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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
Upper bound on products in sequence
tapir1729   11
N a few seconds ago by HamstPan38825
Source: TSTST 2024, problem 7
An infinite sequence $a_1$, $a_2$, $a_3$, $\ldots$ of real numbers satisfies
\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]for every positive integer $n$. Prove that there exists a real number $C$ such that $a_{n} a_{n+1} < C$ for every positive integer $n$.

Merlijn Staps
11 replies
tapir1729
Jun 24, 2024
HamstPan38825
a few seconds ago
Long and wacky inequality
Royal_mhyasd   4
N a minute ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
4 replies
Royal_mhyasd
May 12, 2025
Royal_mhyasd
a minute ago
Consecutive squares are floors
ICE_CNME_4   6
N 3 minutes ago by mszew

Determine how many positive integers \( n \) have the property that both
\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]are consecutive perfect squares.
6 replies
ICE_CNME_4
5 hours ago
mszew
3 minutes ago
perpendicular diagonals criterion for a cyclic quadrilateral
parmenides51   3
N 10 minutes ago by PEKKA
Source: Sharygin 2005 Finals 9.1
The quadrangle $ABCD$ is inscribed in a circle whose center $O$ lies inside it.
Prove that if $\angle BAO = \angle DAC$, then the diagonals of the quadrilateral are perpendicular.
3 replies
parmenides51
Aug 26, 2019
PEKKA
10 minutes ago
functional inequality with equality
miiirz30   3
N 17 minutes ago by genius_007
Source: 2025 Euler Olympiad, Round 2
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that the following two conditions hold:

1. For all real numbers $a$ and $b$ satisfying $a^2 + b^2 = 1$, We have $f(x) + f(y) \geq f(ax + by)$ for all real numbers $x, y$.

2. For all real numbers $x$ and $y$, there exist real numbers $a$ and $b$, such that $a^2 + b^2 = 1$ and $f(x) + f(y) = f(ax + by)$.

Proposed by Zaza Melikidze, Georgia
3 replies
miiirz30
Today at 10:32 AM
genius_007
17 minutes ago
JBMO Shortlist 2023 N6
Orestis_Lignos   4
N 19 minutes ago by MR.1
Source: JBMO Shortlist 2023, N6
Version 1. Find all primes $p$ satisfying the following conditions:

(i) $\frac{p+1}{2}$ is a prime number.
(ii) There are at least three distinct positive integers $n$ for which $\frac{p^2+n}{p+n^2}$ is an integer.

