Riddle Needs Solved....Now!

Oops, I messed up in the above explanation of Sailor’s Delight but I think the gist of it applies. The fifth step should still be 2,0,0,98,0. Sorry 'bout that.

Enlightenment extra credit (four light switches):

Turn on switches 3 and 4 for a few minutes. Turn off switch 3, turn on switch 2, and go to the room. If the lamp is off and cold, it’s switch 1. If it’s on and cold, it’s switch 2. If it’s off and warm, it’s 3. If it’s on and warm, it’s 4.

Actually, though, I should restate the solution… please ignore the previous post.

[spoiler]12; 0,100
123; 1,0,99
1234; 1,0,0,99
12345; 1,0,0,99,0
123456; 1,0,0,0,1,98
123567; 1,0,0,0,1,98,0
12345678; 1,0,0,0,1,0,1,97
123456789; 1,0,0,0,1,0,1,97,0
1234567890; 1,0,0,0,1,0,1,0,1,96

I was very inconsistently applying the thought of “bloodthirstiness”. However, extra credit for bloodthirstiness

12; 0,100
123; 1,0,99
1234; 2,0,0,98
12345; impossible, pirate 5 cannot offer a bid that will quench greed over bloodthirstiness.[/spoiler]

I like panamajack’s Eat and Run answer - I think that’s the one they were looking for.

For the No Vacancy/ies problems, I understand the solutions you gave, but I don’t see why my answer is wrong. I can even give a nerdier solution that lets an infinite number of groups of infinity into the hotel. Put the first group in prime-numbered rooms (2,3,5,…). Put the second group in rooms that are 2 times a prime number (4,6,10,…). Put the 3rd group in rooms that are 3 times a prime number greater than 2(9,15,21,…). Put the nth group in all the rooms that are p(n) times a prime number greater than or equal to p(n), where p(n) is the nth prime(p(n)p(n),p(n)p(n+1),p(n)p(n+2),…). The nth group has one room for each prime, except for the first n-1 primes, which is an infinite # of rooms for any n.
I can’t see why my answer would be wrong, and I can’t see exactly where the other answers have made an incorrect assumption. I’ll even acknowledge that your answer was the one that they were looking for. But I still think I’m right. Look at it this way:[spoiler]Let’s use coins instead of rooms, because I think it’s easier to see it that way. An equivalent problem with coins is, there are an infinite # of coins, and an infinite # of them are heads-up. Is it possible to make one more heads up (no vacancy)? Is it possible to make an infinite # more heads up (no vacancies)? Now, suppose you have an infinite # of heads-up coins, and 0 tails-up coins. Can you make one more coin heads up? No. But this fits the description posed in the problem. Now, you say that you can order all the coins (1,2,3…). I’ll accept that. Then you “move each person up a room” by transferring the state of the nth coin to the n+1th coin. Now the 1st coin is free, and that can be your extra heads. But you really didn’t change anything. None of the coins flipped. So you didn’t add another heads.

To say it another way, all you have is n occupied rooms. Can you fit one more person into a room, without having double-occupancy? You move into room 1, and have that person move into room 2, and that person moves into room 3… If n is finite & each person move to the next highest room, one person will be bumped out of the highest numbered room & he’ll be homeless. Since n is infinite, you don’t have a highest numbered room, so you never have a homeless person, you just have perpetual bumping. There’s always someone out of a room, moving into the next room, so you really haven’t fit anyone extra in.[/spoiler]

Flight Plight is entertaining:

5 planes (I was incorrect earlier), though explaining it without a diagram is tricky. Think of the equator as divided up into six equal segments. Since a fully-fuelled plane can fly halfway around the world, a plane burns one-third of its fuel per segment. The planes are labelled P1 through P5 and the Checkpoints (at the borders between segments) are labelled C1 (home base) through C6. The sequence plays like this, assuming it takes one hour for a plane to travel the length of one segment. The table shows the location of each plane and how much fuel it has after refuels and transfers. “3/3”, “2/3”, “1/3” and “0/3” shows the quantity.



0 hour:
       C1     C2     C3     C4     C5     C6
P1     3/3
P2     3/3
P3     3/3
P4     3/3
P5     3/3
1 hour:
       C1     C2     C3     C4     C5     C6
P1            2/3
P2            2/3
P3            2/3
P4            2/3
P5            2/3
P1 transfers a third of its fuel to P2.
P3 transfers a third of its fuel to P4.
Result:
       C1     C2     C3     C4     C5     C6
P1            1/3
P2            3/3
P3            1/3
P4            3/3
P5            2/3
2 hours:
       C1     C2     C3     C4     C5     C6
P1     0/3
P2                   2/3
P3     0/3
P4                   2/3
P5                   1/3
P1 and P3 refuel.
P2 transfers a third of its fuel to P5.
P4 transfers a third of its fuel to P5.
Result:
      C1     C2     C3     C4     C5     C6
P1    3/3
P2                  1/3
P3    3/3
P4                  1/3
P5                  3/3
3 hours:
      C1     C2     C3     C4     C5     C6
P1           2/3
P2           0/3
P3           2/3
P4           0/3
P5                         2/3

