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jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
Have fun
centslordm   6
N 28 minutes ago by Moubinool
Source: Instagram
Compute \[\prod^\infty_{k=2} \sqrt[2^k]{1 - \frac 1k - \frac 1{k^2} + \frac1{k^3}}.\]
6 replies
centslordm
5 hours ago
Moubinool
28 minutes ago
Can you Tai(y)lor it?
SatisfiedMagma   1
N an hour ago by Alphaamss
Show that
\[\cos(x) \le \frac{\sin^3(x)}{x^3} \]for $x \in (0,\pi/2)$,
1 reply
SatisfiedMagma
6 hours ago
Alphaamss
an hour ago
On coefficients of a polynomial over a finite field
Ciobi_   1
N Today at 1:00 AM by AndreiVila
Source: Romania NMO 2025 12.4
Let $p$ be an odd prime number, and $k$ be an odd number not divisible by $p$. Consider a field $K$ be a field with $kp+1$ elements, and $A = \{x_1,x_2, \dots, x_t\}$ be the set of elements of $K^*$, whose order is not $k$ in the multiplicative group $(K^*,\cdot)$. Prove that the polynomial $P(X)=(X+x_1)(X+x_2)\dots(X+x_t)$ has at least $p$ coefficients equal to $1$.
1 reply
Ciobi_
Apr 2, 2025
AndreiVila
Today at 1:00 AM
Collatz Conjecture
Cpw945   7
N Yesterday at 11:55 PM by maromex
Source: "Collatz Conjecture Confirmed Through Connectivity of Odd and 8mod12 Positive Integers" on viXra
Hello everyone! I am a college student who has created a potential proof for the Collatz Conjecture, which I have posted on Vixra, under the title “Collatz Conjecture Confirmed by Connectivity of Odds and 8mod12 Positive Integers”. It is in Section 2507, under the name Chloe Williams. Feel free to check it out and tell me if my solution idea would work. The link for my paper is down below.

https://vixra.org/abs/2507.0020
7 replies
Cpw945
Tuesday at 8:56 PM
maromex
Yesterday at 11:55 PM
a sufficient condition for nilpotence
CatalinBordea   3
N Tuesday at 4:26 PM by Filipjack
Source: Romanian District Olympiad 2015, Grade XII, Problem 4
Let $ m $ be a non-negative ineger, $ n\ge 2 $ be a natural number, $ A $ be a ring which has exactly $ n $ elements, and an element $ a $ of $ A $ such that $ 1-a^k $ is invertible, for all $ k\in\{ m+1,m+2,...,m+n-1\} . $
Prove that $ a $ is nilpotent.
3 replies
CatalinBordea
Sep 26, 2018
Filipjack
Tuesday at 4:26 PM
A ring whose set of units is finite
Filipjack   3
N Jul 3, 2025 by a_0a
Source: Selected for Romania District Olympiad 2020, grade 12, problem 3
Let $(A,+, \cdot)$ be a ring in which the set of invertible elements is finite. Prove that the following assertions are equivalent:
[list=1]
[*] For any non-invertible element $a \in A,$ there is a non-invertible element $b \in A$ such that $ab=a+b.$
[*] Any non-invertible element of the ring $A$ is nilpotent.
[/list]

Mihai Opincariu
3 replies
Filipjack
Jun 17, 2025
a_0a
Jul 3, 2025
Inequalities with |H|
WakeUp   4
N Jul 3, 2025 by a_0a
Source: Romanian MO 2010 Grade 12
Let $G$ be a finite group of order $n$. Define the set
\[H=\{x:x\in G\text{ and }x^2=e\},\]
where $e$ is the neutral element of $G$. Let $p=|H|$ be the cardinality of $H$. Prove that
a) $|H\cap xH|\ge 2p-n$, for any $x\in G$, where $xH=\{xh:h\in H\}$.
b) If $p>\frac{3n}{4}$, then $G$ is commutative.
c) If $\frac{n}{2}<p\le\frac{3n}{4}$, then $G$ is non-commutative.

Marian Andronache
4 replies
WakeUp
Aug 6, 2012
a_0a
Jul 3, 2025
Endomorphisms on finite groups
Valentin Vornicu   9
N Jul 3, 2025 by a_0a
Source: RMO 2008, Grade 12, Problem 4
Let $ \mathcal G$ be the set of all finite groups with at least two elements.

a) Prove that if $ G\in \mathcal G$, then the number of morphisms $ f: G\to G$ is at most $ \sqrt [p]{n^n}$, where $ p$ is the largest prime divisor of $ n$, and $ n$ is the number of elements in $ G$.

