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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
GCD of terms in a sequence
BBNoDollar   0
25 minutes ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
25 minutes ago
0 replies
Number Theory
fasttrust_12-mn   13
N 31 minutes ago by KTYC
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
13 replies
fasttrust_12-mn
Aug 15, 2024
KTYC
31 minutes ago
GCD of terms in a sequence
BBNoDollar   0
34 minutes ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
34 minutes ago
0 replies
Aime type Geo
ehuseyinyigit   3
N 42 minutes ago by sami1618
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
3 replies
ehuseyinyigit
Yesterday at 9:04 PM
sami1618
42 minutes ago
Putnam 2016 A1
Kent Merryfield   15
N Today at 10:51 AM by anudeep
Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k,$ the integer
\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\](the $j$-th derivative of $p(x)$ at $k$) is divisible by $2016.$
15 replies
Kent Merryfield
Dec 4, 2016
anudeep
Today at 10:51 AM
Determinant is 1
Entrepreneur   2
N Today at 8:27 AM by Entrepreneur
If a determinant is of $n^{\text{th}}$ order, and if the constituents of its first, second, ..., $n^{\text{th}}$ rows are the first $n$ figurate numbers of the first, second, ..., $n^{\text{th}}$ orders respectively, show that it's value is $1.$
2 replies
Entrepreneur
Yesterday at 7:14 PM
Entrepreneur
Today at 8:27 AM
Can cos(√2 t) be expressed as a polynomial in cost?
tom-nowy   1
N Today at 7:17 AM by Aiden-1089
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
Can $\cos ( \sqrt{2}\,  t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
1 reply
tom-nowy
Today at 7:10 AM
Aiden-1089
Today at 7:17 AM
36x⁴ + 12x² - 36x + 13 > 0
fxandi   2
N Today at 7:14 AM by MeKnowsNothing
Prove that for any real $x \geq 0$ holds inequality $36x^4 + 12x^2 - 36x + 13 > 0.$
2 replies
fxandi
Yesterday at 10:02 PM
MeKnowsNothing
Today at 7:14 AM
2024 Putnam A1
KevinYang2.71   20
N Today at 5:50 AM by thelateone
Determine all positive integers $n$ for which there exists positive integers $a$, $b$, and $c$ satisfying
\[
2a^n+3b^n=4c^n.
\]
20 replies
KevinYang2.71
Dec 10, 2024
thelateone
Today at 5:50 AM
2025 OMOUS Problem 4
enter16180   2
N Yesterday at 8:57 PM by Acridian9
Source: Open Mathematical Olympiad for University Students (OMOUS-2025)
Find all matrices $M \in M_{n}(\mathbb{C})$ such that following equality holds

$$
\operatorname{rank}(M)+\operatorname{rank}\left(M^{2023}-M^{2025}\right)=\operatorname{rank}\left(M-M^{2}\right)+\operatorname{rank}\left(M^{2023}+M^{2024}\right)
$$
2 replies
enter16180
Apr 18, 2025
Acridian9
Yesterday at 8:57 PM
Poker hand
Aksudon   1
N Yesterday at 6:32 PM by lucaminiati
Problem: In a standard 52-card deck, how many different five-card poker hands are there of 'two pairs'?

Can someone please explain what is logically wrong with the following solution? (It gives double of the right solution which supposed to be 123552).

13\binom{4}{2}*12\binom{4}{2}*44=247104

Thanks
1 reply
Aksudon
Yesterday at 5:14 PM
lucaminiati
Yesterday at 6:32 PM
Sequence with GCD involved
mathematics2004   3
N Yesterday at 5:54 PM by anudeep
Source: 2021 Simon Marais, A2
Define the sequence of integers $a_1, a_2, a_3, \ldots$ by $a_1 = 1$, and
\[ a_{n+1} = \left(n+1-\gcd(a_n,n) \right) \times a_n \]for all integers $n \ge 1$.
Prove that $\frac{a_{n+1}}{a_n}=n$ if and only if $n$ is prime or $n=1$.
Here $\gcd(s,t)$ denotes the greatest common divisor of $s$ and $t$.
3 replies
mathematics2004
Nov 2, 2021
anudeep
Yesterday at 5:54 PM
Putnam 2000 B2
ahaanomegas   20
N Yesterday at 5:05 PM by reni_wee
Prove that the expression \[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \] is an integer for all pairs of integers $ n \ge m \ge 1 $.
20 replies
ahaanomegas
Sep 6, 2011
reni_wee
Yesterday at 5:05 PM
f(x)f'(x)≥cos, f(∞)=undef. if f is bounded
jasperE3   2
N Yesterday at 4:21 PM by Rohit-2006
Source: VJIMC 2013 1.1
Let $f:[0,\infty)\to\mathbb R$ be a differentiable function with $|f(x)|\le M$ and $f(x)f'(x)\ge\cos x$ for $x\in[0,\infty)$, where $M>0$. Prove that $f(x)$ does not have a limit as $x\to\infty$.
2 replies
jasperE3
May 30, 2021
Rohit-2006
Yesterday at 4:21 PM
Two permutations
Nima Ahmadi Pour   12
N Apr 23, 2025 by Zhaom
Source: Iran prepration exam
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
12 replies
Nima Ahmadi Pour
Apr 24, 2006
Zhaom
Apr 23, 2025
Two permutations
G H J
Source: Iran prepration exam
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Nima Ahmadi Pour
160 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
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ZetaX
7579 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it):

