Player 2 attempting for a common factor

When examining the situation where Player 1 was attempting for a common factor, we found that if $N$ is even, then Player 2 can force a win by pairing the numbers and removing the complement to whatever number Player 1 took. In a similar way, if Player 2 is attempting for a common factor and $N$ is odd, Player 1 can force a win by first removing one of the numbers at either end ($N_{0}$ or $N_{N-1}$), then dividing the remaining values into consecutive pairs and following the same strategy that was described previously. For instance, if the sequence 48, 49, 50, 51, 52, 53, 54 were used, Player 1 could remove 48. That would leave ( 49, 50 ), ( 51, 52 ) and ( 53, 54 ). If Player 2 removes 49, Player 1 would counter with 50, etc...

As a result of this, we need only consider the cases where $N$ is even. As before, we notice that Player 2 gets all of the turns where there are an odd number of values remaining. Using this fact, we can reuse the main theorem, replacing each occurrence of Player 1 with Player 2.

Theorem: Assume that Player 2 is trying to force the last two numbers to have a common factor. If Player 2 has enough turns to remove all values congruent to 1 and 5 mod 6, and leave at least 4 values remaining (i.e. have at least one additional turn), he can force a win.

Proof: All numbers that aren't congruent to 1 or 5 mod 6 are either even, or are congruent to 3 mod 6. Since there are at least 4 values remaining, Player 2 will at some point have a turn where there are exactly 3 numbers left (since he gets all of the odd turns). Since there are 3 numbers left, but only two categories, there must be a pair that are either congruent mod 2 or congruent mod 3, so Player 2 can simply remove the value that isn't part of the pair to win.

All that is left is to calculate minimum values of $N$ as before, and then examine what happens when $N$ is less than these minimums. Unfortunately, it is not possible to recycle the theorem we used to show that if $N_{0}$ was even, Player 1 would win. This is because now only $N/2$ values are even, but Player 1 has $N/2 - 1$ turns. Hence, Player 1 can remove all but 1 of the even numbers, possibly forcing Player 2 to lose. Unfortunately, this leads to both more equations to be solved initially as well as more special cases to be examined later.

To calculate the minimum values of $N$, we will use the same basic formula as before: Number of values congruent to 1 mod 6 + Number of values congruent to 5 + 1 extra turn $\leq$ Number of Player 2 turns. Since this was already described in detail while examining Player 1's situations, the results are presented in Table 3 without explanation.

Table 3: Minimum values for $N$ when Player 2 wants a common factor

$N_{0} \equiv 0 \bmod 6$	$N \equiv 0 \bmod 6$	$2 \times \frac{N}{6} + 1 \leq \frac{N}{2} - 1$
		$12 \leq N$
	$N \equiv 2 \bmod 6$	$2 \times \frac{N-2}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$14 \leq N$
	$N \equiv 4 \bmod 6$	$2 \times \frac{N-4}{2} + 1 + 1 \leq \frac{N}{2} - 1$
		$10 \leq N$
$N_{0} \equiv 1 \bmod 6$	$N \equiv 0 \bmod 6$	$2 \times \frac{N}{6} + 1 \leq \frac{N}{2} - 1$
		$12 \leq N$
	$N \equiv 2 \bmod 6$	$2 \times \frac{N-2}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$14 \leq N$
	$N \equiv 4 \bmod 6$	$2 \times \frac{N-4}{2} + 1 + 1 \leq \frac{N}{2} - 1$
		$10 \leq N$
$N_{0} \equiv 2 \bmod 6$	$N \equiv 0 \bmod 6$	$2 \times \frac{N}{6} + 1 \leq \frac{N}{2} - 1$
		$12 \leq N$
	$N \equiv 2 \bmod 6$	$2 \times \frac{N-2}{6} + 1 \leq \frac{N}{2} - 1$
		$8 \leq N$
	$N \equiv 4 \bmod 6$	$2 \times \frac{N-4}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$10 \leq N$
$N_{0} \equiv 3 \bmod 6$	$N \equiv 0 \bmod 6$	$2 \times \frac{N}{6} + 1 \leq \frac{N}{2} - 1$
		$12 \leq N$
	$N \equiv 2 \bmod 6$	$2 \times \frac{N-2}{6} + 1 \leq \frac{N}{2} - 1$
		$8 \leq N$
	$N \equiv 4 \bmod 6$	$2 \times \frac{N-4}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$10 \leq N$
$N_{0} \equiv 4 \bmod 6$	$N \equiv 0 \bmod 6$	$2 \times \frac{N}{6} + 1 \leq \frac{N}{2} - 1$
		$12 \leq N$
	$N \equiv 2 \bmod 6$	$2 \times \frac{N-2}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$14 \leq N$
	$N \equiv 4 \bmod 6$	$2 \times \frac{N-4}{6} + 2 + 1 \leq \frac{N}{2} - 1$
		$16 \leq N$
$N_{0} \equiv 5 \bmod 6$	$N \equiv 0 \bmod 6$	$2 \times \frac{N}{6} + 1 \leq \frac{N}{2} - 1$
		$12 \leq N$
	$N \equiv 2 \bmod 6$	$2 \times \frac{N-2}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$14 \leq N$
	$N \equiv 4 \bmod 6$	$2 \times \frac{N-4}{6} + 1 + 1 \leq \frac{N}{2} - 1$
		$10 \leq N$

