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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.

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[*]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]
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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
Finding all possible $n$ on a strange division condition!!
MathLuis   10
N 13 minutes ago by justaguy_69
Source: Bolivian Cono Sur Pre-TST 2021 P1
Find the sum of all positive integers $n$ such that
$$\frac{n+11}{\sqrt{n-1}}$$is an integer.
10 replies
MathLuis
Nov 12, 2021
justaguy_69
13 minutes ago
IMO 2012 P5
mathmdmb   123
N 21 minutes ago by SimplisticFormulas
Source: IMO 2012 P5
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.

Proposed by Josef Tkadlec, Czech Republic
123 replies
mathmdmb
Jul 11, 2012
SimplisticFormulas
21 minutes ago
Fixed line
TheUltimate123   14
N 23 minutes ago by amirhsz
Source: ELMO Shortlist 2023 G4
Let \(D\) be a point on segment \(PQ\). Let \(\omega\) be a fixed circle passing through \(D\), and let \(A\) be a variable point on \(\omega\). Let \(X\) be the intersection of the tangent to the circumcircle of \(\triangle ADP\) at \(P\) and the tangent to the circumcircle of \(\triangle ADQ\) at \(Q\). Show that as \(A\) varies, \(X\) lies on a fixed line.

Proposed by Elliott Liu and Anthony Wang
14 replies
TheUltimate123
Jun 29, 2023
amirhsz
23 minutes ago
Computing functions
BBNoDollar   7
N 37 minutes ago by ICE_CNME_4
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
7 replies
BBNoDollar
May 18, 2025
ICE_CNME_4
37 minutes ago
RMO 2024 Q2
SomeonecoolLovesMaths   14
N 44 minutes ago by Adywastaken
Source: RMO 2024 Q2
For a positive integer $n$, let $R(n)$ be the sum of the remainders when $n$ is divided by $1,2, \cdots , n$. For example, $R(4) = 0 + 0 + 1 + 0 = 1,$ $R(7) = 0 + 1 + 1 + 3 + 2 + 1 + 0 = 8$. Find all positive integers such that $R(n) = n-1$.
14 replies
SomeonecoolLovesMaths
Nov 3, 2024
Adywastaken
44 minutes ago
Decimal functions in binary
Pranav1056   3
N an hour ago by ihategeo_1969
Source: India TST 2023 Day 3 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
3 replies
Pranav1056
Jul 9, 2023
ihategeo_1969
an hour ago
Beautiful numbers in base b
v_Enhance   21
N an hour ago by Martin2001
Source: USEMO 2023, problem 1
A positive integer $n$ is called beautiful if, for every integer $4 \le b \le 10000$, the base-$b$ representation of $n$ contains the consecutive digits $2$, $0$, $2$, $3$ (in this order, from left to right). Determine whether the set of all beautiful integers is finite.

Oleg Kryzhanovsky
21 replies
v_Enhance
Oct 21, 2023
Martin2001
an hour ago
Polynomial method of moving points
MathHorse   6
N an hour ago by Potyka17
Two Hungarian math olympians achieved significant breakthroughs in the field of polynomial moving points. Their main results are summarised in the attached pdf. Check it out!
6 replies
MathHorse
Jun 30, 2023
Potyka17
an hour ago
Intertwined numbers
miiirz30   2
N an hour ago by Gausikaci
Source: 2025 Euler Olympiad, Round 2
Let a pair of positive integers $(n, m)$ that are relatively prime be called intertwined if among any two divisors of $n$ greater than $1$, there exists a divisor of $m$ and among any two divisors of $m$ greater than $1$, there exists a divisor of $n$. For example, pair $(63, 64)$ is intertwined.

a) Find the largest integer $k$ for which there exists an intertwined pair $(n, m)$ such that the product $nm$ is equal to the product of the first $k$ prime numbers.
b) Prove that there does not exist an intertwined pair $(n, m)$ such that the product $nm$ is the product of $2025$ distinct prime numbers.
c) Prove that there exists an intertwined pair $(n, m)$ such that the number of divisors of $n$ is greater than $2025$.

