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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
Number of functions satisfying sum inequality
CyclicISLscelesTrapezoid   19
N a few seconds ago by john0512
Source: ISL 2022 C5
Let $m,n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_1,X_2,\ldots,X_m$ be pairwise distinct non-empty, not necessary disjoint subset of $X$. A function $f \colon X \to \{1,2,\ldots,n+1\}$ is called nice if there exists an index $k$ such that \[\sum_{x \in X_k} f(x)>\sum_{x \in X_i} f(x) \quad \text{for all } i \ne k.\]Prove that the number of nice functions is at least $n^n$.
19 replies
CyclicISLscelesTrapezoid
Jul 9, 2023
john0512
a few seconds ago
Inequality em981
oldbeginner   21
N 5 minutes ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
21 replies
1 viewing
oldbeginner
Sep 22, 2016
sqing
5 minutes ago
Find the minimum
sqing   29
N 7 minutes ago by sqing
Source: Zhangyanzong
Let $a,b$ be positive real numbers such that $a^2b^2+\frac{4a}{a+b}=4.$ Find the minimum value of $a+2b.$
29 replies
sqing
Sep 4, 2018
sqing
7 minutes ago
Interesting inequality
sqing   4
N 8 minutes ago by sqing
Source: Own
Let $a,b\geq 0, 2a+2b+ab=5.$ Prove that
$$a+b^3+a^3b+\frac{101}{8}ab\leq\frac{125}{8}$$
4 replies
sqing
an hour ago
sqing
8 minutes ago
nice ecuation
MihaiT   1
N Yesterday at 7:24 PM by Hello_Kitty
Find real values $m$ , s.t. ecuation: $x+1=me^{|x-1|}$ have 2 real solutions .
1 reply
MihaiT
Yesterday at 2:03 PM
Hello_Kitty
Yesterday at 7:24 PM
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
Easy IMO 2023 NT
799786   133
N Apr 24, 2025 by Maximilian113
Source: IMO 2023 P1
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
133 replies
799786
Jul 8, 2023
Maximilian113
Apr 24, 2025
Easy IMO 2023 NT
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2023 P1
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Ywgh1
139 posts
#135 • 1 Y
Y by radian_51
IMO 2023 p1

We claim that the answer is $n=p^k$, where $p$ is a prime.

Let $p_1$ and $p_2$ be the first two primes of $n$. Now let $s$ be the largest number, such that $p_1^s < p_2$

So we have that
\[\frac{n}{p_2},\frac{n}{p_1^s},\frac{n}{p_1^{s-1}}\]are consecutive divisors of $n$.
Hence we must have that.
$$\frac{n}{p_2}|(\frac{n}{p_1^s}+\frac{n}{p_1^{s-1}}) \implies p_1^s|p_2(p_1+1)$$Which is a contradiction, hence $n$ has less than 2 primes.
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ezpotd
1295 posts
#137 • 1 Y
Y by radian_51
I claim the answer is only prime powers. It is easy to see that these work. To see nothing else works, assume for the sake of contradiction some $n$ with more than $1$ prime power worked. Then take the second smallest prime divisor, let it be $q$, and the largest divisor less than it, $p^{k}$, where $k$ can be $1$. We would then have $\frac nq \mid \frac{n}{p^{k - 1}} + \frac{n}{p^k}$, which is obviously not true by taking $\nu_p$ on both sides, the left side has $\nu_p(\frac nq) = \nu_p(n)$, right hand side has $\nu_p({\frac{n}{p^{k - 1}} + \frac{n}{p^k}}) = \nu_p({\frac{n}{p^k}})$, so the left side cannot divide the right.
This post has been edited 1 time. Last edited by ezpotd, Sep 30, 2024, 1:00 AM
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lelouchvigeo
183 posts
#139 • 1 Y
Y by S_14159
Nothing new
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cursed_tangent1434
643 posts
#140
Y by
Trivial NT on the IMO. Finally found some time to actually post my solution from the contest.

We claim that the answer is all positive integers of the form $p^m$ for some prime $p$ and $m\ge 2$. These clearly all work since for all $1\le i \le m+1$ , $d_i=p^{i-1}$. Thus,
\[d_i=p_{i-1}\mid p_i + p_{i+1}=d_{i+1}+d_{i+2}\]quite clearly.