Version 2. Let $p \neq 5$ be a prime number such that $\frac{p+1}{2}$ is also a prime. Suppose there exist positive integers $a <b$ such that $\frac{p^2+a}{p+a^2}$ and $\frac{p^2+b}{p+b^2}$ are integers. Show that $b=(a-1)^2+1$.
4 replies
Orestis_Lignos
Jun 28, 2024
MR.1
19 minutes ago
functional equation with exponentials
produit   7
N 26 minutes ago by GreekIdiot
Find all solutions of the real valued functional equation:
f(\sqrt{x^2+y^2})=f(x)f(y).
Here we do not assume f is continuous
7 replies
produit
6 hours ago
GreekIdiot
26 minutes ago
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   7
N 30 minutes ago by DeathIsAwe
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
7 replies
OgnjenTesic
3 hours ago
DeathIsAwe
30 minutes ago
Computing functions
BBNoDollar   2
N an hour ago by youochange
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
2 replies
BBNoDollar
Yesterday at 10:06 AM
youochange
an hour ago
Serbian selection contest for the IMO 2025 - P1
OgnjenTesic   1
N an hour ago by grupyorum
Source: Serbian selection contest for the IMO 2025
Let \( p \geq 7 \) be a prime number and \( m \in \mathbb{N} \). Prove that
\[\left| p^m - (p - 2)! \right| > p^2.\]Proposed by Miloš Milićev
1 reply
OgnjenTesic
3 hours ago
grupyorum
an hour ago
Serbian selection contest for the IMO 2025 - P5
OgnjenTesic   1
N an hour ago by math90
Source: Serbian selection contest for the IMO 2025
Determine the smallest positive real number $\alpha$ such that there exists a sequence of positive real numbers $(a_n)$, $n \in \mathbb{N}$, with the property that for every $n \in \mathbb{N}$ it holds that:
\[
        a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
    \]Proposed by Pavle Martinović
1 reply
OgnjenTesic
2 hours ago
math90
an hour ago
Some Length Equality
SatisfiedMagma   7
N 2 hours ago by BlueCrate04
Source: RMO 2024/5
Let $ABCD$ be a cyclic quadrilateral such that $AB \parallel CD$. Let $O$ be the circumcenter of $ABCD$ and $L$ be the point on $AD$ such that $OL$ is perpendicular to $AD$. Prove that
\[ OB\cdot(AB+CD) = OL\cdot(AC + BD).\]Proposed by Rijul Saini
7 replies
SatisfiedMagma
Nov 3, 2024
BlueCrate04
2 hours ago
Inspired by lbh_qys
sqing   2
N 2 hours ago by BraveHedgehog91
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a+k-1}{b - 3} + \frac{b+k-1}{3 - a} + \frac{k+2}{a - b} \right)^2 + 2(a^2 + b^2 )\geq6(k+8)$$Where $ k\in N^+.$
2 replies
sqing
May 20, 2025
BraveHedgehog91
2 hours ago
Geomerty
Teacher886699   0
2 hours ago
Source: Geometry plese help Me
To each side aof a convex polygon we assign the maximum area of a triangle contained
in the polygon and having a as one of its sides. Show that the sum of the areas assigned to all
sides of the polygon is not less than twice the area of the polygon.
0 replies
Teacher886699
2 hours ago
0 replies
Sequence with condition on million consecutive terms
jbaca   13
N Apr 26, 2025 by Ilikeminecraft
Source: 2021 Iberoamerican Mathematical Olympiad, P3
Let $a_1,a_2,a_3, \ldots$ be a sequence of positive integers and let $b_1,b_2,b_3,\ldots$ be the sequence of real numbers given by
$$b_n = \dfrac{a_1a_2\cdots a_n}{a_1+a_2+\cdots + a_n},\ \mbox{for}\ n\geq 1$$Show that, if there exists at least one term among every million consecutive terms of the sequence $b_1,b_2,b_3,\ldots$ that is an integer, then there exists some $k$ such that $b_k > 2021^{2021}$.
13 replies
jbaca
Oct 20, 2021
Ilikeminecraft
Apr 26, 2025
Sequence with condition on million consecutive terms
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 Iberoamerican Mathematical Olympiad, P3
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jbaca
225 posts
#1
Y by
Let $a_1,a_2,a_3, \ldots$ be a sequence of positive integers and let $b_1,b_2,b_3,\ldots$ be the sequence of real numbers given by
$$b_n = \dfrac{a_1a_2\cdots a_n}{a_1+a_2+\cdots + a_n},\ \mbox{for}\ n\geq 1$$Show that, if there exists at least one term among every million consecutive terms of the sequence $b_1,b_2,b_3,\ldots$ that is an integer, then there exists some $k$ such that $b_k > 2021^{2021}$.
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blackbluecar
303 posts
#2 • 1 Y
Y by PRMOisTheHardestExam
Notice that if $b_i$ and $b_j$ are consecutive integer terms then $i-j$ is bounded. This implies that there exists some integer $k$ where if $b_i=k=b_j$ and $b_i$ and $b_j$ are consecutive terms with this property, then $i-j$ is bounded. Now, notice that the only thing that decreases the denominator is $a_i=1$ and it does so linearly. And, if $a_i>1$ then the numerator increases exponentially while the denominator grows linearly. Thus, we require an arbitrarily large number of $a_i=1$'s to drag $b_i$ down to $k$, contradicting $i-j$ bounded.
This post has been edited 2 times. Last edited by blackbluecar, Dec 19, 2021, 4:54 PM
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L567
1184 posts
#3 • 1 Y
Y by PRMOisTheHardestExam
Suppose not, and let the $b_i$ be bounded above by $M$. Ignore the first $10000000M$ terms and only consider the rest, so the denominator of the $b_i$'s now only has numbers $> 10000000M$.

