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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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[*]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
Find min and max
lgx57   0
24 minutes ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
24 minutes ago
0 replies
Find min
lgx57   0
28 minutes ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
28 minutes ago
0 replies
Geometry
Lukariman   3
N 2 hours ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
3 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
2 hours ago
III Lusophon Mathematical Olympiad 2013 - Problem 5
DavidAndrade   2
N 2 hours ago by KTYC
Find all the numbers of $5$ non-zero digits such that deleting consecutively the digit of the left, in each step, we obtain a divisor of the previous number.
2 replies
DavidAndrade
Aug 12, 2013
KTYC
2 hours ago
Maximum number of terms in the sequence
orl   11
N 2 hours ago by navier3072
Source: IMO LongList, Vietnam 1, IMO 1977, Day 1, Problem 2
In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.
11 replies
orl
Nov 12, 2005
navier3072
2 hours ago
Combinatorics
P162008   2
N 2 hours ago by cazanova19921
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
2 replies
P162008
3 hours ago
cazanova19921
2 hours ago
USAMO 2003 Problem 1
MithsApprentice   68
N 2 hours ago by Mamadi
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
68 replies
MithsApprentice
Sep 27, 2005
Mamadi
2 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   3
N 3 hours ago by EeEeRUT
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
3 replies
BR1F1SZ
Monday at 9:45 PM
EeEeRUT
3 hours ago
IMO Genre Predictions
ohiorizzler1434   60
N 3 hours ago by Yiyj
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
60 replies
ohiorizzler1434
May 3, 2025
Yiyj
3 hours ago
square root problem
kjhgyuio   5
N 3 hours ago by Solar Plexsus
........
5 replies
kjhgyuio
May 3, 2025
Solar Plexsus
3 hours ago
Diodes and usamons
v_Enhance   47
N 3 hours ago by EeEeRUT
Source: USA December TST for the 56th IMO, by Linus Hamilton
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?

Proposed by Linus Hamilton
47 replies
v_Enhance
Dec 17, 2014
EeEeRUT
3 hours ago
3-var inequality
sqing   1
N 4 hours ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
1 reply
sqing
4 hours ago
sqing
4 hours ago
IMO ShortList 2001, combinatorics problem 3
orl   37
N 4 hours ago by deduck
Source: IMO ShortList 2001, combinatorics problem 3, HK 2009 TST 2 Q.2
Define a $ k$-clique to be a set of $ k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.
37 replies
orl
Sep 30, 2004
deduck
4 hours ago
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N 4 hours ago by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
4 hours ago
Distinct Integers with Divisibility Condition
tastymath75025   16
N Apr 21, 2025 by ihategeo_1969
Source: 2017 ELMO Shortlist N3
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu
16 replies
tastymath75025
Jul 3, 2017
ihategeo_1969
Apr 21, 2025
Distinct Integers with Divisibility Condition
G H J
G H BBookmark kLocked kLocked NReply
Source: 2017 ELMO Shortlist N3
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tastymath75025
3223 posts
#1 • 2 Y
Y by Adventure10, Mango247
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu
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nmd27082001
486 posts
#2 • 1 Y
Y by Adventure10
My solution:i will prove that there is no sequence sastisfied the problem
Let $p|C$ and $t=v_p(C)$ and $b_k=v_p(a_k)$
So we need $b_{k+1}.k<=tk+\sum_{i=1}$
Consider sequence $c_n=\sum_{k=1}^{n}\frac{1}{k}$
We can easily prove that $b_n<=[ta_n]+b_1$
But note that there exist m such that for all $n>m$,$n-b_1>[ta_n]$
Which implies there exist infinitive number i,j:$b_i=b_j$
Let S={(i,j),$b_i=b_j$}
Consider prime $q=!p$ of C
We found that in S there are infinitive (t,s):$v_p(a_t)=v_p(a_s)$ and $v_q(a_t)=v_q(a_s)$
Continue this with all prime divisor of C we are done
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ThE-dArK-lOrD
4071 posts
#3 • 2 Y
Y by Adventure10, Mango247
My lengthly solution
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jun 29, 2018, 8:03 PM
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Wizard_32
1566 posts
#4 • 2 Y
Y by centslordm, PRMOisTheHardestExam
Nice problem. I will give a sketch.
tastymath75025 wrote:
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.
The answer is no. Suppose not for a fixed $C.$ Consider any prime $p.$ Then the problem gives
$$k\nu_p (a_{k+1}) \le k\nu_p(C)+\nu_p(a_1)+\dots+\nu_p(a_k).$$Now we have the following key claim which can be proven by simple induction:
Claim: Let $\text{H}_n$ denote the $n$th harmonic number. Then
$$\nu_p( a_n)-\nu_p(a_1) \le \text{H}_n \nu_p(C).$$
The key hypothesis we need now is that $a_i$ are pairwise distinct. The claim gives $\nu_p(C)=0 \implies \nu_p(a_n) \le \nu_p(a_1).$ In particular $\nu_p(a_m)$ is eventually constant. So ignore primes for which $\nu_p(C)=0.$ This means we only have a finite set of prime factors to worry about for the sequence now.

