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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]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
Posting Guidelines
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
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How do I study for mathcounts?
Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

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
Find the value
sqing   5
N 2 minutes ago by sqing
Source: Own
Let $ a,b $ be real numbers such that $ (a^2 + b^2) (a + 1) (b + 1) =  a ^ 3 + b ^ 3 =2 $. Find the value of $ a b .$

Let $ a,b $ be real numbers such that $ (a^2 + b^2) (a + 1) (b + 1) = 2 $ and $ a ^ 3 + b ^ 3 = 1 $. Find the value of $ a + b .$
5 replies
sqing
Yesterday at 2:29 PM
sqing
2 minutes ago
2-var inequality
sqing   0
33 minutes ago
Source: Own
Let $ a,b\geq  0 ,a+b+ab=2.$ Prove that
$$ (a^2+\frac{27}{5}ab+b^2)(a+1)(b+1) \leq 12 $$$$ (a^2+\frac{11}{2}ab+b^2)(a+1)(b+1) \leq 45(2-\sqrt 3) $$
0 replies
sqing
33 minutes ago
0 replies
circumcenter of ARS lies on AD
Melid   1
N 39 minutes ago by Acrylic3491
Source: own
In triangle $ABC$, let $D$ be a point on arc $BC$ of circle $ABC$ which doesn't contain $A$. $AD$ and $BC$ intersect at $E$. Let $P$ and $Q$ be the reflection of $E$ about to $AB$ and $AC$, respectively. $PD$ intersects $AB$ at $R$, and $QD$ intersects $AC$ at $S$. Prove that circumcenter of triangle $ARS$ lies on $AD$.
1 reply
Melid
5 hours ago
Acrylic3491
39 minutes ago
2-var inequality
sqing   10
N an hour ago by sqing
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq 8$$$$ (a^2+b^2)(a+1)(b+1) \leq 8$$Let $ a,b> 0 ,a^3+ab(a+b)+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$
10 replies
sqing
Yesterday at 1:35 PM
sqing
an hour ago
Overly wordy problems
ZMB038   16
N Today at 6:47 AM by Yiyj
Hey everyone, here we can post questions with way to many extraneous words, that are actually easy.
Try to solve the one above yours.
I'll start:
Click to reveal hidden text
16 replies
ZMB038
May 28, 2025
Yiyj
Today at 6:47 AM
No one here has solved the revised version yet
PikaVee   7
N Today at 6:19 AM by PikaVee
(Thanks Random Stranger for the idea and I will be making it so it is extremely specific to your solution.)
We are playing Pokemon Scarlet and Violet and you are fighting a friend. You and your friend don't have any items at all and the pokemon does not have any held items. Your friend challenges you to a battle because he just said nah I'd win.

You two start the battle using only one Pokemon each which neither of you knows the type of the other. Luckily he had used a level 28 Squirtle and you had used a level 25 Pikachu. Surprisingly both of the Pokemon each have one HP. Your Pikachu has a move set of one single move of Thunder with 10/10 PP and has 1 HP because you forgot to go to the Pokemon center. Your Pikachu also has a bad IV stat in speed with 1/15 and the 252 EV speed stat of the Squirtle combined with a perfect IV stat in speed makes it so it guarantees to always out speed your move. To account for that he made his Squirtle have 1 HP on purpose for absolutely no reason.

After he saw what kind of moves you have and since that person was so cocky and confident that they decided to gamble all their moves with each having an equal chance of being used. Their Squirtle has a move set of Protect 10 PP which has 100% chance of being used and has has the success probability multiplied by 1/3 every time it is being used (Meaning the second time it is being used has a 33% chance of succeeding and a third time it will be 11%. This also ignores the rules of how the move is regularly used by making the 4th move 1/27 instead of it being a guaranteed fail and so on.), Tackle which has a 100% chance to hit having 10/35 PP , Water Gun which has a 100% chance of hitting with 10/25 and Rain Dance with 5/5 PP and 100% chance of being used. If the amount of PP reaches 0 it will be unavailable for the rest of the fight meaning that the probability for each other move to be used goes from 25% all the way to 33%.

