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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
About old Inequality
perfect_square   0
4 minutes ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
1 viewing
perfect_square
4 minutes ago
0 replies
inquality
karasuno   1
N 27 minutes ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
1 viewing
karasuno
an hour ago
sqing
27 minutes ago
Number Theory
karasuno   0
an hour ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
an hour ago
0 replies
Minimum number of values in the union of sets
bnumbertheory   5
N an hour ago by Rohit-2006
Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3
For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
5 replies
bnumbertheory
Oct 14, 2023
Rohit-2006
an hour ago
No more topics!
|B-A| >= 3 for any subsets A,B
mlm95   13
N May 31, 2022 by thczarif
Source: Iran TST 2013: TST 1, Day 1, Problem 2
Find the maximum number of subsets from $\left \{ 1,...,n \right \}$ such that for any two of them like $A,B$ if $A\subset B$ then $\left | B-A \right |\geq 3$. (Here $\left | X \right |$ is the number of elements of the set $X$.)
13 replies
mlm95
Apr 17, 2013
thczarif
May 31, 2022
|B-A| >= 3 for any subsets A,B
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Source: Iran TST 2013: TST 1, Day 1, Problem 2
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mlm95
245 posts
#1 • 2 Y
Y by Adventure10, Mango247
Find the maximum number of subsets from $\left \{ 1,...,n \right \}$ such that for any two of them like $A,B$ if $A\subset B$ then $\left | B-A \right |\geq 3$. (Here $\left | X \right |$ is the number of elements of the set $X$.)
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Math-lover123
304 posts
#2 • 1 Y
Y by Adventure10
What is $A-B$_?
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mlm95
245 posts
#3 • 1 Y
Y by Adventure10
$B-A$ is set of the elements that are in $B$ but not in $A$
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Goutham
3130 posts
#4 • 4 Y
Y by hyperbolictangent, blackwave, Adventure10, Mango247
Let $X_i = \sum _{ j = 0} ^{\left\lfloor\frac{n-i}{3}\right\rfloor} \dbinom{n}{i+3j}, i = 0, 1, 2$ and $M = X_k$ such that $X_k \ge X_{k+1}, X_k \ge X_{k+2}$ where indices are modulo $3$. Let $S_n = \{1, 2, \ldots, n \}$. Consider a set $T$ such that each element of $T$ is of the form $\{\phi, \{a_1\}, \{a_1, a_2\}, \ldots, \{a_1, a_2, \ldots, a_n \} \}$ where $\{a_1, a_2, \ldots, a_n \} = S_n$ and if we have any $K\subset S_n$ with $|K| = k$, then $K$ should occur in $\frac{ \prod_{i = 0} ^{n} \dbinom{n}{i}} {\dbinom{n}{k} }$ elements of $T$. The existence of such a set $T$ can be proven using Hall's Marriage Theorem by considering a set to have $\dbinom{n}{k}$ copies of each subset of $S_n$ of size $k$ and then noticing that there is a matching from subsets of size $k$ and those of size $k+1$ because they can be shown to satisfy Hall's condition and gluing the matchings would give rise to $T$.
Let $K$ be any set satisfying the given conditions. Then \[|K| = \frac{\sum_{A\in K}\dbinom{n}{|A|}\cdot \sum_{P}^{A\in P \in T} 1}{|T|} = \frac{\sum_{P\in T} \sum _{A\in P\cap K } \dbinom{n}{|A|}}{|T|}\le \frac{\sum_{P\in T} M}{|T|} = M\]
Equality holds when
\[K = \bigcup_{j=0}^{\left\lfloor\frac{n-k}{3}\right\rfloor} \{ S| S\subset\{1, 2, \ldots, n\}, |S|=k+3j\}\]
This post has been edited 2 times. Last edited by Goutham, May 28, 2013, 7:05 AM
Reason: Added a few extra explanations thanks to hyperbolictangent
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goodar2006
1347 posts
#5 • 2 Y
Y by Adventure10, Mango247
Proposed by Ali Khezeli
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leader
339 posts
#6 • 2 Y
Y by Adventure10, Mango247
is this correct?
if we call $X_{n}$ the solution of the problem for $n$ then it is enough to find $X_{1},..,X{6}$ and then continue on knowing what sum of binomial coefficents to take(module $3$ either $1,2,3$)
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qwert1234
4 posts
#7 • 3 Y
Y by blackwave, Adventure10, Mango247
Let $A_1,A_2,\dots ,A_r$ be sets which satisfies the conditions of the problem.
Let $f(A_i)=\{ A_i,A_i\cup \{1\},A_i\cup \{2\},\dots ,A_i\cup \{n\},A_i-\{1\},A_i-\{2\},\dots , A_i-\{n\}\}$.
It is easy to see that $|f(A_i)|=n+1$ and
1) $f(A_i)\cap f(A_j)=\varnothing$
or
2) $|A_i-A_j|=1$, $|A_i|=|A_j|$ and so $|f(A_i)\cap f(A_j)|=2$.

