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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:
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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
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   9
N 29 minutes ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
9 replies
OgnjenTesic
May 22, 2025
JARP091
29 minutes ago
DE is tangent to a fixed circle whose radius is half the radius of (O)
parmenides51   1
N 29 minutes ago by TigerOnion
Source: 2017 Saudi Arabia JBMO Training Tests 2
Let $ABC$ be a triangle inscribed in circle $(O)$ such that points $B, C$ are fixed, while $A$ moves on major arc $BC$ of $(O)$. The tangents through $B$ and $C$ to $(O)$ intersect at $P$. The circle with diameter $OP$ intersects $AC$ and $AB$ at $D$ and $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.
1 reply
parmenides51
May 28, 2020
TigerOnion
29 minutes ago
Orthocorrespondent of P on Euler line
Luis González   1
N 44 minutes ago by AuroralMoss
Let $O,G$ and $K$ be the circumcenter, centroid and symmedian point of $\triangle ABC,$ respectively. $P$ is an arbitrary point on Euler line $OG.$ Show that the orthocorrespondent of $P$ WRT $\triangle {ABC}$ falls on $GK.$
1 reply
Luis González
Feb 8, 2025
AuroralMoss
44 minutes ago
JBMO Shortlist 2023 N3
Orestis_Lignos   9
N an hour ago by Just1
Source: JBMO Shortlist 2023, N3
Let $A$ be a subset of $\{2,3, \ldots, 28 \}$ such that if $a \in A$, then the residue obtained when we divide $a^2$ by $29$ also belongs to $A$.

Find the minimum possible value of $|A|$.
9 replies
Orestis_Lignos
Jun 28, 2024
Just1
an hour ago
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   7
N an hour ago by sttsmet
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
7 replies
mathematics2003
Apr 14, 2021
sttsmet
an hour ago
a^2-bc square implies 2a+b+c composite
v_Enhance   41
N an hour ago by cursed_tangent1434
Source: ELMO 2009, Problem 1
Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Evan o'Dorney
41 replies
v_Enhance
Dec 31, 2012
cursed_tangent1434
an hour ago
IMO Shortlist 2008, Geometry problem 2
April   42
N an hour ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
an hour ago
3^n + 61 is a square
VideoCake   25
N an hour ago by endless_abyss
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
25 replies
VideoCake
Monday at 5:14 PM
endless_abyss
an hour ago
Problem 5
blug   3
N an hour ago by Jt.-.
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
3 replies
blug
May 19, 2025
Jt.-.
an hour ago
Inequality with xy+yz+zx=1
Kimchiks926   14
N 2 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
2 hours ago
what number have you memorized perfect squares
ellenssim   3
N 3 hours ago by whwlqkd
Up to what number have you memorized perfect squares, and how often does it help you in solving problems?
3 replies
ellenssim
Today at 3:44 AM
whwlqkd
3 hours ago
Challenge: Make every number to 100 using 4 fours
CJB19   266
N 4 hours ago by Leeoz
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:
$$(4-4)*4*4=0$$$$4-4+4/4=1$$$$4/4+4/4=2$$$$(4+4+4)/4=3$$$$4+(4-4)*4=4$$$$4+4^{4-4}=5$$$$4!/4+4-4=6$$$$4+4-4/4=7$$$$4+4+4-4=8$$
266 replies
CJB19
May 15, 2025
Leeoz
4 hours ago
Last challenge problems in the books
ysn613   10
N 5 hours ago by mdk2013
Algebra
It is known that $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}\dots=\frac{\pi^2}{6}$ Given this fact, determine the exact value of $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}\dots.$$(Source: Mandelbrot)
Counting and Probability
A $3\times3\times3$ wooden cube is painted on all six faces, then cut into 27 unit cubes. One unit cube is randomly selected and rolled. After it is rolled, $5$ out of the $6$ faces are visible. What is the probability that exactly one of the five visible faces is painted? (Source: MATHCOUNTS)
Number Theory(This technically isn't the last problem but the last chapter doesn't have challenge problems)
The integer p is a 50-digit prime number. When its square is divided by 120, the remainder is not 1. What is the remainder?
I didn't include geometry because I haven't taken it yet, feel free to post it
Answer these problems and post what you think is the order of difficulty
10 replies
ysn613
Monday at 7:04 PM
mdk2013
5 hours ago
What's $(-1)^0?$
Vulch   12
N 6 hours ago by Li0nking
What's $(-1)^0?$

