# Three-digit multiplication puzzle, part III: Return of the Hex

Place different three-digit hexadecimal numbers (000-FFF) on each of the seven nodes of the following diagram:

There are six lines in the diagram, on each of which are three of the nodes. On each of those lines, the result of multiplying the numbers on those nodes should end in 001 in hexadecimal. The same thing for the circle, on which are also three of the nodes.

Possible solutions to the previous two questions [hidden in spolier tags, in case you want to solve the previous questions first]

(For part I, the left was basically the only solution. For part II, the right was one solution, but there were others.)

Based on which you might guess

But, you would be wrong, this does not work! Can you find something that does work?

Call the corners $$a, b, c$$, the midpoints of the opposite edges $$d, e, f$$ respectively, and the center $$g$$, so that $$abf \equiv ace \equiv bcd \equiv adg \equiv beg \equiv cfg \equiv def \equiv 1 \pmod{16^3}$$. Then $$a^6 \equiv a^6(bcd)(beg)(cfg)(def) \equiv (abf)^2(ace)^2(adg)^2 \equiv 1 \pmod{16^3}$$. By symmetric reasoning (e.g. we can cycle through $$a \mapsto b \mapsto c \mapsto e \mapsto g \mapsto f \mapsto d \mapsto a$$ seven times, preserving the diagram), we also have $$b^6 \equiv c^6 \equiv d^6 \equiv e^6 \equiv f^6 \equiv g^6 \equiv 1 \pmod{16^3}$$. But there are only four different sixth roots of $$1 \pmod{16^3}$$, namely $$001_{16}, 7FF_{16}, 801_{16}, FFF_{16}$$.