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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 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
9 When Is your Mathcounts State Competition?
MathyMathMan   150
N an hour ago by K1mchi_
When is your Mathcounts State Comp? Mine is on the 11th of March.
150 replies
MathyMathMan
Mar 1, 2023
K1mchi_
an hour ago
1234th Post!
PikaPika999   218
N an hour ago by K1mchi_
I hit my 1234th post! (I think I missed it, I'm kinda late, :oops_sign:)

But here's a puzzle for you all! Try to create the numbers 1 through 25 using the numbers 1, 2, 3, and 4! You are only allowed to use addition, subtraction, multiplication, division, and parenthesis. If you're post #1, try to make 1. If you're post #2, try to make 2. If you're post #3, try to make 3, and so on. If you're a post after 25, then I guess you can try to make numbers greater than 25 but you can use factorials, square roots, and that stuff. Have fun!

1: $(4-3)\cdot(2-1)$
218 replies
PikaPika999
Apr 21, 2025
K1mchi_
an hour ago
Israel Number Theory
mathisreaI   63
N an hour ago by Maximilian113
Source: IMO 2022 Problem 5
Find all triples $(a,b,p)$ of positive integers with $p$ prime and \[ a^p=b!+p. \]
63 replies
mathisreaI
Jul 13, 2022
Maximilian113
an hour ago
0!??????
wizwilzo   64
N an hour ago by K1mchi_
why is 0! "1" ??!
64 replies
wizwilzo
Jul 6, 2016
K1mchi_
an hour ago
100th post! (actually 105)
K1mchi_   0
an hour ago
here’s a math problem!
if the aops forum has 50 users and the users post once a minute, then how many posts will be made in a week? note that Fred and ted are twins and Elizabeth only visits the forum on weekends. assume that the users do not post during the 5 hour school day and 8 hours of sleep, but also that they have no life and only grind AOPS. it is not a leap year

anyway

this is my 105th post and I feel like I’ve grown a lot
not rlly lol
i have remained the same perfect person hahaha

Click to reveal hidden text
0 replies
K1mchi_
an hour ago
0 replies
I need help for British maths olympiads
RCY   1
N an hour ago by Miquel-point
I’m a year ten student who’s going to take the bmo in one year.
However I have no experience in maths olympiads and the best results I have achieved so far was 25/60 in intermediate maths olympiads.
What shall I do?
I really need help!
1 reply
RCY
3 hours ago
Miquel-point
an hour ago
Value of the sum
fermion13pi   0
2 hours ago
Source: Australia
Calculate the value of the sum

\sum_{k=1}^{9999999} \frac{1}{(k+1)^{3/2} + (k^2-1)^{1/3} + (k-1)^{2/3}}.
0 replies
fermion13pi
2 hours ago
0 replies
NT Functional Equation
mkultra42   0
2 hours ago
Find all strictly increasing functions \(f: \mathbb{N} \to \mathbb{N}\) satsfying \(f(1)=1\) and:

\[ f(2n)f(2n+1)=9f(n)^2+3f(n)\]
0 replies
mkultra42
2 hours ago
0 replies
Cyclic sum of 1/((3-c)(4-c))
v_Enhance   22
N 2 hours ago by Aiden-1089
Source: ELMO Shortlist 2013: Problem A6, by David Stoner
Let $a, b, c$ be positive reals such that $a+b+c=3$. Prove that \[18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15. \]Proposed by David Stoner
22 replies
v_Enhance
Jul 23, 2013
Aiden-1089
2 hours ago
x^101=1 find 1/1+x+x^2+1/1+x^2+x^4+...+1/1+x^100+x^200
Mathmick51   6
N 3 hours ago by pi_quadrat_sechstel
Let $x^{101}=1$ such that $x\neq 1$. Find the value of $$\frac{1}{1+x+x^2}+\frac{1}{1+x^2+x^4}+\frac{1}{1+x^3+x^6}+\dots+\frac{1}{1+x^{100}+x^{200}}$$
6 replies
Mathmick51
Jun 22, 2021
pi_quadrat_sechstel
3 hours ago
IMO Shortlist 2014 N5
hajimbrak   60
N 3 hours ago by sansgankrsngupta
Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p -1} + y$ and $x + y^ {p -1}$ are both powers of $p$.

