It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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
Feb 2, 2025
0 replies
I need this. Pls help me
Giahuytls2326   12
N 6 minutes ago by Giahuytls2326
Source: Own
Given $\triangle ABC$ , orthocenter $H$, $E$ is the center of Euler's circle of $\triangle ABC$. $X,Y,Z$ is the midpoint of $AH,BH,CH$. From $X,Y,Z$ draw the tangent of $(E)$, cut a line through $E$ perpendicular to sides $BC, CA,AB$ at $M,N,P$ .Prove that $AM,BN,CP$ concurrent
12 replies
Giahuytls2326
Thursday at 4:59 PM
Giahuytls2326
6 minutes ago
Thanks u!
Ruji2018252   4
N 7 minutes ago by sqing
Let $a,b,c>0$ and $a+b+c=abc$
Find minimum (and prove)
\[P=(ab-1)(bc+1)^6(ca-1)\]
4 replies
Ruji2018252
Yesterday at 2:38 PM
sqing
7 minutes ago
Functional Equation
AnhQuang_67   4
N 7 minutes ago by gordian.knot
Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$x^2f(x)+f(1-x)=2x-x^4, \forall x \in \mathbb{R} $$
4 replies
AnhQuang_67
Yesterday at 4:02 PM
gordian.knot
7 minutes ago
thanks u!
Ruji2018252   1
N 27 minutes ago by sqing
Let $a,b,c\in \mathbb{R},a,b,c\ne 0$ and $a+b+c=0.$ Find minimum (and prove)
\[C=\dfrac{b^2+c^2-a^2}{b^2+c^2}+\dfrac{c^2+a^2-b^2}{c^2+a^2}+\dfrac{a^2+b^2-c^2}{a^2+b^2}\]
1 reply
1 viewing
Ruji2018252
3 hours ago
sqing
27 minutes ago
No more topics!
Every positive integer divides some number of a sequence
freemind   8
N Aug 20, 2020 by VicKmath7
Source: 1st Romanian Master in Mathematics (RMIM) 2008, Bucharest, Problem 3
Let $ a>1$ be a positive integer. Prove that every non-zero positive integer $ N$ has a multiple in the sequence $ (a_n)_{n\ge1}$, $ a_n=\left\lfloor\frac{a^n}n\right\rfloor$.
8 replies
freemind
Feb 9, 2008
VicKmath7
Aug 20, 2020
Every positive integer divides some number of a sequence
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Source: 1st Romanian Master in Mathematics (RMIM) 2008, Bucharest, Problem 3
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freemind
337 posts
#1 • 7 Y
Y by anantmudgal09, tenplusten, Adventure10, megarnie, Mango247, and 2 other users
Let $ a>1$ be a positive integer. Prove that every non-zero positive integer $ N$ has a multiple in the sequence $ (a_n)_{n\ge1}$, $ a_n=\left\lfloor\frac{a^n}n\right\rfloor$.
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Fedor Petrov
520 posts
#2 • 4 Y
Y by rafayaashary1, Adventure10, and 2 other users
Strange problem. Let $ N=N_1N_2$, where $ N_2$ and $ a$ are coprime, and $ N_1$ divides $ a^C$ for suitable positive integer $ C$. Without loss of generality, $ CN_1$ divides $ a^{C}$ (we may take $ C$ being large power of $ a$). Take large prime $ p$ such that $ p-1$ is divisible by $ \varphi(n)$. The existence of such $ p$ is well-known elementary case of the Dirichlet theorem, which may be obtained using the cyclotomic polynomials. Then take $ n=pC$. Then by Fermat's little theorem we have $ [a^{Cp}/(Cp)]=(a^{Cp}-a^C)/(Cp)$, which is divisible by $ N=N_1N_2$.
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hrkksym0
6 posts
#3 • 1 Y
Y by Adventure10
We can solve this problem without the Dirichlet theorem.

There are positive integers $m$ and $M$ such that $GCD(m,M)=1, N\mid a^{m}M$.
We consider $M>1$(it is obvious when $M=1$)
Let $R$ is an integer such that $M^{R}>a^{m+\varphi(M)}$ and
$a^{R} mod \varphi(M), a^{R+1} mod \varphi(M), a^{R+2} mod \varphi(M), ...$
is a periodical sequence(the period is not more than $\varphi(M)$).
Then there exists an integer $0\le i\le\varphi(M)$ and
$a^{a^{R}M^{R}-m-i}\equiv a^{R}\pmod{\varphi(M)}$.
Let $k=a^{R}M^{R}-m-i$ and $n=a^{k}M^{R}$, then

$
\varphi(M^{R+1})=M^{R}\varphi(M)\mid n-k-m-i\\
\Rightarrow a^{n-k-m-i}\equiv 1\pmod{M^{R+1}}\\
\Rightarrow a^{n-k}\equiv a^{m+i}\pmod{a^{m}M^{R+1}}\\
\Rightarrow \lfloor\dfrac{a^{n-k}}{M^{R}}\rfloor=\lfloor\dfrac{a^{n}}{n}\rfloor\\
=\lfloor\dfrac{Ia^{m}M^{R+1}+a^{m+i}}{M^{R}}\rfloor (I\in\mathbb{Z})\\
=Ia^{m}M+\lfloor\dfrac{a^{m+i}}{M^{R}}\rfloor=Ia^{m}M   ( a^{m+i}\le a^{m+\varphi(M)}<M^{R})\\
\Rightarrow N\mid \lfloor\dfrac{a^{n}}{n}\rfloor\\
$
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dinoboy
2903 posts
#4 • 2 Y
Y by Adventure10, Mango247
This solution was found in collaboration with proglote. it is quite similar to Fedor Petrov's solution actually.

