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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
[PMO18 Qualifying] III.3 Functional Equation
Magdalo   2
N an hour ago by jasperE3
Suppose a function $f:\mathbb R\to \mathbb R$ satisfies the following conditions:
\begin{align*}
&f(4xy)=2y[f(x+y)+f(x-y)]\text{ for all }x,y\in\mathbb R\\
&f(5)=3
\end{align*}
Find the value of $f(2015)$.
2 replies
Magdalo
Yesterday at 10:53 AM
jasperE3
an hour ago
[PMO25 Areas, I.13] log + cubic equation
jdcuber13   3
N 2 hours ago by aops-g5-gethsemanea2
Let $a,b,c$ be real numbers with $1 < a < b < c$ that satisfy the equations
\begin{align*}
\log_{a} b + \log_{b} c + \log_{c} a &= 6.5 \\\\
\log_{b} a + \log_{c} b + \log_{a} c &= 5
\end{align*}
Then max{$\log_{a} b, \log_{b} c, \log_{c} a$} can be written in the form $\sqrt{x} + \sqrt{y}$, where $x$, and $y$ are positive integers. What is $x+y$?

Answer Confirmation
3 replies
jdcuber13
Yesterday at 12:44 PM
aops-g5-gethsemanea2
2 hours ago
23rd PMO Qualifying stage #15
Kyj9981   3
N 3 hours ago by aops-g5-gethsemanea2
In the figure below, $BC$ is the diameter of a semicircle centered at $O$, which intersects $AB$ and
$AC$ at $D$ and $E$ respectively. Suppose that $AD = 9$, $DB = 4$, and $\angle ACD = \angle DOB$. Find the
length of $AE$.
IMAGE

Answer
3 replies
Kyj9981
Yesterday at 12:18 PM
aops-g5-gethsemanea2
3 hours ago
[PMO27 Qualifying] I.15
Magdalo   2
N 3 hours ago by aops-g5-gethsemanea2
Given that $\log_25\approx2.321$, how many positive integers $k\leq100$ are there such that there are exactly four powers of 2 with exactly $k$ digits? (Note that 1 is considered a power of 2)
2 replies
Magdalo
Yesterday at 3:48 PM
aops-g5-gethsemanea2
3 hours ago
Linear algebra problem
Feynmann123   1
N Yesterday at 3:51 PM by Etkan
Let A \in \mathbb{R}^{n \times n} be a matrix such that A^2 = A and A \neq I and A \neq 0.

Problem:
a) Show that the only possible eigenvalues of A are 0 and 1.
b) What kind of matrix is A? (Hint: Think projection.)
c) Give a 2×2 example of such a matrix.
1 reply
Feynmann123
Yesterday at 9:33 AM
Etkan
Yesterday at 3:51 PM
Linear algebra
Feynmann123   6
N Yesterday at 1:09 PM by OGMATH
Hi everyone,

I was wondering whether when I tried to compute e^(2x2 matrix) and got the expansions of sinx and cosx with the method of discounting the constant junk whether it plays any significance. I am a UK student and none of this is in my School syllabus so I was just wondering…


6 replies
Feynmann123
Saturday at 6:44 PM
OGMATH
Yesterday at 1:09 PM
Local extrema of a function
MrBridges   2
N Yesterday at 11:36 AM by Mathzeus1024
Calculate the local extrema of the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $(x,y)\mapsto x^4+x^5+y^6$. Are they isolated?
2 replies
MrBridges
Jun 28, 2020
Mathzeus1024
Yesterday at 11:36 AM
Integral
Martin.s   3
N Yesterday at 10:52 AM by Figaro
$$\int_0^{\pi/6}\arcsin\Bigl(\sqrt{\cos(3\psi)\cos\psi}\Bigr)\,d\psi.$$
3 replies
Martin.s
May 14, 2025
Figaro
Yesterday at 10:52 AM
Reducing the exponents for good
RobertRogo   1
N Yesterday at 9:29 AM 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
1 reply
RobertRogo
May 20, 2025
RobertRogo
Yesterday at 9:29 AM
Sequence divisible by infinite primes - Brazil Undergrad MO
rodamaral   5
N Yesterday at 8:01 AM by cursed_tangent1434
Source: Brazil Undergrad MO 2017 - Problem 2
Let $a$ and $b$ be fixed positive integers. Show that the set of primes that divide at least one of the terms of the sequence $a_n = a \cdot 2017^n + b \cdot 2016^n$ is infinite.
5 replies
rodamaral
Nov 1, 2017
cursed_tangent1434
Yesterday at 8:01 AM
Reduction coefficient
zolfmark   2
N Yesterday at 7:42 AM by wh0nix

find Reduction coefficient of x^10

in(1+x-x^2)^9
2 replies
zolfmark
Jul 17, 2016
wh0nix
Yesterday at 7:42 AM
a^2=3a+2imatrix 2*2
zolfmark   4
N Yesterday at 2:44 AM by RenheMiResembleRice
A
matrix 2*2