Let $N$ be a sufficiently large number and consider the first $N$ terms. Every million terms, there is at least one integer, so there are at least $\frac{N}{10^6}$ integers. Since there are only $M$ possibilities for these, some integer appears at least $\frac{N}{10^6M}$ times, so by PHP, we can find two of these that are at most $10^6M$ apart.

Say for the first number, we had the $b_i$ equal to $\frac{x}{y}$. The numbers in between cant be all $1$'s since then the integers cannot be equal. Let $P$ be the product of the numbers in between and $S$ be the sum.

We have $\frac{xP}{y+S} = \frac{x}{y} \implies yP = y+S$. Note that we cant have $P = 1$ because then LHS is too small, so $P \ge 2$. I claim the LHS is now too big. Note that if an $a_i$ in between is $> 1$, then replacing it with $a_i - 1$ only decreases the difference between RHS and LHS.

So, the worst case is when one number is $2$ and others are $1$, but then we want to show $2y > y+S \iff y > S$, which is true since $y$ was at least $10000000M$ and $S$ is at most $1000000M$ since there are those many numbers, so it is indeed impossible for the $b_i$ to be bounded, as claimed. $\blacksquare$

Remark

oops @below, I meant "un"bounded
This post has been edited 1 time. Last edited by L567, Oct 21, 2021, 4:40 PM
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juckter
323 posts
#4 • 3 Y
Y by PRMOisTheHardestExam, guptaamitu1, MatBoy-123
L567's remark wrote:
We can also prove $a_i$ is bounded but this is useless for the problem.

Wrong. Assume FTSOC that $b_i$ is bounded above.

Claim. $a$ is bounded.

Proof. Assume otherwise and choose a large $k$ such that $a_k > \max(a_1, a_2, \dots, a_{k - 1})$. Let $r \ge k - 10^6$ be such that $b_r$ is an integer and let $k = r + t$. Let $P_r = a_1\dots a_r$ and $S_r = a_1 + \dots + a_r$, so $P_r \ge S_r$. Then

\begin{align*}
b_k = \frac{a_1 \dots a_k}{a_1 + a_2 + \dots + a_k} > \frac{P_ra_k}{S_r + ta_k} \ge \frac{S_ra_k}{S_r + ta_k} \ge \frac{S_ra_k}{2\max(S_r, ta_k)} = \min \left(\frac{a_k}{2}, \frac{S_r}{2t}\right)
\end{align*}
Since $S_r \ge k - 10^6$ and both $k$ and $a_k$ can be arbitrarily large, this provides a contradiction. $\square$

Assume now that $a$ is bounded above by $M$. Let $c_1, c_2, \dots$ be the sequence of integers in the sequence $b_1, b_2, \dots$. Notice that the numerators of the $c_i$ are $M$-smooth, i.e. all of their prime factors are at most $M$. Since the $c_i$ are integers it follows that all the denominators must be $M$-smooth too.

On the other hand, by the problem condition the sequence of denominators is strictly increasing, and any two consecutive denominators differ by at most $10^6 M$. This means the sequence of denominators has a positive lower density, which is a contradiction as the $M$-smooth numbers have density $0$.
This post has been edited 1 time. Last edited by juckter, Oct 21, 2021, 4:04 PM
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v_Enhance
6877 posts
#5 • 4 Y
Y by PRMOisTheHardestExam, HamstPan38825, Mathematicsislovely, pog
Let $K = 10^{10^{100}}$ be a huge absolute constant. We consider two cases. The first case is captured in the following claim.

Claim: Suppose that after the first $K$ terms, at least one of every $K$ terms $a_i$ is greater than $1$. Then $b_i$ is unbounded.
Proof. In that case, consider the first $n$ terms for some large $n$. Let $M = \max(a_1, \dots, a_n)$. We have that \[ b_n \ge \frac{ 		2^{\left\lfloor 10^{-6} n - 1 \right\rfloor} M} 	{M+M+\dots+M} 	\ge \frac{1}{n} 		2^{\left\lfloor 10^{-6} n - 1 \right\rfloor} \]since the product of every million terms is at least $2$ and one of the products is actually equal to $M$. The right-hand side is unbounded in $n$, as desired. $\blacksquare$