Examining this table, we see that for all even $N \geq 12$, Player 2 can force a win by removing all of the 1's and 5's. As before, we need to calculate by hand what will happen for values of $N < 12$. These results are presented in Table 4. In every instance, Player 1 will win by removing the numbers specified.

Table 4: Special cases for $N < 12$ where Player 2 wants a common factor

$N = 4$	$N_{0} \equiv 0 \bmod 6$
	$\stackrel{0}{N_{0}} \stackrel{1}{N_{1}} \stackrel{2}{N_{2}} \stackrel{3}{N_{3}}$	Player 1 removes $N_{0}$
	$N_{0} \equiv 1 \bmod 6$
	$\stackrel{1}{N_{0}} \stackrel{2}{N_{1}} \stackrel{3}{N_{2}} \stackrel{4}{N_{3}}$	Player 1 removes $N_{1}$ or $N_{3}$
	$N_{0} \equiv 2 \bmod 6$
	$\stackrel{2}{N_{0}} \stackrel{3}{N_{1}} \stackrel{4}{N_{2}} \stackrel{5}{N_{3}}$	Player 1 removes $N_{0}$ or $N_{2}$
	$N_{0} \equiv 3 \bmod 6$
	$\stackrel{3}{N_{0}} \stackrel{4}{N_{1}} \stackrel{5}{N_{2}} \stackrel{0}{N_{3}}$	Player 1 removes $N_{3}$
	$N_{0} \equiv 4 \bmod 6$
	$\stackrel{4}{N_{0}} \stackrel{5}{N_{1}} \stackrel{0}{N_{2}} \stackrel{1}{N_{3}}$	Player 1 removes $N_{0}$ or $N_{2}$
	$N_{0} \equiv 5 \bmod 6$
	$\stackrel{5}{N_{0}} \stackrel{0}{N_{1}} \stackrel{1}{N_{2}} \stackrel{2}{N_{3}}$	Player 1 removes $N_{1}$ or $N_{3}$
$N = 6$	$N_{0} \equiv 0 \bmod 6$
	$\stackrel{0}{N_{0}} \stackrel{1}{N_{1}} \stackrel{2}{N_{2}} \stackrel{3}{N_{3}} \stackrel{4}{N_{4}} \stackrel{5}{N_{5}}$	Player 1 removes $N_{0}$, $N_{4}$ or $N_{2}$
	$N_{0} \equiv 1 \bmod 6$
	$\stackrel{1}{N_{0}} \stackrel{2}{N_{1}} \stackrel{3}{N_{2}} \stackrel{4}{N_{3}} \stackrel{5}{N_{4}} \stackrel{0}{N_{5}}$	Player 1 removes $N_{5}$, $N_{1}$ or $N_{3}$
	$N_{0} \equiv 2 \bmod 6$
	$\stackrel{2}{N_{0}} \stackrel{3}{N_{1}} \stackrel{4}{N_{2}} \stackrel{5}{N_{3}} \stackrel{0}{N_{4}} \stackrel{1}{N_{5}}$	Player 1 removes $N_{4}$, $N_{0}$
	$N_{0} \equiv 3 \bmod 6$
	$\stackrel{3}{N_{0}} \stackrel{4}{N_{1}} \stackrel{5}{N_{2}} \stackrel{0}{N_{3}} \stackrel{1}{N_{4}} \stackrel{2}{N_{5}}$	Player 1 removes $N_{3}$, $N_{5}$