For everyone who wants to solve the easy part. If the probability that Squirtle will survive turn 1 when simplified is a/b then what is a+b?

Alright so for Squirtle to survive turn one then we try to find out how Squirte will faint at turn one. First of all Pikachu needs to hit Thunder bolt at a 70% chance then get a 25% chance that the Squirtle will use Rain Dance so that the Squirtle will not faint the Pikachu because it didn't attack. Another possible option is for it to choose Water Gun and miss so it would be a 70% * 25% * 5% chance for Pikachu to faint Squirtle. 70% = \frac{7}{10}, 25% = \frac{1}{4}, and 5% = \frac{1}{20} So the complement of what we are trying to find is $\frac{7}{10}*\frac{1}{4}+\frac{7}{10}*\frac{1}{4}*\frac{1}{20}=\frac{7}{10*4}+\frac{7}{10*4*20}=\frac{7}{40}+\frac{7}{800}=\frac{140}{800}+\frac{7}{800}=\frac{147}{800}$. The complement of this would be $1-\frac{147}{800}$ or $\frac{653}{800}$. The final thing we can do is to make sure it is simplified and add the numerator and the denominator which is $653$ and $800$ so $653+800=1453$ This should be final answer. (Mathdash rating 800)

For the very hard question, What is the probability that the Squirtle will win this fight? (This is going to be a very long arithmetic series with a lot of cases. The max amount of turns this fight can have is 11 turns.)
7 replies
PikaVee
May 28, 2025
PikaVee
Today at 6:19 AM
Problem of the day
sultanine   20
N Today at 3:16 AM by EthanNg6
[center]Every day I will post 3 new problems
one easy, one medium, and one hard.
Please hide your answers so others won't be affected
:D :) :D :) :D
20 replies
sultanine
May 23, 2025
EthanNg6
Today at 3:16 AM
A Variety of Math Problems to solve
FJH07   48
N Today at 3:13 AM by EthanNg6
Hi, so people can post different math problems that they think are hard, and I will post some (I think middle school math level) problems so that the community can help solve them. :)
48 replies
FJH07
May 22, 2025
EthanNg6
Today at 3:13 AM
Worst Sillies of All Time
pingpongmerrily   61
N Today at 3:01 AM by shaayonsamanta
Share the worst sillies you have ever made!