The claim clearly gives $\nu_p(a_n/a_1) \le A \log n$ for some constant $A$ and all primes $p.$ Now it is not too hard to see that we can find arbitrarily large intervals over which $\left \lfloor A\log x \right \rfloor$ is constant. Since the set of prime factors of $\{a_k\}$ is finite now, hence by pigeonhole two terms $a_i,a_j$ will have the same $\nu_p$ for all primes $p,$ hence will be equal, a contradiction. $\square$

EDIT: Ok, I finally typed the elaborated version of the sketch: Full Proof
This post has been edited 3 times. Last edited by Wizard_32, Nov 6, 2020, 6:51 AM
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mathlogician
1051 posts
#5
Y by
There is no such sequence for any $C$. For convenience, define $x_n = v_p(a_n)$ for every integer $n$. Furthermore, define $h_n = \frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}$. Finally, let $S = v_p(C)$. The pith of the problem lies in the following claim:

Claim: $x_n \leq x_1 + Sh_n$.

Proof: By strong induction. Remark that the condition tells us $$x_n \leq S + \frac{x_1+x_2+\dots+x_{n-1}}{n-1}.$$
This claim implies that for any prime $p$, there exists an arbitrary long sub-sequence of terms $a_i,a_{i+1},\dots,a_j$ such that $\nu_p(a_i) = \nu_p(a_{i+1}) = \dots = \nu_p(a_j)$ for sufficiently large $i$ and $j$. Note that for primes $p$ in $a_1$ but not $C$, $\nu_p(a_i) \leq \nu_p(a_1)$, so there are only a finite number of possible terms with equal $v_p$ for all primes $p \in C$. Therefore, for sufficiently large $i$ there will be two equal terms, the end.
This post has been edited 1 time. Last edited by mathlogician, Jan 25, 2021, 2:43 PM
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Sprites
478 posts
#6
Y by
tastymath75025 wrote:
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu

Define $\mathbf{H}_n=1+\frac{1}{2}+.............+\frac{1}{n}$
The condition is equivalent to $k\nu_p(a_{k+1}) \le k \nu_p(C)+\sum_{j=1}^k \nu_p(a_j)$
Claim: $\nu_p(a_n)-\nu_p(a_1) \le\mathbf{H}_n \nu_p(C)$
Proof: Obvious by strong induction.
This implies that $\nu_p(a_j)$ is bounded and when $j$ is sufficiently large we will get $\nu_p(a_j)=\nu_p(a_{j+1}),\forall p \implies a_j=a_{j+1} $, contradiction.
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IvoBucata
46 posts
#7
Y by
I'll prove that for no $C$ there exists such a sequence. I'll start with the following claim:

Claim: $v_p(a_{k+1})\leq (1+\frac12 + \cdots +\frac1k)v_p(C) + v_p(a_1)$ for every prime $p$.

Proof : It's not hard to show this by induction.

