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
P6 Geo Finale
math_comb01   7
N 2 minutes ago by GuvercinciHoca
Source: XOOK 2025/6
Let $ABC$ be a triangle with incenter $I$ and excenters $I_A$, $I_B$, $I_C$ opposite to $A,B,C$ respectively. Suppose $BC$ meets the circumcircle of $I_AI_BI_C$ at points $D$ and $E$. $X$ and $Y$ lie on the incircle of $\triangle ABC$ so that $DX$ and $EY$ are tangents to the incircle (different from $BC$). Prove that the circumcircles of $\triangle AXY$ and $\triangle ABC$ are tangent.

Proposed by Anmol Tiwari
7 replies
math_comb01
Feb 10, 2025
GuvercinciHoca
2 minutes ago
A functional equation
super1978   1
N 14 minutes ago by pco
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(y-x)-xf(y))+f(x)=y(1-f(x)) $$for all $x,y \in \mathbb R$
1 reply
super1978
an hour ago
pco
14 minutes ago
Sequences Handout
M11100111001Y1R   4
N 18 minutes ago by MR.1
Source: Own
Hi everyone, I wrote this handout about sequences in NT.
Hope you enjoy!
4 replies
+2 w
M11100111001Y1R
Oct 19, 2022
MR.1
18 minutes ago
[Handout] 50 non-traditional functional equations
gghx   2
N 44 minutes ago by GreekIdiot
Sup guys,

I'm retired. I love FEs. So here's 50 of them. Yea...

Functional equations have been one of the least enjoyed topics of math olympiads in recent times, mostly because so many techniques have been developed to just bulldoze through them. These chosen problems do not fall in that category - they require some combi-flavoured creativity to solve (to varying degrees).

For this reason, this handout is aimed at more advanced problem solvers who are bored of traditional FEs and are up for a little challenge!

In some sense, this is dedicated to the "covid FE community" on AoPS who got me addicted to FEs, people like EmilXM, hyay, IndoMathXdZ, Functional_equation, GorgonMathDota, BlazingMuddy, dangerousliri, Mr.C, TLP.39, among many others: thanks guys :). Lastly, thank you to rama1728 for suggestions and proofreading.

Anyways...
2 replies
gghx
Sep 23, 2023
GreekIdiot
44 minutes ago
No more topics!
2015 solutions for quotient function!
raxu   47
N Feb 14, 2025 by megahertz13
Source: TSTST 2015 Problem 5
Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$.

Proposed by Iurie Boreico
47 replies
raxu
Jun 26, 2015
megahertz13
Feb 14, 2025
2015 solutions for quotient function!
G H J
Source: TSTST 2015 Problem 5
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jj_ca888
2726 posts
#42
Y by
I claim we can construct an infinite sequence of primes $p_i$ for which no $p_i$ divides another $p_j - 1$. We do this inductively. For the base case, say we simply pick $p_1 = 5$ ad $p_2 = 7$. Thereafter, for every $k \geq 3$, we pick $p_k \equiv 2 \pmod {p_1p_2\ldots p_{k-1}}$ which clearly exists by Dirichlet.

Construct this sequence up to a certain $n$. I claim that the $2^n$ possible numbers $X = x_1x_2\ldots x_n$ generated by $x_i \in \{p_i(p_i - 1), (p_i - 1)^2\}$ all satisfy $\phi(X)$ being equal to the same value. Indeed, consider $X_1$ with primes in $S_1$ contributing $x_i$ of the former form and $X_2$ with primes in $S_2 \neq S_1$ contributing $x_i$ of the latter form. Then, since each $p_i$ is relatively prime to each other $p_j - 1$, we have\[\phi(X_1) = \phi(x_1\ldots x_n) = \left(\prod_{p_i \in S_1} \phi(p_i)\right) \phi\left(\prod (p_i - 1)^e\right)\]where $e = 1$ if $p_i \in S_1$ and $2$ otherwise. Similarly, \[\phi(X_2) = \phi(x_1\ldots x_n) = \left(\prod_{p_i \in S_2} \phi(p_i)\right) \phi\left(\prod (p_i - 1)^e\right)\]where $e = 1$ if $p_i \in S_2$ and $2$ otherwise. Divide them and quickly see that the quotient is $1$. Indeed, so taking sufficietly large $n$, we have $2^n > 2015$ numbers for which $\phi$ generates the same value.
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blackbluecar
303 posts
#43
Y by
Interesting!

