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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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 A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

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Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
integer functional equation
ABCDE   150
N a few seconds ago by youochange
Source: 2015 IMO Shortlist A2
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
150 replies
ABCDE
Jul 7, 2016
youochange
a few seconds ago
min A=x+1/x+y+1/y if 2(x+y)=1+xy for x,y>0 , 2020 ISL A3 for juniors
parmenides51   13
N 4 minutes ago by AylyGayypow009
Source: 2021 Greece JMO p1 (serves also as JBMO TST) / based on 2020 IMO ISL A3
If positive reals $x,y$ are such that $2(x+y)=1+xy$, find the minimum value of expression $$A=x+\frac{1}{x}+y+\frac{1}{y}$$
13 replies
parmenides51
Jul 21, 2021
AylyGayypow009
4 minutes ago
Thailand MO 2025 P3
Kaimiaku   3
N 8 minutes ago by EeEeRUT
Let $a,b,c,x,y,z$ be positive real numbers such that $ay+bz+cx \le az+bx+cy$. Prove that $$ \frac{xy}{ax+bx+cy}+\frac{yz}{by+cy+az}+\frac{zx}{cz+az+bx} \le \frac{x+y+z}{a+b+c}$$
3 replies
Kaimiaku
2 hours ago
EeEeRUT
8 minutes ago
Another geo P1
alchemyst_   32
N 10 minutes ago by tilya_TASh
Source: Balkan MO 2022 P1
Let $ABC$ be an acute triangle such that $CA \neq CB$ with circumcircle $\omega$ and circumcentre $O$. Let $t_A$ and $t_B$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $CX$. The line through $C$ parallel to line $AB$ meets $t_A$ at $Z$. Prove that the line $YZ$ passes through the midpoint of the line segment $AC$.

Proposed by Dominic Yeo, United Kingdom
32 replies
alchemyst_
May 6, 2022
tilya_TASh
10 minutes ago
Funny Moments
Dream9   32
N 5 hours ago by valisaxieamc
What's the goofiest thing you've seen happen at a math competition? :starwars:
32 replies
Dream9
Sunday at 8:43 PM
valisaxieamc
5 hours ago
CDR Scores List
MathRook7817   30
N 5 hours ago by DhruvJha
List of all the scores in the matchups put them here
30 replies
MathRook7817
Yesterday at 3:43 PM
DhruvJha
5 hours ago
Favorite Memory at MATHCOUNTS
MathRook7817   37
N 5 hours ago by DhruvJha
Hey guys, what is everyone's favorite memory at any Mathcounts competition?

Mine was arriving at the hotel for the 2024 nats comp.
37 replies
MathRook7817
Yesterday at 12:31 AM
DhruvJha
5 hours ago
2025 MATHCOUNTS State Hub
SirAppel   850
N Today at 2:16 AM by mathkiddus
Previous Years' "Hubs": (2022) (2023) (2024)Please Read

Now that it's April and we're allowed to discuss ...
[list=disc]
[*] CA: 43 (45 44 43 43 43 42 42 41 41 41)
[*] NJ: 43 (45 44 44 43 39 42 40 40 39 38) *
[*] NY: 42 (43 42 42 42 41 40 38 38 38 38 38 38)
[*] TX: 42 (43 43 43 42 42 40 40 38 38 38)
[*] MA: 41 (45 43 42 41)
[*] WA: 41 (41 45 42 41 41 41 41 41 41 40) *
[*]VA: 40 (41 40 40 40)
[*] FL: 39 (42 41 40 39 38 37 37)
[*] IN: 39 (41 40 40 39 36 35 35 35 34 34)
[*] NC: 39 (42 42 41 39)
[*] IL: 38 (41 40 39 38 38 38)
[*] OR: 38 (44 39 38 38)
[*] PA: 38 (41 40 40 38 38 37 36 36 34 34) *
[*] MD: 37 (43 39 39 37 37 37)
[*] AZ: 36 (40? 39? 39 36)
[*] CT: 36 (44 38 38 36 35 35 34 34 34 33 33 32 32 32 32)
[*] MI: 36 (39 41 41 36 37 37 36 36 36 36) *
[*] MN: 36 (40 36 36 36 35 35 35 34)
[*] CO: 35 (41 37 37 35 35 35 ?? 31 31 30) *
[*] GA: 35 (38 37 36 35 34 34 34 34 34 33)
[*] OH: 35 (41 37 36 35)
[*] AR: 34 (46 45 35 34 33 31 31 31 29 29)
[*] NV: 34 (41 38 ?? 34)
[*] TN: 34 (38 ?? ?? 34)
[*] WI: 34 (40 37 37 34 35 30 28 29 29 29) *
[*] HI: 32 (35 34 32 32)
[*] NH: 31 (42 35 33 31 30)
[*] DE: 30 (34 33 32 30 30 29 28 27 26? 24)
[*] SC: 30 (33 33 31 30)
[*] IA: 29 (33 30 31 29 29 29 29 29 29 29 29 29) *
[*] NE: 28 (34 30 28 28 27 27 26 26 25 25)
[*] SD: 22 (30 29 24 22 22 22 21 21 20 20)
[/list]
Cutoffs Unknown