b) Find all the groups in $ \mathcal G$ for which the inequality at point a) is an equality.
9 replies
Valentin Vornicu
Apr 30, 2008
a_0a
Jul 3, 2025
Equivalence of two propositions
Filipjack   1
N Jun 17, 2025 by alexheinis
Source: Selected for Romania District Olympiad 2020, grade 12, problem 1
Let $k$ and $n$ be two positive integers and let $(G, \cdot)$ be a group of order $n.$ Prove that the following assertions are equivalent:

[list=1]
[*] The numbers $k$ and $n$ are coprime.
[*] For any subgroup $H$ of $G,$ the set $\{x \in G: x^k \in H \}$ is included in $H.$
[/list]

Adrian Boțan
1 reply
Filipjack
Jun 17, 2025
alexheinis
Jun 17, 2025
Preferred permutations
pablock   2
N Jun 15, 2025 by GoldenBoy03
Source: 2023 IMC #5
Fix positive integers $n$ and $k$ such that $2 \le k \le n$ and a set $M$ consisting of $n$ fruits. A permutation is a sequence $x=(x_1,x_2,\ldots,x_n)$ such that $\{x_1,\ldots,x_n\}=M$. Ivan prefers some (at least one) of these permutations. He realized that for every preferred permutation $x$, there exist $k$ indices $i_1 < i_2 < \ldots < i_k$ with the following property: for every $1 \le j < k$, if he swaps $x_{i_j}$ and $x_{i_{j+1}}$, he obtains another preferred permutation.

Prove that he prefers at least $k!$ permutations.
2 replies
pablock
Aug 2, 2023
GoldenBoy03
Jun 15, 2025
Find n such that "deg n and no roots => irreducible"
Filipjack   1
N Jun 14, 2025 by alexheinis
Source: Romanian National Olympiad 1995 – Grade 12 – Problem 2
Let $K$ be a finite field. Find the set of integers $n \ge 2$ having the property that any $n$-th degree polynomial in $K[X]$ with no roots in $K$ is irreducible over $K.$
1 reply
Filipjack
Jun 14, 2025
alexheinis
Jun 14, 2025
Number of (infinite) idempotents in a ring
RobertRogo   7
N Jun 7, 2025 by ysharifi
Source: -
(crucial so you don't waste your time) spoiler
Let $R$ be a reversible ring such that $\lvert E(R)  \rvert \geq \lvert \mathbb{N}  \rvert$. Prove that $ \lvert E(R)  \rvert \geq \lvert \mathbb{R} \rvert$.
7 replies
RobertRogo
Jun 7, 2025
ysharifi
Jun 7, 2025
On units in a ring with a polynomial property
Ciobi_   4
N Jun 5, 2025 by KevinYang2.71
Source: Romania NMO 2025 12.1
We say a ring $(A,+,\cdot)$ has property $(P)$ if :
\[
\begin{cases}

\text{the set } A \text{ has at least } 4 \text{ elements} \\
\text{the element } 1+1 \text{ is invertible}\\
x+x^4=x^2+x^3 \text{ holds for all } x \in A
\end{cases}
\]a) Prove that if a ring $(A,+,\cdot)$ has property $(P)$, and $a,b \in A$ are distinct elements, such that $a$ and $a+b$ are units, then $1+ab$ is also a unit, but $b$ is not a unit.
b) Provide an example of a ring with property $(P)$.
4 replies
Ciobi_
Apr 2, 2025
KevinYang2.71
Jun 5, 2025
Reducing the exponents for good
RobertRogo   3
N Jun 2, 2025 by RobertRogo
Source: The national Algebra contest (Romania), 2025, Problem 3/Abstract Algebra (a bit generalized)
Let $A$ be a ring with unity such that for every $x \in A$ there exist $t_x, n_x \in \mathbb{N}^*$ such that $x^{t_x+n_x}=x^{n_x}$. Prove that
a) If $t_x \cdot 1 \in U(A), \forall x \in A$ then $x^{t_x+1}=x, \forall x \in A$
b) If there is an $x \in A$ such that $t_x \cdot 1 \notin U(A)$ then the result from a) may no longer hold.

Authors: Laurențiu Panaitopol, Dorel Miheț, Mihai Opincariu, me, Filip Munteanu
3 replies
RobertRogo
May 20, 2025
RobertRogo
Jun 2, 2025
Possible values of determinant of 0-1 matrices
mathematics2004   4
N May 30, 2025 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
May 30, 2025
Possible values of determinant of 0-1 matrices
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 Simon Marais, A3
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mathematics2004
82 posts
#1 • 1 Y
Y by Tip_pay
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 $A$.
Z K Y
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Isolemma
62 posts
#2 • 1 Y
Y by loup blanc
We introduce some more notations.
Let $\mathcal M_n$ be the set of all $n \times n$ matrices with at most two entries in each row equal to $1$ and all other entries equal to $0$.
Let $D_n$ be the set $\{\det A: A \in \mathcal M_n\}$.
Thus the problem asks for the size of $D_{2021}$.
It is easy to see that $D_1 = \{0, 1\}$ and $D_2 = \{0, 1, -1\}$.
Also for $n \geq 2$, it is clear that $x \in D_n \iff -x \in D_n$ for any $x$, by exchanging any two rows of a matrix.