Clearly, $\varphi(p(p-1))= \varphi((p-1)^2)$. Clearly, by induction, Dirichlet, and CRT we can construct a set of primes $P = \{p_1,p_2,...,p_{\text{big }} \}$ where $p_i-1$ is not divisible by any element in $P$. For any $i$, we can either choose $p_i(p_i-1)$ or $(p_i-1)^2$ and each is independent since they are relatively prime. Clearly, $2^{\text{big}} \geq 2015$. So we are done.
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Sprites
478 posts
#44
Y by
Let $p_1,p_2,p_3,......$ be an increasing sequence of primes and fix a positive integer $k$
Clearly $N=p_1 p_2 p_3.....p_k$ and $x_i=N \cdot (1 - \frac{1}{p_i})$ satisfies $\phi(N)=\phi(x_i)$ since $\frac{\phi(x_i)}{x_i}$$=\prod(1-\frac{q}{p_j}) \cdot \frac{p_i}{p_i-1}$$=$${\phi(N) \over N }\cdot$$ \frac{p_i}{p_i-1}$$=$$\frac{\phi(n)}{x_i}$,as required.
This post has been edited 1 time. Last edited by Sprites, Aug 20, 2021, 5:53 AM
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ChrisWren
323 posts
#45
Y by
$m=2^{20}\cdot 3^9\cdot 5^3\cdot 7^2\cdot 11^2$ works. We can let $\nu_{p}{(n)}$ equal $0$ or $1$ for every prime in $\{13, 17, 19, 23, 29, 31, 37, 41, 43, 61, 67\}$ and adjust the number of factors of $2,3,5,7,11$ in $n$ as needed.

The number $m$ is equivalent to $$15296168853504000$$
This post has been edited 1 time. Last edited by ChrisWren, Dec 1, 2021, 9:08 PM
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megarnie
5488 posts
#46 • 1 Y
Y by GoodMorning
Solved with GoodMorning.

We claim $m=\boxed{2^{41}\cdot 3^{19}}$ works.

Let $S$ be the set of primes $\{5,13,17,37,73,97,109,163,193,577,769\}$. We chose these primes because they are $1$ more than a power of $2$ times a power of $3$. Note that $m=\varphi\left({\prod_{x\in S}x}\right)$.

Claim: There exists a positive integer $x$ with $\varphi(x)=2^a\cdot 3^b$, for all positive integers $a$, nonnegative integers $b$, and $a=b=0$. Furthermore, $x$ has no prime factors except $2$ and $3$.
Proof: Set $x=2^a\cdot 3^{b+1}$. $\blacksquare$

Take any subset $T\subset S$. We have \[\frac{\varphi\left(\prod_{x\in S}x\right)}{\varphi\left(\prod_{x\in T}x\right)}=\varphi\left(\prod_{x\in S, x\not\in T}x\right)\]Note that this value can be written in the form $2^a3^b$, where $a>0$. Setting $n=2^a\cdot 3^{b+1}\cdot \prod_{x\in T}x$ gives $\varphi(n)=m$. Also setting $n=\prod_{x\in S}x$ works.

There are $2^{11}-1=2047$ ways to choose $T$, and one additional way, which gives at least $2048>2015$ different values of $n$.
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CyclicISLscelesTrapezoid
370 posts
#47
Y by
Oops.

We claim that \[m=312 \cdot 408 \cdot 576 \cdot 660 \cdot 936 \cdot 1152 \cdot 1800 \cdot 1320 \cdot 1008 \cdot 1200=149967695669069120274432000000\]works.

