OK, I put this in GQ because there’s probably a simple physics answer. Here’s the riddle:
You have two spheres. They are externally identical: They have the same surface hardness, diameter, color, weight, mass, etc… They will bounce to the same height if dropped.
One of the spheres is uniformly filled with a low-density substance. The other sphere has an interior coating of a high-density substance and the center is hollow (basically, think tennis ball v. golf ball, if they were externally identical)
How do you tell which sphere is which?
I don’t know the answer, so I can’t say if the answer is correct or not
And for those of you who don’t have any idea, but like answering puzzles, here’s a bonus riddle that I do know the answer to: You sit down with the devil at a circular table which is exactly 2 foot in diameter. You each have a limitless supply of U.S. quarters, and you take turns placing them on the table. The only rules are: you can place your quarter anywhere on the table that you like, but you can’t move any quarters that are already on the table, you can’t have any overlap at all with any quarters that are already on the table, and the quarter must stay on the table by itself (i.e., if it is on the edge and falls off it doesn’t count–there’s no wind, glue, etc.). The winner is the last person who can place a quarter according to the above rules. What is the winning strategy to this game?
And a second bonus riddle: 10 prisoners are held in separate cells. On the first day of each month one random prisoner is allowed access to a common area which has nothing in it except a light bulb and light switch. The prisoners can’t see the common area from their cells, can’t be heard by the other prisoners, and have no way of leaving marks for the other prisoners. The common area is left untouched by the guards, except they will clean up any markings and will remove any left items). The warden then makes a deal with the prisoners; if they can tell him when all the prisoners have visited the common area at least once he’ll let them go, but if they’re wrong he will execute them all. What strategy should the prisoners use to determine that they have each visited the common area at least once?
Well, if sawing the balls in half is out of the question, I suppose a high pressure chamber would have different effects on them. I’d think the hollow one would contract more under high pressure, since air/void is very compressible. That being said, I was a horrible science student, so… yeah.
Expanding on adhay’s solution to problem #2, you go first, and you take the center as your first move. After that, you respond to each of the devil’s moves in the way described.
[spoiler]Not sure if this will result in the minimum time, but:
The first prisoner switches the light off.
Each subsequent prisoner will turn the light on only if it’s their first visit to the room when the light is off. (If they enter the room and the light is already on, it doesn’t count.) Otherwise they leave the light alone.
Whenever prisoner #1 visits the room again, he checks the position of the light. If it’s on, he adds a tally and switches it off.
When prisoner #1 has nine tallies, he knows all the other prisoners must have visited the room also.[/spoiler]
As for riddle #3, this may not be the most efficient solution, but:
Whichever prisoner is selected on the first day, designate him The Leader. Every time The Leader comes into the room, if the lightbulb is off, he turns it on. Every time someone else comes into the room, if it’s the first time they’ve seen an on lightbulb, they turn it off; otherwise, they leave it alone. After The Leader has turned the lightbulb on 9 times, the next time The Leader sees an off lightbulb, he tells the warden that they’ve all visited the common area.
The strategy that is easiest to describe while still being reasonably quick is probably the “single counter” strategy described above. My quick mental calculations suggest that this would take about 120 visits, so I don’t think it’s the optimal solution. Possible speed-ups include having a heirarchy of counters, or having a so-called snowball stage – an initial period of, say, 5 visits in which the light is switched on by the first person to enter the common area twice, who then becomes the counter and has a head start in the count, equal to the number of other prisoners who entered the room before his second visit (each of them entering just once). If no one enters twice in that period, the last person becomes the counter, with a head start of 5 (including himself).
I thought that, barring air resistance, rolled objects would all reach the bottom at the same time, that it was analogous to dropping them from a height, or am I mistaken?
I should have made that explicit; before being locked in their cells there’s a period of time when they can talk to each other (and then they never get a chance to talk to each other again until they’re released)
Well, a single bit of data can convey all sorts of messages. Such as “has anyone visited twice yet?” If not, then you know that you have n different visitors so far, where n is the number of months that have passed since the start.