Proposed by Stijn Cambie, Belgium
2 replies
miiirz30
Yesterday at 10:12 AM
Gausikaci
an hour ago
Geometry
shactal   0
an hour ago
Two intersecting circles $C_1$ and $C_2$ have a common tangent that meets $C_1$ in $P$ and $C_2$ in $Q$. The two circles intersect at $M$ and $N$ where $N$ is closer to $PQ$ than $M$ . Line $PN$ meets circle $C_2$ a second time in $R$. Prove that $MQ$ bisects angle $\widehat{PMR}$.
0 replies
shactal
an hour ago
0 replies
Inspired by 2025 KMO
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=0 $ and $ a^2+b^2+c^2+d^2= 6 .$ Prove that $$ -\frac{3}{4} \leq abcd\leq\frac{9}{4}$$Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=6 $ and $ a^2+b^2+c^2+d^2= 18 .$ Prove that $$ -\frac{9(2\sqrt{3}+3)}{4} \leq abcd\leq\frac{9(2\sqrt{3}-3)}{4}$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
RMO 2024 Q1
SomeonecoolLovesMaths   25
N 2 hours ago by Adywastaken
Source: RMO 2024 Q1
Let $n>1$ be a positive integer. Call a rearrangement $a_1,a_2, \cdots , a_n$ of $1,2, \cdots , n$ nice if for every $k = 2,3, \cdots , n$, we have that $a_1 + a_2 + \cdots + a_k$ is not divisible by $k$.
(a) If $n>1$ is odd, prove that there is no nice arrangement of $1,2, \cdots , n$.
(b) If $n$ is even, find a nice arrangement of $1,2, \cdots , n$.
25 replies
SomeonecoolLovesMaths
Nov 3, 2024
Adywastaken
2 hours ago
4 variables
Nguyenhuyen_AG   10
N 2 hours ago by Butterfly
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
10 replies
Nguyenhuyen_AG
Dec 21, 2020
Butterfly
2 hours ago
2025 KMO Inequality
Jackson0423   3
N 2 hours ago by sqing
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
3 replies
Jackson0423
Yesterday at 4:32 PM
sqing
2 hours ago
Sum of whose elements is divisible by p
nntrkien   45
N May 10, 2025 by Jupiterballs
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
45 replies
nntrkien
Aug 8, 2004
Jupiterballs
May 10, 2025
Sum of whose elements is divisible by p
G H J
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
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nntrkien
61 posts
#1 • 5 Y
Y by Davi-8191, Adventure10, Mango247, cubres, and 1 other user
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
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iura
481 posts
#2 • 6 Y
Y by Adventure10, ken3k06, v4913, Mango247, cubres, and 1 other user
This is problem 6 form IMO 1995.There are two different solutions to it:
The first is purely combinatorial: Tahe $ A = \{0..p - 1\}, B = \{p..2p - 1\}$.
For a set $ S$ different from $ A$ and $ B$ denote $ C = S \bigcap A$ ,$ D = S \bigcap B$.
Then the sets $ (C + x) \bigcap D$ form a group of $ p$ sets where all the residues appear. (The set $ A + x = \{(a + x) mod p| a \in A\}$)
So we can couunt easily.
The second uses multisection formula counting the coeff $ x^{pk} y^p$ at the polynomial
$ (1 + y)(1 + xy)\cdots (1 + x^{2p - 1}y)$
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ZetaX
7579 posts
#3 • 5 Y
Y by Adventure10, Mango247, balllightning37, cubres, and 1 other user
Define $X: =\{ 1,2,...,p\},Y: =\{p+1,p+2,...,2p\}$.
See $X$ and $Y$ as a representant system for $\mathbb{Z}/ p \mathbb{Z}$, since not more is necessary (so all identities concerning $X$ or $Y$ must be seen $\mod p$).
For a subset $A \subset X$ and $z \in \mathbb{Z}/ p \mathbb{Z}$ define $A+z: =\{ a+z \in X| a \in A \}$ and similar for a subset $B \subset Y$.
Call $A \subset X$ trivial iff $A=\emptyset$ or $A=X$, similar for $B \subset Y$ again.
Now there are only $4$ subsets $P \subset (X \cup Y)$ such that both $P \cap X$ and $P \cap Y$ are trivial, so call these subsets trival from now on too.
Then define for a nontrivial subset $P \subset (X \cup Y)$ a 'translation':
if $P \cap X$ is nontrivial define $P+z : = ((P\cap X) +z) \cup (P \cap Y)$ and $P+z : = (P \cap X) \cup ((P\cap Y) +z)$ otherwise.
Now it's easy to see that when $z$ goes through all possible residue classes, that also the sum of the elements of $P+z$ goes through all of them, so there is exactly one sum that is divisible by $p$.
So the 'translation' divides the nontrivial subsets into equivalence classes of $p$ sets each, all sets of the same class having same number of elements.

