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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
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!
The game of circulate
Zhero   7
N Apr 20, 2024 by Om245
Source: ELMO Shortlist 2010, C7
The game of circulate is played with a deck of $kn$ cards each with a number in $1,2,\ldots,n$ such that there are $k$ cards with each number. First, $n$ piles numbered $1,2,\ldots,n$ of $k$ cards each are dealt out face down. The player then flips over a card from pile $1$, places that card face up at the bottom of the pile, then next flips over a card from the pile whose number matches the number on the card just flipped. The player repeats this until he reaches a pile in which every card has already been flipped and wins if at that point every card has been flipped. Hamster has grown tired of losing every time, so he decides to cheat. He looks at the piles beforehand and rearranges the $k$ cards in each pile as he pleases. When can Hamster perform this procedure such that he will win the game?

Brian Hamrick.
7 replies
Zhero
Jul 5, 2012
Om245
Apr 20, 2024
The game of circulate
G H J
Source: ELMO Shortlist 2010, C7
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Zhero
2043 posts
#1 • 3 Y
Y by Adventure10, Mango247, buddyram
The game of circulate is played with a deck of $kn$ cards each with a number in $1,2,\ldots,n$ such that there are $k$ cards with each number. First, $n$ piles numbered $1,2,\ldots,n$ of $k$ cards each are dealt out face down. The player then flips over a card from pile $1$, places that card face up at the bottom of the pile, then next flips over a card from the pile whose number matches the number on the card just flipped. The player repeats this until he reaches a pile in which every card has already been flipped and wins if at that point every card has been flipped. Hamster has grown tired of losing every time, so he decides to cheat. He looks at the piles beforehand and rearranges the $k$ cards in each pile as he pleases. When can Hamster perform this procedure such that he will win the game?

Brian Hamrick.
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fwolth
202 posts
#2 • 6 Y
Y by toto1234567890, MathbugAOPS, Adventure10, Mango247, and 2 other users
This problem is amazing. The following is excessively verbose for a very simple idea, but I tried to make everything rigorous.

We'll term a card displaying the number of a pile as pointing to that pile. We claim that it's possible iff there doesn't exist a proper subset of the piles $S$ s.t. each card in $S$ points to a pile in $S$ (a self-contained subset).

To prove necessity, notice that if we do have such an $S$, then since all $|S|k$ of the cards with numbers pointing to $S$, the complement of $S$ contains only cards that point to itself as well, so either way, pile 1 is in a self-contained subset. Then no matter how you order the cards in each pile of that subset, it's impossible to reach a card pointing to a pile outside of that subset from within the subset, so we can't flip over all the cards.

To prove sufficiency, we show the more general result that if we're given $n$ piles with $a_1,a_2,\ldots, a_n$ cards to start s.t. the number $i$ appears exactly $a_i$ times and the first card is to be drawn from any given pile, the result still holds. From this point on, we assume WLOG Hamster is female.

Since the "game" is predetermined, the real fun lies in the rearrangement of piles. Suppose our player lays out the piles in no particular order and lets herself flip whichever card from each pile she wants at each stage, starting from pile 1 as in the real game, and proceeding as according to normal rules. Our desired result is equivalent to her being able to flip all the cards, since once she has done this, she can go back and put the cards in each pile in the order she picked them, and if she can go back and put the cards in such an order, then she can likewise find the according sequence of picks.

Notice that if she is unsuccessful and does end up running out of cards in pile $t$, then she must have picked up $t$ $a_t+1$ times unless $t$ is the pile she first picked from, so the last card picked is the index of the first pile picked from.

We establish an isomorphism complete sequences of picks (complete or not) and $1,2,\ldots,n$ written around a circle s.t.:
a) $t$ at exactly $a_t$ times
b) The numbers directly following occurrences of $t$ around a circle are those in column $t$.
Clearly, if we write a sequence of picks Hamster makes around a circle, they satisfy these conditions (note that the first pick follows the last pick, which again is the index of the first pile). Similarly, with any such sequence, we can pick a starting point and take a card with that value from any pile, then proceed according to the rules; clearly, we never run out of cards until all are chosen by a) and the successive cards according to b) can be chosen according to the rules, by construction. Notice that this gives a symmetry: any sequence of picks can be turned into a circle, which can then be turned back into a sequence starting at any point, meaning it doesn't matter what pile the starting point is.

