3. There are three tables in a room. The left table has $L \mathrm{~kg}$ of chocolate on it, the middle table has $M \mathrm{~kg}$ of chocolate on it, and the right table has $R \mathrm{~kg}$ of chocolate on it, for some positive integers $L, M$ and $R$. People enter the room one at a time. Each person determines the table at which they will sit by determining the table at which they would currently receive the greatest mass of chocolate, assuming that the people at each table would share the chocolate at that table equally. If at least two tables would result in the same share of the chocolate, the person will sit at whichever of these tables is farther to the left.
For example, suppose that $L=5, M=3$ and $R=6$. Person 1 sits at the right table, because their share would be $6 \mathrm{~kg}$ instead of $5 \mathrm{~kg}$ or 3 $\mathrm{kg}$ at the other two tables. Person 2 then sits at the left table, because their share would be $5 \mathrm{~kg}$ instead of $3 \mathrm{~kg}$ at each of the other two tables. The chart to the right shows the possible shares
\begin{tabular}{||c|c||c|c||c|c||}
\multicolumn{2}{||c||}{ Left } & \multicolumn{2}{l|}{ Middle } & \multicolumn{2}{c|}{ Right } \\
\hline 5 & & 3 & & 6 & $\mathrm{P} 1$ \\
5 & $\mathrm{P} 2$ & 3 & & & \\
$\frac{5}{2}$ & & 3 & $\mathrm{P} 3$ & 3 & \\
$\frac{5}{2}$ & & $\frac{3}{2}$ & & 3 & $\mathrm{P} 4$
\end{tabular}
in $\mathrm{kg}$ and the final decision for each of the first four people.
(a) Suppose that $L=5, M=3$ and $R=6$ and that 7 people enter the room in total. Continue the chart above to show where the 7 people sit and explain why each of Person 5, Person 6 and Person 7 make their choice.
(b) Explain why there cannot be positive integers $L, M$ and $R$ for which the first 6 people who enter the room sit at the tables in the following order: Left, Middle, Right, Left, Left, Left.
(c) Suppose that $L=9, M=19$ and $R=25$. Determine, with justification, where Person 2019 will sit.
3. (a) After 4 people, there is 1 person at the Left table, 1 person at the Centre table, and 2 people at the Right table.
Continuing the table, we obtain
\begin{tabular}{|c|c|c|c|c|c|}
\hline \multicolumn{2}{|c|}{ Left } & \multicolumn{2}{|c|}{ Middle } & \multicolumn{2}{|c|}{ Right } \\
\hline 5 & & 3 & & 6 & $\mathrm{P} 1$ \\
\hline 5 & P2 & 3 & & \begin{tabular}{l}
3
\end{tabular} & \\
\hline$\frac{5}{2}$ & & 3 & P3 & 3 & \\
\hline$\frac{5}{2}$ & & $\frac{3}{2}$ & & 3 & $\mathrm{P} 4$ \\
\hline$\frac{5}{2}$ & P5 & $\frac{3}{3}$ & & 2 & \\
\hline$\frac{5}{3}$ & & $\frac{3}{2}$ & & 2 & P6 \\
\hline$\frac{5}{3}$ & P7 & $\frac{3}{2}$ & & $\frac{5}{2}$ & \\
\hline
\end{tabular}

Person 5 sits at the Left table because $\frac{5}{2}=2.5$ is greater than $\frac{3}{2}=1.5$.
Person 6 sits at the Right table because $\frac{5}{3} \approx 1.67$ and $\frac{3}{2}=1.5$ are both less than 2 .
Person 7 sits at the Left table because $\frac{5}{3} \approx 1.67$ is greater than $\frac{3}{2}=1.5$.
Therefore, Person 5 sits at the Left table, Person 6 at the Right table, and Person 7 at the Left table.
(b) Suppose that there are integers $L, M, R$ for which the first 6 people sit as described. We construct a similar chart from the given information, using the information about where the previous people have sat to calculate the shares in each row:
\begin{tabular}{|c|c|c|c|c|c|}
\hline \multicolumn{2}{|c|}{ Left } & \multicolumn{2}{|c|}{ Middle } & \multicolumn{2}{|c|}{ Right } \\
\hline$L$ & P1 & $M$ & & $R$ & \\
\hline$\frac{1}{2} L$ & & $M$ & P2 & $R$ & \\
\hline$\frac{1}{2} L$ & & $\frac{1}{2} M$ & & $R$ & P3 \\
\hline$\frac{1}{2} L$ & P4 & $\frac{1}{2} M$ & & $\frac{1}{2} R$ & \\
\hline$\frac{1}{3} L$ & P5 & $\frac{1}{2} M$ & & $\frac{1}{2} R$ & \\
\hline$\frac{1}{4} L$ & P6 & $\frac{1}{2} M$ & & $\frac{1}{2} R$ & \\
\hline
\end{tabular}

