SELECTION EXAMINATION 2019

We observe that: $P_A\cdot P_B=8!=c$. By symmetry, without loss of generality, we suppose that $P_A\le P_B$, and hence $P_A\le \sqrt{c}$. We write $$P_A+P_B=P_A+\frac{c}{P_A}$$ and $P_A=x$. We consider the function $f(x)=x+\frac{c}{x}$, with $1\le x\le \sqrt{c}$. The function $f(x)$ is strictly decreasing, because for $1\le x<y\le \sqrt{c}$ it follows that $f(x)>f(y)$. Indeed, for $1\le x<y\le \sqrt{c}$, we have $x-y<0$, $xy-c<0$ and $$f(x)-f(y)=x-y+\frac{c(y-x)}{xy}=\frac{(x-y)(xy-c)}{xy}>0.(\ast )$$ Since $x$ is integer and it is not possible to be equal to $\sqrt{c}$, the least possible value is the closest integer to $\sqrt{c}$. We have $\lfloor \sqrt{8!}\rfloor =\lfloor 24\sqrt{70}\rfloor =200$, and so the closest integer is $24\cdot 8=192$. Therefore the least possible value is $$f(192)=192+210=402$$ and it can be approached, for example for the sets $A=\{4,6,8\}$, $B=\{1,2,3,5,7\}$.