Plan ahead for the next school year. Schedule your class today!

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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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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
A problem of functional equation
deltapc   2
N 6 minutes ago by deltapc
Find all function \( f: \mathbb{R} \to \mathbb{R} \) such as \( f(xf(y^2)+f(x))=f(f(x))f(y)^2+x, \forall x,y \in \mathbb{R} \)
2 replies
deltapc
Today at 2:46 AM
deltapc
6 minutes ago
sum (a+b)/(a^2+ab+b^2) <=2 if 1/a+1/b+1/c =3 for a,b,c>0
parmenides51   16
N 33 minutes ago by skellyrah
Source: 2020 Greek JBMO TST p2
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=3$. Prove that
$$\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}\le 2$$When is the equality valid?
16 replies
parmenides51
Nov 14, 2020
skellyrah
33 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   4
N 39 minutes ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
4 replies
Dattier
May 15, 2025
Dattier
39 minutes ago
Max value of function with f(f(n)) < n+50
Rijul saini   3
N an hour ago by Saucepan_man02
Source: India IMOTC Day 3 Problem 2
Let $S$ be the set of all non-decreasing functions $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying $f(f(n))<n+50$ for all positive integers $n$. Find the maximum value of
$$f(1)+f(2)+f(3)+\cdots+f(2024)+f(2025)$$over all $f \in S$.

Proposed by Shantanu Nene
3 replies
Rijul saini
Jun 4, 2025
Saucepan_man02
an hour ago
No more topics!
Combinatorics Problem
Hopeooooo   1
N Oct 13, 2022 by M11100111001Y1R
Source: SRMC 2022 P4
In a language$,$ an alphabet with $25$ letters is used$;$ words are exactly all sequences of $($ not necessarily different $)$ letters of length $17.$ Two ends of a paper strip are glued so that the strip forms a ring$;$ the strip bears a sequence of $5^{18}$ letters$.$ Say that a word is singular if one can cut a piece bearing exactly that word from the strip$,$ but one cannot cut out two such non-overlapping pieces$.$ It is known that one can cut out $5^{16}$ non-overlapping pieces each containing the same word$.$ Determine the largest possible number of singular words$.$
(Bogdanov I.)
1 reply
Hopeooooo
May 23, 2022
M11100111001Y1R
Oct 13, 2022
Combinatorics Problem
G H J
G H BBookmark kLocked kLocked NReply
Source: SRMC 2022 P4
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Hopeooooo
819 posts
#1 • 1 Y
Y by Mango247
In a language$,$ an alphabet with $25$ letters is used$;$ words are exactly all sequences of $($ not necessarily different $)$ letters of length $17.$ Two ends of a paper strip are glued so that the strip forms a ring$;$ the strip bears a sequence of $5^{18}$ letters$.$ Say that a word is singular if one can cut a piece bearing exactly that word from the strip$,$ but one cannot cut out two such non-overlapping pieces$.$ It is known that one can cut out $5^{16}$ non-overlapping pieces each containing the same word$.$ Determine the largest possible number of singular words$.$
(Bogdanov I.)
This post has been edited 1 time. Last edited by Hopeooooo, May 23, 2022, 2:19 PM
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M11100111001Y1R
130 posts
#2 • 3 Y
Y by Hopeooooo, Mahdi.sh, a22886
I don't remember a combinatorics problem harder than this one...

