Given $4$ variables and $5$ pairwise products, find the $6$th pairwise product? 
Consider four positive numbers (not necessarily integers). The pairwise products are $2$, $3$, $4$, $5$, $6$, plus one more number.
What is the 6th product? What are the numbers?

I found this from Quora and I would be interested in a nice solution!
If we name the four numbers $x_1, x_2, x_3, x_4$ and the missing product $p_6$, then all of the possible products are:
$$x_1 x_2,\quad
x_1 x_3,\quad
x_1 x_4,\quad
x_2 x_3,\quad
x_2 x_4,\quad\text{and}\quad
x_3 x_4$$
There are six equations and five unknowns, but I don't know how to assign the six different numbers to each of them.
I understand that the partial products which do not share a common factor (for example, $x_1 x_2$ and $x_3 x_4$) should not be assigned to numbers which do have a common factor, for example $2$ and $4$, or $2$ and $6$, or $3$ and $6$.
 A: Suppose the four numbers are $0<a<b<c<d$  Note:  we can assume they are distinct since no products are repeated.
Let the six products be $P_1≤P_2≤\cdots ≤P_6$.
It is easy to see that these must be of the form $$ab<ac<\{bc,ad\}<bd<cd$$
Where the order of the two terms in the middle is uncertain.
If we assume that $P_6>6$ then we get $$\frac {P_1\times P_6}{P_2}=P_5\implies \frac {2P_6}{3}=6\implies P_6=9$$
But there are no solutions for $(a,b,c,d)$ consistent with this (brute force).
Similarly we can not have $P_1<2$.
If we had $P_2$ as the missing term then we'd get $$\frac {2\times 6}{5}=P_2\implies \boxed {P_2=\frac {12}5}$$
That one works!  Indeed we could have $$a=2\sqrt {\frac 25}\quad b=\sqrt {\frac 52}\quad c= 3\sqrt {\frac25}\quad d=\sqrt {10}$$
I did not try to analyze the other cases, though this would be no harder.
A: You can separate the six products into three pairs with each pair having different factors
$$x_1\cdot x_2\quad \ x_3\cdot x_4\\
x_1\cdot x_3\quad \ x_2 \cdot x_4\\
x_1 \cdot x_4\quad \ x_2 \cdot x_3$$
When we multiply the partial products on each line, we should get the same result.  The only two pairs that have the same product are $2 \cdot 6$ and $3 \cdot 4$, so the product of the last line must also be $12$.  The sixth partial product is $$\frac {12}5$$
Now we can check that the solution works.  By symmetry we can assign the first line $2 \cdot 6,$ the second $3 \cdot 4$ and the last $5 \cdot \frac {12}5$ but we cannot be sure of the order of the last.Then $\frac {x_3}{x_2}=\frac 32.$  If the last is $5 \cdot \frac {12}5$ then $\frac {x_4}{x_2}=\frac 52, \frac {x_1}{x_2}=\frac 54$.  The product of them all is $12$, so we have
$$\frac 32\cdot \frac 52 \cdot \frac 54 x_2^4=12\\x_2=\sqrt{\frac 85}\\
x_1=\frac 54\sqrt {\frac 85}\\x_3=\frac 32\sqrt{\frac 85}\\x_4=\frac 52 \sqrt{\frac 85}$$
If we switch the products in the last line, we get another solution
$$x_2=\sqrt{\frac {10}3}\\
x_1=\sqrt {\frac 65}\\x_3=\sqrt{\frac{15}2}\\x_4=\sqrt{\frac{24}5}$$
We can permute the assignment of the variables at will.
