TITLE: Six real numbers so that product of any five is the sixth one
QUESTION [40 upvotes]: One needs to choose six real numbers $x_1,x_2,\cdots,x_6$ such that the product of any five of them is equal to other number. The number of such choices is
A) $3$
B) $33$
C) $63$
D) $93$

I believe this is not so hard problem but I got no clue to proceed. The work I did till now.
Say the numbers be $a,b,c,d,e, abcde$. Then, $b\cdot c\cdot d\cdot e\cdot abcde=a$ hence $bcde= +-1$.
Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?

REPLY [4 votes]: Even though the question stated real numbers, we can extend this to complex fairly easily.
Let our vector of numbers be represented by $\underline{x}$ and define a new vector $\underline{z}$ such that
$$ \exp(\underline{z}) = \underline{x} $$
Now from the condition in the question, we require
$$ \exp( \left( E - I) \underline{z} \right ) = \exp(\underline{z}) $$
where $E$ is a matrix of ones and $I$ is the identity matrix. To help illustrate this, the $i$th row of the above equation states
$$ \exp \left (\left[\sum_{j=1}^6 z_j \right] - z_i \right) = \exp(z_i) $$
i.e.
$$ \frac{\prod_{j=1}^6 x_j}{x_i} = x_i $$
which is exactly the conditions required in the question.
Now from rearranging the above equation,
$$ \exp( \left( E - 2I) \underline{z} \right ) = \exp(2\pi i \underline{k}) $$
for some vector of integers $\underline{k}$. Equating the powers and using $(E - 2I)^{-1} = \frac{1}{8}(E - 4I)$, gives
$$ \underline{z} = \frac{K \pi i}{4} - \pi i \underline{k}$$
where $K = \sum_{i=1}^6 (\underline{k})_i$ and hence
$$ \underline{x} = \exp \left (\frac{K \pi i}{4} \right) \cdot \exp(\pi i \underline{k})$$
The second term is clearly a vector of 1s and -1s.
The first term can take the following values:

If the total 1s in the second term are even i.e. $K$ is even

$1$
$i$
$-1$
$-i$


If the total 1s in the second term are odd i.e. $K$ is odd

$\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$
$-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$
$-(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$
$-(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$



Finally, consider the subsets of $A = \{-1, 1\}^6$, $A_E \subset A$ containing an even number of 1s and $A_O \subset A$ containing an odd number of 1s.
Clearly $A_E = -A_E$ (multiplying each element in the set by -1) and $A_O = - A_O$, hence the unique combinations of complex numbers is the union of disjoint sets
$$C = A_E \cup iA_E \cup (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) A_O \cup (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) A_O$$
As $|A_E| = |A_O| = 32$, and the above is a union of disjoint subsets, $|C| = 128$ and including the case where all values are $0$, we have the total complex combinations as $129$.
Note
To answer the question, we know that the only subset containing real values is $A_E$ so the total combinations in the real case is $33$.