This problem is proposed by Demetres Christofides, Cyprus.

Obviously some intuition would give that we can make Alice win since forcing primes is basically impossible. This is a case bash basically. We check all cases for Alice less than $8$ and we find that Bob always wins, so we conjecture that $n=8$ is the winning number for Alice and to prove this we check all the possible cases...

If Bob originally picks 3, then how does Alice win?

jonathan12345 wrote: If Bob originally picks 3, then how does Alice win? She picks 2. Now if he picks 1 Alice will have 2+4+6+8, otherwise he will pick 4, then Alice picks 5. If Bob now picks 1 Alice will have 2+5+6+8=21 and if Bob picks 6 Alice will pick 1 and have 1+2+5+8.

For $n=2$ Bob chooses $1$ and $2$ is not a composite number. For $n=3$ Bob chooses $1$, Alice must choose $2$ again which is not composite. For $n=4$ Bob chooses $3$. If Alice chooses $4$ she ends up with $5$ so she should choose $2$. Bob chooses $4$ and $3$ is not composite. For $n=5$ Bob chooses $2$ and conclusion easily follows. (You learned the method ) For $n=6$ Bob chooses $2$ and conclusion follows. (A bit more complicated ) For $n=7$ Bob chooses $2$ and conclusion follows similar to $n=6$. For $n=8$ let's exhaust all the cases. If $B$ chooses $1$ or $8$ it is obvious to see that $A$ wins. Let $B$ choose $2$. Then if $A$ chooses $3$ she wins. Let $B$ choose $3$. $A$ chooses $2$. If $B$ chooses $1$ then $A$ chooses $2+4+6+8$ which is composite. Then $B$ must choose $4$. $A$ chooses $5$. If $B$ chooses $1$ then A chooses $2+5+6+8=21$ which is composite. $B$ must choose $6$. Then $A$ chooses $1+2+5+8=16$ which is composite. Let $B$ choose $4$. $A$ chooses $3$. If $B$ chooses $2$ then $A$ chooses $3+1+6+8=18$ which is composite. $B$ must choose $5$. $A$ chooses $6$. From now on if $B$ chooses $2$, next move $A$ chooses $1$; if $B$ chooses $7$, next move $A$ chooses $8$ so sum of $A$'s choices is $18$ which is composite.

I don't understand why checking n=1,....,8 suffices, can someone explain why this is enough?

Well, checking $n=2,...7$ gives you information that Bob wins for these $n$ and this is not necessary to write in your solution. When you check $n=8$, you see that maybe Alice will win. Since choosing $n$ is part of Alice's strategy, it suffices to do a case bash and thus show that Alice always can "answer" to Bob's choice, such that in the end the number is composite.

VicKmath7 wrote: Well, checking $n=2,...7$ gives you information that Bob wins and it is not necessary to write in your solution. When you check $n=8$, you see that maybe Alice will win. Since choosing $n$ is part of Alice's strategy, it suffices to do a case bash and thus show that Alice always can "answer" to Bob's choice, such that in the end the number is composite. Can you write a complete proof (just exclude the cases n = 1,..., 8), as I am still slightly confused about the "answer" to Bob's choice part?

Ok, here is the official one To say that Alice has a winning strategy means that she can find a number n to form the set A, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such n does not exist, this would mean that Bob has a winning strategy instead. Alice can try first to check the small values of n. Indeed, this gives the following winning strategy for her: she initially picks n = 8 and responds to all possible choices made by Bob as in the list below (in each row the choices of Bob and Alice are given alternatively, starting with Bob): 1 2 3 4 5 6 7 8 2 3 1 4 5 6 7 8 2 3 4 1 5 6 7 8 3 2 1 4 5 6 7 8 3 2 4 5 1 6 7 8 3 2 4 5 6 1 7 8 4 5 3 6 2 1 7 8 4 5 3 6 7 8 2 1 4 5 6 7 3 2 1 8 4 5 6 7 3 2 8 1 4 5 6 7 8 3 2 1 5 4 3 2 1 6 7 8 5 4 3 2 6 7 1 8 5 4 3 2 6 7 8 1 5 4 6 3 2 1 7 8 5 4 6 3 7 8 2 1 6 7 5 4 3 8 2 1 6 7 5 4 8 3 2 1 6 7 8 5 4 3 2 1 7 6 8 5 4 3 2 1 7 6 5 8 4 3 2 1 8 7 6 5 4 3 2 1 In all cases, Alice’s sum is either an even number greater than 2, or else 15 or 21, thus Alice always wins.

VicKmath7 wrote: Ok, here is the official one To say that Alice has a winning strategy means that she can find a number n to form the set A, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such n does not exist, this would mean that Bob has a winning strategy instead. Alice can try first to check the small values of n. Indeed, this gives the following winning strategy for her: she initially picks n = 8 and responds to all possible choices made by Bob as in the list below (in each row the choices of Bob and Alice are given alternatively, starting with Bob): 1 2 3 4 5 6 7 8 2 3 1 4 5 6 7 8 2 3 4 1 5 6 7 8 3 2 1 4 5 6 7 8 3 2 4 5 1 6 7 8 3 2 4 5 6 1 7 8 4 5 3 6 2 1 7 8 4 5 3 6 7 8 2 1 4 5 6 7 3 2 1 8 4 5 6 7 3 2 8 1 4 5 6 7 8 3 2 1 5 4 3 2 1 6 7 8 5 4 3 2 6 7 1 8 5 4 3 2 6 7 8 1 5 4 6 3 2 1 7 8 5 4 6 3 7 8 2 1 6 7 5 4 3 8 2 1 6 7 5 4 8 3 2 1 6 7 8 5 4 3 2 1 7 6 8 5 4 3 2 1 7 6 5 8 4 3 2 1 8 7 6 5 4 3 2 1 In all cases, Alice’s sum is either an even number greater than 2, or else 15 or 21, thus Alice always wins. Oh.... I didn't realize that Alice picks n. Then it's very easy Alice just picks n=8 and wins. I thought that the problem was asking for which n, Alice wins. Sorry for bothering you and thanks for your explanation.

Is there some characterization of all $n$ for which Alice can win?

I do not think this is a mathematical problem it is just case bash and assume the winning number is $10^{2024}-1$ how can the contestant find this ?