Write $N = bc$ where $\gcd(c,a) = \gcd(b,c) = 1$. Let $b|a^k$ and $k$ clearly exists since $b$'s prime factors are purely those of $a$ as well. Then consider $n = a^kp$ for a prime $p \equiv 1 \pmod{\text{ord}_c(a)}$ and $p > \max(c,a^k - k)$ which exists by Dirichlet. It is easy to see that: \[\frac{a^{a^kp}}{a^kp} = \frac{a^{a^kp-k}}{p}\]
Then remark that $a^{a^kp-k} \equiv a^{a^k-k} \pmod{p}$, so it follows \[a_{a^kp} = \frac{a^{a^kp - k} - a^{a^k-k}}{p}\]

It is clear that $a^{a^k(p-1)}-1$ divides the numerator. It follows that as $c|(a^{a^k(p-1)}-1)$ due to the order condition and $\gcd(c,p) = 1$ due to $p > c$ that $c$ divides $a_{a^kp}$. It is also clear that $b$ must divide $a_{a^kp}$ as $a^{a^k-k}$ divides the numerator. Thus we are done as we have shown an arbitrary $N$ divides some term of the sequence.
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JuanOrtiz
366 posts
#5 • 1 Y
Y by Adventure10
I solved it pretty ugly, no Dirichlet. Sorry I used different values. I want $k | floor(a^r/r)$

Instead of $r$ I used $ra^x$, and then I define $P$ very big and then $Q$ such that $ord_Q a=P$ and then $r$ such that $ord_r a= Q$ using two Zsigmondys.

Then I can control $x$ in mod $d=ord_{k_1} a$ (where $k_1$ is biggest divisor of $k$ coprime to $a$), mod $ORD_d a$ (where $ORD$ is like the $ord$ function but without caring about $a$ and the modulus coprime), mod $P$ and mod $Q$ and use Chinese Theorem to finish, choosing $a^{ra^x-x} \equiv a^{c+i}$ (mod $kr$), where $c$ is such that $g | a^c$ ($g=k/k_1$) and $i<d$).

$i$ is useful since $ORD_da$ and $d$ aren't necessarily coprime.
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anantmudgal09
1979 posts
#6 • 1 Y
Y by Adventure10
Probably i am mistaken or this is on the easier end of RMM. Indeed let $p$ be a prime such that $p>a$ and $p$ is congruent to 1 modulo $\phi(N)$.(Clearly by the properties of cyclotomic polynomials infinitely many such $p$ exist or just use Weak Dirichlet's theorem) Now clearly $N$ divides $a_p$
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Shaddoll
688 posts
#7 • 4 Y
Y by guangzhou-2015, canhhoang30011999, tenplusten, Adventure10
Solution
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MNJ2357
644 posts
#8 • 2 Y
Y by Adventure10, Mango247
I'll just post this cuz it's different from the above solutions.
(Actually, #6 and #7 are really nice, but proving that there are infinitely many primes in the $nk+1$ takes a lot of time.)

My main idea is Power Towers.

We'll prove that $n=b_m-1$ works for sufficiently large $m>2$. One can easily check that
$$a_n=\left\lfloor\frac{a^n}n\right\rfloor=\left\lfloor\frac{a^{a^{b_{m-1}}-1}}{a^{a^{b_{m-2}}}-1}\right\rfloor=\frac{a^{a^{b_{m-1}}-1}-a^{a^{b_{m-2}}-1}}{a^{a^{b_{m-2}}}-1}=a^{b_{m-1}-1}\cdot\frac{a^{b_m-b_{m-1}}-1}{a^{b_{m-1}}-1}$$Now let $N=N_1N_2$, where $N_2$ is the largest divisor of $N_2$ such that $(a,N_2)=1$. We'll choose $m$ large enough so that $N_1|a^{b_{m-1}-1}$. Observe that for each $p|N_2$, the sequence $v_p(a^{b_m}-1)$ is bounded, so let $x_p$ be the upper bound. Using the well-known property of power-towers, find some $M$ such that
$$m>M\Longrightarrow \phi(N_2\cdot\prod_{p|N_2}p^{x_p})|{b_m-b_{m-1}}$$, and we're done. :)
This post has been edited 1 time. Last edited by MNJ2357, Feb 8, 2019, 7:49 AM
Reason: (N,N_2)=1 -> (a,N_2)=1
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VicKmath7
1383 posts
#9
Y by
Same as above solutions, posting for storage.
Solution. Take a prime $p$, such that $p>a$ (1) and $p=k.\phi{N}$ (*) and $(p,N)=1$ (**)(existence is guaranteed by Dirichlet`s theorem). By FLT, $a^p-a$ is divisible by $p$ (2).From (1) and (2), we have that $a_p=\lfloor\frac{a^p}{p}\rfloor=\frac {a^p-a}{p}$. Using (*), (**) and Euler`s theorem, we immediately see that $a_p$ is divisible by $N$, which is what we need. $\blacksquare $
This post has been edited 1 time. Last edited by VicKmath7, Aug 20, 2020, 12:06 PM
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