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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.

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0 replies
jlacosta
Apr 2, 2025
0 replies
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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
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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
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9 Did you get into Illinois middle school math Olympiad?
Gavin_Deng   17
N 6 minutes ago by Gavin_Deng
I am simply curious of who got in.
17 replies
Gavin_Deng
Apr 19, 2025
Gavin_Deng
6 minutes ago
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1234th Post!
PikaPika999   182
N 12 minutes 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: $(4-3)\cdot(2-1)$
182 replies
PikaPika999
Apr 21, 2025
K1mchi_
12 minutes ago
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Purple Comet Math Meet Recources
RabtejKalra   2
N 42 minutes ago by Andyluo
I heard that you can take a packet of information to the Purple Comet examination with some formulas, etc. Does anybody have a copy of a guidebook with all the important formulas? I'm just too lazy to write one myself.......
2 replies
RabtejKalra
an hour ago
Andyluo
42 minutes ago
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Mathcounts Nationals Roommate Search
iwillregretthisnamelater   41
N an hour ago by K1mchi_
Does anybody want to be my roommate at nats? Every other qualifier in my state is female. :sob:
Respond quick pls i gotta submit it in like a couple of hours.
41 replies
iwillregretthisnamelater
Mar 31, 2025
K1mchi_
an hour ago
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hard binomial sum
PRMOisTheHardestExam   7
N an hour ago by P162008
Find
\[ \frac{\displaystyle\sum_{k=0}^r \binom nk \binom{n-2k}{r-k}}{\displaystyle\sum_{k=r}^n \binom nk \binom{2k}{2r}\left(\frac{3}{4}\right)^{n-k}\left(\frac{1}{2}\right)^{2k-2r}}\]\
where $n \ge 2r$.
Options: 1/2, 1, 2, none.
7 replies
1 viewing
PRMOisTheHardestExam
Mar 6, 2023
P162008
an hour ago
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Telescopic Sum
P162008   0
an hour ago
Compute the value of $\Omega = \sum_{r=1}^{\infty} \frac{14 - 9r - 90r^2 - 36r^3}{7^r r(r + 1)(r + 2)(4r^2 - 1)}$
0 replies
P162008
an hour ago
0 replies
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Triple Sum
P162008   0
an hour ago
Find the value of

$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{k.2^n + 2m + 1}$
0 replies
P162008
an hour ago
0 replies
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Combinatorial Sum
P162008   0
an hour ago
Compute $\sum_{r=0}^{n} \sum_{k=0}^{r} (-1)^k (k + 1)(k + 2) \binom {n + 5}{r - k}$
0 replies
P162008
an hour ago
0 replies
J
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Combinatorial Sum
P162008   0
an hour ago
Evaluate $\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}}$
0 replies
P162008
an hour ago
0 replies
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Combinatorial Sum
P162008   0
an hour ago
$\frac{\sum_{r=0}^{24} \binom{100}{4r} \binom{100}{4r + 2}}{\sum_{r=1}^{25} \binom{200}{8r - 6}}$ is equal to
0 replies
P162008
an hour ago
0 replies
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binomial sum ratio
thewayofthe_dragon   3
N 3 hours ago by P162008
Source: YT
Someone please evaluate this ratio inside the log for any given n(I feel the sum doesn't have any nice closed form).
3 replies
thewayofthe_dragon
Jun 16, 2024
P162008
3 hours ago
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IMO Shortlist 2013, Combinatorics #3
lyukhson   31
N 3 hours ago by Maximilian113
Source: IMO Shortlist 2013, Combinatorics #3
A crazy physicist discovered a new kind of particle wich he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time.
(i) If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it.
(ii) At any moment, he may double the whole family of imons in the lab by creating a copy $I'$ of each imon $I$. During this procedure, the two copies $I'$ and $J'$ become entangled if and only if the original imons $I$ and $J$ are entangled, and each copy $I'$ becomes entangled with its original imon $I$; no other entanglements occur or disappear at this moment.

Prove that the physicist may apply a sequence of such operations resulting in a family of imons, no two of which are entangled.
31 replies
lyukhson
Jul 9, 2014
Maximilian113
3 hours ago
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A wizard kidnaps 101 people
Leicich   14
N 3 hours ago by MathCosine
Source: Argentina TST 2011, Problem 2
A wizard kidnaps $31$ members from party $A$, $28$ members from party $B$, $23$ members from party $C$, and $19$ members from party $D$, keeping them isolated in individual rooms in his castle, where he forces them to work.
Every day, after work, the kidnapped people can walk in the park and talk with each other. However, when three members of three different parties start talking with each other, the wizard reconverts them to the fourth party (there are no conversations with $4$ or more people involved).

