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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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?
How do I get a problem on the contest page?
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
Easy and Elegant Inequality
zqy648   0
2 minutes ago
Source: 2024 November 谜之竞赛-4
For positive integers \( n \) and positive real numbers \( a_1, a_2, \ldots, a_n \), define
\[[a_1, a_2, \cdots, a_n] = \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\cdots + \frac{1}{a_n}}}}. \]Determine the minimal real number \( C \) such that for any positive real numbers \( x_1, x_2, \cdots, x_{2024} \),
\[[x_1, x_2, \cdots, x_{2024}] - [x_1^2, x_2^2, \cdots, x_{2024}^2] \leq C. \]Proposed by Ruixuan Zhu
0 replies
zqy648
2 minutes ago
0 replies
Tangent to two Fixed Circles
zqy648   0
6 minutes ago
Source: 2024 November 谜之竞赛-1
Given a circle \(\Omega\) and its chord \(AB\), let \(C\) be a moving point on the segment \(AB\). \(D\) and \(E\) lie on \(\Omega\) such that \(\angle ADC = \angle BEC = 90^\circ\), \(AE\) and \(BD\) intersect at point \(F\).

Prove that as \(C\) moves along the open segment \(AB\) (excluding endpoints), the circumcircle of \(\triangle DEF\) is tangent to two fixed circles.

Proposed by Hongdao Chen
0 replies
1 viewing
zqy648
6 minutes ago
0 replies
Bernoulli Site Percolation
zqy648   0
13 minutes ago
Source: 2025 Feb 谜之竞赛-2
Let \( \mathbb{Z}^2 \) denote the set of all integer lattice points in the plane with Cartesian coordinates. A graph \( G = (\mathbb{Z}^2, E) \) is constructed by connecting two lattice points with an edge if their Euclidean distance is $1{}{}{}$. For a positive integer \( n \), define \( f(n) \) as the number of connected subgraphs \( H = (V(H), E(H)) \) of \( G \) satisfying:
\[
\{(0, 0)\} \subseteq V(H) \subseteq \mathbb{Z}^2, \quad |V(H)| = n, \quad E(H) \subseteq E.
\]Prove that there exists a positive constant \( C \) such that for any positive integer \( n \), \(
f(n) \leq C \cdot 7^n.
\)

Proposed by Hanqing Huang and Huankun Guo
0 replies
1 viewing
zqy648
13 minutes ago
0 replies
Prove that KI₁=KI₂
zqy648   0
17 minutes ago
Source: 2025 February 谜之竞赛-6
In acute triangle \( ABC \), let \( H \) be the orthocenter. Denote the incenter of \( \triangle ABH \) by \( I_1 \) and its incircle by \( \omega_1 \); the incenter of \( \triangle ACH \) by \( I_2 \) and its incircle by \( \omega_2 \). Let \( l \) be the common external tangent of \( \omega_1 \) and \( \omega_2 \) closer to vertex \( A \), tangent to \( \omega_1 \) and \( \omega_2 \) at \( P \) and \( Q \), respectively. Lines \( BP \) and \( CQ \) intersect at \( K \). Prove that \( KI_1 = KI_2 \).

Created by Lolochen(Xiuyi Chen)
0 replies
zqy648
17 minutes ago
0 replies
AIME p7???
Soupboy0   9
N Today at 2:33 AM by giratina3
how the heck is this p7


Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}$. Find $(x+1)^{48}$.

source: 2005 AIME II p7
9 replies
Soupboy0
Yesterday at 3:40 AM
giratina3
Today at 2:33 AM
9 Favorite topic
A7456321   230
N Today at 2:28 AM by mpcnotnpc
What is your favorite math topic/subject?

If you don't know why you are here, go binge watch something!

If you forgot why you are here, go to a hospital! :)

If you know why you are here and have voted, maybe say why you picked the option that you picked in a response) :thumbup:

CLICK ON ME YOU KNOW YOU WANT TO

Timeline
230 replies
A7456321
May 23, 2025
mpcnotnpc
Today at 2:28 AM
Private Forum for Geometry Study
fossasor   234
N Today at 2:07 AM by Serenah329
Hi all,

I've recently been trying to improve my skill at competition math, but I consistently struggle with geometry. Since doing math is more fun and motivating with others, I've created the GeoPrepClub. This is a private forum to increase geometry skill and help others increase theirs: we'll have problems, marathons, and much more. This is similar to forums such as "AMC8 Prep Buddies" by PatTheKing, but is focused exclusively on this subject. We welcome all skill levels and hope those with greater mathematical knowledge can assist those lacking in it.

