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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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0 replies
jlacosta
May 1, 2025
0 replies
2023 Japan Mathematical Olympiad Preliminary
parkjungmin   0
7 minutes ago
Please help me if I can solve the Chinese question
0 replies
parkjungmin
7 minutes ago
0 replies
Polynomial divisible by x^2+1
Miquel-point   1
N an hour ago by luutrongphuc
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
1 reply
Miquel-point
Apr 6, 2025
luutrongphuc
an hour ago
the same prime factors
andria   5
N an hour ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
5 replies
andria
Sep 6, 2015
bin_sherlo
an hour ago
A sharp estimation of the product
mihaig   0
an hour ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
an hour ago
0 replies
Integration Bee Kaizo
Calcul8er   63
N 3 hours ago by MS_asdfgzxcvb
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
63 replies
Calcul8er
Mar 2, 2025
MS_asdfgzxcvb
3 hours ago
Japanese high school Olympiad.
parkjungmin   1
N 3 hours ago by GreekIdiot
It's about the Japanese high school Olympiad.

If there are any students who are good at math, try solving it.
1 reply
parkjungmin
Yesterday at 5:25 AM
GreekIdiot
3 hours ago
Already posted in HSO, too difficult
GreekIdiot   0
4 hours ago
Source: own
Find all integer triplets that satisfy the equation $5^x-2^y=z^3$.
0 replies
GreekIdiot
4 hours ago
0 replies
Square on Cf
GreekIdiot   0
4 hours ago
Let $f$ be a continuous function defined on $[0,1]$ with $f(0)=f(1)=0$ and $f(t)>0 \: \forall \: t \in (0,1)$. We define the point $X'$ to be the projection of point $X$ on the x-axis. Prove that there exist points $A, B \in C_f$ such that $ABB'A'$ is a square.
0 replies
GreekIdiot
4 hours ago
0 replies
Japanese Olympiad
parkjungmin   4
N Today at 8:55 AM by parkjungmin
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
4 replies
parkjungmin
Saturday at 6:51 PM
parkjungmin
Today at 8:55 AM
ISI UGB 2025 P3
SomeonecoolLovesMaths   11
N Today at 8:21 AM by Levieee
Source: ISI UGB 2025 P3
Suppose $f : [0,1] \longrightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x) | \leq f(x)$ for all $x \in [0,1]$, then show that $f(x) = 0$ for all $x$.
11 replies
SomeonecoolLovesMaths
Yesterday at 11:32 AM
Levieee
Today at 8:21 AM
D1020 : Special functional equation
Dattier   4
N Today at 7:57 AM by Dattier
Source: les dattes à Dattier
1) Are there any $(f,g) \in C(\mathbb R,\mathbb R_+)$ increasing with
$$\forall x \in \mathbb R, f(x)(\cos(x)+3/2)+g(x)(\sin(x)+3/2)=\exp(x)$$?

2) Are there any $(f,g) \in C(\mathbb R,\mathbb R_+)$ increasing with
$$\forall x \in \mathbb R, f(x)(\cos(x)+3/2)+g(x)(\sin(x)+3/2)=\exp(x/2)$$?
4 replies
Dattier
Apr 24, 2025
Dattier
Today at 7:57 AM
Mathematical expectation 1
Tricky123   1
N Today at 6:57 AM by navier3072
X is continuous random variable having spectrum
$(-\infty,\infty) $ and the distribution function is $F(x)$ then
$E(X)=\int_{0}^{\infty}(1-F(x)-F(-x))dx$ and find the expression of $V(x)$

Ans:- $V(x)=\int_{0}^{\infty}(2x(1-F(x)+F(-x))dx-m^{2}$

How to solve help me
1 reply
Tricky123
Yesterday at 9:51 AM
navier3072
Today at 6:57 AM
Tough integral
Martin.s   0
Today at 4:00 AM
$$\int_0^{\pi/2}\ln(\tan(\theta/2))
\;\frac{4\cos\theta\cos(2\theta)}{4\sin^4\theta+1}\,d\theta.$$
0 replies
Martin.s
Today at 4:00 AM
0 replies
Minimum value
Martin.s   3
N Yesterday at 5:24 PM by Martin.s
What is the minimum value of
$$
\frac{|a + b + c + d| \left( |a - b| |b - c| |c - d| + |b - a| |c - a| |d - a| \right)}{|a - b| |b - c| |c - d| |d - a|}
$$over all triples $a, b, c, d$ of distinct real numbers such that
$a^2 + b^2 + c^2 + d^2 = 3(ab + bc + cd + da).$

