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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
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0 replies
jlacosta
Mar 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
Problem about Euler's function
luutrongphuc   0
5 minutes ago
Prove that for every integer $n \ge 5$, we have:
$$ 2^{n^2+3n-13} \mid \phi \left(2^{2^{n}}-1 \right)$$
0 replies
luutrongphuc
5 minutes ago
0 replies
IMO Shortlist 2009 - Problem N4
April   12
N 7 minutes ago by asdf334
Find all positive integers $n$ such that there exists a sequence of positive integers $a_1$, $a_2$,$\ldots$, $a_n$ satisfying: \[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\] for every $k$ with $2\leq k\leq n-1$.

Proposed by North Korea
12 replies
April
Jul 5, 2010
asdf334
7 minutes ago
China Team Selection Test 2015 TST 1 Day 2 Q1
sqing   6
N 16 minutes ago by sttsmet
Source: China Hangzhou
Prove that : For each integer $n \ge 3$, there exists the positive integers $a_1<a_2< \cdots <a_n$ , such that for $ i=1,2,\cdots,n-2 $ , With $a_{i},a_{i+1},a_{i+2}$ may be formed as a triangle side length , and the area of the triangle is a positive integer.
6 replies
sqing
Mar 14, 2015
sttsmet
16 minutes ago
Cool Number Theory
Fermat_Fanatic108   4
N 27 minutes ago by BR1F1SZ
For an integer with 5 digits $n=abcde$ (where $a, b, c, d, e$ are the digits and $a\neq 0$) we define the \textit{permutation sum} as the value $$bcdea+cdeab+deabc+eabcd$$For example the permutation sum of 20253 is $$02532+25320+53202+32025=113079$$Let $m$ and $n$ be two fivedigit integers with the same permutation sum.
Prove that $m=n$.
4 replies
Fermat_Fanatic108
3 hours ago
BR1F1SZ
27 minutes ago
China Mathematical Olympiad 1993 problem5
jred   3
N 35 minutes ago by iStud
Source: China Mathematical Olympiad 1993 problem5
$10$ students bought some books in a bookstore. It is known that every student bought exactly three kinds of books, and any two of them shared at least one kind of book. Determine, with proof, how many students bought the most popular book at least? (Note: the most popular book means most students bought this kind of book)
3 replies
jred
Sep 23, 2013
iStud
35 minutes ago
x and o game, in an infinite grid of regular triangles
parmenides51   5
N an hour ago by Lil_flip38
Source: Norwegian Mathematical Olympiad 2017 - Abel Competition p3b
In an infinite grid of regular triangles, Niels and Henrik are playing a game they made up.
Every other time, Niels picks a triangle and writes $\times$ in it, and every other time, Henrik picks a triangle where he writes a $o$. If one of the players gets four in a row in some direction (see figure), he wins the game.
Determine whether one of the players can force a victory.
IMAGE
5 replies
parmenides51
Sep 3, 2019
Lil_flip38
an hour ago
BMN is equilateral iff rectangle ABCD is square
parmenides51   4
N an hour ago by Tsikaloudakis
Source: 2004 Romania NMO SL - Shortlist VII-VIII p8 https://artofproblemsolving.com/community/c3950157_
Consider a point $M$ on the diagonal $BD$ of a given rectangle $ABCD$, such that $\angle AMC = \angle  CMD$. The point $N$ is the intersection point between $AM$ and the parallel line to $CM$ that contains $B$. Prove that the triangle $BMN$ is equilateral if and only if $ABCD$ is a square.

Valentin Vornicu
4 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
an hour ago
Loop of Logarithms
scls140511   10
N an hour ago by SomeonecoolLovesMaths
Source: 2024 China Round 1 (Gao Lian)
Round 1

1 Real number $m>1$ satisfies $\log_9 \log_8 m =2024$. Find the value of $\log_3 \log_2 m$.
10 replies
scls140511
Sep 8, 2024
SomeonecoolLovesMaths
an hour ago
Proving a kite
Bugi   4
N 2 hours ago by ali123456
Source: Serbian JBTST 3, Day 2
Let $ ABCD$ be a convex quadrilateral, such that

$ \angle CBD=2\cdot\angle ADB, \angle ABD=2\cdot\angle CDB$ and $ AB=CB$.