Since there are exactly $\binom{2p}{p}$ subsets of the desired order $p$ and $2$ of them are trivial, we get that there are $\frac{\binom{2p}{p}-2}{p}+2$ such subsets.

(note that this technique trivially generalises to other types of subsets and all sets of type $\{1,2,...,kp\}$)
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pluricomplex
390 posts
#4 • 3 Y
Y by Adventure10, Mango247, cubres
Thanks for nice work ZetaX (is it true ? :D )!
It's a very famous problem ! It seems that there's another way to count this by using polynomial (or useing generator function) . Do you know who creat that nice solution? It's just my wondering about the exactly author of that nice solution which i saw on The Mathematics and Youth Magazine (Vietnam Magazine) in 1996!
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Bluesea
59 posts
#5 • 3 Y
Y by Adventure10, Mango247, cubres
can you write the second solution in a slightly way iura
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iura
481 posts
#6 • 11 Y
Y by duanby, Davi-8191, huricane, HolyMath, Scrutiny, Adventure10, cubres, and 4 other users
Consider the polynomial $f(x,y)=(1+xy)(1+x^2y)\ldots(1+x^{2p-1}y)$, the $k$-th factor refering to whether $k$ is present in the set or not.

Then the monomial $x^ky^l$ will refer to the set having $l$ elements and sum of elements $k$, so we need to compute the sum of coefficients of $x^{kp}y^p$ of $f$ to find the answer.

To do this, we need to compute the sum of coefficients $x^{pk}y^{pl}$ and substract 2, since there are two terms for $l\neq 1$:$1, x^{p(2p-1)}y^{2p}$.

To compute the sum of coefficients of $x^{pk}y^{pl}$ we use Multisection formula that says us that it equals

$\frac{1}{p^2}\sum_{i,j}f(w^i,w^j)$($w$ is $p$th root of unity).

Now if $w^i$ is not $1$, $f(w^i,w^j)=\prod(1+w^{ik}w^j)=\prod (1+w^k)^2$ (every $w^m$ will be written twice as $w^{ki}w^j$). This equals $\prod (-1-w^k)^2=g(-1)^2=((-1)^p-1)^2=4$ ($g(x)=x^p-1$ is the polynomial with roots $w^i$). Since there are $p-1$ choices for $w^i$, $p$ choices for $w^j$ we get $4p(p-1)$.

Finally, if $w^i=1$, $f(1,w^j)=(1+w^j)^{2p}$. To evaluate $\sum_{j=0}^{p-1} (1+w^j)^{2p}$, note that it equals $p$ times the coefficients of $x^p$ in the polynomial $(1+x)^{2p}$ by the same multisection formula, so it equals $p\binom{2p}{p}$.

So our total sum is $\frac 1{p^2} (4p(p-1)+p\binom{2p}{p}) =\frac{\binom{2p}{p}-2}p+4$.
Substracting 2 we get the desired answer.
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Bluesea
59 posts
#7 • 3 Y
Y by Adventure10, Mango247, cubres
iura wrote:
.


To compute the sum of coefficients of $x^{pk}y^{pl}$ we use Multisection formula that says us that it equals

$\frac{1}{p^2}\sum_{i,j}f(w^i,w^j)$($w$ is $p$th root of unity).