* means that CDR is official in that state.

Notes

For those asking about the removal of the tiers, I'd like to quote Jason himself:
[quote=peace09]
learn from my mistakes
[/quote]

Help contribute by sharing your state's cutoffs!
850 replies
SirAppel
Apr 1, 2025
mathkiddus
Today at 2:16 AM
9 What competitions do you do
VivaanKam   31
N Today at 2:00 AM by nmlikesmath

I know I missed a lot of other competitions so if you didi one of the just choose "Other".
31 replies
VivaanKam
Apr 30, 2025
nmlikesmath
Today at 2:00 AM
How to get a 300+ on the NWEA MAP MATH test (URGENT)
nmlikesmath   27
N Today at 1:58 AM by nmlikesmath
I have 4 days till this test, I'm wondering how do I get a 300+ and what do I need to know, thank you.
27 replies
nmlikesmath
May 3, 2025
nmlikesmath
Today at 1:58 AM
9 Mathcounts poll
AndrewZhong2012   17
N Today at 1:22 AM by tikachaudhuri
Last year I got 14th at States and 3rd in Chapter
This year I got 91.5 on AMC 10A.
Do you guys think I can get to nationals this year? I am in PA. This year is also my last chance.

17 replies
AndrewZhong2012
Jan 7, 2025
tikachaudhuri
Today at 1:22 AM
The daily problem!
Leeoz   158
N Today at 12:25 AM by giratina3
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! :)

Please hide solutions and answers, hints are fine though! :)

Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!

Past Problems!
158 replies
Leeoz
Mar 21, 2025
giratina3
Today at 12:25 AM
A glass is filled with milk. Two-thirds of its content is poured out and replace
Vulch   5
N Today at 12:23 AM by giratina3
A glass is filled with milk. Two-thirds of its content is poured out and replaced with water. If this process of pouring out two-thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is
5 replies
Vulch
May 10, 2025
giratina3
Today at 12:23 AM
system linear equation with substitution
Miranda2829   3
N Today at 12:18 AM by giratina3
5x-3y=-22
6x+4y=-34

whats the steps by using substitution in this question?

