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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
Factorial Divisibility
Aryan-23   47
N a minute ago by ezpotd
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
$$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$
47 replies
Aryan-23
Jul 9, 2023
ezpotd
a minute ago
2-var inequality
sqing   3
N 25 minutes ago by sqing
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq 8$$$$ (a^2+b^2)(a+1)(b+1) \leq 8$$Let $ a,b> 0 ,a^3+ab(a+b)+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$
3 replies
1 viewing
sqing
an hour ago
sqing
25 minutes ago
Infinite number of sets with an intersection property
Drytime   8
N 41 minutes ago by math90
Source: Romania TST 2013 Test 2 Problem 4
Let $k$ be a positive integer larger than $1$. Build an infinite set $\mathcal{A}$ of subsets of $\mathbb{N}$ having the following properties:

(a) any $k$ distinct sets of $\mathcal{A}$ have exactly one common element;
(b) any $k+1$ distinct sets of $\mathcal{A}$ have void intersection.
8 replies
Drytime
Apr 26, 2013
math90
41 minutes ago
Factorials divide
va2010   37
N an hour ago by ND_
Source: 2015 ISL N2
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
37 replies
va2010
Jul 7, 2016
ND_
an hour ago
IMC 1994 D2 P1
j___d   13
N Yesterday at 11:20 PM by krigger
Let $f\in C^1[a,b]$, $f(a)=0$ and suppose that $\lambda\in\mathbb R$, $\lambda >0$, is such that
$$|f'(x)|\leq \lambda |f(x)|$$for all $x\in [a,b]$. Is it true that $f(x)=0$ for all $x\in [a,b]$?
13 replies
j___d
Mar 6, 2017
krigger
Yesterday at 11:20 PM
D1039 : A strange and general result on series
Dattier   0
Yesterday at 10:33 PM
Source: les dattes à Dattier
Let $f \in C([0,1];[0,1])$ bijective, $f(0)=0$ and $(a_k) \in [0,1]^\mathbb N$ with $ \sum \limits_{k=0}^{+\infty} a_k$ converge.

Is it true that $\sum \limits_{k=0}^{+\infty} f(a_k)\times f^{-1}(a_k)$ converge?
0 replies
Dattier
Yesterday at 10:33 PM
0 replies
Aproximate ln(2) using perfect numbers
YLG_123   5
N Yesterday at 8:55 PM by ei_killua_
Source: Brazilian Mathematical Olympiad 2024, Level U, Problem 1
A positive integer \(n\) is called perfect if the sum of its positive divisors \(\sigma(n)\) is twice \(n\), that is, \(\sigma(n) = 2n\). For example, \(6\) is a perfect number since the sum of its positive divisors is \(1 + 2 + 3 + 6 = 12\), which is twice \(6\). Prove that if \(n\) is a positive perfect integer, then:
\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]where the sums are taken over all prime divisors \(p\) of \(n\).
5 replies
YLG_123
Oct 12, 2024
ei_killua_
Yesterday at 8:55 PM
Quadruple Binomial Coefficient Sum
P162008   4
N Yesterday at 8:40 PM by vmene
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
4 replies
P162008
Thursday at 8:04 PM
vmene
Yesterday at 8:40 PM
IMC 1994 D1 P5
j___d   5
N Yesterday at 5:39 PM by krigger
a) Let $f\in C[0,b]$, $g\in C(\mathbb R)$ and let $g$ be periodic with period $b$. Prove that $\int_0^b f(x) g(nx)\,\mathrm dx$ has a limit as $n\to\infty$ and
$$\lim_{n\to\infty}\int_0^b f(x)g(nx)\,\mathrm dx=\frac 1b \int_0^b f(x)\,\mathrm dx\cdot\int_0^b g(x)\,\mathrm dx$$
b) Find
$$\lim_{n\to\infty}\int_0^\pi \frac{\sin x}{1+3\cos^2nx}\,\mathrm dx$$
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 5:39 PM
2023 Putnam A2
giginori   21
N Yesterday at 3:32 PM by pie854
Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2 n$; that is to say, $p(x)=$ $x^{2 n}+a_{2 n-1} x^{2 n-1}+\cdots+a_1 x+a_0$ for some real coefficients $a_0, \ldots, a_{2 n-1}$. Suppose that $p(1 / k)=k^2$ for all integers $k$ such that $1 \leq|k| \leq n$. Find all other real numbers $x$ for which $p(1 / x)=x^2$.
21 replies
giginori
Dec 3, 2023
pie854
Yesterday at 3:32 PM
Putnam 2019 A1
awesomemathlete   33
N Yesterday at 3:25 PM by cursed_tangent1434
Source: 2019 William Lowell Putnam Competition
Determine all possible values of $A^3+B^3+C^3-3ABC$ where $A$, $B$, and $C$ are nonnegative integers.
33 replies
awesomemathlete
Dec 10, 2019
cursed_tangent1434
Yesterday at 3:25 PM
IMC 1994 D1 P2
j___d   5
N Yesterday at 3:11 PM by krigger
Let $f\in C^1(a,b)$, $\lim_{x\to a^+}f(x)=\infty$, $\lim_{x\to b^-}f(x)=-\infty$ and $f'(x)+f^2(x)\geq -1$ for $x\in (a,b)$. Prove that $b-a\geq\pi$ and give an example where $b-a=\pi$.
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 3:11 PM
A Construction in Multivariable Analysis
MrOrange   0
Yesterday at 2:11 PM
Source: Garling's A COURSE IN MATHEMATICAL ANALYSIS
Construct a continuous real valued function \( f \) on \( \mathbb{R}^2 \) for which
\[
\lim_{R \to \infty} \int_{\|x\|_2 \leq R} f(x) \, dx = 0
\]and for which
\[
\lim_{R \to \infty} \int_{\|x\|_\infty \leq R} f(x) \, dx \text{ does not exist.}
\]
0 replies
MrOrange
Yesterday at 2:11 PM
0 replies
Possible values of determinant of 0-1 matrices
mathematics2004   4
N Yesterday at 1:56 PM by loup blanc
Source: 2021 Simon Marais, A3
Let $\mathcal{M}$ be the set of all $2021 \times 2021$ matrices with at most two entries in each row equal to $1$ and all other entries equal to $0$.
Determine the size of the set $\{ \det A : A \in M \}$.
Here $\det A$ denotes the determinant of the matrix $A$.
4 replies
mathematics2004
Nov 2, 2021
loup blanc
Yesterday at 1:56 PM
2024 IMO P1
EthanWYX2009   104
N Yesterday at 1:29 AM by SYBARUPEMULA
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
104 replies
EthanWYX2009
Jul 16, 2024
SYBARUPEMULA
Yesterday at 1:29 AM
2024 IMO P1
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 IMO P1
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EthanWYX2009
872 posts
#1 • 41 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, vadava_lx
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
1135 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.
This post has been edited 4 times. Last edited by NO_SQUARES, Jul 16, 2024, 2:46 PM
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bin_sherlo
734 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
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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
578 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 • 25 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, corgi61
$\alpha$ 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
840 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
167 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
167 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
38 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
167 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
460 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
167 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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H
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a