Prove that quadrilateral $ ABCD$ is a kite.
4 replies
Bugi
May 31, 2009
ali123456
2 hours ago
Inequality
Marinchoo   6
N 2 hours ago by sqing
If $abc=1$ prove that $8(a^3+b^3+c^3) \geq 3(a^2+bc)(b^2+ac)(c^2+ab)$
6 replies
Marinchoo
Apr 28, 2020
sqing
2 hours ago
Interesting inequality
sqing   4
N 2 hours ago by sqing
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
4 replies
sqing
4 hours ago
sqing
2 hours ago
Orthocentre is collinear with two tangent points
vladimir92   42
N 2 hours ago by AshAuktober
Source: Chinese MO 1996
Let $\triangle{ABC}$ be a triangle with orthocentre $H$. The tangent lines from $A$ to the circle with diameter $BC$ touch this circle at $P$ and $Q$. Prove that $H,P$ and $Q$ are collinear.
42 replies
vladimir92
Jul 29, 2010
AshAuktober
2 hours ago
Problem 4
den_thewhitelion   3
N 2 hours ago by DensSv
Source: Second Romanian JBMO TST 2016
We have a 4x4 board.All 1x1 squares are white.A move is changing colours of all squares of a 1x3 rectangle from black to white and from white to black.It is possible to make all the 1x1 squares black after several moves?
3 replies
den_thewhitelion
Jun 15, 2016
DensSv
2 hours ago
Find the period
Anto0110   2
N 2 hours ago by YaoAOPS
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
2 replies
Anto0110
Yesterday at 7:37 PM
YaoAOPS
2 hours ago
Integer inequality
ayan.nmath   11
N Jul 12, 2024 by AshAuktober
Source: Indian TST 2019 Practice Test 1 P1
Let $a_1,a_2,\ldots, a_m$ be a set of $m$ distinct positive even numbers and $b_1,b_2,\ldots,b_n$ be a set of $n$ distinct positive odd numbers such that
\[a_1+a_2+\cdots+a_m+b_1+b_2+\cdots+b_n=2019\]Prove that
\[5m+12n\le 581.\]
11 replies
ayan.nmath
Jul 17, 2019
AshAuktober
Jul 12, 2024
Integer inequality
G H J
G H BBookmark kLocked kLocked NReply
Source: Indian TST 2019 Practice Test 1 P1
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ayan.nmath
643 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $a_1,a_2,\ldots, a_m$ be a set of $m$ distinct positive even numbers and $b_1,b_2,\ldots,b_n$ be a set of $n$ distinct positive odd numbers such that
\[a_1+a_2+\cdots+a_m+b_1+b_2+\cdots+b_n=2019\]Prove that
\[5m+12n\le 581.\]
This post has been edited 2 times. Last edited by ayan.nmath, May 22, 2020, 3:23 PM
Reason: See #7, thanks @MathPassionForever.
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ayan.nmath
643 posts
#2 • 2 Y
Y by Adventure10, Mango247
Originally the problem had a typo, it missed the word "positive".
This post has been edited 1 time. Last edited by ayan.nmath, Jul 17, 2019, 1:04 PM
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Pluto1708
1107 posts
#3 • 5 Y
Y by AlastorMoody, amar_04, L567, Adventure10, Mango247
:Sigh: how many more typo's? :D
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Pluto1708
1107 posts
#4 • 2 Y
Y by Adventure10, Mango247
Partial Solution
This post has been edited 1 time. Last edited by Pluto1708, Jul 17, 2019, 1:55 PM
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Math-wiz
6107 posts
#5 • 2 Y
Y by amar_04, Adventure10
Anyone with a solution? Maybe Ankoganit with the official solution?
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Physicsknight
635 posts
#6 • 1 Y
Y by Adventure10
Take $m=1,n=31$ Then $b_1,b_2,\hdots,b_{31}=1,3,\hdots,61\implies b_1+b_2+\hdots+b_{31}=(1+61)\cdot\frac {31}{2}=961$
$a_1=2019-961=1058$
$5m+21n>581$
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MathPassionForever
1663 posts
#7 • 1 Y
Y by Math-wiz
Indeed, the problem statement has a typo. The original quantity was $5m+12n$.
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RudraRockstar
606 posts
#8
Y by
Redacted
This post has been edited 1 time. Last edited by RudraRockstar, Mar 3, 2021, 8:52 PM
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ayan_mathematics_king
1527 posts
#9
Y by
As @above mentioned we have,
$m(m+1)+n^2\le 2019$.
or, $(m+\frac 12)^2+ n^2 \le 2019+ \frac 14$.
All numbers $m,n$ that satisfy the above equation lie inside the circle $(x+\frac 12)^2 +y^2 \le 2019+\frac 14$
Just note that interesting points of the above circle and the straight line $5x+12y=581$ are $(13.617,42.74),(20.33,39.94)$(approx).
Let's assume for the sake of contradiction that there exist integers $m,n$ which lie inside the circle but also satisfy $5m+12n >581$.
Then this point must lie on the minor area bounded by the circle and the straight line.
But observe that,
for $n=40$,$m\le 19$,(since (m,n) must lie inside the circle)
for $n=41$,$m\le 17$
for $n=42$,$m\le 15$