3 replies
Martin.s
Oct 17, 2024
Martin.s
Yesterday at 5:24 PM
2024 IMO P1
EthanWYX2009   103
N Apr 27, 2025 by ashwinmeena
Source: 2024 IMO P1
Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$)

Proposed by Santiago Rodríguez, Colombia
103 replies
EthanWYX2009
Jul 16, 2024
ashwinmeena
Apr 27, 2025
2024 IMO P1
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 IMO P1
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EthanWYX2009
865 posts
#1 • 40 Y
Y by David-Vieta, ATGY, MathIQ., holdmyquadrilateral, aaaa_27, mathhotspot, Sylvestra, NO_SQUARES, ehuseyinyigit, vsamc, GeoKing, MathPerson12321, scannose, pingupignu, centslordm, Ziyadel, lpieleanu, kamatadu, KAME06, aidan0626, EpicBird08, Sedro, Rounak_iitr, LeonidasTheConquerer, aqwxderf, Eka01, AlexCenteno2007, Math_.only., MUKESH_KUMAR, mathematicsy, KhaiMathAddict, farhad.fritl, numbertheory97, zaf123, Roh_Hoi_Chan, axsolers_24, Atilla, xytan0585, cubres, NicoN9
Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$)

Proposed by Santiago Rodríguez, Colombia
This post has been edited 2 times. Last edited by EthanWYX2009, Jul 19, 2024, 5:33 AM
Reason: change to original text
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NO_SQUARES
1124 posts
#3 • 7 Y
Y by crocodilepradita, GeoKing, KhaiMathAddict, Roh_Hoi_Chan, cubres, Dhruv777, piano1900
Very nice problem!
I think that answer is all even $\alpha$. They are clearly works. Now let $2 \not | \alpha$.
WLOG $0<\alpha<2$, since we can translate it on $2$.
Let $$\frac{\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor}{n}=f(n)/n.$$Then $\frac{f(n+1)}{n+1}-\frac{f(n)}{n}=\frac{n\lfloor{\alpha(n+1)\rfloor}-f(n)}{n(n+1)}$ is integer and so $n(n+1)|n\lfloor\alpha(n+1)\rfloor-f(n)$. But $0<^*RHS<2n(n+1)$ and so $g(n)=RHS=n(n+1)$.
But now we can look at $g(n+1)$ and get $\lfloor{\alpha(n+2) \rfloor}=2+\lfloor (n+1)\alpha \rfloor$ for all $n$, which is impossible for $0<\alpha <2$. (We can use induction on $n$ to get $\lfloor \alpha (n+1) \rfloor =2n$ and thus $\frac{2n-1}{n+1}< \alpha <\frac{2n+1}{n+1}$ for all $n$).
(*) Here we used the fact that $\alpha>1$. In other case $0<g(n)<n(n+1)$ for large $n$, which is impossible.
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bin_sherlo
728 posts
#4 • 10 Y
Y by Collatz09, GeoKing, sansgankrsngupta, bartu_kalkan, farhad.fritl, KhaiMathAddict, Roh_Hoi_Chan, cubres, Dhruv777, piano1900
Answer is all even integers. If $\alpha=2k,$ then $\frac{n(n+1)}{2}.2k=n(n+1)k$ is divisible by $n$. Let's show that other real numbers does not satisfy the condition. If $\alpha$ satisfies, then $\alpha+2$ also satisfies hence we can assume that $0\leq \alpha<2$. Let $\alpha\neq 0\iff 0<\alpha<2$. If $\alpha<1,$ denote $m$ as $\lceil \frac{1}{\alpha}\rceil=m\iff (m-1)\alpha<1\leq m\alpha$ where $2\leq m\in Z^+$. Since $m|\lfloor \alpha \rfloor+...+\lfloor (m-1)\alpha \rfloor+\lfloor m\alpha \rfloor=\lfloor m\alpha \rfloor$ But $1\leq \lfloor m\alpha \rfloor<m$ which is impossible. Also $\alpha=1$ doesn't hold which can be seen by taking $n=2$. Thus, $1<\alpha<2$. Let $\alpha=1+\beta$ where $0<\beta<1$. We have $n|\frac{n(n+1)}{2}+\sum{\lfloor i\beta} \rfloor$ Let $\lceil \frac{1}{\beta}\rceil =k$ We have $k|\frac{k(k+1)}{2}+\lfloor k\beta \rfloor=\frac{k(k+1)}{2}+1$ since $(k-1)\beta<1$ yields $k\beta<\beta+1<2$. So $2k|k^2+k+2\implies k=1,2$. $\frac{1}{\beta}>1\implies k=2$. Let's show that if $\lfloor m\beta \rfloor=m-1$ for all $1\leq m\leq N,$ then $\lfloor (m+1)\beta \rfloor=m$ for $m\geq 3$.