We can find arbitrarily large intervals over which $(1+\frac12 + \cdots +\frac1k)v_p(C) + v_p(a_1)$ is constant, but note that $a_{k+1}$ can take only values which are divisors of $a_1C^{X}$ where $X=\lceil (1+\frac12 + \cdots +\frac1k)v_p(C)\rceil $. Now at some point $d(a_1C^{X})$ is going to be smaller than the length of these intervals because $1+\frac12 + \cdots +\frac1k$ grows very slowly, so the $a_i$'s can't be distinct over these intervals.
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guptaamitu1
656 posts
#8
Y by
Here's a different proof (we will bound the $v_p$'s differently without the harmonic number).
We will show for any $C$, such a sequence does not exist. Assume contrary. Fix $C$ and such a sequence. Call a prime nice if it divides some $a_i$. Note all nice primes must divide $C \cdot a_1$, in particular number of nice primes is finite. Fix a $c > 1$ such that $c > v_p(C)$ for all primes $p$. Fix any nice prime $p$. Let $b_i = v_p(a_i)$. Then we know that
$$k b_{k+1} \le kc + b_1 + b_2 + \cdots + b_k ~~ \forall ~ k \ge 2 \qquad \qquad (1)$$Intuitively, we will show that $b_i$'s grow slowly. Call an $i \ge 2$ a peak if $b_i > b_{i-1},\ldots,b_1$. Let $i_1+1 < i_2 + 1< i_3 + 1 < \cdots$ be all the peaks.


Claim 1: For all $n \ge 1$ we have
$$\frac{i_1 + i_2 + \cdots + i_n}{i_n} \le c ~ \iff ~ \frac{i_1 + \cdots + i_{n-1}}{i_n} \le c-1$$
Proof: Define $$f(m) = (b_m - b_1) + (b_m - 2) + \cdots + (b_m - b_{m-1})~ \forall ~ m \ge 2$$. Note $(1)$ is equivalent to
$$ \frac{f(k)}{k-1} \ge c  \qquad \qquad (2) $$Observe that for any $n \ge 1$ that
\begin{align*}
f(i_{n + 1} + 1) &= \sum_{j=1}^{i_{n+1}}\left( b_{i_{n+1} + 1} - b_j \right) = \sum_{j= i_n + 1}^{i_{n+1}} (b_{i_{n+1} + 1} - b_j) + \sum_{j=1}^{i_n} (b_{i_{n+1} + 1} - b_j) \\ &\ge  (i_{n+1} - i_n)(b_{i_{n+1} + 1} - b_{i_n + 1}) + \sum_{j=1}^{i_n} \bigg( (b_{i_{n+1} + 1} - b_{i_{n} + 1}) + (b_{i_n + 1} - b_j) \bigg) \\
&= i_{n+1}(b_{i_{n+1} + 1} - b_{i_n + 1}) + f(i_n + 1) \ge i_{n+1} + f(i_n + 1)
\end{align*}Now note $f(i_1 + 1) \ge i_1$. So it follows that
$$f(i_n + 1) \ge i_1 + i_2 + \cdots + i_n ~~ \forall ~ n \ge 1$$Plugging $k = i_n + 1$ in $(2)$ and using above implies our claim. $\square$


Claim 2: $\exists$ an $\alpha > 1$ such that $i_k \ge \alpha^{k-1} ~ \forall ~ k \ge c+2$.