If this sounds interesting to you, sing up below and I'll let you know once I've added you. Although the forum may not have much now, that's because I've only just released it, and I hope once I build a community, It will be a very useful and motivating space for those interested in improving their geometry. The link is
here.

I look forward to seeing you all in the forum!
234 replies
fossasor
Jun 10, 2025
Serenah329
Today at 2:07 AM
Bogus Proof Marathon
pifinity   7732
N Today at 2:06 AM by melloncandy
Hi!
I'd like to introduce the Bogus Proof Marathon.

In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution :P

Use classic Marathon format:
[hide=P#]a1b2c3[/hide]
[hide=S#]a1b2c3[/hide]


Example posts:

P(x)
-----
S(x)
P(x+1)
-----
Let's go!! Just don't make it too hard!
7732 replies
pifinity
Mar 12, 2018
melloncandy
Today at 2:06 AM
Puzzles for Fun!!!
Yoonyoung11   21
N Today at 2:04 AM by DynamicFox52
IF YOU SOLVE ONE, ADD ANOTHER ONE PLS:
1)There are three colored boxes: Yellow, Black, and Orange. Each box contains 2 envelopes. Each envelope contains cash - two of them contain Rs. 250000 each, two of them contain Rs. 150000 each, and the remaining two contain Rs. 100000 each.
There is one statement written on the cover of each box.
* Yellow Box: Both a yellow box and an orange box contain Rs. 100000 each.
* Black Box: Both a black box and a yellow box contain Rs. 250000 each.
* Orange Box: Both an orange box and a black box contain Rs. 150000 each.
Only one of the above 3 statements is true, and the corresponding box contains the maximum amount.
Can you tell which box contains the maximum amount and how much?
2)Stranded on a deserted island, Harry Puttar is left with only a 40 litres container of milk. To conserve his milk he decides that on the first day he will drink one litre of milk and then refill the container back up with water. On the 2nd day he will drink 2 litres and refill the container. On the 3rd day he will drink 3 litres and so on... By the time all the milk is gone, how much water has he drunk?
21 replies
Yoonyoung11
Jun 29, 2025
DynamicFox52
Today at 2:04 AM
Foil
Silverfalcon   29
N Today at 1:54 AM by booking
$(1+x^2)(1-x^3)$ equals

$ \text{(A)}\ 1 - x^5\qquad\text{(B)}\ 1 - x^6\qquad\text{(C)}\ 1+ x^2 -x^3\qquad \\ \text{(D)}\ 1+x^2-x^3-x^5\qquad \text{(E)}\ 1+x^2-x^3-x^6 $
29 replies
Silverfalcon
Oct 21, 2005
booking
Today at 1:54 AM
Exponent of 1 and -1
Silverfalcon   31
N Today at 12:13 AM by melloncandy
$(-1)^{5^2} + 1^{2^5} =$

$\textbf{(A)}\ -7 \qquad
\textbf{(B)}\ -2 \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ 1 \qquad
\textbf{(E)}\ 57$
31 replies
Silverfalcon
Oct 22, 2005
melloncandy
Today at 12:13 AM
9 Square roots
A7456321   34
N Yesterday at 10:27 PM by A7456321
Me personally I only have $\sqrt2=1.414$ memorized but I'm sure there are people out there with more!
update: i recently learned $\sqrt3=1.732$

CLICK ON ME YOU KNOW YOU WANT TO
34 replies
A7456321
May 29, 2025
A7456321
Yesterday at 10:27 PM
Troll Problem
giratina3   12
N Yesterday at 10:06 PM by ManolidisS
If $\frac{a}{a - 1} = \frac{b^2 + 2b - 1}{b^2 + 2b - 2}$, then what does $a$ equal in terms of $b$?