Proposed by Belgium
60 replies
hajimbrak
Jul 11, 2015
sansgankrsngupta
3 hours ago
n variables with n-gon sides
mihaig   0
3 hours ago
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
0 replies
mihaig
3 hours ago
0 replies
4 variables with quadrilateral sides
mihaig   3
N 4 hours ago by mihaig
Source: VL
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$4\left(abc+abd+acd+bcd\right)\geq3\left(a+b+c+d\right)+4.$$
3 replies
mihaig
Today at 5:11 AM
mihaig
4 hours ago
Calculate the distance of chess king!!
egxa   5
N 4 hours ago by Tesla12
Source: All Russian 2025 9.4
A chess king was placed on a square of an \(8 \times 8\) board and made $64$ moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called $\emph{pleasant}$ if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
5 replies
egxa
Apr 18, 2025
Tesla12
4 hours ago
HELPMEPLEASE
czala   19
N Mar 31, 2025 by AbhayAttarde01
One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted 74 heads and 196 legs. How many humans and horses were there?
A.
37 humans and 98 horses
B.
24 horses and 50 humans
C.
31 horses and 74 humans
D.
24 humans and 50 horses



SOMEONE HELP MEEEEE
19 replies
czala
Dec 23, 2022
AbhayAttarde01
Mar 31, 2025
HELPMEPLEASE
G H J
G H BBookmark kLocked kLocked NReply
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czala
7 posts
#1
Y by
One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted 74 heads and 196 legs. How many humans and horses were there?
A.
37 humans and 98 horses
B.
24 horses and 50 humans
C.
31 horses and 74 humans
D.
24 humans and 50 horses



SOMEONE HELP MEEEEE
Z K Y
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TetraFish
1088 posts
#2 • 3 Y
Y by Inaaya, AoPS_0829, eagles2018
alr lets play this cheap. If there were all horses then there are $74 * 4 = 296 \text{ legs}$. This is an overcount by $296-196 = 100 \text{ legs}$. So each time we replace a human in the spot of a horse $2$ legs are lost. So there must be $50$ humans to get this down by $100$. So the correct answer is $\boxed{B}$

edit: Were u in a live test?
This post has been edited 2 times. Last edited by TetraFish, Dec 23, 2022, 6:12 PM
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Inaaya
290 posts
#3 • 1 Y
Y by Mango247
bruh this is a system of equations...
sol
oops sohum beat it to me and in a orzer way :love:
This post has been edited 1 time. Last edited by Inaaya, Dec 23, 2022, 6:13 PM
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TetraFish
1088 posts
#4 • 2 Y
Y by Inaaya, Mango247
Inaaya wrote:
bruh this is a system of equations...
sol
oops sohum beat it to me and in a orzer way :love:

cheap tricks are fun when doing mathcounts problems.
Z K Y
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FaThEr-SqUiRrEl
4721 posts
#5
Y by
czala wrote:
One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted 74 heads and 196 legs. How many humans and horses were there?
A.
37 humans and 98 horses
B.
24 horses and 50 humans
C.
31 horses and 74 humans
D.
24 humans and 50 horses



SOMEONE HELP MEEEEE

If this is homework (a safe assumption), please ask your teacher or a classmate for help.
Z K Y
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AlexanderRuan
13 posts
#6
Y by
you could also just plug in all the answer choices and see which one works
Z K Y
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TetraFish
1088 posts
#7
Y by
AlexanderRuan wrote:
you could also just plug in all the answer choices and see which one works
even more orz stratergy
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apex304
524 posts
#8
Y by
SohumUttamchandani wrote:
Inaaya wrote:
bruh this is a system of equations...
sol
oops sohum beat it to me and in a orzer way :love:

cheap tricks are fun when doing mathcounts problems.