Can anyone help me to complete this idea?
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gev_gev
31 posts
#8 • 3 Y
Y by blackwave, Adventure10, Mango247
I think this must be right
First, note that the group of all subsets with number of elements $3i+1$ ($n$ is even) or $3i+2$ ($n$ is odd) satisfy the requirements of the problem, the number of the subsets in this group is $\Sigma(\frac{n}{3i+1})$ or $\Sigma(\frac{n}{3i+2})$ one of which equals $\frac{2^{n}-(-1)^{n}}{3}$.
We will prove by induction on $n$ that the maximum number of subsets is $\frac{2^{n}-(-1)^{n}}{3}$.
The case $n=4$ is obvious.
We suppose that we have the answer for $n-1$, and we will prove for $n$.
Let's split the group of subsets satisfying the requirements of the problem into $2$ subgroups, on of which contains all subsets which do not contain the element ${1}$ and the other subgroup contains the rest of the subsets. The first subgroup has at most $\frac{2^{n-1}-(-1)^{n-1}}{3}$ by induction, the second subgroup has at most $\frac{2^{n-1}-(-1)^{n-1}}{3}+1$ elements (we can consider subsets $A-\{1\}$ instead of $A$ if $A \neq \{1\}$, then add to the number of that subsets $1$ for the subset $\{1\}$) by induction. This gives the result in case $n$ is odd, still for even $n$ more tricks must be done.
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DVDthe1st
339 posts
#9 • 8 Y
Y by gev_gev, hyperbolictangent, blackwave, K.N, Ankoganit, mijail, Adventure10, Mango247
A faster way to see it is as follows:

Let $S_n$ be the maximum.

We note that $S_1=1,S_2=1$.

Now we show that $S_{n+2}=2^n+S_n$, from which we can easily finish by induction.

Consider a family of sets satisfying the condition for $n+2$. Note for every $A\subseteq \{1,2,...,n\}$, at most one of $A,A\cup\{n+1\},A\cup\{n+1,n+2\}$ is in the family. For the remaining sets (which contain $n+2$ and exclude $n+1$), we can pick at most $S_n$ more sets. Hence our conclusion follows.

Of course, it still remains to show that the maximum can always be attained, but this has been demonstrated in previous posts.
This post has been edited 1 time. Last edited by DVDthe1st, Sep 16, 2015, 8:38 AM
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yunxiu
571 posts
#10 • 3 Y
Y by blackwave, Adventure10, Mango247
${S_2} = 2$so we have ${S_n} = \frac{{{2^{n + 1}} + 3 + {{( - 1)}^n}}}{6}$.
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yugrey
2326 posts
#11 • 3 Y
Y by blackwave, Adventure10, Mango247
Err, another way is to show by straight induction you can split the $2^n$ subsets into the floor of $\frac {2^n} {3}$ chains of $3$ elements and then the last is either a singleton or a chain of $2$, depending on the parity of $n$. The induction step is more or less "clone and combine."
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blackwave
202 posts
#12 • 2 Y
Y by Adventure10, Mango247
Can someone help me explain the idea of GOUTHAM ? ( I don't understand much his solution)
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Wonder-Cop
1 post
#13 • 2 Y
Y by Adventure10, Mango247
Can anyone give a more detailed solution for me please? Thank you so much!
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thczarif
36 posts
#14
Y by
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Are you sure that for $n$ the answer is $\frac{2^{n}-(-1)^{n}}{3}$ ?

For $n=4$ we can get $6$ subsets whereas you predicted to have at most $5$. Here are the $6$ subsets, $\{1,2\},\{2,3\},\{1,3\},\{1,4\},\{2,4\},\{3,4\}$. It seems to have at most $\left \lceil \dfrac{2^n}{3} \right \rceil $ subsets instead. Sorry, in advance if I am wrong somewhere.
This post has been edited 1 time. Last edited by thczarif, May 31, 2022, 9:38 PM
Reason: hided the quoted part
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