Six real numbers so that product of any five is the sixth one

Here's an argument which extends to $n$ real numbers quite nicely! We can start by noting that $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$ And by commutativity, we get $x_i^2 = x_j^2$ which implies that all the magnitudes are equal.

Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we conclude that $L=0$ or $1$.

Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s

If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that $$\frac{\prod_{i=1}^6 x_i}{1} = 1 \text{ and } \frac{\prod_{i=1}^6 x_i}{-1} = -1$$ Now we can count configurations. There will be $$\sum_{i=0}^3 \binom{6}{2i} = 2^5$$ possibilities. And finally, we have $1+32 = 33$.

Well, if there are an even number of $1$'s and $-1$'s, then the property holds, so that's $$\binom{6}{0}+ \binom{6}{2}+ \binom{6}{4}+ \binom{6}{6}=32$$ And then there’s the case that they’re all zero. Thus, there are $33$ total cases.

Just to make the solution more explicit, the product of all six values is the square of any one of them, so they're all $\pm k$ for some $k$, and $k^6=k^2\implies k\in\{0,\,1\}$. We get one $k=0$ solution and, because we require a non-negative product, one $k=1$ solution for each choice of six $\pm 1$s with an even number of $-1$s. Famously, the numbers of even- and odd-sized subsets of a finite non-empty set are equal, and so the total number of solutions is $1+\frac12\cdot2^6$, i.e. B).