	$N_{0} \equiv 4 \bmod 6$
	$\stackrel{4}{N_{0}} \stackrel{5}{N_{1}} \stackrel{0}{N_{2}} \stackrel{1}{N_{3}} \stackrel{2}{N_{4}} \stackrel{3}{N_{5}}$	Player 1 removes $N_{2}$, $N_{0}$
	$N_{0} \equiv 5 \bmod 6$
	$\stackrel{5}{N_{0}} \stackrel{0}{N_{1}} \stackrel{1}{N_{2}} \stackrel{2}{N_{3}} \stackrel{3}{N_{4}} \stackrel{4}{N_{5}}$	Player 1 removes $N_{1}$, $N_{5}$
$N= 10$	$N_{0} \equiv 0 \bmod 6$
	$\stackrel{0}{N_{0}} \stackrel{1}{N_{1}} \stackrel{2}{N_{2}} \stackrel{3}{N_{3}} \stackrel{4}{N_{4}} \stackrel{5}{N_{5}} \stackrel{0}{N_{6}} \stackrel{1}{N_{7}}$	Player 1 removes $N_{0}$, $N_{6}$, $N_{2}$
	$N_{0} \equiv 1 \bmod 6$
	$\stackrel{1}{N_{0}} \stackrel{2}{N_{1}} \stackrel{3}{N_{2}} \stackrel{4}{N_{3}} \stackrel{5}{N_{4}} \stackrel{0}{N_{5}} \stackrel{1}{N_{6}} \stackrel{2}{N_{7}}$	Player 1 removes $N_{5}$, $N_{1}$, $N_{7}$
	$N_{0} \equiv 4 \bmod 6$
	$\stackrel{4}{N_{0}} \stackrel{5}{N_{1}} \stackrel{0}{N_{2}} \stackrel{1}{N_{3}} \stackrel{2}{N_{4}} \stackrel{3}{N_{5}} \stackrel{4}{N_{6}} \stackrel{5}{N_{7}}$	Player 1 removes $N_{2}$, $N_{0}$, $N_{6}$
	$N_{0} \equiv 5 \bmod 6$
	$\stackrel{5}{N_{0}} \stackrel{0}{N_{1}} \stackrel{1}{N_{2}} \stackrel{2}{N_{3}} \stackrel{3}{N_{4}} \stackrel{4}{N_{5}} \stackrel{5}{N_{6}} \stackrel{0}{N_{7}}$	Player 1 removes $N_{1}$, $N_{3}$, $N_{7}$
$N= 10$	$N_{0} \equiv 4 \bmod 6$
	$\stackrel{4}{N_{0}} \stackrel{5}{N_{1}} \stackrel{0}{N_{2}} \stackrel{1}{N_{3}}... ...stackrel{4}{N_{6}} \stackrel{5}{N_{7}} \stackrel{0}{N_{8}} \stackrel{1}{N_{9}}$	Player 1 removes $N_{2}$, $N_{8}$, $N_{0}$
		Remove $N_{6}$ if $N_{1} \equiv 0 \bmod 5$
		Remove $N_{4}$ if $N_{9} \equiv 0 \bmod 5$
	$N_{0} \equiv 5 \bmod 6$
	$\stackrel{5}{N_{0}} \stackrel{0}{N_{1}} \stackrel{1}{N_{2}} \stackrel{2}{N_{3}}... ...stackrel{5}{N_{6}} \stackrel{0}{N_{7}} \stackrel{1}{N_{8}} \stackrel{0}{N_{9}}$	Player 1 removes $N_{1}$ and $N_{7}$
		Remove $N_{5}$, $N_{2}$ if $N_{0} \equiv 0 \bmod 5$
		Remove $N_{9}$, $N_{3}$ or $N_{5}$ if $N_{9} \equiv 0 \bmod 5$