Mine was probably on the 2024 MathCounts State Target Round Problem 8, where I wrote my answer as a fraction instead of a percent, which cost me a trip to Nationals that year.
61 replies
pingpongmerrily
May 30, 2025
shaayonsamanta
Today at 3:01 AM
AMC 8 info
VivaanKam   5
N Today at 3:00 AM by shaayonsamanta
Hi I will be attending the AMC 8 contest in 2026. How does it work? time? number of questions? points? scoring?
5 replies
VivaanKam
Today at 1:11 AM
shaayonsamanta
Today at 3:00 AM
MathDash help
Spacepandamath13   11
N Today at 1:27 AM by Yiyj
AkshajK ORZ by the way invited me to do MathDash a few months ago and I did try it one day but haven't done it much after (Sorry). Now, I'm getting back into it and finding the format kind of weird. When selecting certain problem type sometimes it lets me pick immediately, other times not. Any fixes?
11 replies
Spacepandamath13
May 29, 2025
Yiyj
Today at 1:27 AM
Challenge: Make every number to 100 using 4 fours
CJB19   274
N Today at 12:30 AM by AllenHou
I've seen this attempted a lot but I want to see if the AoPS community can actually do it. Using ONLY 4 fours and math operations, make as many numbers as you can. Try to go in order. I'll start:
$$(4-4)*4*4=0$$$$4-4+4/4=1$$$$4/4+4/4=2$$$$(4+4+4)/4=3$$$$4+(4-4)*4=4$$$$4+4^{4-4}=5$$$$4!/4+4-4=6$$$$4+4-4/4=7$$$$4+4+4-4=8$$
274 replies
CJB19
May 15, 2025
AllenHou
Today at 12:30 AM
DHR Amc8?
imsuper   139
N Today at 12:23 AM by Moon_settler
What do yall think the DHR this year will be? Will 22 be enough?
139 replies
imsuper
Jan 30, 2025
Moon_settler
Today at 12:23 AM
Combo Bash
DhruvJha   5
N Yesterday at 11:01 PM by EthanNg6
Devin and Cowen are playing a game where they take turns flipping a biased coin. The coin lands on heads with probability 2/3 and tails with probability 1/3. Devin goes first. On each turn, the current player flips the coin repeatedly until the coin lands tails. For each heads flipped, the player gains 1 point and continues flipping. If the coin lands tails, their turn ends, and the other player takes their turn. The first player to reach 3 points wins the game immediately. What is the probability that Devin wins the game? Express your answer as a common fraction in lowest terms.
5 replies
DhruvJha
May 27, 2025
EthanNg6
Yesterday at 11:01 PM
Parity and sets
betongblander   7
N Apr 22, 2025 by ihategeo_1969
Source: Brazil National Olympiad 2020 5 Level 3
Let $n$ and $k$ be positive integers with $k$ $\le$ $n$. In a group of $n$ people, each one or always
speak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set $A$, what is the parity of people who speak to
true? ”, where $A$ is a subset of size $ k$ of the set of $n$ people. The answer can only
be $even$ or $odd$.
a) For which values of $n$ and $k$ is it possible to determine which people speak the truth and
which people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
7 replies
betongblander
Mar 18, 2021
ihategeo_1969
Apr 22, 2025
Parity and sets
G H J
G H BBookmark kLocked kLocked NReply
Source: Brazil National Olympiad 2020 5 Level 3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
betongblander
144 posts
#1
Y by
Let $n$ and $k$ be positive integers with $k$ $\le$ $n$. In a group of $n$ people, each one or always
speak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set $A$, what is the parity of people who speak to
true? ”, where $A$ is a subset of size $ k$ of the set of $n$ people. The answer can only
be $even$ or $odd$.
a) For which values of $n$ and $k$ is it possible to determine which people speak the truth and
which people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
This post has been edited 14 times. Last edited by betongblander, Jun 7, 2021, 8:08 PM
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v_Enhance
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#3 • 3 Y
Y by HamstPan38825, Yuuhhuuuuuuu, PHSH
The answer is that the task is possible exactly when $k$ is even in which case exactly $n$ questions are needed.
Treat each person as an element in ${\mathbb F}_2$, where $1$ for truth and $0$ for liar.
If you ask person $p$ about the set $A$, their response is \[ (p+1) + \sum_{a \in A} \pmod 2. \]So in other words, a query amounts to sampling a set of either $k-1$ elements ($p \in A$) or $k+1$ elements $(p \notin A$), and taking the sum.
Now, if $k$ is odd, the task is impossible, because replacing every $x \mapsto x+1$ changes no responses.
On the other hand, when $k$ is even, the following $n$ queries suffice:
  • Query $(x_1 + \dots + x_k) - x_i$ for $i=1,\dots,k$ By summing, one gets the value of $(k-1)(x_1 + \dots + x_k)$, and hence knows $x_i $ for $1 \le i \le k$.
  • Query $(x_1 + \dots + x_{k-2}) + x_i$ for $i = k+1, \dots, n$. This gets $x_i$ for $i \ge n$.
Moreover, at least $n$ queries are necessary because there are $2^n$ possible final answers for Arnaldo, and each query has two possible responses.
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john0512
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#5
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The answer is that it is not possible when $k$ is odd, and it takes $n$ turns when $k$ is even.

