# 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 $$6$$.

• Yes, I've seen this myself - I don't think any of the replies there is correct! Apr 7 '20 at 13:01
• Are these meant to be integers? But: The only way to get a prime $p$ as a product is $1\times p$. Thus, since $2,3,5$ are on the list of products, we need to have $1,2,3,5$ on the list of numbers, hence, that's the full list. Which does not work since $10,15$ are not among the products.
– lulu
Apr 7 '20 at 13:08
• What is the source of the problem? I expect some information has been omitted. If it's online, please provide a link.
– lulu
Apr 7 '20 at 13:10
• @lulu good point, not necessarily integers. I edited my question. Apr 7 '20 at 13:10
• "I understand the partial products which do not share a common factor"... since you aren't working exclusively with integers, "factors" are now meaningless as everything is a factor of everything else (except zero). Apr 7 '20 at 13:11

## 3 Answers

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.

• Stunning wave of thought! Really astounding idea!!! Apr 7 '20 at 14:02

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 $$\bf p$$.

Suppose the four numbers are $$0 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

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.