A game involving a set $A={1,...,n}$ where the goal for one player is for the addition of numbers she chooses to be composite and other opposite. I was just doing the following problem:

Alice and Bob play the following game: Alice picks a set $A=\{1,2,\ldots,n\}$ for some natural number $n\ge2$. Then, starting with Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each step should be distinct from all the already chosen numbers and should differ by $1$ from an already chosen number. Alice wins if the sum of all the numbers that she has chosen is composite. Otherwise, Bob wins. Decide which player has a winning strategy.

I state that $k$ is the sum of the numbers which Bob picked and $l$ is the sum of the numbers that Alice picked.
We have that $k+l=\frac{(n+1)\cdot n}2$ is an invariable so $2l=n^2+n-2k$, $l=\frac{n^2-n-2k}2$
If Bob originially picked $x$ then we have that:
$k=x+(x+1\text{ or }x-1)+(x+1\text{ or }x-1\text{ or }x-2\text{ or }x+2)$
And this is as far as I got. Could you please show me how to finish off the question using the maximum amount of what I have already written?
 A: Alice wins.  I think if the problem has a solution she must win because I don't see how you could prove that Bob wins all numbers.  There could be a large number where it just works out that Alice wins.  The task is to prove that Alice can win.  The easiest way is to find some small $n$ that she can win.
The outcome of a particular game is determined when the first person picks $1$ or $n$ because there are no choices left.
Alice wins with $n=8$.
If Bob picks an end, Alice gets $16$ or $20$
If Bob picks $2$, Alice picks $3$, the other of $1,4$ from Bob and gets $18$ or $21$
If Bob picks $3$, Alice picks $2$.  Bob picking $1$ gives Alice $20$, so he must pick $4$.  Alice picks $5$, Bob $6$, Alice $1$ and gets $16$.
If Bob picks $4$, Alice picks $3$.  Bob picking $2$ lets Alice pick $1$ and get $18$.  Bob picking $5$ lets Alice pick $6$ and get $18$.
If Bob picks $5$, Alice picks $6$.  Bob $4$ and Alice $3$ give $18$, Bob $2$ and Alice $1$ give $18$
If Bob picks $6$ we have Alice $7$, Bob $5$, Alice $4$, Bob $3$, Alice $8$ getting $20$
If Bob picks $7$ Alice picks $6$, Bob $5$ Alice $8$ gets $18$.
All of the cases I gave Bob only one choice lose immediately after the other because it is an end.