P1 transfers a third of its fuel to P2
P3 transfers a third of its fuel to P4
Result:
      C1     C2     C3     C4     C5     C6
P1           1/3
P2           1/3
P3           1/3
P4           1/3
P5                         2/3
4 hours:
      C1     C2     C3     C4     C5     C6
P1    0/3
P2    0/3
P3    0/3
P4    0/3
P5                                1/3

P1 refuels and will now travel in the *opposite direction* around the world, going to C6 instead of C2.

5 hours:
      C1     C2     C3     C4     C5     C6
P1                                       2/3
P2    0/3
P3    0/3
P4    0/3
P5                                       0/3

P1 transfers a third of its fuel to P5
Result:
      C1     C2     C3     C4     C5     C6
P1                                       1/3
P2    0/3
P3    0/3
P4    0/3
P5                                       1/3

By the sixth hour, P1 and P5 reach C1, wth P5 having circled the Earth.


I think Lazy Susan is the only riddle we haven’t answered yet.

Step 1: turn over opposite corners.
Step 2: turn over any glasses that are next to each other.
Step 3: turn over opposite corners.
Step 4: turn over any glass.
Repeat steps 1, 2, and 3.
I don’t know if this is the best solution. I think it is though.

The original question, about the locks, has a practical application in cryptography. A lot of advanced codes have as their key an enormous (say 100-200 digits) number. So you want to send one of these gigantic numbers to someone so they can decode your message. You take this number, and you have an enormous prime number (of which there are lots and lots), unknown to your correspondent. You multiply these numbers together, to get a product that has maybe 200-400 digits. You send this number to your buddy.

Your buddy multiplies this number by ANOTHER enormous prime and sends the result back to you. You factor out your prime number and send back to your buddy. The idea here is that anyone can read this number in transmission and unless they know one of your two huge prime numbers, they can’t decode the original number.

Oh, I should mention that these numbers are large enough that it would take a computer many many months to figure out what your prime might be. Or at least that was the case when I heard about this, maybe five years ago. Technology has probably simplified this quite a bit since then, but I’m not sure. Still it’s pretty interesting stuff. To an English major that is.

Here is my answer for Flight Plight, using a similar strategy to Bryan but getting a different result.

[spoiler]I think it is possible using only 4 planes.
0 hour:

P1 3/3 (C1)
P2 3/3 (C1)
P3 3/3 (C1)
P4 3/3 (C1)

1 hour:
P1 2/3 (C2)
P2 2/3 (C2)
P3 3/3 (C1)
P4 3/3 (C1)

P1 transfers a third of its fuel to P2.
Result:
P1 1/3 (C2)
P2 3/3 (C2)
P3 3/3 (C1)
P4 3/3 (C1)

2 hours:
P1 0/3 (C1)
P2 2/3 (C3)
P3 3/3 (C1)
P4 3/3 (C1)

P1 refuels.
Result:
P1 3/3 (C1)
P2 2/3 (C3)
P3 3/3 (C1)
P4 3/3 (C1)

3 hours:
P1 2/3 (C6)
P2 1/3 (C4)
P3 2/3 (C6)
P4 3/3 (C1)
P1 transfers a third of its fuel to P3
Result:
P1 1/3 (C6)
P2 1/3 (C4)
P3 3/3 (C6)
P4 3/3 (C1)

4 hours:
P1 0/3 (C1)
P2 0/3 (C5)
P3 2/3 (C5)
P4 3/3 (C1)
P1 refuels.
P3 transfers a third of its fuel to P2.
Result:
P1 3/3 (C1)
P2 1/3 (C5)
P3 1/3 (C5)
P4 3/3 (C1)
5 hours:
P1 2/3 (C6)
P2 0/3 (C6)
P3 0/3 (C6)
P4 2/3 (C6)
P1 transfers a third of its fuel to P2
P4 transfers a third of its fuel to P3
Result:
P1 1/3 (C6)
P2 1/3 (C6)
P3 1/3 (C6)
P4 1/3 (C6)
By the sixth hour, all planes reach C1, wth P2 having circled the Earth.

[/spoiler]

The typical answer to the tractor problem is that

they waited until Winter and drove it over the frozen lake.That’s probably the answer he’s looking for but it doesn’t tie in real well with his storyline.

Holy shit, Nighttime. That’s one spoiler.

I believe the answer to the original question can be summed up more easily by saying your friend cuts the locks off when they get there.