Bumppppppp

Answer is 6 Let $k\geq 5$ be the number of cups left. I will show that we can destroy one more cup as long as $k\geq 5$. Strategy of $A$ is that $A$ tells $B$ to put $\left\lfloor\frac{k}{2}\right\rfloor$ to the cup that they think has the most pebbles in it, and put $k-\left\lfloor\frac{k}{2}\right\rfloor$ to the cup that they think has the second most pebbles in it, according to the info that $B$ gives previous round. Note that $B$ switches cups, so $A$ can not be certain about a cup having the most pebbles in it. Let $x_1\geq x_2\geq\cdots\geq x_k$ be the numbers $B$ said previous round. That means $A$ tells $B$ to put the pebbles to the cups that they think have $x_1$ and $x_2$ pebbles. There are 4 cases: Case 1. $B$ did not touch the cups with $x_1$ and $x_2$ pebbles. Then new situation is $(x_1+\left\lfloor\frac{k}{2}\right\rfloor-1,x_2+k-\left\lfloor\frac{k}{2}\right\rfloor-1,x_3-1,\cdots,x_k-1)$ Case 2. $B$ switches the cups with $x_1$ and $x_2$ pebbles. Then new situation is $(x_1+k-\left\lfloor\frac{k}{2}\right\rfloor-1,x_2+\left\lfloor\frac{k}{2}\right\rfloor-1,x_3-1,\cdots,x_k-1)$ Case 3. $B$ switches the cups with $x_1$ and $x_3$ (WLOG) pebbles. Then new situation is $(x_1-1,x_2+k-\left\lfloor\frac{k}{2}\right\rfloor-1,x_3+\left\lfloor\frac{k}{2}\right\rfloor-1,\cdots,x_k-1)$ Case 4. $B$ switches the cups with $x_2$ and $x_3$ (WLOG) pebbles. Then new situation is $(x_1+\left\lfloor\frac{k}{2}\right\rfloor-1,x_2+1,x_3+k-\left\lfloor\frac{k}{2}\right\rfloor-1,\cdots,x_k-1)$ Claim. In all 4 cases, the total number of pebbles in the cups with most and second most pebbles either increases or does not change. If it does not change, then the number of pebbles in the cup with the most pebbles increases. Proof. Let $a=\left\lfloor\frac{k}{2}\right\rfloor-1$, and $b=k-\left\lfloor\frac{k}{2}\right\rfloor-1$. Since $k\geq 5$, $a\geq 1$ and $b\geq 2$. For case 1 case 2, $x_1+x_2\rightarrow x_1+x_2+k-2$, so it increases. For case 3, $(x_1,x_2,x_3)\rightarrow (x_1-1,x_2+b,x_3+a)$. There are 6 subcases for the order of these. If $x_1-1\geq x_2+b\geq x_3+a$, then $(x_1-1)+(x_2+b)>x_1+x_2$ because $b\geq 2$. If $x_1-1\geq x_3+a\geq x_2+b$, then $(x_1-1)+(x_3+a)>x_1+x_2$ because $x_3-x_2\geq b-a>1-a$. For other subcases, total of top two cups also increases. For case 4, $(x_1,x_2,x_3)\rightarrow (x_1+a,x_2-1,x_3+b)$. There are 6 subcases for the order of these. Similar to case 3, since $a\geq 1$, total of top two cups either increases or does not change. If it stays the same, then $x_1\rightarrow x_1+a$, so top cup increases. Therefore, with this strategy, pebbles can be concentrated in the top two cups and it works until $k=4$, so $A$ can guarantee to destroy 6 cups.