Thus, we may focus on the case where $a_N = a_{N+1} = \dots = a_{N+K-1} = 1$ for some $N \ge K$. Look at two indices $b_{N+i}$ and $b_{N+j}$ with $|j-i| \le 10^6$ and such that $b_i$ and $b_j$ are integers, as promised by the problem condition. We may write \[ b_i = \frac{X}{Y+i}, \qquad b_j = \frac{X}{Y+j} \]where $Y = a_1 + \dots + a_N \ge N \ge K$. Then \[ X \ge \operatorname{lcm} (Y+i, Y+j) 	= \frac{(Y+i)(Y+j)}{\gcd(Y+i, Y+j)} 	\ge \frac{1}{10^6} (Y+i)(Y+j) \]which means that \[ b_j \ge \frac{Y+j}{10^6} > \frac{K}{10^6} \]as needed.
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SnowPanda
186 posts
#6
Y by
Solution
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guptaamitu1
656 posts
#7
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Assume contrary. Let $\text{one million} = x$ and $2021^{2021} = y$. Note
$$ \frac{a_{n+k}}{a_n} = a_{n+1} \cdots a_{n+k} \cdot \frac{a_1 + a_2 + \cdots + a_n}{a_1 + \cdots + a_{n+k}} $$Observe decreasing any of $a_1,\ldots,a_{k+n}$ would decrease the above value and increasing them would increase it (this fact will be used several times). By assumption the sequence $\{b_i\}_{i \ge 1}$ is bounded above by $y$. Fix a large $M$. Pick a $N > M$ such that $b_N$ is the max possible integer among $b_M,b_{M+1},\ldots$. We have two cases:

Case 1: All of $a_{N+1},\ldots,a_{N+x}$ equal $1$. Pick a $1 \le r \le x$ such that $a_{N+r} \in \mathbb Z$. Then both
$$ b_N = \frac{a_1a_2 \cdots a_N}{a_1 + \cdots + a_N} ~~,~~ b_{N+r} = \frac{a_1 \cdots a_N}{a_1 + \cdots + a_N + r} $$are integers.
In particular $a_1a_2 \cdots a_N$ is divisible by both $a_1 + \cdots + a_N, a_1 + \cdots + a_N + x$. Thus,
$$ b_N \ge \frac{\text{lcm}(a_1 + \cdots + a_N, a_1 + \cdots + a_N+r)}{a_1 + a_2 + \cdots + a_N} \ge \frac{a_1 + a_2 + \cdots + a_N + r}{r} \ge \frac{M}{x}  $$But as $M$ is large, so $b_N > y$, contradiction. $\square$

Case 2: $a_{N+s} > 1$ for some $1 \le s \le x$. Then,
$$ \frac{b_{N+s}}{b_N} = a_{N+1} \cdots a_{N+s} \cdot \frac{a_1 + \cdots + a_N}{a_1 + \cdots + a_{N+s}} \ge 2 \cdot \frac{N}{N+s+1} \ge \frac{2M}{M+s+1} \ge \frac{2M}{M+x+1} $$Now pick a $1 \le t \le x$ such that $b_{N + s + t} \in \mathbb Z$. Then,
$$ \frac{b_{N+s+t}}{b_{N+s}} = a_{N+s+1} \cdots a_{N+s+t} \cdot \frac{a_1 + \cdots + a_{N+s}}{a_1 + \cdots + a_{N+s+t} } \ge \frac{N+s}{N+s+t} \ge \frac{M+s}{M+s+t} \ge \frac{M}{M+t} \ge \frac{M}{M+x} $$Thus it follows that
$$ \frac{b_{N+s+t}}{b_N} \ge \frac{2M}{M+x+1} \cdot \frac{M}{M+x} $$As $M$ is large, so $b_{N+s+t} > b_N$, which contradicts the maximality assumption on $b_N$. $\blacksquare$
This post has been edited 1 time. Last edited by guptaamitu1, Feb 7, 2022, 6:55 PM
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megarnie
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#8
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Suppose that $b_i\le 2021^{2021}$ for all $i$.

