Blackboard Game

The whole game is about remainders modulo $4$. Nothing else about the numbers matters.

Bob needs Alice's chosen number $a$ and his number $b$ to satisfy:

$a+b \equiv 3 \pmod 4$

That means the possible residue pairings are:

• $0$ pairs with $3$
• $3$ pairs with $0$
• $1$ pairs with $2$
• $2$ pairs with $1$

So the blackboard is really just four piles: numbers congruent to $0,1,2,3$ modulo $4$.

What happens in one full round?

Suppose Alice chooses residue $0$. Then Bob must choose residue $3$. Net effect: one $0$ and one $3$ disappear.

Suppose Alice chooses residue $3$. Then Bob must choose residue $0$. Same net effect: one $0$ and one $3$ disappear.

So for the pair $(0,3)$, a completed round always removes one from both piles. Same deal for $(1,2)$.

Bob loses immediately if Alice chooses from a pile whose matching pile is empty. For example, if there is a $0$ left but no $3$, Alice picks that $0$, Bob has no legal response, and Bob gets cooked.

Now look at a pair, say $(0,3)$.

• If the counts are equal, every time this pair is used, both counts decrease together until both hit zero. No leftover appears.
• If the counts are unequal, the larger pile will eventually have leftovers after the smaller pile is exhausted. Then Alice can pick one of those leftovers and Bob has no answer.

Therefore Bob can only win if both paired groups are perfectly balanced:

$c_0 = c_3 \quad \text{and} \quad c_1 = c_2$

where $c_i$ is the number of values from $0$ to $n-1$ with remainder $i$ modulo $4$.

For consecutive numbers starting at $0$, the residues repeat like this:

$0,1,2,3,0,1,2,3,\dots$

All four residue counts are equal exactly when $n$ is divisible by $4$.

So:

• If $n \equiv 0 \pmod 4$, Bob wins.
• Otherwise, Alice wins.

That's the entire trick. The game looks interactive, but Bob's move is basically forced by residue pairing, so the answer collapses to one modulo check.