\[\frac{m+1}{2}|m+1|\frac{(m+1)(m+2)}{2}+\frac{(m-1)m}{2}+\lfloor (m+1)\beta \rfloor\equiv \frac{m^2-m+m^2+3m+2}{2}+\lfloor (m+1)\beta \rfloor\equiv 1+\lfloor (m+1)\beta \rfloor\]Since $m+1|2+2\lfloor (m+1)\beta \rfloor$ and $m-1\leq \lfloor (m+1)\beta \rfloor<m+1,$ we get $\lfloor (m+1)\beta \rfloor=m$ which completes our induction.
So $\lfloor n\beta \rfloor=n-1$ for all $n$ but this is impossible since it gives that $n\beta>n-1\iff \beta>\frac{n-1}{n}=1-\frac{1}{n}$ where $n$ goes to infinity. Therefore $\alpha \not \in (0,2)$ as desired.$\blacksquare$
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Supertinito
45 posts
#5 • 75 Y
Y by holdmyquadrilateral, ATGY, crocodilepradita, bin_sherlo, ehuseyinyigit, silouan, Stuffybear, Davi Medeiros, Sylvestra, L567, vuanhnshn, Jalil_Huseynov, Tintarn, dxd29070501, GeoKing, meowme, Patrik, MathPerson12321, InternetPerson10, azzam2912, denistusk, Seicchi28, pingupignu, centslordm, Orthogonal., CyclicISLscelesTrapezoid, Assassino9931, KST2003, ihatemath123, aidan0626, EpicBird08, timon92, talkon, justJen, Gato_combinatorio, IndoMathXdZ, Sedro, Jack_w, tricky.math.spider.gold.1, ohiorizzler1434, MS_Kekas, CahitArf, aqwxderf, khina, ESAOPS, RobertRogo, Marcus_Zhang, study1126, eduD_looC, andlind, Mufara07, Natrium, APZZ-NHR, Funcshun840, KhaiMathAddict, Samujjal101, Genius132, DensSv, Nomad_from_QZ, zaf123, Roh_Hoi_Chan, namanrobin08, axsolers_24, Atilla, stillwater_25, Alex-131, cubres, TheBaiano, Dhruv777, NicoN9, NonoPL, SatisfiedMagma, SteppenWolfMath, manomanguinha, NeonMaths
I’m really excited to have a P1 on IMO for the second year in a row. This problem was proposed by me Santiago Rodríguez from Colombia. I genuinely hope that you enjoy my problem as much as I do.
Story of how I came up with the problem
This post has been edited 4 times. Last edited by Supertinito, Aug 31, 2024, 9:42 PM
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BR1F1SZ
577 posts
#6 • 4 Y
Y by Jack_w, KhaiMathAddict, Roh_Hoi_Chan, cubres
@above Thanks!!! It's a beautiful problem!! :coolspeak:
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chronondecay
145 posts
#7 • 24 Y
Y by crocodilepradita, azzam2912, ihatemath123, Pluto1708, Wikiliks, clarkculus, peter_ib, ehuseyinyigit, mastermind.hk16, Math_.only., cappucher, Vahur, ACGNmath, andlind, on1921379, Natrium, Turker31, KhaiMathAddict, Roh_Hoi_Chan, RobertRogo, cubres, TheBaiano, D4N13LCarpenter, ektorasmiliotis
$\alpha$ any even integer works, because $2(1+2+\cdots+n)=n(n+1)$. We claim these are the only solutions.

If $\alpha$ is not an even integer, we can add any even integer to $\alpha$ without changing whether the condition holds (due to the same identity), so we may assume that $-1<\alpha\le1$.