Claim: There exist arbitrarily large groups of consecutive $a_i$ all equal to $1$.
Proof: Suppose this was not the case. Then for any arbitrarily large $k$, we have that at least one of $k$ consecutive terms in the sequence $(a_i)$ is greater than $1$.

Fix a $k=C> 69^{420^{665^{1434}}}$ satisfying that condition. Notice that for any $n$ at least $n$ of $a_1, a_2, \ldots, a_{n\cdot C}$ are greater than $1$, which implies that \[b_{n\cdot C} \ge \frac{2^n}{a_1 + a_2 + \cdots + a_{n\cdot C}}\]Notice that the function $\frac{kx}{k+x}$ for fixed $k$ is increasing over the interval $(1, \infty)$, which implies that \[b_{n\cdot C} \ge \frac{2^n}{n\cdot 2 + (C-n)\cdot 1} = \frac{2^n}{n(C+1)},\]which exceeds $2021^{2021}$ for large $n$. $\square$


Now we have for any $M$, there exist $a_i, a_{i+1}, \ldots, a_{i+M-1}$ all equal to $1$.

Fix a large $M > 665^{1434^{7776}}$ for which $a_i = a_{i+1} = \cdots = a_{i+M-1} = 1$, where $i> 665^{1434^{7776}}$ is a positive integer.

Notice that there exist $x<y$ strictly between $i$ and $i+M - 1$ such that $y-x\le 10^6$ and $b_x, b_y\in \mathbb{Z}$.

Let $x+d = y, a_1a_2 \cdots a_x = P, a_1 + a_2 \cdots + a_x = S$. We have \[b_x = \frac{P}{S}, b_y = \frac{P}{S+d}\]This implies that $\mathrm{lcm}(S, S+d)\mid P$, so \[P\ge \frac{S(S+d)}{\gcd(S, S+d)} \ge \frac{S(S+d)}{10^6},\]which implies that $b_x \ge \frac{S+d}{10^6}$, which exceeds $2021^{2021}$ because $S\ge x$ is a sufficiently large positive integer, contradiction.
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IAmTheHazard
5003 posts
#9 • 1 Y
Y by megarnie
We prove that $b_i$ is unbounded which clearly finishes. Suppose for the sake of contradiction that the $b_i$ are all bounded above by some positive integer $M$.
First note that $S_n:=a_1+\cdots+a_n$ is clearly unbounded, so $P_n:=a_1\ldots a_n$ must be as well, since it's bounded below by $S_n$ at least once in every million consecutive terms. We now prove the central claim.

Claim: There cannot exist one trillion consecutive terms $a_{n+1}=\cdots=a_{n+10^{12}}=1$ for $n$ sufficiently large.
Proof: We will first prove that there cannot exist one billion consecutive terms $a_{n+1}=\cdots=a_{n+10^9}=1$ if $b_n$ is an integer, if $n$ is sufficiently large, so $S_n\geq GM$, where $G$ is Graham's number. Suppose that we have $P_n=kS_n$, so $k\leq M$. Then if $a_{n+1}=\cdots=a_{n+10^9}=1$, we have
$$b_{n+10^9}=\frac{kS_n}{S_n+10^9}>k-1 \iff kS_n>kS_n-S_n+10^9k-10^9 \iff S_n>10^9k-10^9,$$which is true. Thus $b_i \in (k-1,k)$ for all $n+1 \leq i \leq n+10^9$, so we have $10^9$ consecutive noninteger $b_i$: illegal. To finish, note that if $a_{n+1}=\cdots=a_{n+10^{12}}=1$ we can find some $1 \leq i \leq 10^6$ such that $b_i$ is an integer, so applying this lemma implies that there are at most $10^6+10^9$ consecutive one terms at a time, which clearly implies the desired claim.