Hint 1
Hint 2
Hint 3
12 replies
giratina3
Friday at 8:28 PM
ManolidisS
Yesterday at 10:06 PM
The 24 Game, but with a twist!
PikaPika999   280
N Yesterday at 7:34 PM by PreciseScorpion58
So many people know the 24 game, where you try to create the number 24 from using other numbers, but here's a twist:

You can only use the number 24 (up to 5 times) to try to make other numbers :)

the limit is 5 times because then people could just do $\frac{24}{24}+\frac{24}{24}+\frac{24}{24}+...$ and so on to create any number!

honestly, I feel like with only addition, subtraction, multiplication, and division, you can't get pretty far with this, so you can use any mathematical operations!
280 replies
PikaPika999
Jul 1, 2025
PreciseScorpion58
Yesterday at 7:34 PM
Decimal functions in binary
Pranav1056   3
N May 23, 2025 by ihategeo_1969
Source: India TST 2023 Day 3 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
3 replies
Pranav1056
Jul 9, 2023
ihategeo_1969
May 23, 2025
Decimal functions in binary
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Source: India TST 2023 Day 3 P1
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Pranav1056
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#1 • 4 Y
Y by GeoKing, CahitArf, Supercali, Siddharth03
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
This post has been edited 2 times. Last edited by Pranav1056, Jul 9, 2023, 6:22 AM
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Supercali
1263 posts
#2 • 2 Y
Y by Om245, thepassionatepotato
A story about this problem: It was originally meant to be D3 P3 (basically the hardest problem in the TSTs), but a few days before the test, some of us found an easier solution while trying. Hence the problem had to be demoted to D3 P1. D4 P3 at that time, which was a very hard geo, was shifted to D3 P3 (I think it was more suitable for that position anyway), and we had to use the shortlist for D4 P3. Anyway, I think this is a very cute problem.

Here is the solution that we found:

For $n\in\mathbb N$, let $d(n)$ denote the number of $1$'s in the binary representation of $n$. Let $P(x,y)$ denote the statement that $f(x)+y$ and $f(y)+x$ have the same number of $1$'s in their binary representation.

Claim 1: For any $y,n \in \mathbb{N}$ with $2^n>f(y)$, $f(2^n-f(y))+y$ is a power of two.
Proof: $P(2^n-f(y),y)$ gives us that $2^n$ and $f(2^n-f(y))+y$ have the same number of $1$'s, and the former has exactly one $1$, so $f(2^n-f(y))+y$ has exactly one $1$, from which the claim follows. $\blacksquare$

Claim 2: $f(y+2^k)-f(y)$ is a power of two for any $k \geq 0$ and $y \geq 2^k$.
Proof: Choose an $n$ such that $n>1000+\log_2(10+|f(y+2^k)-f(y)|)$. By Claim 1, $f(2^n-f(y))=2^t-y >0$ for some $t$ with $t \geq k+1$. Therefore $P(2^n-f(y),y+2^k)$ gives $$d(2^n-f(y)+f(y+2^k))=d(2^t+2^k)=2$$since $t \geq k+1$. If $f(y)=f(y+2^k)$, then LHS is $d(2^n)=1$, contradiction! If $f(y)>f(y+2^k)$, and $f(y)-f(y+2^k)$ has $m<\log_2(10+|f(y+2^k)-f(y)|)$ digits, then $$d(2^n-f(y)+f(y+2^k)) \geq n-m-1 \geq 999>2$$since $2^n-f(y)+f(y+2^k)$ starts with at least $n-m-1$ ones, contradiction! Therefore $f(y+2^k)>f(y)$, and since $n$ is bigger that the number of digits in $f(y+2^k)-f(y)$, there is no carry-over, so
$$2=d(2^n-f(y)+f(y+2^k))=1+d(f(y+2^k)-f(y))$$which gives us $f(y+2^k)-f(y)$ is a power of $2$, as required. $\blacksquare$


