Now I've got it

transparency!

Makes more sense with the "when you're with me don't beguile" line.

With my refraction answer, I was interpreting that line as a warning not to make a funhouse mirror

On the right track, for sure. Close, very close, but no cigar . . .

luminescence. Although I don't remember ever seeing a glowing muntin, I can easily imagine that such a thing exists.

refraction. But I'm not sure how alpha and gamma would fit... I don't know if gamma waves and alpha particles can even be refracted. Unless there's a naming convention where alpha is the incident angle and gamma = alpha/n is the refracted angle?

sorry, just saw the "RIGHT" answer, lol....I told you I'm new.

Lol, no biggie, I was a newbie too not too long ago

Welcome to Brainden

spedhead got this one on the nose

welcome to brainden!

please use spoilers when giving answers, not a big deal but one of our little customs

I boast leaves pruned and neatly arranged

Along with the seasons, their colors have changed

Attracting creatures with an outward display fair

They carry my seed far, hungering the fruit that I bear

Attacked by worms, rarely suffering damage dire

What I fear is consumption of me and my kindred by fire

With modernization my place in the world's shrinking

What am I now, surely this won't need too much thinking

Pineapple, slang for... GRENADE!

Then we would ALL be Shakee-ing. Nice riddle from the one and only; it was a blast. And I promise to cut out the puns now.

My tipoff was from looking up pomegranate to follow the Persephone and Hades trail, and then finding out its French name

Awww, this one doesn't spell out anything, even when I put the unique set of letters that must be used in the solution into an anagram program.

Maybe I made a mistake.

Let's see what the other puzzlers get.

[1000 people in a circle]

Congratulations guys you got it correctly .....

General case when there are n person standing in a circle.

Lemma 1 : If n = 2^k then the last man standing is the person which carries the sword initially.

In this case its person number 1. (See proof below)

Lemma 2 : In first round after x (x < n/2) person have been killed the sword is with 2x+1 person.

Now when n is not equal to 2^k then we can write n = 2^q + a, where a < 2^q .

Now as soon as a people are killed, there are 2^q person left in the circle and now the sword is with person no 2a+1 (from lemma 2).

So from lemma 1 we can say that the last man standing will be person number 2a+1.

where , a = n -2^q

Now since, 2^q < n < 2^q+1

taking log and solving for q we get,

q = floor(log₂ n)

that is the general formula.

Proof of Lemma 1 :

It can be proved via induction.

whenever there are even number of person ,then after round 1, the person 1 will get back the sword.

To prove : If n = 2^k then the last man standing is the person which carries the sword initially.

In this case its person number 1.

for n=1 it is trivial..

assume it is true for n = m

for n = m+1 ,total number of person = 2^m+1 .(an even number)

So, after the first round,there are 2^m person left and the sword will be with person number 1.

which reduces to the case of 2^m person that we had already assumed to be true.

Hence the lemma is proved.

Neat! Much more elegant proof, hardly uses any math at all (which is a good thing).

[1000 people in a circle]

This is definitely not the most eloquent proof ever, but...

Define S(n) as the solution to the problem if you have n swordsmen. That is, you're asking us to find S(1000).

Suppose n is even. Then after one round, all of the even numbered people would be dead and you would have swordsman #1 holding the sword again. The remaining people would be numbered {1, 3, 5, 7, ...} and the size of the group would be n/2. So, if you knew what S(n/2) was, then you could figure out what S(n) was. Suppose S(n/2) was person #3: then looking at the set above, we would see that the person who was #3 in the group of size n/2 would have been person #5 in the group of size n. In general, person #i in the small group would have been person #(2i-1) from the big group. So S(n) = S(n/2) x 2 - 1. Remember, all this is assuming that n is even.

Suppose n is odd. Then after one round, all of the even numbered people will still be dead, and as the last (odd numbered) person takes the sword he will kill #1 and person #3 will start off with the sword on the next round. The remaining people will be numbered {3, 5, 7, 9, ...} and the size of the group would be (n-1)/2. So this time, if you knew what S(n/2-1) was, you could figure out what S(n) was. Person #i in the small group would have been person #(2i+1) from the big group. So S(n) = S([n-1]/2) x 2 + 1. That's if n is odd.