Now for large $n$ such that $S_n \geq \mathrm{TREE}(3)M$ (note that $\mathrm{TREE}(3)\gg G$), if we have $a_{n+1}=\cdots=a_{n+k}=1$ and $a_{n+k+1}=c>1$, we have
$$b_{n+k+1}=\frac{cP_n}{S_n+c+k}\geq \frac{1.001P_n}{S_n}=1.001b_n \iff cS_n\geq 1.001(S_n+c+k) \iff S_n \geq \frac{1.001c+1.001k}{c-1.001}\geq 1.001+\frac{1.001k+1.001^2}{c-1.001}.$$By our claim, $k\leq M$, hence the last inequality is certainly true as $c\geq 2$. Thus we can find a subsequence of the $b_i$ which grows exponentially, which contradicts our initial assumption. This clearly gives us some $b_k>2021^{2021}$, so we're done. $\blacksquare$
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mathlogician
1051 posts
#10
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The goal is to prove that $(b_i)$ is unbounded; suppose otherwise. Now we first prove that we can never have an infinitely long string of $1$s, since otherwise for infinitely many consecutive $b_i$ we have $b_i < 1$ as $a_1a_2\cdots a_i$ is bounded whilst $a_1+a_2+\cdots + a_i$ is unbounded. Now take a very large $n$ and $k$ such that some term between $a_n$ and $a_{n+k}$ is greater than $1$ and with $b_n \geq b_k$; we can do this since $(b_i)$ is bounded. We get that $S \left( a_{n+1}a_{n+2}\dots a_{n+k} \right) \leq a_{n+1} + a_{n+2} + \dots + a_{n+k}$ where $S = a_1 + a_2 + \dots + a_n$. Suppose that $t$ of the $a_i$ between $a_n$ and $a_{n+k}$ are greater than $1$. Then if $t = 1$ then with sufficiently high $S$ we note that $S(a_i -1) > 10^6-1 + a_i$ which is a contradiction. Otherwise if $t>1$ we can just induct down on $t$ by noticing $a_i > 1$ contributes a lot more to the left-hand side than the right-hand side, done.
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Bigtaitus
75 posts
#11 • 1 Y
Y by Vahe_Arsenyan
We assume FTSOC that $b_k\leq 2021^{2021}$ $\forall k.$ Now the key claim is the following:

Claim. For all $a\in\mathbb{N}$ $\exists b s.t. a_{b+i}=1 \forall i\in \{1,2,\dots, a\}.$

Proof. For some $a_i$ with $i$ arbitrarely big, let $c_1,c_2,...,c_k$ be the $a_j\geq 2$ for $j\leq i.$ Now: $$b_i=\frac{c_1\cdots c_k}{c_1+\cdots + c_k+(i-k)}.$$If we can show that $(i-k)>ck$ has a solution $k$ $\forall c,$ then the claim is proven by $PHP.$
In order to show this, we claim that $b_i$ would dicrease if we swap some $a_l\to a_l-1f.$ For the sake of simplicity we'll replace $a_1\to a_1-1.$ This is equivalent to: $$\frac{(c_1-1)c_2\cdots c_k}{c_1-1+c_2+\cdots c_k}\leq \frac{c_1\cdots c_k}{c_1+c_2+\cdots c_k}$$which is trivially true by expanding. So, iterating this procces until all $c_i's$ are $2$ we get that $$b_i\leq \frac{2^{k}}{2k+(i-k)}\leq 2021^{2021},$$which ends the claim.
\blacksquare

This means that we can get a $d$ such that there exist $d_1,d_2,...,d_m<10^6$ satisfying that all $b_{d}, b_{d+d_1}, b_{d+d_1+d_2}, \dots, b_{d+d_1+\cdots +d_m}$ are all integers, and $a_d,a_{d+1}, \dots a_{d+d_1+\cdots +d_k}=1.$ Let $f(i)=a_1+a_2+\cdots a_i.$ Now we have that $lcm(f(d),f(d+d_1),\dots f(d+d_1+\cdots +d_m))\mid a_1\cdot a_2\cdots a_d.$ Bounding the quotient of this $lcm$ and $f(d)$ is trivial, and ends the problem immediately.
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HamstPan38825
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#12
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Let $\{x_i\}$ be the maximal subsequence of $\{b_i\}$ that consists of only integers. (For convenience, the indices of $\{x_i\}$ will match those of $\{b_i\}$ and thus not be always consecutive.) I claim that for every $x_i$, there exists an $r$ such that $x_{i+r} > x_i$, which will imply the result.

