Let’s write 6 equations that must be satisfied and label them A to F:
A: x₁x₂x₃x₄x₅=x₆
B: x₁x₂x₃x₄x₆=x₅
C: x₁x₂x₃x₅x₆=x₄
D: x₁x₂x₄x₅x₆=x₃
E: x₁x₃x₄x₅x₆=x₂
F: x₂x₃x₄x₅x₆=x₁
If we take any pair of these, say B and E, and divide one by the other we get: x₂/x₅=x₅/x₂, so x₂²=x₅².
If we do this for all pairs we end up with x₁²=x₂²=x₃²=x₄²=x₅²=x₆².
So all the numbers have the same magnitude. The only difference is their sign.
If the magnitude of each is x then each number is ±x. So the arrangement of plus or minus signs is what determines how many arrangements there can be. The only x such that x⁵=x is x=1 or 0. But zero cannot be signed. However, 0 for each x is a solution. If one of the x’s is 1 all of the other x’s can be 1 also. But if there are an even number of -1’s (that is, 2 or 4) the product of 5 of them will be 1, equal to the selected x.
If we switch to the binary system where 0 represents + and 1 represents -, then we have 63 combinations of plus and minus, but we only want those with an even number of 1’s. This is sometimes referred to as parity, and there are 32 combinations of bits (binary digits) with even parity. So we have 32 arrangements for the x’s with their signs that satisfy the requirements plus the all-zero solution, making 33 in all, answer b.
