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This pirate will always be pirate $p_{n}$, and he will need strictly more than $\\floor*{\\dfrac{n}{2}}$ votes to stay alive; in other words, he will need at least $\\ceil*{\\dfrac{n+1}{2}}$ votes to stay alive. However, since he will always vote for his proposition, we will need at least $\\ceil*{\\dfrac{n-1}{2}}$ other votes to stay alive.\n+\n+%%%%%%%%%%%%%%%%%%\n+%%%%%%% 1 %%%%%%%%\n+%%%%%%%%%%%%%%%%%%\n+\\paragraph{Block 1, $n = 1$.}\n+When pirate $p_{1}$ is alone, he will take the 10 gold pieces for himself.\n+\n+%%%%%%%%%%%%%%%%%%\n+%%%%%%% 2 %%%%%%%%\n+%%%%%%%%%%%%%%%%%%\n+\\paragraph{Block 2, $n = 2$.}\n+When the crew has two pirates, pirate $p_{2}$ will need strictly more than 1 vote, or at least 2 votes, to stay alive. Hence he will need his vote, and 1 extra vote. Putting everything together, we will need the vote of pirate $p_{1}$.\n+He can propose 10 gold pieces to pirate $p_{1}$ and 0 for himself, which seems like a proposition pirate $p_{1}$ will vote for. However, if pirate $p_{1}$ has the chance to throw pirate $p_{2}$ overboard, he will favor this option, since it does not affect his gold revenue.\n+This is why we wrote 0 and 10 in red on Table \\ref{tab:pirates5}, because we could think that this is the final division of the gold, but in fact this situation will never happen: pirate $p_{1}$ will never vote for the proposition of pirate $p_{2}$, hence pirate $p_{2}$ will always die, and we will have $n = 2 \\rightarrow n = 1$.\n+\n+%%%%%%%%%%%%%%%%%%\n+%%%%%%% 3 %%%%%%%%\n+%%%%%%%%%%%%%%%%%%\n+\\paragraph{Block 3, $n = 3$.}\n+When the crew has three pirates, pirate $p_{3}$ will need strictly more than 1 vote, or at least 2 votes, to stay alive. Hence he will need his vote, and 1 extra vote. Putting everything together, we will need the vote of pirate $p_{1}$ or the vote of pirate $p_{2}$.\n+Pirate $p_{2}$ will accept anything, because if pirate $p_{3}$ dies, then we are back to $n = 2$, and in that case, pirate $p_{2}$ dies as well.\n+Hence, pirate $p_{3}$ can simply propose 0 to $p_{2}$, 0 to $p_{1}$ and 10 for himself. Pirate $p_{1}$ will not vote for that proposition, but pirate $p_{2}$ will clearly vote for that proposition, hence we have found the repartition.\n+\n+%%%%%%%%%%%%%%%%%%\n+%%%%%%% 4 %%%%%%%%\n+%%%%%%%%%%%%%%%%%%\n+\\paragraph{Block 4, $n = 4$.}\n+When the crew has four pirates, pirate $p_{4}$ will need strictly more than 2 votes, or at least 3 votes, to stay alive. Hence he will need his vote, and 2 extra votes. Putting everything together, from the three pirates who are beneath him, he can only afford to loose one vote, not more.\n+It is impossible to make pirate $p_{3}$ happy, because he will always prefer to kill pirate $p_{4}$ and go back to block 3, where he wins 10. Hence, we will never have the vote of pirate $p_{3}$. However, in the previous case, pirates $p_{1}$ and $p_{2}$ both receive 0 gold coins, hence we can easily win their votes by giving them 1 coin each.\n+Hence, pirate $p_{4}$ will propose 1 to $p_{1}$, 1 to $p_{2}$, 0 for $p_{3}$ and 8 for himself. Pirate $p_{3}$ will not vote for that proposition, but pirates $p_{1}$ and $p_{2}$ will clearly vote for that proposition, hence we have found the repartition.\n+\n+%%%%%%%%%%%%%%%%%%\n+%%%%%%% 5 %%%%%%%%\n+%%%%%%%%%%%%%%%%%%\n+\\paragraph{Block 5, $n = 5$.}\n+When the crew has five pirates, pirate $p_{5}$ will need strictly more than 2 votes, or at least 3 votes, to stay alive. Hence he will need his vote, and 2 extra votes. Putting everything together, from the four pirates who are beneath him, he can only afford to loose two votes, not more.\n+It is difficult to make pirate $p_{4}$ happy, because he will always prefer to kill pirate $p_{5}$ and go back to block 4, where he wins 8: the only way to make pirate $p_{4}$ happy is to give him 9 or 10 coins, but then there are almost no coins left for pirate $p_{5}$. Hence, pirate $p_{5}$ will never have the vote of pirate $p_{4}$. However, in the previous case, pirate $p_{3}$ received 0 gold coins, hence we can easily win his vote by giving him 1 coin. We can also see that, in the previous case, pirate $p_{1}$ received 1 gold coin, hence we can easily win his vote by giving him 2 coins. The same can be said about pirate $p_{2}$, and since since pirate $p_{5}$ is highly intelligent and ferociously logical, he will, if his life is not threatened, always act to maximize his gain. This means that we will not give 2 coins to pirate $p_{1}$ and 2 other coins to pirate $p_{2}$, because pirate $p_{5}$ only needs the vote of one of these two pirates. This is where the extra rule is important: since he has the choice between two divisions of gold with the same amount for him, pirate $p_{5}$ will choose \\textit{the one that gives the highest amount to the highest ranked pirates}, i.e., pirate $p_{2}$. This rule is really important because it allows to eliminate multiple solutions, and it makes the solution unique.\n+Hence, pirate $p_{5}$ will propose 0 to $p_{1}$, 2 to $p_{2}$, 1 for $p_{3}$, 0 for $p_{4}$ and 7 for himself. Pirates $p_{4}$ and $p_{1}$ will not vote for that proposition, but pirates $p_{2}$ and $p_{3}$ will clearly vote for that proposition, hence we have found the repartition.\n+\n+\\paragraph{Putting everything together.} The final decisions regarding the division of the gold are given on Table \\ref{tab:pirates5}. On this table, each horizontal block shows the division of the gold, in function of the value of $n$, where $n$ is the number of pirates in the crew.\n+\n+\n+\\begin{tabular}[h!]{|lllll||r|}\n+\\hline\n+&         &           &                    &                     & $n$ \\\\ \\hline\n+&         &           &                    & $p_{1}$             & 1   \\\\\n+&         &           &                    & 10                  &     \\\\ \\hline\n+&         &           & $p_{2}$            & $p_{1}$             & 2   \\\\\n+&         &           & \\textcolor{red}{0} & \\textcolor{red}{10} &     \\\\ \\hline\n+&         & $p_{3}$   & $p_{2}$            & $p_{1}$             & 3   \\\\\n+&         & 10        &  0                 & 0                   &     \\\\ \\hline\n+& $p_{4}$ & $p_{3}$ & $p_{2}$              & $p_{1}$             & 4   \\\\\n+& 8       &  0        & 1                  & 1                   &     \\\\ \\hline\n+$p_{5}$ & $p_{4}$ & $p_{3}$   & $p_{2}$            & $p_{1}$             & 5   \\\\\n+7       &  0      & 1         & 2                  &  0                  &     \\\\ \\hline\n+\\end{tabular}\n+\\captionof{Gold divisions for $n$ pirates, with $1 \\leq n \\leq 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