Claim 2 gives us $f(y+1)-f(y)=2^{t(y)}$ for some $t(y)$, for all $y$. But for $y \geq 2$, $f(y+2)-f(y)$ is also a power of two $\implies$ $2^{t(y)}+2^{t(y+1)}$ is a power of two, which is only possible if $t(y)=t(y+1)$ for all $y \geq 2$. Therefore $f(y+1)-f(y)$ is a constant power of two for all $y \geq 2$, say $2^k$. This gives us $f(y)=2^ky+c$ for some constant $c$, for all $y \geq 2$. Putting this in Claim 1, we get
$$2^{k+n}-(2^{2k}-1)y-(2^k-1)c$$is a power of two for any $y \geq 2$ and any sufficiently large $n$. This is only possible if, for all $y \geq 2$,
$$(2^{2k}-1)y+(2^k-1)c=0$$$$\iff (2^k-1)((2^k+1)y+c)=0$$which can only hold for all $y \geq 2$ if $2^k=1$, i.e., $f(y)=y+c$ for all $y \geq 2$. But Claim 1 for $y=1$ and large $n$ gives
$$2^n-f(1)+1+c$$is a power of two for all sufficiently large $n$, which is only possible if $f(1)=1+c$. Therefore the only solutions are
$$\boxed{f(x)=x+c \ \ \forall x \in \mathbb{N}}$$where $c$ is a non-negative integer.
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i3435
1351 posts
#3 • 2 Y
Y by GeoKing, Om245
$P(2^n-f(x),x)$ means $x=2^k-f(2^n-f(x))$ for some $k$, for all $x,n$. $P(2^a+2^k-f(x),x)$ means that $f(2^a+2^k-f(x))+x$ is of the form $2^c+2^d$, where $c\neq d$ when $a\neq k$ and $c=d$ when $a=k$. Replacing $x$ in the previous equation with $2^n-f(x)$, $f(2^a+x)+2^n-f(x)$ either has one or two ones in its binary representation. If you make $n$ large, we get that $f(2^a+x)-f(x)$ is a power of two for all $a,x$. In the same manner as the previous post, you can get $f(x)=2^kx+c$ for some $k,c$. $P(x,2^kx)$ means $2^{k+1}x+c$ and $(2^{2k}+1)x+c$ have the same number of $1$'s in their binary representation. If $x$ is a large power of $2$, then the second one will have one more $1$ than the first one unless $k=0$. Thus $k=0$ and $f(x)=x+c$, which works.
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ihategeo_1969
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#4
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Let $P(x,y)$ denote the assertion. Call $2$ such numbers quirky.

$P(2^n-f(y),y)$ gives us $f(2^n-f(y))+y=2^{g(n,y)}$ for any $n>f(y)$ where $g: \mathbb{N}^2 \to \mathbb N$ is a function. See that $g(n,y)$ is unbounded.

Claim: $f(x+2^\ell)-f(x)$ is a power of $2$ for any $\ell \ge 0$ and $x \ge 2^\ell$.
Proof: $P(x+2^\ell,2^n-f(x))$ gives us that $f(x+2^\ell)+2^n-f(x)$ and $2^{g(n,x)}+2^\ell$ are quirky.

Now $2^{g(n,x)}>2^\ell$ so $f(x+2^\ell)+2^n-f(x)=2^{g_1(n,x,\ell)}+2^{g_2(n,x,\ell)}$ where $g_1$, $g_2: \mathbb{N}^3 \to \mathbb Z _{\ge 0}$ and $g_1(n,x,\ell) \neq g_2(n,x,\ell)$. Fix $x$ and $\ell$ and we will abuse some notation by letting $g_i(n,x,\ell)=g_i(n)$ because I am lazy. So we have \begin{align*}
& 2^{g_1(n)}+2^{g_2(n)}-2^n \text{ is constant} \\
\implies & 2^{g_1(n)}+2^{g_2(n)}+2^m=2^{g_1(m)}+2^{g_2(m)}+2^n
\end{align*}Say $m>ng_1(n)g_2(n)$ and $g_1(n)$, $g_2(n)$, $m$ are all distinct and so is $g_1(m)$, $g_2(m)$. If $n=g_2(m)$ then LHS have $3$ $1$'s in their binary representation but RHS has atmost $2$.

Now as $n \neq m$ so $n \in \{g_1(n),g_2(n)\}$ and hence we get that $2^{g_1(n)}+2^{g_2(n)}-2^n$ is a power of $2$ and so we are done. $\square$

Choose $\ell=0$ and $1$ and easily get that $f(x+1)-f(x)$ is a constant power of $2$ for $x \ge 2$. By a bit case bash we get that that must be $1$. And similarly we get that $f(2)-f(1)=1$ as well.

Hence the only solution is $\boxed{f(x) \equiv x+c \text{ } \forall \text{ }x \in \mathbb{N}}$ where $c \ge 0$; which obviously works.
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