \pmod{2021} \iff p^{n + 1} \equiv p \pmod{2021}. \]Let $g_{43}$ and $g_{47}$ be arbitrary primitive roots of $43$ and $47$ respectively, by CRT and Dirichlet we can find a prime $p$ such that $p \equiv g_{43} \pmod{43}$ and $p \equiv g_{47} \pmod{47}$ which forces $\text{lcm}(43 - 1, 47 - 1) = \text{lcm}(42, 46) \mid n$.

On the other hand, if $\gcd(p - 1, 2021) \ne 1$, we have $p \equiv 1 \pmod{43}$ or $p \equiv 1 \pmod{47}$. Assume the former, plugging in $a = p$ gives
\[1 + p + p^2 + \dots + p^n \equiv n + 1 \equiv 1 \pmod{43} \iff 43 \mid n. \]A similar argument for $47$ means $47 \mid n$. Altogether, we get $43, 47 \mid n$.

From both of these cases we easily obtain $\text{lcm}(42, 43, 46, 47) \mid n$, it is easy to show sufficiency for $n = \text{lcm}(42, 43, 46, 47)$ (just generalize the form of $a$, use multiplicity, and re-iterate through the previous arguments) so we are done.
This post has been edited 1 time. Last edited by blueprimes, Dec 2, 2024, 3:03 AM
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AshAuktober
1011 posts
#97
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Obtain from various values of $a$ that \[2021\phi(2021)\mid n\]is necessary,and then that it is sufficient,yielding 125.
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Magnetoninja
277 posts
#98
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Straightfoward for its placement:

Let $a=p_1^{e_1}*p_2^{e_2}*\cdots{p_k^{e_k}}$. $a^n=p_1^{ne_1}*p_2^{ne_2}*\cdots{p_k^{ne_k}} \Longrightarrow \sigma{(a^n)}=\prod_{j=1}^{k}{\sum_{i=0}^{e_j}{(p_j^i)}}=\prod_{j=1}^{k}{\frac{p_j^{ne_j+1}-1}{p_j-1}}$. Let prime $p|2021$. By induction, if $\sigma{(a_n)}\equiv 1\pmod{p}$, we also need $\sigma{(aq^{e_q})^n}=\prod_{j=1}^{k}{\sum_{i=0}^{e_j}{(p_j^i)}}*\frac{q^{ne_q+1}-1}{q-1} \equiv 1\pmod{p}$ so $\frac{q^{ne_q+1}-1}{q-1}\equiv 1\pmod{p}$ for all $q, e_q$. The base case is $a=1$, which is true. Now, if $q\neq{1}\pmod{p}$, then $q^{ne_1+1}-1\equiv q-1\pmod{p} \Longrightarrow q^{ne_q}=(q^{e_q})^n\equiv 1\pmod{p}$. By Fermat's Little Theorem, we need $p-1|n$ to satisfy this congruence for all $q, e_q$. If $q\equiv 1\pmod{p}$, then $(1+q+q^2\cdots{+q^{ne_q}})\equiv \underbrace{1+1\cdots{1}}_{ne_q+1}\equiv ne_q+1\equiv 1\pmod{p} \Longrightarrow n\equiv 0\pmod{p}$. Therefore $n(n-1)|p$. In our case $p=43, 47$, so we get $\text{lcm}{(42, 43, 46, 47)}=2*3*7*23*43*47$, giving us $\boxed{125}$.
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megahertz13
3194 posts
#99
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Notice that the condition is equivalent to \[\sigma(p^n)\equiv 1\pmod {2021}\]for primes $p$ (this is necessary since $a$ can be a prime, and sufficient since $\sigma$ is multiplicative). Since $2021=43\cdot 47$ and \[\sigma(p^n)=1+p+\dots+p^n=\frac{p^{n+1}-1}{p-1},\]we want \[\frac{p^{n+1}-1}{p-1}\equiv 1\pmod {43}\]and \[\frac{p^{n+1}-1}{p-1}\equiv 1\pmod {47}.\]
Consider only the first condition.

Case 1: If $p-1\equiv 0\pmod {43} \implies p\equiv 1\pmod {43}$, then $\sigma(p^n)\equiv (n+1)\pmod {43}$. However, we need $\sigma(p^n)\equiv 1\pmod {43}$, so $n\equiv 0\pmod {43}$.

Case 2: If $p-1\ne 0\pmod {43} \implies p\ne 1\pmod {43}$, then we have \[p^{n+1}-1\equiv p-1\pmod {43}.\]This implies that $p^{n+1}-p$ is a multiple of $43$. If $p=43$, then the result is always true, so we can ignore this case as it provides no information. Otherwise, we have $p^n\equiv 1\pmod {43}$ for all prime $p$. There exists a $0<g<43$ that is a primitive root $\pmod {43}$ (well-known), and by Dirichlet there exists a prime $p$ that is congruent to $g\pmod {43}$. Thus $n\equiv 0\pmod {42}$.

Similarly, we can find that $n\equiv 0\pmod {47}$ and $n\equiv 0\pmod {46}$, so the answer is \[43+2+3+7+47+23=\boxed{125}.\]
This post has been edited 2 times. Last edited by megahertz13, May 22, 2025, 4:54 PM
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