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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
$$fracprod_i=1^6 x_i1 = 1 text and fracprod_i=1^6 x_i-1 = -1$$
Now we can count configurations. There will be
$$sum_i=0^3 binom62i = 2^5$$
possibilities. And finally, we have $1+32 = 33$.

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 $k^6=k^2implies kin0,,1$. 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 $1+frac12cdot2^6$, i.e. B).

Even though the question stated real numbers, we can extend this to complex fairly easily.

Note we require

$$ exp( left( E - I) boldsymbolz right ) = exp(boldsymbolz) $$

where $exp(boldsymbolz) = boldsymbolx$ (the vector of numbers to multiply), $E$ is a matrix of ones and $I$ is the identity matrix.

Now,

$$ exp( left( E - 2I) boldsymbolz right ) = exp(2pi i boldsymbolk) $$

for some vector of integers $boldsymbolk$. Equating the powers and using $(E - 2I)^-1 = frac18(E - 4I)$, gives

$$ boldsymbolz = fracK pi i4 - pi i boldsymbolk$$

where $K = sum_i=1^6 (boldsymbolk)_i$ and hence

$$ boldsymbolx = exp left (fracK pi i4 right) cdot exp(pi i boldsymbolk)$$

The second term is clearly a vector of 1s and -1s.
The first term can take the following values:

1. 
   

   If the total 1s in the second term are even i.e. (even number of even $k_i$),

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


2. 
   

   If the total 1s in the second term are odd i.e. (odd number of even $k_i$),

   
   
   
   • $frac1sqrt2 + frac1sqrt2i$
   
   
   • $-frac1sqrt2 + frac1sqrt2i$
   
   
   • $-(-frac1sqrt2 + frac1sqrt2i)$
   
   
   • $-(frac1sqrt2 + frac1sqrt2i)$
   
   

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

$$C = A_E cup iA_E cup (frac1sqrt2 + frac1sqrt2i) A_O cup (-frac1sqrt2 + frac1sqrt2i) A_O$$

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 $33$.

HINT:

It is obvious that you need to have the numbers absolute value equal to $1$ (besides one special case but we get to that later). The question is how to set the signs of these $1$s. The product will be negative if you have an odd number of negative signs in it and positive whenever you have an even number of negative signs. Since you are to determine signs for an even number of $1$s if you decide to give an even number of them a negative sign you have one of the following two situations

• the $1$ that is NOT in the product has negative sign. In this case there are a $texteven-1=textodd$ number of negative signs distributed along the other five $1$s which result in $-1$ so the condition is fullfilled




• the $1$ that in NOT in the product has positive sign. In this case there are a $texteven-0=texteven$ number of negative signs distributed along the five other $1$s which result in $+1$ so the condition is once again fullfilled.

So your question translates to in how many ways can you distribute an even number of negative signs among the six $1$s. This is solved in the easiest way by considering the number of ways to distribute $0,2,4$ and $6$ negative signs.

Finally the one solution not having to deal with $1$s is the one dealing with all zeros.

Hope this helped