When $k$ is odd, he cannot distinguish between everyone telling the truth and everyone lying (since in both cases, all queries will result in the answer being "odd").

There are $2^n$ possible configurations, so if he asks at most $n-1$ questions, by Pidgeonhole there exists some string of answers that corresponds to more than one possible configuration, hence he cannot determine the configuration, which shows the lower bound.

When $k$ is even, we claim that he can simply query the same set $A$ each time and ask each person once. First, he asks each person in $A$ about $A$ and record the number of "even" responses and "odd" responses. These are the numbers of truth-tellers and liars in $A$ in some order. However, since $|A|$ is even, these two numbers are either both odd or both even. If they are both odd, then the people that said odd are truth-tellers and the people that said even are liars, and vice versa for the both even case. In both cases, he can determine the type of each person in $A$. Furthermore, he knows the true answer to his question at this point, so he can simply ask the same question to the remaining $n-k$ people to determine their type, hence done.
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bryanguo
1032 posts
#6
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Hmm...this problem hurts my head

For odd $k,$ the process is impossible. For even $k,$ the process requires $n$ queries.

For odd $k,$ observe that a string of all $T$'s and all $F$'s cannot be distinguished, since every query results with the answer ``odd."

For even $k,$ we first observe that $n$ operations are necessary: At each step, we at most determine if a set of $k$ people has an odd number of $T$'s or $F$'s, which halves our possibilities. Thus, to get from $2^n$ possibilities to $1$ unique string, we require at least $n$ queries.

For upper bound, since $k$ is even, in any arbitrary set of $k$ people, the parity of the truthtellers and liars must be the same. Querying everyone in the set of $k$ people, we count number of ``evens" and ``odds" we obtain as answers. If the parity of the number of people who said ``odd" lines up with the number of ``odds" we counted, then these people are the truthtellers (similarly for even), and we know the rest are liars. Arnaldo can continue this process until there are $q<k$ people remaining, from which $q$ queries is sufficient to determine who are liars and truthtellers--by taking a set of $k$ people and querying the $q$ people of which we don't know who are truthtellers or liars.
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HamstPan38825
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#7
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For odd $k$, Arnaldo cannot distinguish between all truth-tellers or all liars, as he will receive an answer of ``odd" every time.

For even $k$, I claim that at least $n$ queries are required. $n$ queries are sufficient because Arnaldo may fix a group of $k$ people and ask all $n$ people the same question for those $k$ people. Among the $k$ people themselves, suppose $a$ people reply ``odd" and $b$ people reply ``even". Then $a$ and $b$ must be the same parity, and whichever response matches that common parity comes from truth-tellers. Correspondingly Arnaldo may determine the truthfulness of all $n$ people.

To see that $n-1$ questions cannot work, notice that upon asking $n$ questions, one to every member of the group, we receive $2^n$ possible combinations of responses. By the above discussion, every response corresponds to a unique distribution of truth-tellers and liars, and every distribution of truth-tellers and liars yields a unique set of responses. Hence, upon only asking $n-1$ questions, there will be at least $2$ possibilities for the distribution of liars.
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cursed_tangent1434
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#8 • 1 Y
Y by mathematical717
Solved with mathematical717. Very interesting problem. We faced some unnecessary difficulty with the bound, but we were just clowning around. Let a positive integer $k$ be called $n$-detectable if it possible to determine which people speak the truth and which people always lie. Further, the type of a person is whether he speaks the truth or lies.

(a) We claim that the answer is all even integers $k$ (and all $n\ge k$ for each $k$). First of all note that if $k$ is odd and all the $n$ people are of the same type, Arnaldo has no way of knowing which type it is since irrespective of which set of $k$ people and which person Arnaldo selects to question, the reply will always be 'even'. So he gains no new information, and will never know the liars and truth-tellers exactly.