many thanks
3 replies
Miranda2829
Yesterday at 1:56 AM
giratina3
Today at 12:18 AM
Divisibility on 101 integers
BR1F1SZ   5
N May 8, 2025 by Grasshopper-
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
5 replies
BR1F1SZ
Aug 9, 2024
Grasshopper-
May 8, 2025
Divisibility on 101 integers
G H J
G H BBookmark kLocked kLocked NReply
Source: Argentina Cono Sur TST 2024 P2
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BR1F1SZ
577 posts
#1 • 2 Y
Y by cubres, Rounak_iitr
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
This post has been edited 2 times. Last edited by BR1F1SZ, Jan 27, 2025, 5:01 PM
Z K Y
The post below has been deleted. Click to close.
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KAME06
159 posts
#2 • 1 Y
Y by cubres
Let $a_k$ the largest of the 101 numbers.
-Supose that $a_k>201$. First we are going to proof by induction on $n$ that $a_{k-n}=a_k-n$ for all $n=0, 1, 2, ...100$ (Is easy to see that $a_k, a_{k-1}, ..., a_{k-100}$ are all the numbers $a_1, a_2, ...., a_{101}$ where k isn't necessarily 1. Just think them as modulo 101).
Case base: Note that $a_k \mid a_{k-1} +1 \Rightarrow a_k \le a_{k-1}+1 \le a_k+1$, so $a_{k-1}=a_k$, or $a_{k-1}=a_k -1$, but $a_k$ and $a_k+1$ are coprimes so $a_{k-1}=a_k -1$.
Inductive step: If $a_{k-t}=a_k-t$ for some $0\le t \le 99$ we have that: $a_{k-t}=a_k-t\mid a_{k-(t+1)} +1 \Rightarrow a_k-t\le a_{k-(t+1)} +1\le a_k +1$, so $a_{k-(t+1)}$ has all this possibilities: $a_k-(t+1), a_k-t, a_k-(t-1), a_k -(t-2), ..., a_k$.
If $a_{k-(t+1)}=a_k-(u)$, $0\le u \le t$, we have that: $a_{k-t}=a_k-t \mid a_{k-(t+1)} +1= a_k-(u) +1 \Rightarrow a_k-t \mid(a_k-u+1)- ( a_k-t)=t-u+1>0$, (definition of u)$ \Rightarrow a_k-t \le t-u+1 \Rightarrow a_k \le 2t -u +1 \le 2*100 - 0 +1=201$. Contradiction. So that implies $a_{k-(t+1)}=a_k-(t+1)$, end of the induction.
Then, we know that $a_{k-100}=a_k-100$, so we have that: $a_{k-100}=a_k-100 \mid a_k +1 \Rightarrow a_k-100 \mid (a_k+1) - (a_k - 100)=101 \Rightarrow a_k - 100 \le 101 \Rightarrow a_k \le 201$ Contradiction.
-We conclude that $a_k\le 201$. Consider the next configuration: $a_i=100 +i$ for all $i=1, 2, ...101$. Note that:
For all $1, 2, ..., 100$, $a_i+1=100+i+1=a_{i+1}$, so $a_i+1$ is a multiple of $a_{i+1}$.
$a_{101}+1=202= 101*2=a_1 * 2$ so $a_{101}+1$ is a multiple of $a_1$.
-The answer is 201.
This post has been edited 2 times. Last edited by KAME06, Nov 13, 2024, 1:23 AM
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lian_the_noob12
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#4 • 2 Y
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$\color{green} \boxed{\textbf{SOLUTION}}$
We have, $i=a_1 \le a_{101}+1\le a_{100}+2 \cdots \le a_{2} + 100 \le a_{1} + 100=i+100$

So $a_k  \in [i, i+100]$
As every number have to be different so $(a_1,a_2,...a_{101})$ must be a combination of $(i,i+1,i+2...,i+100)$

Let the are not consecutive,
$a_{k-1} = i + m$ and $a_k = i+n$
We have $i+n | i+m+1 \implies n < m+1 \implies n-1 < m$ But $m,n$ are not equal so $n < m$

Now,
$ i+n \le (m-n) +1$
And $102 \ge (m-n) +1 \ge i+n$
Which gives the maximum number is $102$

But while taking them consecutive, $a_k = i + (k-1)$
Its easy to see all $k \le 100$ does work now for
$i=a_1| a_{101}+1=i+101|101$ gives $a_1=101,...a_{101}=201$
So $\boxed{201}$
This post has been edited 2 times. Last edited by lian_the_noob12, Jan 10, 2025, 3:37 PM
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ClassyPeach
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#5 • 1 Y
Y by cubres
It says, 101 numbers. But if i=101 then we need a_102
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BR1F1SZ
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#6 • 1 Y
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ClassyPeach wrote:
It says, 101 numbers. But if i=101 then we need a_102

The indices are taken module $101$, so $a_{102}=a_1$.
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Grasshopper-
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#7 • 1 Y
Y by cubres
lian_the_noob12 wrote:
$\color{green} \boxed{\textbf{SOLUTION}}$
We have, $i=a_1 \le a_{101}+1\le a_{100}+2 \cdots \le a_{2} + 100 \le a_{1} + 100=i+100$

So $a_k  \in [i, i+100]$
As every number have to be different so $(a_1,a_2,...a_{101})$ must be a combination of $(i,i+1,i+2...,i+100)$

Let the are not consecutive,
$a_{k-1} = i + m$ and $a_k = i+n$
We have $i+n | i+m+1 \implies n < m+1 \implies n-1 < m$ But $m,n$ are not equal so $n < m$

Now,
$ i+n \le (m-n) +1$
And $102 \ge (m-n) +1 \ge i+n$
Which gives the maximum number is $102$

But while taking them consecutive, $a_k = i + (k-1)$
Its easy to see all $k \le 100$ does work now for
$i=a_1| a_{101}+1=i+101|101$ gives $a_1=101,...a_{101}=201$
So $\boxed{201}$

Why every number has to be different?
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