Proof: We will only use each $i_n \ge 1$ and Claim 1 to prove this. Observe $i_1 + \cdots + i_{c+1} > c+1$, since all of them cannot be $(1)$ (by Claim 1 for $n=c+1$). Now choose a very small $\alpha > 1$ such that:
\begin{align*}
i_1 + i_2 + \cdots + i_{c+1} \ge 1 + \alpha + \cdots + \alpha^c \\
\frac{1}{\alpha} + \frac{1}{\alpha^2} + \cdots + \frac{1}{\alpha^c} \ge c-1
\end{align*}We show this $\alpha$ works. Suppose for some $t \ge c+1$ it holds that $i_1 + \cdots + i_t \ge 1 + \alpha + \cdots + \alpha^{t-1}$. We will show $i_{t+1} \ge \alpha^t$ (note this would imply our claim by induction). $k=t+1$ in $(3)$ gives
$$i_t \ge \frac{i_1 + \cdots + i_t}{c-1} \ge \frac{1 + \alpha + \cdots + \alpha^{t-1}}{c-1} = \frac{\alpha^t - 1}{(c-1)(\alpha - 1)}  $$So it suffices to show
\begin{align*}
 \frac{\alpha^t - 1}{(c-1)(\alpha - 1)} \ge \alpha^t \iff (c-1)(\alpha-1) \le \frac{\alpha^t - 1}{\alpha^t} = 1 - \frac{1}{\alpha^t} \iff (c-1)(\alpha -1) + 1 - \frac{1}{\alpha^t} \le 0
\end{align*}Indeed,
\begin{align*}
(c-1) (\alpha -1) + 1 - \frac{1}{\alpha^t} &= (\alpha -1) \left( (c-1) - \left( \frac{1}{\alpha} + \frac{1}{\alpha^2} + \cdots + \frac{1}{\alpha^{t-1}} \right) \right) \\ &\le (\alpha -1) \left(c-1 -\left( \frac{1}{\alpha} + \frac{1}{\alpha^2} + \cdots + \frac{1}{\alpha^c} \right) \right)  \le 0
\end{align*}This proves our claim. $\square$


Claim 3 (Key Result): For large $t$, all of $b_1,b_2,\ldots,b_{\alpha^t}$ are at most $c(t+1)$.

Proof: Observe that $b_{i_{n+1} + 1} - b_{i_n + 1} \le c ~ \forall ~ n \ge 1$. As $b_{i_1 + 1} = b_2 \le b_1 + c$, so it follows $$b_{i_m + 1} \le b_1 + cm ~~ \forall ~ m \ge 1$$Now $i_{t+1} + 1 \ge \alpha^t+1$, thus all of $b_1,b_2,\ldots,b_{\alpha^t-1}$ are $\le b_{i_t + 1} \le b_1 + tc$, and $b_1 + tc \le t(c+1)$ for large $t$. This proves our claim. $\square$


Now let $p_1,p_2,\ldots,p_k$ be all the nice primes, and $\alpha_1,\alpha_2,\ldots,
\alpha_k > 1$ be any numbers for which Claim 3 is true. Let $\alpha = \min(\alpha_1,\alpha_2,\ldots,\alpha_k)$. For a large $t$, look at the numbers
$$ a_1,a_2,\ldots,a_{\alpha^t} $$By Claim 3 we know that for any $p_i$ adic valuation of any of these numbers is $\le c(t+1)$. It follows number of distinct numbers between them is at most
$$ \bigg(c(t+1) + 1 \bigg)^k$$As all $a_i$'s are distinct, so this forces
$$ \bigg( c(t+1) + 1 \bigg)^k \ge \alpha^t \qquad \text{for all large } t $$But this is a contradiction as exponential functions grow faster than polynomial functions. This completes the proof. $\blacksquare$


Motivation: The problem isn't even true if $a_i$'s are not given to be distinct, so we somehow had to prove that. So we basically had to prove the $v_p$'s don't grow fast. Now $(1)$ was only interesting for peaks. So it was natural to consider peaks. Now after getting Claim 1, I was sure we only have to use Claim 1 and ignore all other conditions on peaks, which along with $b_{i_{n+1} + 1} - b_{i_n + 1} \le c$ would give us the sequence $\{b_k\}_{k \ge 1}$ doesn't grow fast. So we only had to prove Claim 2. Basically, we conjecture $i_k \ge \alpha^{k-1}$ for all $k$ (for some $\alpha > 1$). We then find the sufficient conditions on $\alpha$ for which we can prove this by induction. The two equations in proof of Claim 2 were precisely those. Now we had some problems like it might happen $i_1 = i_2 = 1$, but that was easy to fix by showing $i_k \ge \alpha^{k-1}$ for large $k$.
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IAmTheHazard
5001 posts
#9 • 1 Y
Y by centslordm
The answer is no such $C$. Let $p$ be some prime, and let $\nu_p(C)=c$, $x_i=\nu_p(a_i)$ for $i \geq 1$. Viewing the divisibility condition in $p$-adic terms only, it is equivalent to
$$kx_{k+1} \leq kc+x_1+\cdots+x_k.$$Let $(H_n)$ denote the sequence of harmonic numbers. The crux of the problem is the following:

Claim: $x_n-x_1 \leq cH_{n-1}$.
Proof: Shifting $(x_i)$ doesn't modify the truth of the condition, so WLOG let $x_1=0$. We now use strong induction:
\begin{align*}
(n-1)c+x_1+\cdots+x_{n-1}&\leq(n-1)c+c\left(\left(\frac{1}{1}\right)+\left(\frac{1}{1}+\frac{1}{2}\right)+\cdots+\left(\frac{1}{1}+\cdots+\frac{1}{n-2}\right)\right)\\
&=c\left(1+(n-2)+\frac{n-2}{1}+\frac{n-3}{2}+\cdots+\frac{1}{n-2}\right)\\
&=c\left(\frac{n-1}{1}+\frac{n-1}{2}+\cdots+\frac{n-1}{n-2}+\frac{n-1}{n-1}\right)\\
&=c(n-1)H_{n-1},
\end{align*}so $(n-1)x_n \leq c(n-1)H_{n-1} \implies x_n \leq cH_{n-1}$ as desired.

The claim implies that $\nu_p(a_n)$ is zero if $p \nmid a_1C$, $O(1)$ if $p \nmid C$ but $p \mid a_1$, and $O(\log n)$ otherwise. Suppose there are $a$ distinct primes dividing $C$ and $b$ distinct primes dividing $a_1$ but not $C$. Then there are $O(1^b(\log n)^a)\sim O((\log n)^a)$ choices for the value of $a_n$we have $O(1)$ options for $\nu_p(a_n)$ if $p \mid a_1$ and $p \nmid C$, and $O(\log n)$ options for $\nu_p(a_n)$ if $p \mid C$. But $n$ dominates $O((\log n)^a)$, so by Pigeonhole for sufficiently large $n$ there must exist two non-distinct elements of the sequence: contradiction. $\blacksquare$
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PianoPlayer111
40 posts
#10 • 2 Y
Y by Mango247, Mango247
The anwser is, that there is no sequnce, which satisfies the problem.
Proof:
Now choose any $a_{k+1} < a_{k+2}$.
First of all we see, that $a_{k+1}^k \mid C^k a_1 a_2 ... a_k$ $\Rightarrow$ $k \cdot v_p(a_{k+1}) \le k \cdot v_p(C) + v_p(a_1) + v_p(a_2) + ... +v_p(a_k)$ $\Rightarrow$ $k \cdot (v_p(a_{k+1} - v_p(C)) \le v_p(a_1) + v_p(a_2) + ... +v_p(a_k)$ $(1)$. Similarly and because of $a_{k+1} < a_{k+2}$, we get $(k+1) \cdot (v_p(a_{k+2} - v_p(C)) \le v_p(a_1) + v_p(a_2) + ... + v_p(a_{k+1})$ $(2)$, for every prime dividing both sides of the equations. Now we calculate $(2) - (1)$, which is equivalent after some boring basics in arithmetic to $(k+1) \cdot [v_p(a_{k+2}) - v_p(a_{k+1})] \le v_p(C)$ $\Rightarrow$ $v_p((\frac{a_{k+2}}{a_{k+1}})^{k+1}) \le v_p(C)$ $\Rightarrow$ $(\frac{a_{k+2}}{a_{k+1}})^{k+1} \mid C$ $\Rightarrow$ $(\frac{a_{k+2}}{a_{k+1}})^{k+1} \le C$. But this inequality is not true for a large $k$, hence we get a contradiction and we are done :ninja:
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aaabc123mathematics
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#11
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PianoPlayer111 wrote:
{a_{k+1}})^{k+1} \le C$. But this inequality is not true for a large $k$, hence we get a contradiction and we are done  :ninja:[/quote] 
Why ?If {a_{k+1}})^{k+1} \le C$. isn't very big ,how can explain the numer is larger than C?
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oty
2314 posts
#12
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Nice problem
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DongerLi
22 posts
#13
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How do the $\nu_p$'s grow? Define $S$ as the set of prime divisors of $a_1C$. Note that the given sequence has no prime factors outside of $S$. Let prime $p \in S$ be arbitrary. Denote $b_n = \nu_p(a_n)$ for all positive integers $n$. The given condition rewrites as:
\[kb_{k + 1} \leq k\nu_p(C) + b_1 + b_2 + \cdots + b_k.\]Define $s_n = \frac{b_1 + b_2 + \cdots + b_n}{n}$ as the average of the first $n$ elements of our new sequence. Adding $k(b_k + \cdots + b_1)$ to both sides of this inequality and dividing through by $k(k + 1)$ lends us a useful inequality.
\[\frac{b_{k+1}+\cdots + b_1}{k+1} \leq \frac{1}{k+1} \nu_p(C) + \frac{b_k + \cdots + b_1}{k}.\]Applying this inequality repeatedly gives us the following inequality.
\[s_k \leq \left(\frac{1}{k} + \cdots + \frac{1}{2}\right)\nu_p(C) + s_1.\]Plugging back into the original inequality and bounding the harmonic series gives us:
\[b_{k+1} \leq \nu_p(C) + s_k \leq \left(\frac{1}{k} + \cdots + \frac{1}{1}\right)\nu_p(C) + b_1 \leq \ln(k+1) + b_1.\]Something is seriously wrong, and we are very happy about it. Since the above is true for all primes $p$, there are $O((\ln k)^{|S|})$ different possibilities for the values of $a_1, \dots, a_k$. Since
\[O((\ln k)^{|S|}) < k\]as $k$ approaches infinity, there must exist two terms in the sequence that are equal.
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thdnder
198 posts
#15
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Since $a_{k + 1}^k \mid C^k a_{1}a_{2}\dots a_{k}$ and $C$ is constant, so the set of primes dividing an element in $(a_{n})_{n \ge 1}$is finite. Let $p$ be a arbitrary prime dividing one of $a_{1}, a_{2}, \dots$. Then the divisibility condition becomes $k\nu_{p}(a_{k + 1}) \le k\nu_{p}(C) + \nu_{p}(a_{1}) + \nu_{p}(a_{2}) + \dots + \nu_{p}(a_{k})$. Now consider the following claim:

Claim:
For $n \ge 2$, we have $\nu_{p}(a_{n}) \le \nu_{p}(a_{1}) + \nu_{p}(C)(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n - 1})$.

Proof:

Apply strong induction on $n$. Base case is clear. For the induction step, $\nu_{p}(a_{n + 1}) \le \nu_{p}(C) + \frac{1}{n}(\nu_{p}(a_{1}) + \nu_{p}(a_{2}) + \dots + \nu_{p}(a_{n})) \le \nu_{p}(C) + \nu_{p}(a_{1}) + \frac{1}{n}(\nu_{p}(C)((\frac{1}{1}) + (\frac{1}{1} + \frac{1}{2}) + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3}) + \dots + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}))) = \nu_{p}(a_{1}) + \nu_{p}(C)(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n})$. $\blacksquare$

Since the set of primes dividing an element in $(a_{n})_{n \ge 1}$is finite, let $a$ be a number of elements in the set of primes dividing an element in $(a_{n})_{n \ge 1}$. Then taking large $N$, we see that there are most $(O(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{N})^k) = O((\ln N)^k) < N$ distinct values in $a_{1}, a_{2}, \dots, a_{N}$, a contradiction. Therefore there are no such $C$. $\blacksquare$
This post has been edited 1 time. Last edited by thdnder, Sep 29, 2023, 2:24 PM
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Mathandski
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#16
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Subjective Rating (MOHs) $       $
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cursed_tangent1434
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#17
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The first observation is that the set of all primes dividing some term of $(a_i)$ is finite as any prime dividing $a_i$ for $i >1$ must divide $Ca_1$ by induction due to the given divisibility condition. Now, we prove our key bound.