Now, we show that no other composite $n$ work. Say $n$ has atleast two distinct primes divisors, of which the smallest two are $p<q$. Note that we have $d_i = p^{i-1}$ for all $1\le i \le r$ for some $r \ge 2$ and $d_{r+1}=q$ as the divisors are arranged in increasing order. Now, if $n$ satisfies the desired condition we must have,
\begin{align*}
d_{m-r} &\mid d_{m-r+1} + d_{m-r+2}\\
\frac{n}{d_{r+1}} & \mid \frac{n}{d_r} + \frac{n}{d_{r-1}}\\
\frac{n}{q} & \mid \frac{n}{p^{r-1}} + \frac{n}{p^{r-2}}\\
p^{r-1} & \mid q + pq\\
p &\mid q+pq\\
p& \mid q
\end{align*}which is a very clear contradiction since $q>p$ are both primes. Thus, there are no such $n$ of this form, which finishes the proof.
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EVKV
71 posts
#141
Y by
Clear answer is n =p^c where p is a prime
Assume that n can have more than 1 prime factor and assume some prime q also divides n
Well you get that p|q contradiction.
Too easy for P1 tho
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maths_enthusiast_0001
133 posts
#142 • 1 Y
Y by radian_51
My first NT after decades..... :D
IMO 2023 P1 wrote:
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
All numbers of the form $p^l$ where $p$ is a prime and $l>1$ work evidently. We now show these are the only ones.
Let $m$ denote the number of divisors of $n$. Then we have $d_{i}d_{m+1-i}=n$. Thus,
$$d_{i} \mid d_{i+1}+d_{i+2}$$$$\implies \frac{n}{d_{m-i+1}} \mid \frac{n}{d_{m-i}}+\frac{n}{d_{m-i-1}}$$$$\implies d_{m-i}d_{m-i-1} \mid d_{m-i+1}(d_{m-i}+d_{m-i-1})$$$$\implies \boxed{d_{i}d_{i+1} \mid d_{i+2}(d_{i}+d_{i+1})}  [\because i \mapsto m-i-1]$$Now $i=1$ gives, $d_{1}d_{2} \mid d_{3}(d_{1}+d_{2}) \implies d_{2} \mid d_{3}$ since $d_{1}=1$. Also, $d_{2}$ is a prime number (say $p$). If $d_{3}$ has any prime factor $q$(say) other than $p$ then $d_{1}<q<d_{3};q \neq d_{2}$ which is a blatant contradiction. Thus $\boxed{d_{3}=p^{2}}$ as it can not be $p^e$ where $e \geq 3$ otherwise $d_{2} < p^{2} < d_{3}$ again, a contradiction. Now by an easy strong induction and similar argument we can conclude that $d_{i+1}=p^{i}$. Thus $n=p^{l}$ where $l>1$ and $p$ is a prime number. $\blacksquare$ ($\mathcal{QED}$)
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Rohit-2006
245 posts
#143 • 1 Y
Y by radian_51
Suppose $n$ has a prime divisors $\geq 2$.

Say the least ones are $p$ and $q$ and $p$ be the minimum.

Let the multiplicity of $p$ be $m$.

At the $(k+1)$-th step, $k \leq m$:

$p^{k-1} < q$, so $q$ can write

$p^{k-1} \mid p^k + q \quad ( \rightarrow \leftarrow)$.

Hence, $n = p^t$, $t \in \mathbb{N}$ and $p$ is prime.
This post has been edited 2 times. Last edited by Rohit-2006, Jan 17, 2025, 10:34 AM
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iStud
268 posts
#144 • 1 Y
Y by radian_51
Note that since $d_{k-2}\mid d_{k-1}+d_k$ and $d_{k-2}\mid n=d_k$, then $d_{k-2}\mid d_{k-1}$. Recall that $d_{k-2}=\frac{n}{d_3}$ and $d_{k-1}=\frac{n}{d_2}$, so $\frac{\frac{n}{d_2}}{\frac{n}{d_3}}\in\mathbb{N}\Longleftrightarrow\frac{d_3}{d_2}\in\mathbb{N}\Longleftrightarrow d_2\mid d_3$. But we also have $d_2\mid d_3+d_4$, so we must have $d_2\mid d_4\Longleftrightarrow\frac{\frac{n}{d_2}}{\frac{n}{d_4}}\in\mathbb{N}\Longleftrightarrow\frac{d_{k-1}}{d_{k-3}}\in\mathbb{N}\Longleftrightarrow d_{k-3}\mid d_{k-1}$. Again, we know that $d_{k-3}\mid d_{k-2}+d_{k-1}$ so we must have $d_{k-3}\mid d_{k-2}$. Repeating the process, eventually we'll end with $1=d_1\mid d_2\mid d_3\mid\dots\mid d_{k-1}\mid d_k=n$. Notice that for any natural number $n$, the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{q}$ for prime $2\le p<q$ or $n,\frac{n}{p},\frac{n}{p^2}$ for prime $p\ge 2$. If the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{q}$, then $\frac{\frac{n}{p}}{\frac{n}{q}}\in\mathbb{N}\Longleftrightarrow\frac{q}{p}\in\mathbb{Z}\Longleftrightarrow p\mid q$, contradiction. So the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{p^2}$ for prime $p\ge 2$. In the similar spirit, we end up showing that $n=p_{k-1}$, which indeed works for any prime $p\ge 2$. Done. $\blacksquare$
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cubres
119 posts
#145
Y by
Storage - grinding IMO problems
This post has been edited 1 time. Last edited by cubres, Feb 3, 2025, 9:44 PM
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megahertz13
3194 posts
#146
Y by
The answer is $n=p^k$ for a prime number $p$ and a positive integer $k$. These clearly work, and we prove that all other $n$ fail. Assume the contrary, and let the smallest prime factors of $n$ be $a$ and $b$.

