I’ve always heard it that you have a square house with all walls facing south. Less nit-picky.
In Woeg’s solution, replace the 0 on the first cube with a 3.
How about up to 54?
Moving thread from IMHO to The Game Room.
Actually, it IS for a game.
Not that I’m admitting that I play, mind you mumble mumble mumble event horizon of geekery mumble mumble my brothers made me mumble…
…But my brothers want me to finish the layout for their new Dungeons and Dragons adventure, and they want it yesterday, and I had this lovely riddle prepared as the centerpiece. Even sign up and the OOTS forum to ask for some advice. Unfortunately, my brother also went to the OOTS forum to browse, and now I’m out a riddle. :smack: I need something that’s moderately difficult and references a physical object.
Huh?
If you start less than a mile from the North Pole and walk south, you won’t cross the pole. If you start less than a mile from the South Pole and try to walk a mile south, you can’t.
That’s not what the centroid is at all. It is the intersection of all such lines that will divide a shape into equal areas. That doesn’t mean any line through the centroid must divide it that way.
Moreover, it’s not clear whether this is a 2D problem or not; 3-dimensional missing pieces complicate the problem and even the ‘horizontal cut’ wouldn’t work.
There need to be restrictions on the missing piece for a simple solution to work.(No, I haven’t proven that but I suspect it’s true.)
There are other solutions in which you circumnavigate the South Pole a number of times. You can start one mile north of a circle 1/2 mile in circumference around the pole, 1/3 mile in circumference, 1/4 mile in circumference, etc.
OK – The cake problem:
Take the cake and attach a plumb line to it somehow. Hang plumb line such that the line and the cake are free-standing. Cut cake along the imaginary continuation of the plumb line through it. You will now have two equal portions of cake, even adjusting for vagaries in the density and thickness of the cake, and in the 3-D shape of the missing piece!
If you use the pan to hold the cake and attach the plumb line, there may be a slight risk of slightly more than half the pan being on one side and slightly more than half the cake on the other, but given the convenience it’s probably worth the small error… 
What’s with the “not to remove it from the pan”?
How else would you cut it?
Why wouldn’t you remove it from the pan?
I think he meant that the answer isn’t: Remove it from the pan and cut it horizontally.
One I swiped from a Piers Anthony book:
You have a balance scale, and twelve coins. The twelve coins are, to all outward appearances, identical. However, one of the coins is definitely a counterfeit. The only way it differs from the other coins is by weight. The counterfeit may be lighter than the other coins, or it may be heavier; you only know that the weight is different. You cannot determine the weight by touch.
In no more than three weighings on the balance scale, find the counterfeit. Note: there is no “trick”, only a method.
In this box is the method I know:
[spoiler]First weighing: 1 2 3 4 against 5 6 7 8. Note which side is heavier and which is lighter.
Second weighing: 1 2 3 5 against 4 9 10 11. Again, note which side is heavier and lighter. This is where the gimmick lies: swapping one coin from each side of the scale to the other.
Now, let’s talk about the possibilities, of which there are only three. If the same side is heavier or lighter in the second weighing, then you know that the counterfeit must be 1, 2, or 3, and you know whether the counterfeit is heavy or light. Just weigh 1 against 2 for your final weighing to find the counterfeit.
If the opposite side is heavy or light for the second weighing, you know that the counterfeit must be either 4 or 5, because those coins switched sides, but you don’t know yet if the counterfeit is heavy or light. Weigh 4 against a known good coin to find the counterfeit. If they unbalance, it’s 4; if they balance, it’s 5.
And if the second weighing results in a balance, you know that the counterfeit is one of the coins you removed (6, 7, or 8), and you know whether the counterfeit is heavy or light. For the third weighing, weigh 6 against 7.
And of course there was an additional possibility at the first weighing that it would result in a balance. If weighing number one balances, there’s no need to swap coins; just make the second weighing 1 2 3 against 9 10 11, and go from there.
Note that it’s possible to find the counterfeit without determining whether it’s heavier or lighter in certain situations. However, that doesn’t matter; the only problem the puzzle poses is “find the counterfeit”, not determining whether it’s a heavy or light counterfeit. Simply weighing the coins as groups of four will not find the counterfeit in all situations, but the method of switching places with two coins for the second weighing works no matter where the counterfeit is.[/spoiler]
I can’t picture what you’re saying at all. The cake is floating in the air? On its side? What is the plumb line attached to? Please clarify.
On its side, with the line attached somewhere along its edge (or the edge of the pan)
This is basically a usage of the technique for finding the centroid of a planar figure – where you do this twice (hang the object from two different points) and the intersection of the two lines is the centroid. This technique is most famously used, e.g., in order to find “The Heart of Texas” by using a cardboard cut-out of the state’s shape.
We don’t even need the centroid here, so just one hanging is enough – gravity says you’ll have equal masses on either side of the imaginary plumb line continuation through the object [the cake in this case] – put the cake back flat, and cut normally along the line where the plumb-line would have left a mark had you continued it
There are a few practical problems here, of course - cakes aren’t rigid bodies; attaching something isn’t easy to do; the remaining cake portions may not be connected. If frosting is considered part of the ‘cake’, then the question of what ‘equal’ means is muddled.
Except there are only two turns made, so he’s at the North Pole.
If there’s ANY sort of question regarding the color of a bear, the answer is “White, because it’s a polar bear.” Other standard answers to these questions involve ice, skydiving, or some relative (brother, sister, etc.).
Cube A: 1, 2, 3, 4, 6, 0
Cube B: 1, 2, 3, 7, 8, 0
Both the “2” and the “6” can be reversed to be the “5” and the “9”, respectively.
There were only two turns made in my example as well, although you are correct that I misstated. The man goes in 3 directions (South, East, North); he doesn’t make 3 turns.
But of course, I made it clear that because a bear is involved, it is probably at the north pole.
Start at a distance just far enough north of the south pole that if you walk a mile south and then due east, you will make a full circle around the pole in a distance of one mile. Then return north to your starting point.
It also works if your loop around the pole is 1/2 mile long, 1/3, 1/4 …
Definitely no polar bears there.
Hmmm…this doesn’t work - 2 and 5 are mirror opposites, not…transpositional(?). Maybe the 1s and 7 can somehow be combined…
You don’t need two zeros – one of the stipulations was that you can represent single-digit numbers as single digits in this case:
1, 2, 3, 4, 5, 6
1, 2, 3, 7, 8, 0
The zero only needs to be paired with the 1, 2, 3, 4 on the other cube. 6 and 9 interchange – you can get to 43 this way