A perfect circle with an infinite number of doors encloses a hexadecagon with one door per side. The Hexadecagon encloses two octagons with one door per side. Each octagon is enclosed by a cube with one door per side. All doors can be exited or entered only once, except for the cube's doors which may be entered or exited freely.

How many times can you pass from outside the circle, into the hexadecagon, into both of the octagons, into the cube, and back outside the circle. (Keep in mind the rules concerning the doors).

Now what is the cubed root of your answer. Now take this number and multiply by 2 to the power of 2. The cubed root of the number is your final answer.

The first 5 members to answer correctly will receive a coupon for $75 off their next order.

Cheers,

Chris @ Alpha

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EDIT: the cube is inside the octagon.

An infinite amount of times.

The wording is somewhat confusing - is one octagon inside the other? Are there two cubes or just one that's inside both octagons? When you say a door can only be exited or entered once - do you mean it can be exited once AND entered once, or exited OR entered but not both? Are you looking for the total number of combinations or the amount of times it can be done with one passing? lol..

Edit: Couldn't send a PM, but given what I understand from your question, I believe the answer is 2.

For what I understand cube is the 3-dimensional figure and this is about 2-dimensional figures, then you possibly meant square?

If there is a single square inside both octagons, then I'd say that the answer is 2.
But indeed I'm not sure I understood the task the way you meant it.

I suck at math, so I say 2

An explanation is necessary for why you choose the number.

1.85

The wording is a bit confusing

ok.. stop... I am drawing a picture to accurately represent the scenario. After I post this in a few minutes, we will start over.

And if every door can be entered OR exited only once, but not both, then the answer is 1.85 according to my calculator ^^

Guess I will be wrong anyway giving two answers, but lots of options make me lose my sleep
Now finally heading off to bed.

Good luck to anyone else trying this!

Suppose each door can be exited once AND entered once. The cube (square) has unlimited entries so we don't even need to count its existence. We consider the inner octagon: each of its doors can be exited and entered once and there are eight doors, hence eight total entries and exits. Every time you enter the octagon you are able to enter the cube, and when you exit you can leave the circle (since each shape outside the octagon has >= 8 doors). Hence we have that the maximum number of passes is eight. We then have ((8^(1/3))*4)^(1/3), giving the answer as 2.

Suppose instead each door can be exited once OR entered once. The answer (by the same method as above) is now 4, and the answer is then ((4^(1/3))*4)^(1/3) = 1.8517...