Since Person 1 sits at the Left table, then their potential share at the Left table is at least as large as it is at the Middle and Right tables.
Therefore, $L \geq M$ and $L \geq R$.
Since Person 2 sits at the Middle table, then their potential share at the Middle table is larger than that at the Left table (if they were equal, they would choose Left) and at least as large as it is at the Right table.
Therefore, $M>\frac{1}{2} L$ and $M \geq R$.
Since Person 6 sits at the Left table, then their potential share at the Left table is at least as large as it is at the Middle and Right tables.
Therefore, $\frac{1}{4} L \geq \frac{1}{2} M$ and $\frac{1}{4} L \geq \frac{1}{2} R$.
The inequality $\frac{1}{4} L \geq \frac{1}{2} M$ is equivalent to saying $\frac{1}{2} L \geq M$.
But this gives $M>\frac{1}{2} L$ and $\frac{1}{2} L \geq M$.
These two inequalities cannot both be true.
This means that we have a contradiction and so there are no values of $L, M, R$ that would give the seating pattern shown.
(c) Since $L=9$, the 9th person to sit at the Left table would be getting a share of $1 \mathrm{~kg}$. Since $M=19$, the 19th person to sit at the Middle table would be getting a share of $1 \mathrm{~kg}$. Since $R=25$, the 25 th person to sit at the Right table would be getting a share of $1 \mathrm{~kg}$. We note also that each additional person who sits at a table causes the share per person at that table to decrease.
Since each person sits at the table that maximizes their current share of chocolate, then there cannot be 10 people at the Left table before there are 19 people at the Middle table and 25 people at the Right table, because the 10th person at the Left table would get a larger share by sitting at the Middle or Right table.
Similarly, there cannot be 20 people at the Middle table before there are 9 people at the Left table and 25 people at the Right table, and there cannot be 26 at the Right table before there are 9 people at the Left table and 19 people at the Middle table.
In other words, we must have 9 people at the Left table, 19 people at the Middle table, and 25 people at the Right table before there are more than 9,19 and 25 people at the Left, Middle and Right tables, respectively.
At this point, there are $9+19+25=53$ people in total that have been seated.
The largest multiple of 53 less than 2019 is $38 \times 53=2014$.
We note also that $38 \times 9=342$ and $38 \times 19=722$ and $38 \times 25=950$.
The 342nd person to sit at the Left table would be getting a share of $\frac{9}{342} \mathrm{~kg}=\frac{1}{38} \mathrm{~kg}$.
The 722nd person to sit at the Middle table would be getting a share of $\frac{19}{722} \mathrm{~kg}=\frac{1}{38} \mathrm{~kg}$. The 950th person to sit at the Right table would be getting a share of $\frac{25}{950} \mathrm{~kg}=\frac{1}{38} \mathrm{~kg}$.
Using a similar argument to that above, we cannot have more than 342 people at the Left table before there are 722 people and 950 people at the Middle and Right tables. Similarly, we cannot have more than 722 people at the Middle table before there are 342 people and 950 people at the Left and Right tables, or more than 950 people at the Right table before there are 342 people and 722 people at the Left and Middle tables.
In other words, once 2014 people have been seated, there are 342 people at the Left table, 722 people at the Middle table, and 950 people at the Right table.
To determine at which table Person 2019 sits, we first determine where Person 2015, Person 2016, Person 2017, and Person 2018 sit:
\begin{tabular}{|c|c|c|c|c|c|}
\hline \multicolumn{2}{|l|}{ Left } & \multicolumn{2}{|l|}{ Middle } & \multicolumn{2}{|l|}{ Right } \\
\hline$\frac{9}{343} \approx 0.026239$ & & $\frac{19}{723} \approx 0.026279$ & & $\frac{25}{951} \approx 0.026288$ & P2015 \\
\hline$\frac{9}{343} \approx 0.026239$ & & $\frac{19}{723} \approx 0.026279$ & P2016 & $\frac{25}{952} \approx 0.026261$ & \\
\hline$\frac{9}{343} \approx 0.026239$ & & $\frac{19}{724} \approx 0.026243$ & & $\frac{25}{952} \approx 0.026261$ & P2017 \\
\hline$\frac{9}{343} \approx 0.026239$ & & $\frac{19}{724} \approx 0.026243$ & P2018 & $\frac{25}{953} \approx 0.026233$ & \\
\hline$\frac{9}{343} \approx 0.026239$ & P2019 & $\frac{19}{725} \approx 0.026207$ & & $\frac{25}{953} \approx 0.026233$ & \\
\hline
\end{tabular}

Therefore, Person 2019 sits at the left table.