Let $ (G, + )$ be a finite abelian group of order $ n$.
Let also $ a_1,a_2,...,a_{n - 1} \in G$ be arbitrary.
Then there exist pairwise distinct $ b_1,b_2,...,b_{n - 1} \in G$ and pairwise distinct $ c_1,c_2,...,c_{n - 1} \in G$ such that $ a_k = b_k + c_k$ for $ k = 1,2,...,n - 1$.

[Moderator edit: The Ars Combinatoria paper is:
F. Salzborn, G. Szekeres, A problem in Combinatorial Group Theory, Ars Combinatoria 7 (1979), pp. 3-5.]
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epitomy01
240 posts
#3 • 3 Y
Y by Dan37kosothangnao, Adventure10, Mango247
so could someone post a proof of either the problem or its generalization?
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spdf
136 posts
#4 • 1 Y
Y by Adventure10
you can find the proof in file shortlist 2005 which has been posted by orl
The main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\sum a_{i}\equiv 0(mod n)$ and $\sum b_{i}\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\neq\ b_{i}(modn)$ and $a_{j}\neq\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
i will come back with detail if you need
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ZetaX
7579 posts
#5 • 1 Y
Y by Adventure10
Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper...
It's a bit different from the ISL one.
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keira_khtn
485 posts
#6 • 2 Y
Y by Adventure10, Mango247
I think you didnt keep promise,Zetax :lol: Please post it here and now!
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bodom
123 posts
#7 • 2 Y
Y by Adventure10, Mango247
to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so :)
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ZetaX
7579 posts
#8 • 2 Y
Y by Adventure10, Mango247
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).
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arandomperson123
430 posts
#9 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).

that is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?
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ThE-dArK-lOrD
4071 posts
#10 • 16 Y
Y by Lam.DL.01, Mosquitall, nmd27082001, Arc_archer, MathbugAOPS, iceillusion, Aryan-23, magicarrow, k12byda5h, gabrupro, Mop2018, Adventure10, Mango247, bhan2025, CyclicISLscelesTrapezoid, winniep008hfi
Since it's almost twelve years without complete solution, here's the official solution:

Suppose there exists permutations $\sigma$ and $\tau$ of $[n]$ for some sequence $\{ a_i\}_{i\in [n]}$ so that $a_i\equiv_n \sigma (i)+\tau (i)$ for all $i\in [n]$.
Given a sequence $\{ b_i\}_{i\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\{ a_i\}_{i\in [n]}$ in only two positions, say $i_1$ and $i_2$.
We want to construct permutations $\sigma'$ and $\tau'$ of $[n]$ so that $b_i\equiv_n \sigma' (i) +\tau' (i)$ for all $i\in [n]$.
Recall that $b_i\equiv a_i\pmod{n}$ for all $i\in [n]$ that $i\neq i_1,i_2$.
Construct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\geq 2$, define $i_{k+1}\in [n]$ to be the unique integer satisfy $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$.
Let (clearly exists) $p<q$ are the indices that $i_p=i_q$ with minimal $p$, and then minimal $q$.

If $p>2$. This means $i_j\not\in \{ i_1,i_2\} \implies \sigma (i_j) +\tau (i_j) \equiv_n b_{i_j}$ for all $j\in \{ p,p+1,...,q\}$.
Summing the equation $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$ for $k\in \{ p,p+1,...,q-1\}$ gives us
$$\sum_{j=p-1}^{q-2}{\sigma (i_j) } +\sum_{j=p+1}^{q}{\tau (i_j)} \equiv_n\sum_{j=p}^{q-1}{b_{i_j}} \implies \sigma (i_{p-1}) +\sigma (i_p) +\tau (i_{q-1}) +\tau (i_q) \equiv_nb_{i_p}+b_{i_{q-1}}.$$Plugging $i_p=i_q$ and use $\sigma (i_p) +\tau (i_p)\equiv_n b_{i_p}$ gives us $\sigma (i_{p-1}) +\tau (i_{q-1})\equiv_n b_{i_{q-1}} \equiv_n \sigma (i_{q-1})+\tau (i_{q-1})$.
Hence, $\sigma (i_{p-1}) \equiv_n \sigma (i_{q-1})\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$.

