Japan Junior Mathematical Olympiad

Let us denote by $S_A$ and $P_A$ the sum and product of the 3 numbers chosen by $A$, respectively, and by $S_B$ and $P_B$ those for $B$. Then the condition of the problem states that $S_A=P_B$ and $S_B=P_A$ hold. We then have $P_B=S_A\le 9+9+9\le 27$. In view of the fact that for any pair of real numbers $a$ and $b$ satisfying $a\le b$ the inequality $ab>(a-1)(b+1)$ holds, we see that if $S_B\ge 14$ then $P_B\ge 1\times 4\times 9=36$. Therefore, we must have $S_B\le 13$ since $P_B\le 27$. But this implies that $P_A\le 13$, which yields, by the same argument as above, $S_A\le 11$. We then have $P_B\le 11$. But since 11 cannot be a product of 3 positive integers, we must have $P_B\le 10$, and $S_A\le 10$. Similarly, we must have $S_B=P_A\le 10$. There are only 14 triples of positive one-digit positive integers whose sum and product are both less than or equal to 10, which are listed below: $$\begin{gathered} (1,2,5),(1,3,3),(1,1,8),(1,2,4),(2,2,2),(1,1,7),(1,1,6) \\ (1,2,3),(1,1,5),(1,1,4),(1,2,2),(1,1,3),(1,1,2),(1,1,1). \end{gathered}$$ Thus, it is enough to decide the combination of triples of numbers chosen by $A$ and $B$ from the list above so as to satisfy the condition of the problem. Suppose we denote the choice by $A$ of the triple $(x,y,z)$ by $A(x,y,z)$ and similarly for $B$ by $B(x,y,z)$, then we see that $$\begin{gathered} A(1,2,5),B(1,1,8);A(1,3,3),B(1,1,7);A(2,2,2),B(1,1,6);A(1,2,3),B(1,2,3) \\ A(1,1,8),B(1,2,5);A(1,1,7),B(1,3,3);A(1,1,6),B(2,2,2) \end{gathered}$$ are the only combinations satisfying the requirement, and therefore, we conclude that the answer we seek is 7.