Case 1: $\alpha=1$. This fails because $1+2=3$ is not a multiple of 2.

Case 2: $0<\alpha<1$. Let $k>1$ be the smallest integer such that $k\alpha\ge1$; note that $k\alpha=(k-1)\alpha+\alpha<1+1=2$. Hence taking $n=k$, the sum is $0+0+\cdots+0+1=1$, which is not a multiple of $n$.

Case 3: $-1<\alpha<0$. Let $k>1$ be the smallest integer such that $k\alpha<-1$; note that $k\alpha=(k-1)\alpha+\alpha>(-1)+(-1)=-2$. Hence taking $n=k$, the sum is $(-1)+(-1)+\cdots+(-1)+(-2)=-n-1$, which is not a multiple of $n$.

Hence the only $\alpha$ which work are the even integers.
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ehuseyinyigit
835 posts
#8 • 4 Y
Y by KhaiMathAddict, zaf123, Roh_Hoi_Chan, cubres
Floor functions are becoming popular in international contests nowadays. It's interesting. Nice problem @Supertinito.
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Patrik
86 posts
#9 • 3 Y
Y by KhaiMathAddict, Roh_Hoi_Chan, cubres
Let $\alpha = a + b$ where $a = \lfloor \alpha \rfloor$ and $b = \alpha - \lfloor \alpha \rfloor$.
Then we get that $n$ divides
$$\frac{a}{2}(n(n + 1)) + \lfloor 2b \rfloor + \dots \lfloor nb \rfloor$$
Then if $a$ is even we have $n \mid \sum_{m = 2}^n \lfloor mb \rfloor$ from here we see clearly that $\lfloor mb \rfloor = 0$ for all $m$. This of course implies $b = 0$ therefore $\alpha = a = 2k$.

Now suppose $a$ is odd. Then we have that $\lfloor 2b \rfloor = 1$ and $n \mid \sum_{m = 2}^n \lfloor mb \rfloor$ for all odd $n$.

From this inductively we get $mb \geq \lfloor mb \rfloor = m - 1$ a contradiction as we know $b < 1$ on RHS is linear function of higher slope.
This post has been edited 3 times. Last edited by Patrik, Jul 16, 2024, 2:19 PM
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shanelin-sigma
165 posts
#10 • 4 Y
Y by Wikiliks, KhaiMathAddict, Roh_Hoi_Chan, cubres
Notice that $\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$
is the number of integer points in [1,n] within the interval (excluding boundaries) between y=nx and y=0
full solution (hand writting)
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shanelin-sigma
165 posts
#11 • 3 Y
Y by KhaiMathAddict, Roh_Hoi_Chan, cubres
What's the genre of the problem? Is it A or N?
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Funcshun840
22 posts
#12 • 2 Y
Y by Roh_Hoi_Chan, cubres
shanelin-sigma wrote:
What's the genre of the problem? Is it A or N?

It’s algebra as P2 is NT
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shanelin-sigma
165 posts
#13 • 3 Y
Y by Roh_Hoi_Chan, cubres, NicoN9
Funcshun840 wrote:
It’s algebra as P2 is NT

But I think this problem is more like an NT problem
since it reminds me some steps when proving quadratic reciprocity

Besides I think P3 is also an NT problem
NNN this year?
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CrazyInMath
457 posts
#14 • 2 Y
Y by Roh_Hoi_Chan, cubres
shanelin-sigma wrote:
Funcshun840 wrote:
It’s algebra as P2 is NT

But I think this problem is more like an NT problem
since it reminds me some steps when proving quadratic reciprocity

Besides I think P3 is also an NT problem
NNN this year?

If 2020 IMO P6 can be a G, I don't see how this can't be an A
Also P3 clearly is C, I don't see any number theory elements in it.
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vsamc
3789 posts
#15 • 3 Y
Y by KevinYang2.71, Roh_Hoi_Chan, cubres
Solution
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shanelin-sigma
165 posts
#16 • 2 Y
Y by Roh_Hoi_Chan, cubres
CrazyInMath wrote:
If 2020 IMO P6 can be a G, I don't see how this can't be an A
: D
CrazyInMath wrote:
Also P3 clearly is C, I don't see any number theory elements in it.
QAQ
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G
H
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a