For any $A \in \mathcal M_{n - 1}$, the matrix $\begin{pmatrix}1 & \\ & A\end{pmatrix}$ belongs to $\mathcal M_n$ and has the same determinant as $A$. This shows that $D_{n - 1} \subseteq D_n$.

Let $A$ be a matrix in $\mathcal M_n$ such that $\det A \in D_n \setminus D_{n - 1}$.
If there is one row of $A$ with no $1$, then the determinant is $0$, which is already in $D_{n - 1}$, contradicting our assumption.
If there is one row of $A$ with only one $1$, then removing the row and column containing that $1$ from $A$ will give a matrix in $\mathcal M_{n - 1}$, which has the same determinant as $A$. This means that $\det A$ belongs to $D_{n - 1}$, contradicting our assumption.
It follows that each row of $A$ contains exactly two $1$'s.

Similarly, if there is one column of $A$ with no $1$ or only one $1$, then $\det A$ belongs to $D_{n - 1}$, which is impossible.
As the total number of $1$'s is equal to $2n$, each column of $A$ also contains exactly two $1$'s.

Moreover, we know that no two rows of $A$ are identical, otherwise the determinant would be zero.
Thus we have a simple graph, whose vertices are the columns of $A$, and each row defines an edge connecting the two columns on which this row has a $1$.
We have seen that each vertex of this graph has degree $2$. Thus the graph is a disjoint union of several circles.
This means that, up to permutation of rows and columns, the matrix $A$ can be written as a block diagonal matrix, whose diagonal blocks are all of the form $$\begin{pmatrix}1 & 1 & & \\& 1 & 1 &  \\ & & \ddots & \ddots &  \\ & & & 1 & 1\\1& & & & 1\end{pmatrix}.$$Note that each block has size at least $3 \times 3$.
It is easy to see that for a block of size $k$, the determinant of the block is $(-1)^k + 1$, which is $0$ or $2$ according to whether $k$ is even or odd.
Thus all blocks are of odd size. That is, we can write $n = k_1 + \cdots + k_r$ with all $k_i \geq 3$ odd, and the determinant of $A$ is, up to sign, equal to $2^r$.

We summerize the above discussion as follows: for $n \geq 2$, we have $$D_n = \{0, \pm 1\} \cup \bigcup_{m = 3}^n \{\pm 2^r: m = k_1 + \cdots + k_r, k_i \geq 3, 2\nmid k_i\}.$$From here, it is easy to see that $D_{2021} = \{0\} \cup \{\pm 2^r: 0 \leq r \leq 673\}$.
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stanislavssss2025
1 post
#3
Y by
I believe, you are wrong. 2021 is divided into blocks of size at least 3. If one of the blocks is of even parity, the whole thing is zero. Thus consider when all k are odd. But then, it is not true that number of blocks can be any value from 0 to 673.

It can't be any even value, as then we have even number of odd blocks = 2021, which can't be true. However, we can show that any odd number of blocks can be achieved. Indeed, substract 1 from the length of each block. Now they are all even, and all >= 2. And 2021 turns into 2021-odd = even number. And it is obvious, that we can achieve any even number as a sum of even numbers >= 2 (Just take all 2's and one number of what's left). Thus, the correct answer is 0 and +- 2^(2r+1), where r goes from 0 to (673-1)/2 = 336.
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Isolemma
62 posts
#4
Y by
stanislavssss2025 wrote:
I believe, you are wrong.

Note that $D_n$ also contains $D_{n - 1}$. The division into blocks is only relevant for those in $D_n \backslash D_{n - 1}$.
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loup blanc
3624 posts
#5
Y by
The key of the Isolemma's proof is "We have seen that each vertex of this graph has degree $2$. Thus the graph is a disjoint union of several circles."
cf here
https://math.stackexchange.com/questions/3749686/let-g-be-a-graph-such-that-all-its-vertices-have-degree-2-prove-that-g-is-a
In fact, the elements of ${D_n}^{+}=\{x\in D_n;x\geq 2\}$ are generated by the block diagonal matrices containing some of the following matrices on its diagonal:
$I_k$, $I_l+P_l$ -where $l\geq 3$ and $P_l$ is the matrix of the cycle $(1,l,l-1,\cdots,2,1)$- .
For example, if $n=11$:
$I_3+P_3,I_8$ gives $2$
$I_3+P_3,I_3+P_3,I_5$ gives $4$
$I_3+P_3,I_3+P_3,I_3+P_3,I_2$ gives $8$.
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