(It may be a silly question,but still I want to know it's value)
12 replies
Vulch
Oct 26, 2024
Li0nking
6 hours ago
random problem i just thought about one day
ceilingfan404   27
N Apr 29, 2025 by PikaPika999
i don't even know if this is solvable
Prove that there are finite/infinite powers of 2 where all the digits are also powers of 2. (For example, $4$ and $128$ are numbers that work, but $64$ and $1024$ don't work.)
27 replies
ceilingfan404
Apr 20, 2025
PikaPika999
Apr 29, 2025
random problem i just thought about one day
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ceilingfan404
1143 posts
#1 • 4 Y
Y by aidan0626, e_is_2.71828, Exponent11, PikaPika999
i don't even know if this is solvable
Prove that there are finite/infinite powers of 2 where all the digits are also powers of 2. (For example, $4$ and $128$ are numbers that work, but $64$ and $1024$ don't work.)
This post has been edited 1 time. Last edited by ceilingfan404, Apr 20, 2025, 7:55 PM
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huajun78
75 posts
#2 • 1 Y
Y by PikaPika999
well as the number gets bigger, there are more digits, so it's less likely that ALL the digits will be a power of 2 (1, 2, 4, 8).

for the first 20 powers of 2 after $2^{10}$ ($2^{11}$ to $2^{30}$), none of them satisfy the condition (I tested all of them), so it's very unlikely that numbers with even more digits will.

I don't know how to prove this but that fact suggests that there are only a finite number.
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vincentwant
1445 posts
#3 • 1 Y
Y by PikaPika999
if the number is greater than 512 then the last four digits must be 2112, 4112, 8112, 2224, 4224, 8224, 1424, 1824, 2144, 4144, 8144, 1184, 2128, 4128, 8128, 1248, 2448, 4448, 8448, 2848, 4848, 8848, 2288, 4288, 8288, 1488, 1888

dont think this helps
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Yummo
299 posts
#4 • 1 Y
Y by PikaPika999
@above, what about 1024?
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vincentwant
1445 posts
#5 • 1 Y
Y by PikaPika999
Yummo wrote:
@above, what about 1024?

0 is not a power of 2
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e_is_2.71828
222 posts
#6 • 1 Y
Y by PikaPika999
ceilingfan404 wrote:
i don't even know if this is solvable
Prove that there are finite/infinite powers of 2 where all the digits are also powers of 2. (For example, $4$ and $128$ are numbers that work, but $64$ and $1024$ don't work.)

I won't look into it completely, but we can start somewhere. We'll see if it is possible to "generate" a formula for these numbers. So let $n$ be a $k$-digit number such that $n=a_ka_{k-1}...a_2a_1a_0$. Then $n=10^ka_k+10^{k-1}a_k-1...+10a_1+a_0$, and note for all $i$ $a_i=2^b$, for some $b$. So, $n=10^k \cdot 2^{b_k}+10^{k-1}\cdot 2^{b_{k-1}}+...+10\cdot 2^{b_1}+2^{b_0}$. From there we need also $n=2^c$ for some $c$, and presumably we can take the largest $b_i$, factor it out, and we need the remaining sum to also be a power of $2$. Someone can try working it out from here, I think I started it off well enough.
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wangzrpi
159 posts
#7
Y by
See
https://math.stackexchange.com/questions/2238383/how-many-powers-of-2-have-only-0-or-powers-of-2-as-digits
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e_is_2.71828
222 posts
#8
Y by
Definitely not middle school math
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e_is_2.71828
222 posts
#10 • 1 Y
Y by PikaPika999
K1mchi_ wrote:
e_is_2.71828 wrote:
Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue


"This problem is unlikely to have a simple proof, because the following holds:

Theorem. For any k, there exists a power of 2 whose first k digits and last k digits are all either 1 or 2.
Proof. We begin with looking at the last digits, taking 2nmod10k. For sufficiently large n, 2n≡0(mod2k). Since 2 is a primitive root modulo 5 and modulo 52, it is a primitive root modulo 5k for any k (Wikipedia), so we can have 2n≡b(mod10k) for any b such that b≡0(mod2k).