Choose $n=a_1 \cdot a_2 \cdots a_{11}$, where
\begin{align*}
a_1 &\in \{1,2\}\\
a_2 &\in \{313,3 \cdot 157\}\\
a_3 &\in \{409,5 \cdot 103\}\\
a_4 &\in \{577,7 \cdot 97\}\\
a_5 &\in \{661,11 \cdot 67\}\\
a_6 &\in \{937,13 \cdot 79\}\\
a_7 &\in \{1153,17 \cdot 73\}\\
a_8 &\in \{1801,19 \cdot 101\}\\
a_9 &\in \{1321,23 \cdot 61\}\\
a_{10} &\in \{1009,29 \cdot 37\}\\
a_{11} &\in \{1201,31 \cdot 41\}
.\end{align*}Since there are $2^{11}>2015$ choices for $n$, we are done.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Apr 5, 2022, 7:16 PM
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john0512
4167 posts
#48
Y by
Am I getting trolled?

We claim that $m=2^{10000}3^{10000}$ works.

Consider the eleven primes $$5,17,257,65537$$$$7,19,163$$$$13,37,109$$$$73.$$Each of these primes has the property that the number 1 less than it contains no prime factors other than possibly 2 or 3. We will generate 2048 different $n$ for which $\phi(n)=2^{10000}3^{10000}$ in the following manner:

Take any subset of the aformentioned 11 primes (which there are 2048 of), multiply all of those primes, and then multiply that result by 6. Call this intermediate result $I$. Clearly, $\phi(I)=2^a3^b$ for $a,b<10000.$ Then, since $I$ is already a multiple of 6, we then have $$\phi(I\cdot 2^{10000-a}3^{10000-b})=2^{10000}3^{10000},$$which gives us a solution for each of the 2048 values of $I$, so we are done
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trying_to_solve_br
191 posts
#49
Y by
Wait a minute..
Just take primes $p_1,p_2,...,p_2015$ such that $p_i$ does not divide $p_j - 1$ for all $i,j$, this is possible by Dirichilet.

Now, take all the prime divisors $q_i$ of $p_i-1$ for all $i$ and let $C$ be a big integer. Take numbers $p_i.q_1^{a_1}.q_2^{a_2}...q_j^{a_j}$, choosing $a_i$ in the following way: noticing $q_i\neq p_j$ for all $i,j$ by our choice, we choose $a_i$ such that $v_{q_i}(p_i-1)+a_i=C$, which solves the problem as all the v_q's of the numbers are equal, and all the $q-1$ all cancel out.
This post has been edited 1 time. Last edited by trying_to_solve_br, Jul 5, 2023, 12:22 AM
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minusonetwelth
224 posts
#50
Y by
For a positive integer $n=p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$, we have that $\phi(n)=\left(p_1^{\alpha_1}-p_1^{\alpha_1-1}\right)\cdot\ldots\cdot \left(p_k^{\alpha_k}-p_k^{\alpha_k-1}\right)$. Note that if we have $\phi(n)=x$ has at least $a$ solutions and $\phi(n)=y$ has at least $b$ solutions, then $\phi(n)=xy$ has at least $ab$ solutions as long as each of the $a$ solutions are coprime to each of the $b$ solutions. Letting the $a$ solutions for the first equation be $\{a_1,\ldots,a_a\}$ and the $b$ solutions for the second equation be $\{b_1,\ldots,b_b\}$, we see that this is true because if $a_i$ and $b_j$ are coprime, then $\phi(a_i\cdot b_j)=\phi(a_i)\cdot\phi(b_j)$. Therefore, finding enough numbers $m$ for which $\phi(n)=m$ has more than one solution will do the trick. For this,, just pick $\left\lceil\log_2(2015)\right\rceil=11$ pairs of paiwise distinct primes $p,q$ such that $r=(p-1)(q-1)+1$ is prime, and which are each also pairwise distinct from those constructed numbers $r$ (just do small cases, sorry :(). As then $\phi(r)=\phi(pq)=(p-1)(q-1)$, we get what we want.
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HamstPan38825
8794 posts
#51
Y by
I feel like I have some kind of obsession with Dirichlet now :/