But each of these contradicts out assumption that $5m+12n >581$
Hence,
$m(m+1)+n^2 \le 2019 \implies 5m+12n\le 581$ for integer $m,n$
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L567
1184 posts
#10 • 1 Y
Y by anurag27826
Observe that $a_1 + a_2 + ... + a_m \ge 2 + 4 + .. + 2m = m(m+1) = m^2 + m$

Also, $b_1 + b_2 + ... + b_n \ge 1 + 3 + 5 + ... + 2n-1 = n^2$.

So, we get that $n^2 + m^2 + m \le 2019$

Now, by Cauchy, we have that

$(169)(2019) \ge 169(n^2 + m^2 + m) = 169(m^2 + n^2) + 169m = (5^2 + 12^2)(m^2 + n^2) + 169m \ge (5m+12n)^2 + 169m$

And so we have that $(5m+12n)^2 \le 169(2019-m) \implies 5m+12n \le 13 \sqrt{2019-m} \le 13 \sqrt{2019}$, which gives that $5m+12n \le 584$

Now, we just need to show that it isnt possible for $5m+12n$ to be any of $584,583,582$. For this, just find a general solution and use the fact that $m^2 + n^2 + m \le 2019$. Also, $n$ needs to be odd as otherwise the whole LHS would end up being even.

So, since $5m+12n \le 584$ and we proved that it cant be any of $582,583,584$, we must have that $5m+12n \le 581$
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lifeismathematics
1188 posts
#11
Y by
clearly $a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots+b_{3}+\cdots+b_{n} \geqslant m(m+1)+n^2$ since all are distinct , now it implies $n^2+m^2+m \leqslant 2019 \implies n^2+m^2 \leqslant 2019-m$ , now $169(2019-m) \geqslant 169(m^2 + n^2)\stackrel{\text{C-S}}{\geqslant} (5m + 12n)^2 \implies 5m + 12n \leqslant 13\sqrt{2019-m} \leqslant 584$

We will show that $5m+12n=584,583,582$ is not possible.

  • $5m+12n=584$ , clearly $n$ has to be odd and $n \equiv 2 \pmod 5$ which really doesn't give any such $m$ and $n$ to be working.
  • $5m+12n=583$ $n \equiv 4 \pmod 5$ which also doesn't give any such $m$ and $n$
  • $5m+12n=582$ $n \equiv 1 \pmod 5$ which also doesn't works.

Hence $5m+12n \leqslant 581$. $\blacksquare$
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AshAuktober
925 posts
#12
Y by
Observe that $m(m+1) + n^2 \le 2019 \implies (m+\frac 12) ^2 + n^2 \le \frac{8077}{4},$ so
$$2025 >\frac{8077}{4} \ge (m+ \frac 12)^2 + n^2 \ge \frac{(5m + \frac 52 + 12n)^2}{13^2}$$$$\implies 5m + \frac 52 + 12n < 585 \implies 5m + 12n \le 582.$$Now show $5m + 12n \ne 582$ as above. $\square$
This post has been edited 3 times. Last edited by AshAuktober, Jul 12, 2024, 3:22 PM
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