Answer: $2\times5^{17}.$
Official Solution: Let the alphabet of the letters $a_1,a_2,\cdots,a_{25}.$ By a piece we always mean a piece of the strip containing exactly $17$ consecutive letters; Different piece may contain the same word. Say that a piece is singular if the word it contain is such. We start with constructing an example containing $N= 2\times5^{17}$ singular words. Define a word $W=a_1,a_2,\cdots,a_{17};$ This will be the word having $k=5^{16}$ non-overlapping copies on the strip. There exist exactly $k=25^8$ possible $8-$letter sequences, consisting of letters $a_{18},a_{19},\cdots,a_{25};$ Put them onto the strip in an arbitrary order, separating each two sequences by an instance of $W.$ Each segment of the strip containing one $8-$sequence mentioned above (and no other letters) will be referred as a part. Notice that the strip contains exactly $(8+17)k=5^{18}$ letters. Clearly, the obtained strip contains $k$ non-overlapping copies of $W.$ Now we show that any piece containing a whole part is singular moreover, that the word it contains is met on no other piece. Since a part can be situated in a piece at $10$ different positions (starting from the $1-$st, from the $2-$nd, \cdots, or from the $10-$th letter of a piece), we will get that there are at least $N=10\times 5^{16}$ singular words. Consider an arbitrary piece $p$ containing a a word $P.$ Either this piece contains a unique non empty prefix which coincides with some suffix of $W,$ or there is no such prefix only in this case we will say that such prefix is empty. Let $b$ be the length of the defined prefix. Define similarly a suffix of $P$ which coincides with a prefix of $W,$ and denote its length by $e.$ Notice that the defined prefix and suffix do not overlap whenever $P\not =W$ (if $P=W,$ we have $b=c=17$). If the piece contains no whole part, then $max\{b,e\}>9.$ If the piece contains a part then $b+e=9$ and $0\le b,e\le9.$ Thus, piece $p$ contains a part iff $max\{b,e\}\le9,$ and in this case the position of the part at $P$ (and hence the position of $p$ at the strip) is uniquely determined. Therefore, in this case $P$ is met only one piece $p,$ so this piece is singular. We have proven that the constructed example works. It remains to prove that the number of singular words cannot exceed $N.$ Enumerate the positions in the strip successively by $1,2,\cdots,5^{18}$ (the numeration is cyclic module $5^{18}$). Let $p_i$ denote the piece starting at positions $i,$ and let $P_i$ be the word on that piece. Let $n_1,\cdots,n_k$ be a positive such that the pieces $p_{n_1},p_{n_2},\cdots,p_{n_k}$ are pairwise disjoint and contain the same word $W$ (from the problem statement). Clearly, those pieces are not singular. For $i=1,2,\cdots,8$ and $1\le s \le k,$ we say that a piece $p_{n_{s+1}}$ is a rank $i$ follower, while $p_{n_{s-1}}$ is a rank $i$ predecessor. All these pieces (followers and predecessors) are distinct; Moreover. Followers of a fixed rank are pairwise disjoint, and the same holds for predecessors. We will show that $among\ 8\times 5^{16}$ followers of all ranks, at most $5^{16}$ pieces are singular (we will call this statement a quoted claim in the future); By symmetry, the same bound hold for predecessors. This will yield that there are at least $5^{16}+7 \times 5^{16}+7  \times 5^{16}=3\times 5^{17}$ non-singular pieces, which implies the desired bound. Thus, we are left to prove the claim quoted above. For any rank $i$ follower $p_{n_s+i}$ define its tail as its suffix of length $i$ (the tail consists of all letters which do not lie in $p_{n_s};$ We regard a tail as a sequence of letters). We show by induction on $m=0,1,\cdots,8$ that for every sequence $U$ consisting of $(8-m)$ letters, there are no more than $25^m$ followers whose tails contain $U$ as a prefix. The desired claim is obtained by setting $m=8.$ The base case $m=0$ is obvious; If a follower with tail $U$ is singular, then there is only one such follower. Let us perform the inductive step. If there is no singular follower whose tail is $U,$ then every singular follower's tail starting with $U$ starts in fact with some word of the form $U_{a_i}.$ For every $i=1,2,\cdots,25,$ there are at most $25^{m-1}$ such followers, by the inductive hypothesis. So the total number of such followers does not exceed $25\times 25^{m-1}=25^m,$ as desired. Finally, if there is a singular follower $p_{n_s+8-m}$ whose tail is $U,$ then such follower is unique. Therefore, all followers of larger ranks whose tails start with $U$ correspond to the same copy $p_{n_s}$ of $W.$ Then the number of such followers (including $p_{n_s+8-m }$ itself) is at most $m+1 \le 25^m,$ as desired again.
This post has been edited 2 times. Last edited by M11100111001Y1R, Oct 13, 2022, 2:10 PM
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