a) Find out whether it is possible that, after some time, all of the kidnapped people belong to the same party. If the answer is yes, determine to which party they will belong.
b) Find all quartets of positive integers that add up to $101$ that if they were to be considered the number of members from the four parties, it is possible that, after some time, all of the kidnapped people belong to the same party, under the same rules imposed by the wizard.
14 replies
Leicich
Aug 29, 2014
MathCosine
3 hours ago
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PAMO Problem 4: Perpendicular lines
DylanN   11
N 3 hours ago by ATM_
Source: 2019 Pan-African Mathematics Olympiad, Problem 4
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
11 replies
DylanN
Apr 9, 2019
ATM_
3 hours ago
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divisible by 111
aria123   7
N Apr 20, 2025 by aria123
How many 6-digit natural numbers (with distinct digits) can be formed using the digits 2, 3, 4, 5, 6, and 7 that are divisible by 111?
7 replies
aria123
Apr 1, 2025
aria123
Apr 20, 2025
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divisible by 111
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Reveal topic tags
number theory
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aria123
38 posts
aria123
#1 h  Apr 1, 2025, 3:16 PM
Y by
How many 6-digit natural numbers (with distinct digits) can be formed using the digits 2, 3, 4, 5, 6, and 7 that are divisible by 111?
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aria123
38 posts
aria123
#2 h  Apr 7, 2025, 10:57 AM
Y by
aria123 wrote:
How many 6-digit natural numbers (with distinct digits) can be formed using the digits 2, 3, 4, 5, 6, and 7 that are divisible by 111?
Please help me!!!
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ethan2011
297 posts
ethan2011
#3 h  Apr 7, 2025, 11:22 AM • 1 Y
Y by aria123
Use the fact that 1000 is 1 mod 111 to come up with a divisibility rule(kind of like proving the divisibility rules for 9 and 11).
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Apple_maths60
24 posts
Apple_maths60
#4 h  Apr 7, 2025, 1:26 PM • 1 Y
Y by aria123
For divisibility by 111 check divisibility by 3 and 37 as 3×37 =111
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fruitmonster97
2479 posts
fruitmonster97
#5 h  Apr 7, 2025, 1:34 PM • 1 Y
Y by aria123
ethan2011 wrote:
Use the fact that 1000 is 1 mod 111 to come up with a divisibility rule(kind of like proving the divisibility rules for 9 and 11).

orz!!!

if num is abcdef then 111|abcdef implies 111|(abc+def) so we just test!

cases on abc+def:
444: 1
555: 8
666: 27
777: 64
888: 125
999: 216
1110: 125
1221: 64
1332: 27
1443: 8
1554: 1

add them up to get (1+...+216)+(125+...+1)=(1+...+6)^2+(1+...+5)^2=21^2+15^2=441+225=666!

edit: if someone says that the ans is 666 not 666! im going to lose it, i was using ! to show excitement not factorial
This post has been edited 1 time. Last edited by fruitmonster97, Apr 7, 2025, 1:35 PM
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aria123
38 posts
aria123
#6 h  Apr 15, 2025, 9:18 AM
Y by
fruitmonster97 wrote:
ethan2011 wrote:
Use the fact that 1000 is 1 mod 111 to come up with a divisibility rule(kind of like proving the divisibility rules for 9 and 11).

orz!!!

if num is abcdef then 111|abcdef implies 111|(abc+def) so we just test!

cases on abc+def:
444: 1
555: 8
666: 27
777: 64
888: 125
999: 216
1110: 125
1221: 64
1332: 27
1443: 8
1554: 1

add them up to get (1+...+216)+(125+...+1)=(1+...+6)^2+(1+...+5)^2=21^2+15^2=441+225=666!

edit: if someone says that the ans is 666 not 666! im going to lose it, i was using ! to show excitement not factorial

Thank you! But a, b, c, d, e, f are distinct and in {2, 3, 4, 5, 6, 7} so the answer is 48.
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vincentwant
1349 posts
vincentwant
#7 h  Apr 18, 2025, 2:43 AM • 1 Y
Y by aria123
the number is divisible by 111 iff the first three digits + the last three digits is divisible by 111. this sum is divisible by 9, so trying 333, 666, 999, 1332, 1665 gives that 999 is the only possible sum, and this is achieved when the first and fourth digits add to 9, the second and fifth digits add to 9, and the third and sixth digits add to 9. thus the answer is 3!*2^3=48.
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aria123
38 posts
aria123
#8 h  Apr 20, 2025, 3:06 AM
Y by
vincentwant wrote:
the number is divisible by 111 iff the first three digits + the last three digits is divisible by 111. this sum is divisible by 9, so trying 333, 666, 999, 1332, 1665 gives that 999 is the only possible sum, and this is achieved when the first and fourth digits add to 9, the second and fifth digits add to 9, and the third and sixth digits add to 9. thus the answer is 3!*2^3=48.

Thank you!!!
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