(b) We now show that all even integers $k$ are $n$-detectable for all $n\ge k$ with $n$ being the minimum number of moves required to determine which people speak the truth and which people lie. Our algorithm for detecting the liars and the truth-tellers exactly within $n$ moves is as follows.

Consider a random set of $k$ people among the available $n$. Now, ask each and every person of this group the question concerning this group of $k$ people. Then, each person will answer 'even' or 'odd'. Arnaldo then separates them into two groups based on there response. It is easy to see that all the people in each group are of the same type, and two people from the two separate groups must be of different types. Now, after separating into groups, both the groups are of even size, or both groups are of odd size (since $k$ is even). Thus, Arnaldo knows the parity of the size of the set of truth-tellers among this set of $k$ people. Hence, he can exactly distinguish (based on which answer they provided) which group among the separated two consists of only truth-tellers. Now, if this group has atleast one truth-teller Arnaldo picks one of them as his buddy. If all of them are liars he picks one of them as his anti-buddy and negates what ever answer he provides to a question Arnaldo asks and considered him as his buddy.

Using a buddy, Arnaldo can determine the type of all the other people as follows. Arnaldo selects the $k-1$ non-buddy people in his initial set of people. Then, he considers the set of people formed by adding each new person among the available $n$ in turn. Then, he asks from his buddy how many people speak the truth. Since Arnaldo knows the type of all but one person in this group, Arnaldo can then determine the type of the additional temporary member of the group. He repeats this process with each of the $n-k$ left over people, and finishes his task in exactly $n$ steps.

To see why Arnaldo needs $n$ steps, note that if Arnaldo can finish in at most $n-1$ steps, Arnaldo has to uniquely distinguish a set of $2^n$ possible states (each person has two possible types) using $2^{n-1}$ answer sequences. Since $2^n > 2^{n-1}$ there exists atleast two possible states for some answer sequence for $n-1$ queries, making it impossible for Arnaldo to distinguish between the two. Thus, he will always need atleast $n$ queries to determine which people speak the truth and which people always lie, and we are done.
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quantam13
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#9
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I claim that the task is only possible when $k$ is even and in that case, the minimum number of questions is $n$.

When $k$ is odd, to see that the desired task is impssible, notice that we cant differentiate between all people being truth tellers and all people being liars.

Now when $k$ is even, I give a strategy with $n$ questions which is clearly the minimum as there are $2^n$ possible asignements of truth tellers/liars to the $n$ people.

Firstly take $k$ players and use $k$ queries on each of them about the set of those $k$ players. Some work mod 2 can give that this reveals the identity of all $k$ of the players. Now for the rest of the $n-k$ players, to figure them out within $n-k$ queries, we find out the players one by one by asking them about the $k$ players whose status we already know which tells us the identity of that player.
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ihategeo_1969
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#10
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Nice? Oh god pls make me better at combo. Let $f(n,k)$ denote our required (with $\infty$ meaning it ain't possible). Then \[\boxed{f(n,k) = \begin{cases} \infty \text{ if $k$ is  odd} \\ n \text{ if $k$ is even} \end{cases}}\]To see why $k$ is odd fails, see that atmost we can just make $2$ groups and we know one group are liars and one group are truth tellers (just ask everyone the same question on some set of $k$ people). But since $k$ is odd, the parity of wise truth seekers and devilish liars are always different so we can never know for sure, which is which.

For $k$ even, we atleast need $n$ questions as there are $2^n$ possiblities and each question just ``halves" the ones into possible sequences and not at all possible sequences atmost hence we need $\log_2(2^n)=n$ questions atleast.

To see why it is all we need, just fix some group of $k$ people and ask everyone the same question. Now the parity of wise truth seekers and devilish liars in the $k$ set of people is same, and based on what answers the $k$ people gave we know what that parity is and hence whichever people are saying that parity are the truth seekers and others are naughty liars and so done.
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