TITLE: Solution to simple mathematical game
QUESTION [28 upvotes]: Consider the following game (that I made up). Two players each attempt to name a target number. The first player begins by naming 1. On each subsequent turn, a player can name any larger number that differs from the current number by a divisor of the current number, including itself. For example, if the current number were 6, the next number named could be any one of 7, 8, 9, or 12. A player that names the target number wins. A player forced to name a number larger than the target number loses.
If the target number is odd, the first player can win trivially by always naming an odd number, because the difference between the current number and the target number is even, and no odd number has an even divisor. The thing I can't seem to figure out is whether the first player also has a winning strategy for every even number greater than 6, and what that strategy is if so. Obviously, one can determine the optimum course of play for any target number by working backward, and doing so seems to suggest the first player can always win, but I haven't been able to understand why.
Is there any way to prove that the first player can always win for any number greater than 6?

REPLY [5 votes]: I think that every even target greater than $6$ is indeed a first player win. I have no proof but present evidence and speculation on what might lead to a proof. 
For every even target at least up to $10000$ there is a  strategy which is  $2k \rightarrow 2k+1$ except for a small list of exceptions.
For example

Target $T=9580$

Strategy: The other player will always state even numbers. Our response should be $2k \rightarrow 2k+1$ with these four exceptions $$4786 \rightarrow 7179,\ 7184 \rightarrow 7633,\ 9100 \rightarrow 9105,9572 \rightarrow 9574,\ 9578\rightarrow 9580$$


Let me justify that. Consider the sets $$W=\{9574,9580\}$$ $$L=\{4787,7185,9101,9573,9575,9579\}$$
These are the even numbers which it is a winner to state and the odd numbers which it is a loser to state. If the opponent never states a member of $W,$ they can't win. But those are $T=2^2\cdot 5\cdot 479$ and $9574=2\cdot 4787 .$
If we never state the odd numbers $9579,9575,9101,7185=T-1,T-5,T-479,T-2395$ or $4787=9574-4787,$ then they can't state anything from $W$ , unless perhaps we state an even number. We could state $9580$ and win. And if we state $9574=2\cdot 4787$ they can only reply $9576$ to which we can reply $9577 \notin L.$ So we just need responses other than $2k \rightarrow 2k+1$ for $2k+1 \in L.$ So preferably $2k \rightarrow 2k+t\lt T$ where $t \gt 1$ is an odd divisor of $k$ and $2k+t \notin L$. And we have done this for four of the five cases.For $9573-1=9572=2^2\cdot 2393$ there is no such $t$ to use. Fortunately, $9572 \rightarrow 9574$ allows us to state a winner.
OPTIONAL DIGRESSION: Here are three more examples

target $T=2^j$:

Strategy: $2k\rightarrow 2k+1$ except $2^k-2 \rightarrow 2^k.$ So $W=\{2^k\}\ \ L=\{2^k-1\}.$ 

target $T=2p$ for $p$ prime but not a Fermat prime $p=2^s+1.$

Strategy:  $2k \rightarrow 2k+1$ except $2p-2 \rightarrow 2p$ and $p-1 \rightarrow p-1+t$ where $t \gt 1$ is the next smallest odd divisor. So $W=\{2p\}\ \ L=\{p,2p-1\}.$


Since $p-1$ is not a power of $2,$ there is such an odd divisor.

target $T=2p$ for $p$ a Fermat prime $p=2^s+1:$

Strategy:  $2k \rightarrow 2k+1$ except $2p-2 \rightarrow 2p$ and $p-3 \rightarrow p-3+t$ where $t\gt 1$ is the next smallest odd divisor (so also $t \gt 3.$). here $W=\{p-1,2p\}\ \ L=\{p,2p-3\}.$


Since $p-1=2^s$ it is actually an even winner as the only responses are $p$ and even losers. 
END OF DIGRESSION
Question which arise are:

How do we come up with the (or a) valid strategy?
How can we rule out $1 \in L?$
How large might the set of exceptions be?


The methods for an even target $T=2n$ is to start $W=\{2n\}$ and $L=\{\}$ where $W$ will be the even winners and $L$ will be the odd losers. 
While there are unexamined members of $W \cup L$, consider the largest such.

If it is $2k \in W$, then add $2k-t$ to $L$ where $t$ ranges over the odd divisors of $k$. Then pick the next unexamined member.
If it is  $2f+1 \in L$ then either find a good response to $2f$ or, if there is none,  add $2f$ to $W$: Consider $2f+t$ where $t$ goes through the divisors of $2f$, As soon as you get an even winner in $W$ or and odd number not in $L$, add that to your list of exceptional rules. If none of those happen, add $2f$ to $W.$ Then pick the next unexamined member.

Stop when everything currently in $W \cup L$ has been examined.

The exceptions, given an even target $T \gt 6$, are $W,$ the even numbers it is a winner to state and $L$ the odd numbers it is a loser to state. For $\mathcal{T}$ a set of even targets (such as all even integers greater than $6$), One way to prove that every $T \in \mathcal{T}$ is a first player win is to show that every $T \in \mathcal{T}$ has smallest exception $e \gt 1$.
A stronger result might be easier. I would conjecture that $e \gt \frac{T}{4}.$ 
For even $6 \lt T \lt 10000$ there are $114$ cases where $\frac{e}{T} \lt \frac{15}{41}.$ Here are the first few pairs $[T,e]$ if we sort according to increasing ratio $\frac{e}{T}$
$$[114,31],[10,3],[30,9],[462,151],[130,43],[450,151],[1254,421],[2058,691],[1250,421],[3654,1231],[2050,691],[2530,853],[3650,1231],[4930,1663],[5730,1933],[8454,2851],[8450,2851],[9250,3121],[9570,3229]$$
So $\frac{e}{T} \gt \frac14$ (up to $T=10000$) and $\frac{e}{T} \gt \frac13$ for $T \gt 130.$ On the other hand, there are $T$ on the edge of the examined range with $\frac{e}{T} \lt 0.337$ 

There are $12$ cases for $4 \leq T \leq 10000$ with $2$ exceptions, The powers of $2.$
However there are $1285$ with exceptional set of size $3$, all cases of $T=2p$ and some, but not all, $T=2^sp$ for larger $s.$
Here are the first few counts: $$[2, 12], [3, 1285], [4, 28], [5, 609], [6, 187], [7, 60], [8, 525].$$
 The largest are $72,75,80,83,84,94,97$ which occur once each.
Here is a cumulative plot

So just under $\frac23$ of the cases have exceptional sets of size $12$ or less and over $90\%$ have exceptional set of size $25$ or less. 

You can just say that the target is the largest number which can be stated and the first player without a move loses. I didn't consider the case of first player without a move wins, but it should be largely similar.