A^2=3A+2i
A^3=mA+Li


i means identity matrix,

find constant m ، L
4 replies
zolfmark
Feb 23, 2019
RenheMiResembleRice
Yesterday at 2:44 AM
Find solution of IVP
neerajbhauryal   3
N Yesterday at 12:47 AM by MathIQ.
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
3 replies
neerajbhauryal
Sep 23, 2014
MathIQ.
Yesterday at 12:47 AM
Primes and automorphisms
CatalinBordea   2
N Saturday at 5:07 PM by a_0a
Source: Romanian District Olympiad 2016, Grade XII, Problem 3
Let be a group $ G $ of order $ 1+p, $ where $ p $ is and odd prime. Show that if $ p $ divides the number of automorphisms of $ G, $ then $ p\equiv 3\pmod 4. $
2 replies
CatalinBordea
Oct 5, 2018
a_0a
Saturday at 5:07 PM
Another binomial coefficients sum
aether01   8
N Apr 27, 2025 by soryn
Prove that $$\sum_{k=0}^{n}{{\left(-1\right)}^{k}\binom{n}{k} \binom{2n-k}{n-k}} = 1$$
8 replies
aether01
Mar 3, 2022
soryn
Apr 27, 2025
Another binomial coefficients sum
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aether01
5 posts
#1 • 1 Y
Y by HWenslawski
Prove that $$\sum_{k=0}^{n}{{\left(-1\right)}^{k}\binom{n}{k} \binom{2n-k}{n-k}} = 1$$
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alexheinis
10624 posts
#2 • 2 Y
Y by HWenslawski, fungarwai
The sum equals $\sum_0^n (-1)^k {n\choose k} p(x-k)$ where $p(x):={x\choose n}$ and $x=2n$. This sum equals $D^n p(x)$ where $D$ is the difference operator. Since $p$ has degree $n$ it is well-known that $D^np(x)=n! a$ where $a$ is the leading cft of $p$. Since $a=1/n!$, the sum equals 1, as stated.
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BackToSchool
1640 posts
#3
Y by
alexheinis wrote:
The sum equals $\sum_0^n (-1)^k {n\choose k} p(x-k)$ where $p(x):={x\choose n}$ and $x=2n$. This sum equals $D^n p(x)$ where $D$ is the difference operator. Since $p$ has degree $n$ it is well-known that $D^np(x)=n! a$ where $a$ is the leading cft of $p$. Since $a=1/n!$, the sum equals 1, as stated.

I just saw lots of new notations without actual prove process.
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alexheinis
10624 posts
#6 • 1 Y
Y by fungarwai
@above: it's not my fault that you don't know the theory, since it is well-known and basic.
The notations might be new to you but they are standard. So you're wrong when you say that there is no actual proof.
Let $p$ be a polynomial. The difference polynomial $Dp$ is defined with $Dp(x)=p(x)-p(x-1)$.
With induction one shows $D^n(p)(x)=\sum_{k=0}^n (-1)^k {n\choose k} p(x-k)$.
If $\deg(p)=s$ then $\deg(Dp)=s-1$ and the leading cft is multiplied by a factor $s$.
In particular, if $\deg (p)=n$ then $D^n (p)$ is a constant polynomial, and it's value is $n! a$, where $a$ is the leading cft of $p$.
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BackToSchool
1640 posts
#7
Y by
@above,
A well proof should be a friendly one, no matter it is well-known or new to the reader.
Anyway, thanks for explanation.
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hopeless404
588 posts
#8 • 1 Y
Y by fungarwai
Consider functions
$F(x) = \sum_{k\ge0} (-1)^k\binom{n}{k} x^k = (1-x)^n$
and the well-known $G_m(x) = \sum_{k\ge0} \binom{m+k}{k}x^k = \frac{1}{(1-x)^{m+1}}$
Then $$F(x)G_m(x) = \sum_{n\ge0} \sum_{k=0}^n (-1)^k\binom{n}{k} \binom{m+n-k}{n-k} x^n = \frac{(1-x)^n}{(1-x)^{m+1}}$$Equating the coefficients of $x^n$ gives
\begin{align*}
[x^n] F(x)G_n(x) &= [x^n] \frac{1}{1-x}\\
\implies \sum_{k=0}^n (-1)^k\binom{n}{k} \binom{2n-k}{n-k} &= 1
\end{align*}
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P162008
211 posts
#9 • 1 Y
Y by soryn
Cute
This post has been edited 6 times. Last edited by P162008, Apr 28, 2025, 3:21 PM
Reason: Typo
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fungarwai
865 posts
#10 • 3 Y
Y by Taiharward, soryn, P162008
This is a special case of the formula $\boxed{\displaystyle \sum_{i=0}^k (-1)^{k-i}\binom ni \binom {m-1+k-i}{k-i}=\binom {n-m}k}$ introduced in Summation with binomial coefficient

The proof is similar to #8 in this post.

$\displaystyle \sum_{k=0}^n (-1)^{n-k}\binom nk \binom {(n+1)-1+n-k}{n-k}=\binom {-1}n$

$\displaystyle \sum_{k=0}^n (-1)^{k}\binom nk \binom {2n-k}{n-k}=(-1)^n\binom {-1}n=1$

Applying difference operator as #2 is also a clever thought in this case.
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soryn
5348 posts
#11
Y by
Very, very nice!
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