Claim : For all primes $p \mid Ca_1$ and positive integers $k$,
\[\nu_p(a_k) \le \nu_p(a_1)+\nu_p(C)\left(\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k-1}\right)\]
Proof : This is a simple calculation. Note that the condition implies that for all $k \ge 1$,
\[(k-1)\nu_p(a_k)\le (k-1)\nu_p(C) + \nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-1})\]Now replacing $\nu_p(a_2),\dots , \nu_p(a_{k-1})$ inductively we obtain,
\begin{align*}
        \nu_p(a_k) & \le \nu_p(C) + \frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-1})}{k-1}\\
        & \le \nu_p(C) + \frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-2})}{k-1} + \frac{\nu_p(C)+\frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(a_{k-2})}{k-2}}{k-1}\\
        &= \nu_p(C) + \frac{\nu_p(C)}{k-1} + \frac{\nu_p(a_1)+\nu_p(a_2)+\dots + \nu_p(k-2)}{k-2}\\
        & \vdots\\
        & \le \nu_p(C)+\frac{\nu_p(C)}{k-1} + \frac{\nu_p(C)}{k-2}+\dots + \frac{\nu_p(C)}{2} + \frac{\nu_p(a_1)}{2} +\frac{\nu_p(a_1)}{2}\\
        &= \nu_p(a_1) + \nu_p(C) \left (\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k-1}\right)
    \end{align*}which proves the claim.

Note that we may not have $a_k < k$ for all sufficiently large $k$. This is because if $a_k <k$ for all $k \ge N$ then for all $k \ge M=\max(a_1,a_2,\dots , a_{N-1})$, $a_k <M$ but this is a set of $M$ distinct positive integers bounded above by $M-1$ which is a clear contradiction.

However, revisiting our claim, this means for all $k$ the number of distinct positive integers that are possible for $a_1,a_2,\dots , a_k$ is
\[ \prod_{i=1}^r \left(\left \lfloor \nu_{p_i}(a_1) + \nu_{p_i}(C)H_{k-1}\right \rfloor +1\right) = O\left(\frac{1}{1}+\frac{1}{2}+\dots + \frac{1}{k}\right)^{r} = O(\ln k)^r < k\]for all sufficiently large $k$ where $p_1,p_2,\dots , p_r$ is the set of all primes dividing $Ca_1$. But this is a contradiction to what we noted above so the desired is impossible for all $C>1$ and we are done.
This post has been edited 1 time. Last edited by cursed_tangent1434, Apr 22, 2025, 9:12 AM
Reason: minor details
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ihategeo_1969
232 posts
#18
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We will prove $\boxed{\text{no such sequence exists}}$.

See that $\text{rad}(a_1a_2 \dots) \mid \text{rad}(a_1C)$; hence we can write each $a_i$ as $p_1^{e_1} \dots p_t^{e_t}$ for fixed primes $p_1$, $\dots$, $p_t$ ($t$ is finite).

Claim: $\nu_p(a_n) \le O( \log n)$ for each $p \in \{p_1,\dots,p_t\}$.
Proof: See that \[\nu_p(a_n) \le \nu_p(C)+\frac{\nu_p(a_1)+\dots+\nu_p(a_{n-1})}{n-1}\]By simple induction we have that \begin{align*}
\nu_p(a_n) & \le \nu_p(a_1)+\nu_p(C) \left(1+\dots+\frac{1}{n-1} \right) 
 \le \nu_p(a_1)+ \nu_p(C) \left(1+\int_1^{n-1} \frac{dx}{x} \right) 
 = \nu_p(a_1)+\nu_p(C) \left(1+\log(n-1) \right)=O(\log n) \end{align*}As required. $\square$

Now assign a sequence $(\nu_{p_1}(a_n), \dots, \nu_{p_t}(a_n))$ to each $a_n$. Now fix some large $N$ and look at the number of distinct sequences among $a_1$, $\dots$, $a_N$ which is \[O(\log N)^t \ll N \]And hence by PHP, two the sequences must coincide and hence their corresponding numbers are same which is a contradiction.
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