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Please don't post again and again like this. It's a kind of spamming.

I think u are not enough smart to solve the problem so u r saying these kind of rambling

jrc1729 wrote: Please don't post again and again like this. It's a kind of spamming. I have to agree with jrc1729. Why the urgency? thegreatp.d wrote: they have any player in the tournament that has lost all of $k$'s. English is my native (first) language, and I cannot understand this statement. Can you post the question exactly how it was written?

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Okay - I was wondering because last time you posted a link to a forum where the BDMO problems were written in English. Do you mean to say that $k \le n$ is $n$-good if there exists a tournament such that for any $k$ players, there exists a player who has beaten the remaining players? (This would be my best guess, but it doesn't make much sense for $k=2$ since any tournament would be $2$-good under this definition). Mathematics requires precise language, especially on quantifiers like "any/every", "exists", etc. so I cannot answer the question currently. Either you or someone can help me interpret (which would be appreciated) or post a link to the problems.

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Based on my (very) limited grasp of Bengali, and with the help of Google Translate, I'm reading: ahmedittihad wrote: A tournament is played between $n$ people. Everyone plays with everyone else, and no game ends in a draw. A number $k$ is said to be $n$-good if there exists such a tournament in which there is, for every $k$ people, a player who has lost to all of them. a) Prove that $n\geq 2^{k+1}-1$ b) Give all $n$ for which $2$ is $n$-good. The result appeared originally in an an article by Paul Erdős. The following is a replication of his argument: For part a) we will proceed by induction; the base case of $k=1$ is trivial. Assume that the statement holds for $k=m-1$, we will show that this implies the corresponding claim for $k=m$. Suppose, for the sake of contradiction, the opposite—that there exists such a tournament with $n\leq 2^{m+1}-2$. Observe that there necessarily exists a player in this tournament (call her Ruba) who has won at most $\left\lfloor\tfrac{1}{2}(n-1)\right\rfloor=2^m-2$ matches. There are two cases: either (i) Ruba has won at least $m-1$ matches, or (ii) she has won strictly fewer than $m-1$. In the former case, consider the set $L$ of $\l$ (by definition at least $m-1$ at at most $2^m-2$) players whom Ruba has defeated. For any group of $m-1$ players in $L$, we know that there exists some other player (call him Munna) in the tournament who has lost to each of these $m-1$ and to Ruba. Furthermore, and again by definition, Munna is a member of $L$. It follows that $\l$ is $m-1$-good, and thus that $\l\geq 2^m-1$, a contradiction. Alternatively, suppose that $\l<m-1$. Choose some $m-1-\l$ players not Ruba or members of $L$ and add them to $L$ to form a set $L'$ of $m-1$ players. By hypothesis, there is some player in the tournament (say, Topu) who has lost to all of the members of $L'$ as well as to Ruba. Topu is clearly a member of $L$, but $L\subset L'$ (and he has surely not lost to himself!), again a contradiction $\Box$ As for part b), we claim that $n\geq 7$ is necessary and sufficient. The former half is immediate from the inequality from a). Consider first the $n=7$ case. Label the $7$ players $p_0,p_1,\dots,p_6$, and say $p_i\to p_j$ if player $i$ loses to player $j$. Observe that the tournament whose game's results are: $$p_0\to p_1\to p_2\to p_3\to p_4\to p_5\to p_6\to p_0$$$$p_0\to p_2\to p_4\to p_6\to p_1\to p_3\to p_5\to p_0$$$$p_0\to p_4\to p_1\to p_5\to p_2\to p_6\to p_3\to p_0$$Has the desired property (that for every two players there is a third who has lost to both). This is motivated by the fact that $\{1,2,4\}-\{1,2,4\}$ is the complete set of residues modulo $7$. To obtain a construction for $n>7$, simply add $n-7$ players $q_7,q_8,\dots, q_{n-1}$ so that each game between a $q$ and a $p$ is won by the $q$. (The games between the $q$s can be assigned arbitrarily—they don't really matter). It is again fairly easy to check that this satisfies the problem statement, as desired $\blacksquare$

Correct question is posted here. http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=4237&p=20754#p20754