TITLE: Given $4$ variables and $5$ pairwise products, find the $6$th pairwise product?
QUESTION [9 upvotes]: 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 $6$.

REPLY [4 votes]: We have the linear system
$$
\left( {\matrix{
   1 & 1 & 0 & 0  \cr 
   1 & 0 & 1 & 0  \cr 
   1 & 0 & 0 & 1  \cr 
   0 & 1 & 1 & 0  \cr 
   0 & 1 & 0 & 1  \cr 
   0 & 0 & 1 & 1  \cr 
 } } \right)\left( {\matrix{
   {\ln x_{\,1} }  \cr 
   {\ln x_{\,2} }  \cr 
   {\ln x_{\,3} }  \cr 
   {\ln x_{\,4} }  \cr 
 } } \right) = \left( {\matrix{
   {\ln p_{\,1} }  \cr 
   {\ln p_{\,2} }  \cr 
   {\ln p_{\,3} }  \cr 
   {\ln p_{\,4} }  \cr 
   {\ln p_{\,5} }  \cr 
   {\ln p_{\,6} }  \cr 
 } } \right)\quad  \Rightarrow \quad {\bf M}\;{\bf y} = {\bf p}
$$
The rank of the coefficient matrix is $4$, and so  the rank of the augmented matrix
shall be not greater than $4$.
This pose a constraint on the values and on the order of the $p_k$.
We can have a better look to the situation if we perform a Gaussian elimination on  the augmented matrix.
We do that by performing a LU decomposition of $\bf M$ and then right-multiplying by the inverse of the L component,
and arrive to
$$
\left( {\matrix{
   1 & 1 & 0 & 0  \cr 
   0 & { - 1} & 1 & 0  \cr 
   0 & 0 & { - 1} & 1  \cr 
   0 & 0 & 0 & 2  \cr 
   0 & 0 & 0 & 0  \cr 
   0 & 0 & 0 & 0  \cr 
 } } \right)\left( {\matrix{
   {\ln x_{\,1} }  \cr 
   {\ln x_{\,2} }  \cr 
   {\ln x_{\,3} }  \cr 
   {\ln x_{\,4} }  \cr 
 } } \right) = \left( {\matrix{
   1 & 0 & 0 & 0 & 0 & 0  \cr 
   { - 1} & 1 & 0 & 0 & 0 & 0  \cr 
   0 & { - 1} & 1 & 0 & 0 & 0  \cr 
   { - 1} & { - 1} & 2 & 1 & 0 & 0  \cr 
   0 & 1 & { - 1} & { - 1} & 1 & 0  \cr 
   1 & 0 & { - 1} & { - 1} & 0 & 1  \cr 
 } } \right)\left( {\matrix{
   {\ln p_{\,1} }  \cr 
   {\ln p_{\,2} }  \cr 
   {\ln p_{\,3} }  \cr 
   {\ln p_{\,4} }  \cr 
   {\ln p_{\,5} }  \cr 
   {\ln p_{\,6} }  \cr 
 } } \right)
$$
It is evident that the system is solvable, and has a unique solution, if and only if the $p_k$'s satisfy the last two equations, i.e. if
$$
\left\{ \matrix{
  \ln p_{\,2}  + \ln p_{\,5}  = \ln p_{\,3}  + \ln p_{\,4}  \hfill \cr 
  \ln p_{\,1}  + \ln p_{\,6}  = \ln p_{\,3}  + \ln p_{\,4}  \hfill \cr}  \right.\quad  \Rightarrow \quad p_{\,1} p_{\,6}  = p_{\,2} p_{\,5}  = p_{\,3} p_{\,4} 
$$
and this gives a general method to solve this type of problems.
With the data you give, the only way to have two couples with the same product is $2 \cdot 6 = 3 \cdot 4$, and therefore
$$
2 \cdot 6 = {{12} \over 5} \cdot 5 = 3 \cdot 4
$$
which means
$$
{\bf p}^{\,T}  = \left( {\ln 2,\ln \left( {{{12} \over 5}} \right),\ln 3,\ln 4,\ln 5,\ln 6} \right)
$$
Then the upper four equations are readily solved to give
$$
{\bf y} = {1 \over 2}\left( {\matrix{
   {\ln {6 \over 5}}  \cr 
   {\ln {{10} \over 3}}  \cr 
   {\ln {{24} \over 5}}  \cr 
   {\ln {{15} \over 2}}  \cr 
 } } \right)\quad  \Rightarrow \quad {\bf x} = \left( {\matrix{
   {\sqrt {{6 \over 5}} }  \cr 
   {\sqrt {{{10} \over 3}} }  \cr 
   {\sqrt {{{24} \over 5}} }  \cr 
   {\sqrt {{{15} \over 2}} }  \cr 
 } } \right)
$$
and in fact
$$
\eqalign{
  & \sqrt {{6 \over 5}} \sqrt {{{10} \over 3}}  = 2\quad \sqrt {{6 \over 5}} \sqrt {{{24} \over 5}}  = {{12} \over 5}\quad \sqrt {{6 \over 5}} \sqrt {{{15} \over 2}}  = 3  \cr 
  & \sqrt {{{10} \over 3}} \sqrt {{{24} \over 5}}  = 4\quad \sqrt {{{10} \over 3}} \sqrt {{{15} \over 2}}  = 5\quad \sqrt {{{24} \over 5}} \sqrt {{{15} \over 2}}  = 6 \cr} 
$$
However, swapping the values of $p_1$ and $p_6$ we get a different quadruple as result
$$
{\bf x} = \left( {\matrix{
   {\sqrt {{{18} \over 5}} }  \cr 
   {\sqrt {10} }  \cr 
   {\sqrt {{8 \over 5}} }  \cr 
   {\sqrt {{5 \over 2}} }  \cr 
 } } \right)
$$
which checks as well.
I did not check for the other allowed permutations of $\bf p$.