So, we've $p\in \{ 1,2\}$. Let $p'=3-p$. Define the desired permutations $\sigma'$ and $\tau'$ as follows:
$$\sigma' (i_l)=\begin{cases} 
\sigma (i_{l-1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\sigma (i_{q-1}), & \text{ if } l=1
\end{cases} ,\tau' (i_l)= \begin{cases} 
\tau (i_{l+1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\tau (i_{p'}), & \text{ if } l=1
\end{cases}  $$and $\sigma' (i) =\sigma (i),\tau' (i)=\tau (i)$ for the rest $i\in [n]$ that $i\not\in \{ i_1,i_2,...,i_{q-1}\}$.
Note that the reason we choose $\tau (i_{p'})$ is just to not use $\tau (i_p)=\tau (i_{(q-1)+1})$ more than one time.
This construction gives us $\sigma' (i)+\tau' (i)\equiv_n b_i$ for all $i\in [n]$ except when $i=i_1$.
But since both $\sigma'$ and $\tau'$ are permutations of $[n]$, we've $\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$.
This guarantee that $\sigma' (i) +\tau' (i)\equiv_n b_i$ when $i=i_1$ too. This prove the validity of permutations we constructed.
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jan 16, 2018, 3:07 PM
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mathleticguyyy
3217 posts
#11
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The case with $n$ prime is also resolved in this paper
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Bataw
43 posts
#13
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any other solutions ?
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Zhaom
5124 posts
#16
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:star_struck:

Plot $\left(b_k,c_k\right)$ for $k=1,2,\cdots,n$ on the coordinate plane with coordinates modulo $n$. We want to show that it is possible to choose the $n$ points plotted to have $\left\{b_1+c_1,b_2+c_2,\cdots,b_n+c_n\right\}$ be any multiset of $n$ elements $\pmod{n}$ such that sum of the elements is $0\pmod{n}$. Note that the $n$ points plotted can be any $n$ points not in the same row or column.

Instead, we will prove that we can choose $n-1$ points not in the same row or column such that the multiset $S$ of $x+y$ for all of the points $(x,y)$ can be any multiset of $n-1$ elements $\pmod{n}$. Then, we can choose the unique point not in the same row or column as the $n-1$ points to obtain $n$ points for the original statement, as the sum of the coordinates of all $n$ points is necessarily $2\cdot\left(0+1+\cdots+(n-1)\right)\equiv(n-1)n\equiv0\pmod{n}$.

We will construct the $n-1$ points in the following way.

We start with $n-1$ arbitrary points not in the same row or column. We will change the $n-1$ points in such a way that we can replace any element of $S$ by any residue $\pmod{n}$. Suppose we change the element in $S$ corresponding to a point $P$. Then, we will shift $P$ horizontally until the sum of the coordinates of $P$ is the desired value $\pmod{n}$. Now, we might have two points in the same column. We will repeatedly perform the following operation.

Note that there are exactly $n-1$ distinct rows taken up by the $n-1$ points. If the last point moved was $P$ and it is in the same column as $Q$, we will move $Q$ on the line through $Q$ with slope $-1$ until $Q$ occupies the previously empty row.

It suffices that this cannot last forever.

Claim. This operation is a reflection over a fixed line.

Proof. Suppose that this operation moves $P$ to a point $P'$, then on the next step it moves $Q$ to a point $Q'$. We see that $\overline{PP'}$ and $\overline{QQ'}$ both have slope $-1$. Now, note that $P'$ and $Q$ must be in the same row. Furthermore, after the operation moves $P$ to $P'$, the row containing $P$ must be empty, meaning that $Q$ must be moved to that row. This implies that $P$ and $Q'$ are in the same row. Therefore, we see that $PP'QQ'$ is a cyclic isosceles trapezoid and in particular the perpendicular bisectors of $\overline{PP'}$ and $\overline{QQ'}$ are the same, proving the claim.

Now, assume for the sake of contradiction that the operation lasts forever. We will label the $n-1$ points to distinguish them. Now, there are finitely many states of the $n-1$ points, so eventually the operation must loop through a cycle of states. In this cycle, start at an arbitrary state. Now, since at least $1$ point must be moved and then return back to its original position, consider the point $P$ which returns to its original position the earliest. When $P$ was moved, the next point $Q$ which was moved must have occupied the original row of $P$. However, if $P$ was moved back to its original position, then the original row of $P$ must be vacant, so $Q$ must have been moved. However, this implies that $Q$ was moved back to its original position before $P$ by the claim, a contradiction, so we are done.
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