Yeah thats from the Comp Math for Middlle School book. Id reccomend that for AMC 8 Im doing it rn and it is so helpful.
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TetraFish
1088 posts
#9
Y by
apex304 wrote:
SohumUttamchandani wrote:
Inaaya wrote:
bruh this is a system of equations...
sol
oops sohum beat it to me and in a orzer way :love:

cheap tricks are fun when doing mathcounts problems.

Yeah thats from the Comp Math for Middlle School book. Id reccomend that for AMC 8 Im doing it rn and it is so helpful.

yep thats where I originally learn't that. Its a more dynamic way of doing these, systems are foolproof but boring lol
Z K Y
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buddy2007
2071 posts
#10
Y by
czala wrote:
One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted 74 heads and 196 legs. How many humans and horses were there?
A.
37 humans and 98 horses
B.
24 horses and 50 humans
C.
31 horses and 74 humans
D.
24 humans and 50 horses



SOMEONE HELP MEEEEE
AlexanderRuan wrote:
you could also just plug in all the answer choices and see which one works

Even better:
Easily eliminate answers A and C since they dont sum to 74, check B/D and then get your answer(B)
This post has been edited 1 time. Last edited by buddy2007, Dec 23, 2022, 9:01 PM
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TetraFish
1088 posts
#11
Y by
@above
for d 50*4 alrdy fails
so its immeidate b
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The_dynamic_tacocat
51 posts
#12 • 3 Y
Y by Mango247, Mango247, Mango247
SohumUttamchandani wrote:
@above
for d 50*4 alrdy fails
so its immeidate b

speed solving! :-D
Z K Y
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Amkan2022
2009 posts
#13
Y by
Rather than system of equations, sometimes, you have to manipulate the answer choices for a more optimal solution. 74 heads, so A and C are automatically eliminated.
That leaves B and D - 50*4 already exceeds the number of legs, so you can know it's B
This post has been edited 1 time. Last edited by Amkan2022, Dec 25, 2022, 3:10 AM
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sleepypuppy
2366 posts
#14
Y by
FaThEr-SqUiRrEl wrote:
czala wrote:
One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted 74 heads and 196 legs. How many humans and horses were there?
A.
37 humans and 98 horses
B.
24 horses and 50 humans
C.
31 horses and 74 humans
D.
24 humans and 50 horses



SOMEONE HELP MEEEEE

If this is homework (a safe assumption), please ask your teacher or a classmate for help.

yeah no hw questions here plz
This post has been edited 1 time. Last edited by sleepypuppy, Dec 25, 2022, 9:54 PM
Reason: plz
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TetraFish
1088 posts
#15
Y by
Lolz yeah in the rules it said that specifically “we arent here to do your hw”
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Zestra
431 posts
#17
Y by
“It’s elementary, Watson” - Sherlock Holmes
Attachments:
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karthikram0706
89 posts
#18
Y by
Zestra wrote:
“It’s elementary, Watson” - Sherlock Holmes

Totally option c :thumbup: *being sarcastic
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Zestra
431 posts
#19
Y by
karthikram0706 wrote:
Zestra wrote:
“It’s elementary, Watson” - Sherlock Holmes

Totally option c :thumbup: *being sarcastic

:oops_sign: I admit I wasn’t paying enough attention... I had meant Option B.
This post has been edited 1 time. Last edited by Zestra, Dec 28, 2022, 3:10 AM
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HannaC
7 posts
#24
Y by
It’s ovs B
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AbhayAttarde01
1484 posts
#25
Y by
74 heads means that there are 74 humans and horses altogether
we rule out A and C from that info
if all 74 organisms stood on 2 feet (including horses) that results in 148. however, 196-148=48, so there are 2 legs unaccounted for for each horse, so there are 48/2=24 horses and 50 humans
$\boxed{B}$ will be your answer
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