This is possible to accomplish with only 1 and 2 as digits. We start with b1=2 for k=1, and extend bk−1≡0(mod2k−1) to bk≡0(mod2k) by the rule:

If bk−1≡0(mod2k), take bk=2⋅10k−1+bk−1.
If bk−1≡2k−1(mod2k), take bk=10k−1+bk−1.
(This works because 10k−1≡2k−1(mod2)k.)

There is a unique sequence of digits ending …211111212122112 that we obtain in this way; reversed, it is A023396 in the OEIS.

To make sure that 2n ends in bk, there will be some condition along the lines of
n≡c(modϕ(5k))
or n=c+n′ϕ(5k) for some n′. From there, getting the first k digits to be 1 or 2 is easy along the lines of a recently popular question. We might as well aim for the sequence 111…111k, because we can. To do this, we want
log101.11…1<{(c+n′⋅ϕ(5k))log102}<log101.11…2
where {x} denotes the fractional part of x. This translates into a condition of the form
{n⋅log102ϕ(5k)}∈Ik
for some interval Ik, which we know is possible because α=log102ϕ(5k) is irrational, and therefore the sequence {α},{2α},{3α},… is dense in [0,1].

This concludes the proof.

Instead of the digits {1,2} we could have used the digits {1,4} or {1,8} and given a similar proof; if we multiply the solution to one of these by 2 or 4, we get a power of 2 whose first and last digits come from the set {2,4} or {2,8} or {4,8}. (We can't do this with just the set {0,1} or {0,2} or {0,4} or {0,8}, because eventually we can rule these out by a modular condition.)

It's of course still almost certain that there's no large power of 2 entirely made from the digits {0,1,2,4,8}, but you'd have to say something about the "middle digits" of such a power, which is much harder."

From the stack exchange.
This post has been edited 1 time. Last edited by e_is_2.71828, Apr 24, 2025, 6:01 PM
Reason: Added
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Andrew2019
2324 posts
#12 • 2 Y
Y by e_is_2.71828, Demetri
K1mchi_ wrote:
e_is_2.71828 wrote:
Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

it would be crazy if someone who has only done the amc 8 and sold on it says others have a skill issue
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maromex
219 posts
#13
Y by
There is a related question: Does the base-$3$ expression of $2^n$ always have a digit equal to $2$ for sufficiently large $n$? If I recall correctly, this problem is unsolved.

The problem discussed in this topic seems similar to this question, and I don't see why it would be solvable with currently known techniques.
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e_is_2.71828
222 posts
#14 • 1 Y
Y by mithu542
I wouldn't listen to someone who can't even spell figure ...
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maxamc
585 posts
#15
Y by
e_is_2.71828 wrote:
I wouldn't listen to someone who can't even spell figure ...

K1mchi_ is always right 100000 aura.
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K1mchi_
155 posts
#16
Y by
Andrew2019 wrote:
K1mchi_ wrote:
e_is_2.71828 wrote:
Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

it would be crazy if someone who has only done the amc 8 and sold on it says others have a skill issue

slander

i just dont do competitive math

hate me if u like
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MathPerson12321
3795 posts
#17 • 2 Y
Y by e_is_2.71828, mithu542
#11
why dont u?
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K1mchi_
155 posts
#18
Y by
MathPerson12321 wrote:
#11
why dont u?

just quote me


i have better things to do with my time than math rn

i’ll do u the service of enlightenment if i ever find time
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Aaronjudgeisgoat
910 posts
#19
Y by
K1mchi_ wrote:
MathPerson12321 wrote:
#11
why dont u?

just quote me


i have better things to do with my time than math rn

i’ll do u the service of enlightenment if i ever find time

you only have 105 posts, but i feel like ive seen you everywhere
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MathPerson12321
3795 posts
#22
Y by
@bove stop trying to say ur better
do i see mop quals trying to bring me down? no
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RollingPanda4616
266 posts
#24 • 1 Y
Y by PikaPika999
K3mchi_ wrote:
MathPerson12321 wrote:
@bove stop trying to say ur better
do i see mop quals trying to bring me down? no

so? im not trying to bring u down u still bring urself down bc ur very sensitive

dont we celebrate intelligence in our society?