Suppose for the sake of contradiction that there exists an $n$ such that $\{x_i\}$ is constant at $c$ for all $i \geq n$. Then, for $k > \ell > n$, we have
\begin{align*}
a_1a_2\cdots a_k &= c(a_1+a_2+\cdots+a_k) \\
a_1a_2\cdots a_\ell &= c(a_1+a_2+\cdots+a_\ell) \\
\implies a_{\ell+1}a_{\ell+2}\cdots a_k(a_1+a_2+\cdots+a_\ell) &= a_1+a_2+\cdots + a_k.
\end{align*}Let $N = a_1+a_2+\cdots + a_\ell$. Then suppose $a_{\ell+1}+a_{\ell+2}+\cdots + a_k = m \cdot N$, such that $a_{\ell+1}a_{\ell+2}\cdots a_n > m \cdot N - 10^6$ as $k - \ell < 10^6$. It follows that $$Na_{\ell + 1}a_{\ell+2} \cdots a_k > N(m \cdot N - 10^6) > (m+1)N = a_1+a_2+\cdots+ a_k,$$which is an obvious contradiction.
This post has been edited 1 time. Last edited by HamstPan38825, Dec 21, 2023, 11:47 PM
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bin_sherlo
733 posts
#13
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Suppose that $\{b_i\}_{i=1}^{\infty}$ is bounded above by $C$. Since there are finitely many positive integers $b_i$ can take, there are infinitely many pairs where $b_i=b_j$. Assume that the value $T$ is taken infinitely many times. Let $T=b_{m_1}=\dots=b_{m_r}=\dots$
\[\frac{a_1\dots a_k}{a_1+\dots+a_k}=T=\frac{a_1\dots a_k\dots a_{k+l}}{a_1+\dots+a_{k+l}}\implies a_{k+1}\dots a_{k+l}=1+\frac{a_{k+1}+\dots+a_{k+l}}{a_1+\dots+a_k}\]Note that $a_1+\dots+a_k|a_{k+1}+\dots+a_{k+l}$ hence $k\leq a_{k+1}+\dots+a_{k+l}$. We see that
\[\frac{a_{k+1}+\dots+a_{k+l}}{l}\leq a_{k+1}\dots a_{k+l}=1+\frac{a_{k+1}+\dots+a_{k+l}}{a_1+\dots+a_k}\leq 1+\frac{a_{k+1}+\dots+a_{k+l}}{k}\]We pick $l\leq 10^6$ and $k$ sufficiently large (since there are infinitely many $b_i\in \mathbb{Z}^+$ and consecutive ones have difference at most $10^6$ we can apply this).
\[1\geq (a_{k+1}+\dots+a_{k+l})(\frac{1}{l}-\frac{1}{k})\geq k(\frac{1}{l}-\frac{1}{k})\geq \frac{k}{l}-1>1\]Which gives a contradiction as desired.$\blacksquare$
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Ilikeminecraft
658 posts
#15
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Suppose $b_n$ is bounded above. Let $A = 143414341434$ be a fixed constant.
Claim: Let $N$ be any integer. Then, there exists a value $k$ such that $a_k = a_{k + 1} = \dots = a_{k + N - 1} = 1.$
Proof: Suppose $c_1, \dots, c_k$ are the values greater than 1 in the set $a_1, a_2, \dots, a_m,$ where $m$ is some very big number. Then, \[b_m = \frac{c_1\dots c_k}{c_1 + \dots + c_k + (m - k)}.\]Observe that $\frac{x\cdot j}{x + j}$ is an increasing function over $x \in (1, \infty)$ for a fixed $j > 0.$ Thus, if we were to smooth the value of $b_m$ by forcing all of the $c_i = 2,$ we get \[b_m \geq  \frac{2^m}{2m + k - m} \geq 2021^{2021}\]for sufficiently large $m.$ Thus, the claim must be true.

We now prove that this is a contradiction. Assume $b_n$ is bounded above by $B.$ We know there exists $k$ such that $a_{k + 1} = a_{k + 2} = \dots = a_{AB + k} = 1.$ Clearly, at least $B + 1$ values in the set $\{b_{k + 1}, b_{k + 2}, \dots, b_{k + AB + k}\}$ must be integers. However, clearly, the values $b_{k + i} > b_{k + j}$ for $i > j.$ Thus, this is impossible.
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