.
how can you prove that?What is miultisection formula?
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iura
481 posts
#8 • 7 Y
Y by pavel kozlov, Adventure10, guptaamitu1, Mango247, cubres, and 2 other users
If $w$ is a $n$th root of unity then $\sum_{i=0}^{n-1} w^k=0$ if $w\neq 1$ and $\sum_{i=0}^{n-1} w^k=n$ if $w=1$. By using this result, we can prove the Multisection Formula (which holds also for polynomials of more variables) :

$f(x)=\sum a_ix^i$, then $\sum_{i\equiv k\pmod m} a_ix^i=\frac 1n (\sum_{i=0}^{n-1} f(w^ix) w^{-ik})$. Particularly

$\sum_{i \equiv k\pmod m} a_i=\frac 1n(\sum_{i=0}^{n-1}f(w^i)w^{-ik})$
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Bluesea
59 posts
#9 • 3 Y
Y by Adventure10, Mango247, cubres
thank,i understand it now
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mathmanman
1444 posts
#10 • 9 Y
Y by siddigss, Polynom_Efendi, Supercali, Illuzion, Imayormaynotknowcalculus, Adventure10, cubres, and 2 other users
Another (wonderful) solution :

We let $w$ be the p-th root of unity.
We thus have :
\[ \prod_{k=1}^{2p} (x - w^k) = (x^p - 1)^2 = x^{2p} - 2x^p + 1. \]
We now define the quantity :
\[ t(w) = \sum_{1 \leq j_1 < j_2 < \ldots < j_p \leq 2p} w^{j_1}w^{j_2} \ldots w^{j_p}. \]
So we have $t(w) = 2$. Besides, if we write $t(w) = \sum a_jw^j$, then $a_j$ represents the number of sub-sets ${j_1, j_2, \ldots, j_p}$ of ${1, 2, \ldots, 2p}$ such that $j_1 + j_2 + \ldots + j_p \equiv j \pmod p$, and we can write :
\[ (a_0 - 2) + a_1w + \ldots + a_{p-1}w^{p-1} = 0. \]
Since the minimal polynomial of $w$ on $\mathbb{Q}[X]$ is $1 + x + \ldots + x^{p-1}$, it follows that :
\[ a_0 - 2 = a_1 = a_2 = \ldots = a_{p-1}. \]
Besides, $a_0 + a_1 + \ldots + a_{p-1} = \binom {2p}{p}$, we conclude immediately that :
\[ a_0 = \frac 1p \left\{\binom {2p}{p} - 2 \right \} + 2. \]
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bilarev
200 posts
#11 • 5 Y
Y by Polynom_Efendi, Imayormaynotknowcalculus, Adventure10, Assassino9931, cubres
mathmanman wrote:
Another (wonderful) solution
mathmanman is this the solution of Nikolai Nikolov from Bulgaria for which he get a special prize?
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mathmanman
1444 posts
#12 • 5 Y
Y by Polynom_Efendi, Imayormaynotknowcalculus, Adventure10, Mango247, cubres
Yes, exactly. :)
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bilarev
200 posts
#13 • 3 Y
Y by Adventure10, Mango247, cubres
iura wrote:
$\frac 1{p^2} (4p(p-1)+p\binom{2p}{p}) =\frac{\binom{2p}{p}-2}p+4$.
This is not true...$\frac 1{p^2} (4p(p-1)+p\binom{2p}{p}) =\frac{\binom{2p}{p}-4}p+4$. ;)
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bilarev
200 posts
#14 • 4 Y
Y by Adventure10, Adventure10, Mango247, cubres
iura wrote:
Consider the polynomial $f(x,y)=(1+xy)(1+x^2y)\ldots(1+x^{2p-1}y)$, the $k$-th factor refering to whether $k$ is present in the set or not.
I think that we have to consider the polynomial $f(x,y)=(1+xy)(1+x^2y)\ldots(1+x^{2p}y)$...am I right?
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me@home
2349 posts
#15 • 3 Y
Y by Adventure10, Mango247, cubres
Sorry to bump this....

but when I solve it I keep getting $\left\lceil \frac{{2p \choose p}}{p}\right\rceil$, is this the same answer?
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