We now proceed by strong induction. When $n=1,\sum a_i=1$, then we just have the one possible sequence, which clearly works. Now suppose we have a configuration $C$ of $n$ piles with $a_1,\ldots,a_n$ cards with no self-contained subset, and assume that we can find a complete sequence of picks for any arrangement of cards according to the rules with piles and pile sizes identical to any proper subset of $C$. Pick a card from a given pile (doesn't matter by symmetry) and then keep picking at random according to constraints until it's impossible to pick any longer.

If this sequence is complete, we're done. Else, partition $C$ into a configuration with the cards picked $C_1$ and a configuration with those unpicked, $C_2$, preserving the pile number of cards in both configurations. Our sequence from above then must be a complete sequence on $C_1$. But consider $C_2$. Each card valued $i$ not picked in the complete sequence on $C_1$ means that one card from column $i$ was not chosen in that sequence. Thus, the number of occurrences of $i$ in $C_2$ is equal to the number of cards in pile $i$, so by the inductive hypothesis there exists a complete sequence on $C_2$ as well.

Lemma: We can "glue" together complete sequences on $C_1$ and $C_2$ to obtain a complete sequence on $C$.
Proof: From the card-pile duality demonstrated from the circle isomorphism (each card value is followed by a column with that index, including the last card), if they share no values, that means they consist of disjoint sets of piles. But this violates the self-contained subset rule, contradiction.

Thus, assume that $C_1$ and $C_2$ contain only at least common value, or equivalently that there's a pile $i$ they both contain cards from. But then in the construction of $C_1$, when the last value of $i$ is encountered, simply pick one of the values in that pile part of $C_2$ in the original construction. Then follow $C_2$'s cycle until the last value of $i$ encountered in $C_2$'s complete sequence is encountered. By the circle isomorphism, it's clear that this will entail all of $C_2$, since $i$ precedes that value picked in column $i$. Now pick the original value from $C_1$ in column $i$ and complete $C_1$ in the same way as before. The result follows.

Therefore, by strong induction, Hamster wins unless there's a self-contained subset.
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Kayak
1298 posts
#3 • 6 Y
Y by lifeisgood03, FISHMJ25, RudraRockstar, karitoshi, Adventure10, Mango247
My solution is most probably wrong since it looks simple for an ELMO C7 :huh: Can anybody please check it ?

Solution
This post has been edited 1 time. Last edited by Kayak, Apr 27, 2018, 10:34 AM
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FISHMJ25
293 posts
#4 • 3 Y
Y by karitoshi, Adventure10, Mango247
Kayak wrote:
My solution is most probably wrong since it looks simple for an ELMO C7 :huh: Can anybody please check it ?

Solution

It looks correct and i think it is realy nice.
Z K Y
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K_Mirrorheart
57 posts
#5 • 2 Y
Y by Adventure10, Mango247
Well this is an easy C7, at least much easier than 2011's A1, which proved to be a catastrophe. Because the question nearly explicitly asks for a Eulerian Circuit and the condition of existing an EC in directed graph is quite well-known.
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KI_HG
157 posts
#6
Y by
Can someone explain the rule, I don't quite get it. If every card has been flipped, then there must exist a pile that all the cards have been flipped. Isn't it?
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signifance
140 posts
#7
Y by
redacted
This post has been edited 1 time. Last edited by signifance, Jan 13, 2024, 4:06 AM
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Om245
163 posts
#9 • 1 Y
Y by GeoKing
Consider graph $G$ with piles as vertex of $G$ with vertex set $V$ = ${V1,V_2,\cdots,V_k}$. .Let we assume $G$ is connected. Now note that indegree and outdegree of each vertex is $k$. (we consider multiple edges)
hence degree of each vertex is even. It well know if all vertex of graph are even then there exists Eulerian circuit
Now if there $i$ number card in $i$th pile, place it at top as it doesn't matter.
If eulerian circuit is $V{a1} \rightarrow V{a2} \rightarrow \cdots \rightarrow V{al}$. Then rearrange card in that manner. (note as $V{ai} \rightarrow V{a{i+1}}$ there exit card of number $a{i+1}$ in $V{a_i}$ by construction of graph). hence we will have our arrgement.

Now note if $G$ is not connected it is not posibble to reach all piles.
So for $G$ to be connected, for any $m$ piles ${V{b1},V{b2},\cdots,V{b_m}}$ we should not have only number $b_1,b_2,\cdots,b_m$ in these plies.
else there will no edge connected this set of $m$ plies with other.
In cards sence we should not have set of plies which contain only there piles index number

Hopefully not wrong
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