Pick $2015$ primes $p \equiv 1 \pmod {10}$ by Dirichlet, and let $n$ be the LCM of $(p-1)^2$ for these primes multiplied by $10^{1000}$. Notice that we have $$\phi(n) = \phi\left((p-1) \cdot \frac n{p-1}\right) = \phi\left( p \cdot \frac n{p-1}\right)$$for each of these primes $p$ because $p-1$ divides $\frac n{p-1}$. It follows that by varying across all $p$ we can construct $2015$ solutions to $\phi(k) =\phi(n)$, as needed.
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ryanbear
1055 posts
#52
Y by
Let $n=30^{1434143414341434}$.
Let $p$ be the product of any subset of $\{ 7, 11, 13, 17, 31, 37, 41, 61, 97, 151, 181\}$ besides $1$. Note that there are $2047$ values of $p$.
Note that $\phi(\frac{pn}{\phi(p)})=\phi(p)*\phi(\frac{n}{\phi(p)})=\phi(n)$, since for all $p$, $\phi(p)$ is the product of all the prime factors of $p$ minus one, which only has prime factors $2,3,$ or $5$, so it divides $n$ and dividing by $\phi(p)$ does not impact the prime factors and only divides the final value by $\phi(p)$, which is then multiplied back.
All values of $p$ work, which is more than $2015$ values.
This post has been edited 1 time. Last edited by ryanbear, Jun 7, 2024, 12:43 AM
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ezpotd
1241 posts
#53
Y by
Consider the $11$ pairs of primes $(3,5), (7,13), (31,61) ,(37,73) ,( 57,113), (79,147) , (97,193) , (139,277) , (157, 313) , ( 199,397), (211, 421)$. The $2048$ numbers generated from picking one prime from each pair and taking the product have Totient functions differing by powers of $2$, so it suffices to multiply each of those numbers by a power of $2$, and we are done.
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lelouchvigeo
171 posts
#54
Y by
Nice Application of Dirichlet.Sketch
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HamstPan38825
8794 posts
#55
Y by
Let $p_1, p_2, \dots, p_N$ be the first $N$ primes for some sufficiently large $n$, and consider
\[m=\phi(p_1p_2 \cdots p_N) = (p_1-1)(p_2-1)\cdots(p_N-1).\]The claim is that for all $N/2 < k < N$, we can construct an $n$ such that $p_k \nmid n$, $p_i \mid n$ for all $i \neq k$, and $\phi(n) = m$. Clearly these $n$ are distinct, so they exhibit our desired examples.

To do this, let $q_1, q_2, \dots, q_\ell$ be the maximal prime powers that divide $p_k - 1$. By construction, for each $i$ there exists an index $j < k$ such that $q_i$ is equal to the $a_i$th power of $p_j$. Then taking
\[n = (p_k - 1)\prod_{i \neq k} p_i\]works for each $k$.

Remark: The construction for $n$ is oddly simplistic by number theory standards.
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megahertz13
3144 posts
#56
Y by
The value $$m=(2-1)(3-1)(5-1)(7-1)\dots(k-1)$$works, where $k$ is the $2015$th prime number. The values of $n$ that work are of the form $$n=2\cdot 3\cdot5\cdot7\dots k\cdot\frac{p-1}{p}$$where $p$ is one of the first $2015$ primes. To prove that these work, notice that all the prime factors of $p-1$ are included in the other primes, so $$\varphi(n)=(p-1)\varphi\bigg(\frac{2\cdot3\cdot5\dots k}{p}\bigg)=(p-1)\frac{\varphi(2)\varphi(3)\varphi(5)\dots\varphi(k)}{\varphi(p)}=m$$since $\varphi(p)=p-1$.
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