Given four positive numbers: $a, b, c, d$. Among the products $a b, a c, a d, b c, b d, c d$, we know the values of five of them, which are 2, 3, 4, 5, and 6. What is the value of the sixth product?
I. solution. Let's assume the value of $a b$ is missing. We know that $a d \cdot b c = a c \cdot b d = a b \cdot c d = a b c d$. Since among the products, $a c$ and $b d$, as well as $a d$ and $b c$ can also be found, they must yield equal values when multiplied. Let's write down all the numbers that can be formed from the pairwise products of 2, 3, 4, 5, 6: $6, 8, 10, 12, 12, 15, 18, 20, 24, 30$. Only 12 appears twice, so $a d \cdot b c = a c \cdot b d = 2 \cdot 6 = 3 \cdot 4 = 12$.

Since 5 remains without a pair, $c d = 5$. Therefore,

$$
a b = \frac{a b c d}{c d} = \frac{12}{5}
$$

is the sixth product.

II. solution. Let the unknown value of the product be $a b$. Then

$$
a c \cdot a d \cdot b c \cdot b d \cdot c d = \frac{a^{3} b^{3} c^{3} d^{3}}{a b} = 720, \quad \text { so } \quad a b = \sqrt{\frac{720}{c^{3} d^{3}}}
$$

We know the values of $a c$ and $b d$, and $a c \cdot b d = a b \cdot c d$, so $a b \cdot c d$ can be obtained as the product of two known values:

$$
a b \cdot c d \in \{6 ; 8 ; 10 ; 12 ; 15 ; 18 ; 20 ; 24 ; 30\} = A, \quad c d \in \{2 ; 3 ; 4 ; 5 ; 6\}
$$

If $c d = 2$, then

$$
a b = \sqrt{\frac{720}{c^{3} d^{3}}} = \sqrt{90}, \quad c d \cdot a b = 2 \sqrt{90} = 6 \sqrt{10} \notin A \quad \Rightarrow \quad c d \neq 2
$$

If $c d = 3$, then

$$
a b = \sqrt{\frac{720}{c^{3} d^{3}}} = \sqrt{\frac{80}{3}}, \quad c d \cdot a b = 3 \sqrt{\frac{80}{3}} = \sqrt{240} \notin A \quad \Rightarrow \quad c d \neq 3
$$

If $c d = 4$, then

$$
a b = \sqrt{\frac{720}{c^{3} d^{3}}} = \sqrt{\frac{45}{4}}, \quad c d \cdot a b = 4 \sqrt{\frac{45}{4}} = \sqrt{180} \notin A \quad \Rightarrow \quad c d \neq 4
$$

If $c d = 5$, then

$$
a b = \sqrt{\frac{720}{c^{3} d^{3}}} = \sqrt{\frac{144}{25}} = \frac{12}{5}, \quad c d \cdot a b = 5 \frac{12}{5} = 12 \in A \quad \Rightarrow \quad c d \text { can be } 5
$$

If $c d = 6$, then

$$
a b = \sqrt{\frac{720}{c^{3} d^{3}}} = \sqrt{\frac{10}{3}}, \quad c d \cdot a b = 6 \sqrt{\frac{10}{3}} = \sqrt{120} \notin A \quad \Rightarrow \quad c d \neq 6
$$

Therefore, the value of $c d$ can only be 5. If

$$
a = \frac{2}{\sqrt{\frac{10}{3}}}, \quad b = \frac{4}{\sqrt{\frac{10}{3}}}, \quad c = \sqrt{\frac{10}{3}}, \quad d = \frac{5}{\sqrt{\frac{10}{3}}}
$$

then $a b = \frac{12}{5}, b c = 4, a c = 2, c d = 5, a d = 3, b d = 6$. Therefore, we can provide four numbers such that the conditions of the problem are satisfied. The value of the sixth product is $\frac{12}{5}$.