Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B=X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A}+P_{B}$ and find all possible equality cases.
For the maximum, we use the fact that $\left(P_{A}-1\right)\left(P_{B}-1\right) \geqslant 0$, to get that $P_{A}+P_{B} \leqslant P_{A} P_{B}+1=11!+1$. Equality holds if and only if $A=\{1\}$ or $B=\{1\}$.

For the minimum observe, first that $P_{A} \cdot P_{B}=11!=c$. Without loss of generality let $P_{A} \leqslant P_{B}$. In this case $P_{A} \leqslant \sqrt{c}$. We write $P_{A}+P_{B}=P_{A}+\frac{c}{P_{A}}$ and consider the function $f(x)=x+\frac{c}{x}$ for $x \leqslant \sqrt{c}$. Since

$$
f(x)-f(y)=x-y+\frac{c(y-x)}{y x}=\frac{(x-y)(x y-c)}{x y}
$$

then $f$ is decreasing for $x \in(0, c]$.

Since $x$ is an integer and cannot be equal with $\sqrt{c}$, the minimum is attained to the closest integer to $\sqrt{c}$. We have $\lfloor\sqrt{11!}\rfloor=\left\lfloor\sqrt{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11}\right\rfloor=\lfloor 720 \sqrt{77}\rfloor=6317$ and the closest integer which can be a product of elements of $X$ is $6300=2 \cdot 5 \cdot 7 \cdot 9 \cdot 10$.

Therefore the minimum is $f(6300)=6300+6336=12636$ and it is achieved for example for $A=\{2,5,7,9,10\}, B=\{1,3,4,6,8,11\}$.

Suppose now that there are different sets $A$ and $B$ such that $P_{A}+P_{B}=402$. Then the pairs of numbers $(6300,6336)$ and $\left(P_{A}, P_{B}\right)$ have the same sum and the same product, thus the equality case is unique for the numbers 6300 and 6336. It remains to find all possible subsets $A$ with product $6300=2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7$. It is immediate that $5,7,10 \in A$ and from here it is easy to see that all posibilities are $A=\{2,5,7,9,10\},\{1,2,5,7,9,10\},\{3,5,6,7,10\}$ and $\{1,3,5,6,7,10\}$.

Alternative Solution by PSC. We have $P_{A}+P_{B} \geqslant 2 \sqrt{P_{A} P_{B}}=2 \sqrt{11!}=1440 \sqrt{77}$. Since $P_{A}+P_{B}$ is an integer, we have $P_{A}+P_{B} \geqslant\lceil 1440 \sqrt{77}\rceil=12636$. One can then follow the approach of the first solution to find all equality cases.

Remark by PSC. We can increase the difficulty of the alternative solution by taking $X=\{1,2, \ldots, 9\}$. Following the first solution we have $\lfloor\sqrt{9!}\rfloor=\lfloor 72 \sqrt{70}\rfloor=602$ and the closest integer which can be a product of elements of $X$ is $2 \cdot 4 \cdot 8 \cdot 9=576$. The minimum is $f(576)=576+630=1206$ achieved by $A=\{1,2,4,8,9\}$ and $B=\{3,5,6,7\}$. For equality, the set with product 630 must contain 5 and 7, either 2 and 9 or 3 and 6, and finally it is allowed to either contain 1 or not.

Our alternative solution would give $P_{A}+P_{B} \geqslant\lceil 144 \sqrt{70}\rceil=1205$. One would then need to find a way to show that $P_{A}+P_{B} \neq 1205$. To do this we can assume without loss of generality that $5 \in A$. Then the last digit of $P_{A}$ is either 5 or 0. In the first case the last digit of $P_{B}$ would be 0 and so $P_{B}$ would also be a multiple of 5 which is impossible. The second case is analogous.

The computation of the expressions here might be a bit simpler. For example $9!=362880$ so one expects $\sqrt{9!}$ to be slightly larger than 600.