Question
Carolyn and Paul are playing a game starting with a list of the integers $1$ to $n.$ The rules of the game are:
$\bullet$ Carolyn always has the first turn.
$\bullet$ Carolyn and Paul alternate turns.
$\bullet$ On each of her turns, Carolyn must remove one number from the list such that this number has at least one positive divisor other than itself remaining in the list.
$\bullet$ On each of his turns, Paul must remove from the list all of the positive divisors of the number that Carolyn has just removed.
$\bullet$ If Carolyn cannot remove any more numbers, then Paul removes the rest of the numbers.
For example, if $n=6,$ a possible sequence of moves is shown in this chart:
\begin{tabular}{|c|c|c|}
\hline
Player & Removed \# & \# remaining \\
\hline
Carolyn & 4 & 1, 2, 3, 5, 6 \\
\hline
Paul & 1, 2 & 3, 5, 6 \\
\hline
Carolyn & 6 & 3, 5 \\
\hline
Paul & 3 & 5 \\
\hline
Carolyn & None & 5 \\
\hline
Paul & 5 & None \\
\hline
\end{tabular}Note that Carolyn can't remove $3$ or $5$ on her second turn, and can't remove any number on her third turn.
In this example, the sum of the numbers removed by Carolyn is $4+6=10$ and the sum of the numbers removed by Paul is $1+2+3+5=11.$
Suppose that $n=6$ and Carolyn removes the integer $2$ on her first turn. Determine the sum of the numbers that Carolyn removes.
$\bullet$ Carolyn always has the first turn.
$\bullet$ Carolyn and Paul alternate turns.
$\bullet$ On each of her turns, Carolyn must remove one number from the list such that this number has at least one positive divisor other than itself remaining in the list.
$\bullet$ On each of his turns, Paul must remove from the list all of the positive divisors of the number that Carolyn has just removed.
$\bullet$ If Carolyn cannot remove any more numbers, then Paul removes the rest of the numbers.
For example, if $n=6,$ a possible sequence of moves is shown in this chart:
\begin{tabular}{|c|c|c|}
\hline
Player & Removed \# & \# remaining \\
\hline
Carolyn & 4 & 1, 2, 3, 5, 6 \\
\hline
Paul & 1, 2 & 3, 5, 6 \\
\hline
Carolyn & 6 & 3, 5 \\
\hline
Paul & 3 & 5 \\
\hline
Carolyn & None & 5 \\
\hline
Paul & 5 & None \\
\hline
\end{tabular}Note that Carolyn can't remove $3$ or $5$ on her second turn, and can't remove any number on her third turn.
In this example, the sum of the numbers removed by Carolyn is $4+6=10$ and the sum of the numbers removed by Paul is $1+2+3+5=11.$
Suppose that $n=6$ and Carolyn removes the integer $2$ on her first turn. Determine the sum of the numbers that Carolyn removes.
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