alt alert
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RollingPanda4616
266 posts
#26 • 2 Y
Y by PikaPika999, e_is_2.71828
hey

yes
$~~~~~~$
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valisaxieamc
506 posts
#27 • 3 Y
Y by RollingPanda4616, PikaPika999, e_is_2.71828
Bro imagine making alts cause you fear that aops is going to ban you
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RollingPanda4616
266 posts
#28 • 1 Y
Y by PikaPika999
valisaxieamc wrote:
Bro imagine making alts cause you fear that aops is going to ban you

:rotfl:

anyway let's get this thread back on track

I think you might need to break up the digits and use the prime factorization. (like a 3 digit number $abc$ would be broken down into $a \cdot 2^2 5^2 + b \cdot 2^1 5^1 + c$ and since a, b,c are powers of 2, you could just look at the 5s?) idk how to continue though.
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maromex
219 posts
#29
Y by
I'll say this again:
maromex wrote:
There is a related question: Does the base-$3$ expression of $2^n$ always have a digit equal to $2$ for sufficiently large $n$? If I recall correctly, this problem is unsolved.

The problem discussed in this topic seems similar to this question, and I don't see why it would be solvable with currently known techniques.

Unless a problem about digits has good reason to be solvable with currently known techniques, it's probably not solvable, even if the answer seems obviously true/false at first.

Here's another unsolved problem related to the topic of this thread: For $n > 86$, does $2^n$ always have a $0$ in base $10$?
This post has been edited 1 time. Last edited by maromex, Apr 26, 2025, 7:16 PM
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PikaPika999
2407 posts
#30
Y by
K1mchi_ wrote:
e_is_2.71828 wrote:
Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

but if the forum is literally called msm, then shouldn't it be msm? plus, if it is harder than msm, there are high school math and college math and high school olympiads, and it could've been placed there?

k1mchi_

not nice
valisaxieamc wrote:
Bro imagine making alts cause you fear that aops is going to ban you

lol imo aops should use ip bans
This post has been edited 1 time. Last edited by PikaPika999, Apr 27, 2025, 11:09 PM
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PikaPika999
2407 posts
#31 • 2 Y
Y by RollingPanda4616, Pengu14
K3mchi_ wrote:
MathPerson12321 wrote:
@bove stop trying to say ur better
do i see mop quals trying to bring me down? no

so? im not trying to bring u down u still bring urself down bc ur very sensitive

dont we celebrate intelligence in our society?

1. True intelligence shines through clarity and simplicity, not overcomplication.
2. Intelligence isn’t just about flaunting knowledge—it’s also about understanding, humility, and connection.
3. True intelligence lies not in power over others, but in empowering those around us.
4. Creativity/intelligence isn’t just about thinking outside the box—it’s about reshaping the box entirely.
5. Leadership isn’t a title—it’s the trust you earn and the influence you wield wisely.
6. Intelligence is not in the answers we give, but in the questions we dare to ask.
7. Intelligence grows when we challenge our own assumptions, not just those of others.
8. The hallmark of intelligence is recognizing that there’s always more to learn.
9. Intelligence flourishes in collaboration, not isolation.
This post has been edited 2 times. Last edited by PikaPika999, Apr 27, 2025, 11:15 PM
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valisaxieamc
506 posts
#33 • 1 Y
Y by PikaPika999
I completely agree with PikaPika but like RollingPanda said, we probably should get back on topic. I mean the kimchi dude is finally leaving us alone and hopefully getting a life so I'll take it as a win
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fake123
93 posts
#34
Y by
bro why are you guys raging over some random kid why can't you just ignore him
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PikaPika999
2407 posts
#35
Y by
fake123 wrote:
bro why are you guys raging over some random kid why can't you just ignore him

we're not raging over "some random kid" who can be ignored (sorry if this sounds harsher than it is)

they start flamewars on multiple different threads. This is how my 1000th post thread got locked :furious

also, they created multiple different alts, which is explicitly said to be against the rules (probably because of getting postbanned from this sheriff

sry if this sounds harsher than i meant to be
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