Case 1: $a < b < a^2$. Then we have the smallest three divisors are $1, a, b$, so the largest three are $\frac{n}{b}, \frac{n}{a}, n$. However, we need $$\frac{n+\frac{n}{a}}{\frac{n}{b}}=\frac{1+\frac{1}{a}}{\frac{1}{b}}=b+\frac{b}{a}$$to be an integer, but $b/a$ is clearly not an integer. There are no cases here.

Case 2: $a<a^2<\dots<a^k<b$ for $k>1$. We know that $$\frac{b+a^k}{a^{k-1}}=a+\frac{b}{a^{k-1}}$$must be an integer, but this is clearly impossible.

This concludes the problem.

Note: I almost messed up and disregarded Case 1.
This post has been edited 1 time. Last edited by megahertz13, Feb 11, 2025, 10:46 PM
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Ilikeminecraft
658 posts
#147
Y by
I claim $n$ is a perfect power. Assume not. Let $p < q$ be the smallest two prime divisors of $n.$ Let $\frac nq,\frac n{p^c}$ be the two consecutive divisors. Hence, we get $\frac nq \mid \frac n{p^c} + \frac n{p^{c - 1}},$ so $p\mid p + q,$ or $p \mid q.$ All perfect powers work.
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ray66
46 posts
#148
Y by
Let $p$ and $q$ be the smallest two prime factors of $n$. Then $\frac{n}{q} | n\frac{1+p}{p}$ so $\frac{q(p+1)}{p} \in \mathbb{Z}$. But that means $p | q$, impossible. So $n=p^k$ and only has 1 distinct prime factor. All $n$ of this form work for $k \ge 2$.
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Jupiterballs
58 posts
#149
Y by
Let the smallest prime dividing $n$ be $p$
Then $d_{k-1} = \frac{n}{p}$
Which gives that $d_{k-2} \mid d_{k-1} + d_{k} = \frac{n}{p} + n = \frac{n}{p}(p+1)$

Now, we form cases:-
Case-1) $d_{k-2} \mid \frac{n}{p}$
implying that $d_{k-2} = \frac{n}{p^2}$ (easy to see why)

Case-2) $v_p (d_{k-2}) = v_p (n)$, or $d_{k-2}$ does not divide $n$
Which means that we need another prime $p$
Then this implies that $gcd(p, p+1) = p$

Which is absurd, therefore $d_{k-2} = \frac{n}{p^2}$
Reiterating this gives us that $d_1 = \frac{n}{p^{k-1}}$
or $n = p^{k-1}$, which works clearly
q.e.d
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anudeep
202 posts
#150
Y by
We claim $n$ of the form $p^{\alpha}$ with $p$ being a prime and $\alpha\in\mathbb{Z}_{\ge 2}$ are the only solutions (easy to check why these work).
Now we shall show that these are the the only solutions. Suppose this is not true. Let $p,q$ be the smallest two distinct prime divisors of $n$ with $p<q$. Let $j$ be the largest integer such that $p^j\lvert n$ and $p^j<q$ (assume $j>1$ as the case $n=pq$ is trivial). One may notice that $d_j=p^{j-1}$, $d_{j+1}=p^j$ and $d_{j+2}=q$. By the given condition it must be that,
$$p^{j-1}\lvert (p^j+q).$$From the above we say $p^{j-1}\lvert q$, which is ridiculous and hence the claim. $\square$
This post has been edited 2 times. Last edited by anudeep, Apr 20, 2025, 6:30 AM
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Maximilian113
575 posts
#151
Y by
\begin{align*}
d_{k-2} | d_{k-1}+d_k, d_{k-2} | d_k \implies d_{k-2} | d_{k-1} \implies d_2 | d_3. \\
d_2 | d_3, d_2 | d_3+d_4 \implies d_2 | d_4 \implies d_{k-3} | d_{k-1}. \\
d_{k-3} | d_{k-2}+d_{k-1} \implies d_{k-3} | d_{k-2}.
\end{align*}Hence applying this argument in general yields $d_{x} | d_{x+1} \implies d_{x-1} | d_x.$ So $d_1 | d_2 | d_3 | \cdots | d_{k-1} | d_k.$ So if some prime $p | n$ where $p$ is minimal we have $d_1=1, d_2=p \implies p | d_i \, \forall i \geq 2.$ So there is no other prime $q \neq p$ dividing $n.$

Hence $n=p^k$ for some prime $p$ and positive integer $k > 1.$ It is easy to see that this works.
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