
7. Consider the set $X=\{1,2,3, \ldots, 9,10\}$. Find two disjoint nonempty subsets $A$ and $B$ of $X$ such that

(a) $A \cup B=X$;

(b) $\operatorname{prod}(A)$ is divisible by $\operatorname{prod}(B)$, where for any finite set of numbers $C, \operatorname{prod}(C)$ denotes the product of all numbers in $C$;

(c) the quotient $\operatorname{prod}(A) / \operatorname{prod}(B)$ is as small as possible.


Solution: The prime factors of the numbers in set $\{1,2,3, \ldots, 9,10\}$ are $2,3,5,7$. Also only $7 \in X$ has the prime factor 7 . Hence it cannot appear in $B$. For otherwise, 7 in the denominator would not get canceled. Thus $7 \in A$.

Hence

$$
\operatorname{prod}(A) / \operatorname{prod}(B) \geq 7
$$

The numbers having prime factor 3 are $3,6,9$. So 3 and 6 should belong to one of $A$ and $B$, and 9 belongs to the other. We may take $3,6 \in A, 9 \in B$.

Also 5 divides 5 and 10. We take $5 \in A, 10 \in B$. Finally we take $1,2,4 \in A, 8 \in B$. Thus

$$
A=\{1,2,3,4,5,6,7\}, \quad B=\{8,9,10\}
$$

so that

$$
\frac{\operatorname{prod}(A)}{\operatorname{prod}(B)}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{8 \cdot 9 \cdot 10}=7
$$

Thus 7 is the minimum value of $\frac{\operatorname{prod}(A)}{\operatorname{prod}(B)}$. There are other possibilities for $A$ and $B$ : e.g., 1 may belong to either $A$ or $B$. We may take $A=\{3,5,6,7,8\}, B=$ $\{1,2,4,9,10\}$.