Now take the case of S(1000)

S(1000) = S(500) x 2 - 1

S(500) = S(250) x 2 - 1

S(250) = S(125) x 2 - 1

S(125) = S(62) x 2 + 1

S(62) = S(31) x 2 - 1

S(31) = S(15) x 2 + 1

S(15) = S(7) x 2 + 1

S(7) = S(3) x 2 + 1

S(3) = S(1) x 2 + 1

S(1) = 1 (obviously)

So, S(3)=3, S(7)=7, S(15)=15, S(31)=31, S(62)=61, S(125)=123, S(250)=245, S(500)=489, S(1000)=977

Now to simplify that some to make this a general case solution. It would actually be easier if we write the numbers in binary, like S(11001).

Doubling any binary number is done by simply adding a zero to the end (just like multiplying by 10 in base 10). So if you know S(11001), then S(110010) = S(11001) x 2 - 1.

Likewise, multiplying by two and adding one is done by simply adding a one to the end, so if you know S(11001), then S(110011) = S(11001) x 2 + 1.

Now for any number we can start building it from S(1)=1. To get to S(11001) we would do:

S(1) = 1

S(11) = S(1)x2+1 = (in binary) (1)x2+1 = 10+1 = 11

S(110) = S(11)x2-1 = (in binary) (11)x2-1 = 110-1 = 101

S(1100) = S(110)x2-1 = (in binary) (101)x2-1 = 1010-1 = 1001

S(11001) = S(1100)x2+1 = (in binary) (1001)x2+1 = 10010+1 = 10011

Notice what's happening: S(n) will always end in a 1 at the rightmost spot (er, be odd). That's because it's always calculated as S(previous)x2+1 or S(previous)x2-1.

Adding a 0 to the end of the n in S(n) entails performing the operation S(n)x2-1 on an odd binary number, call it abcde...xyz1. Doubling it gives you abcde...xyz10, and then subtracting 1 from it makes it abcde...xyz01. It's exactly the same as putting a zero in the next-to-last spot.

On the other hand, adding a 1 to the end of the n in S(n) entails performing the operation S(n)x2+1. In binary, multiplying by two and adding 1 is just the same as adding a 1 to the end of the number. Using the notation above, abcde...xyz1 goes to abcde...xyz11, you could also say it's the same as adding a 1 to the next-to-last spot.

Now consider the arbitrary number S(1abcdefg) where abcdefg are binary digits

S(1) = 1

S(1a) = a1

S(1ab) = ab1

S(1abc) = abc1

etc.

Since 2^n in binary is 100000... with n zeros, MGH's step 4 of taking the remainder after dividing by the largest available 2^n is just the same as removing the leading 1, leaving you with the abcdefg

Then MGH's step 5 of multiplying by 2 and adding one is the same as adding 1 to the end of the binary number, giving you abcdefg1, the solution above.

(Jeez that was tough. Good problem.)

Unlike diamonds that will scorch or burn when heated, zirconium can be melted into a glass bead. Then solve a puzzle with crystalline geometry to facet the stone such that it will produce a brilliant display when struck by light.

* Edit: grammar

A burnt tree with moss all over it wouldn't be totally bizarre. Moss can in fact most readily grow on a tree that had been dead for a long time, say, one that had been in a forest fire or struck by lightning. And a dead tree would easily fall, either spontaneously or if someone were to try to climb it.

Like Royal said, the weird thing is that they ask you where you live instead of where you are. The Gorden, MS seems important since your sidebar thing says that isn't where you're from. It looks like Gordon is a TINY town (basically in name only) out in the middle of nowhere. Neither Google maps nor Mapquest know of Cherry Blvd there, so I'm assuming that the street isn't important.

lol I had tried arranging "John Dou" as a possible anagram for a clue.

And I was trying to figure out why they were doing an autopsy to determine the cause of death when the guy had a pitchfork sticking out of his back

How did the Easter Bunny get the headline?

Sounds more like the work of a cute psycho bunny...

Is this the right forum for this anymore, I think the new puzzles forum was split. This should probably go in New Word Puzzles

For some reason I'm thinking Ender

So that's where all the WAIs went.

I'd say Magna Carta

[Eat Poop You Cat V]

Jumping in...

1. andromeda

2. Zerep

3. GMaster479

4. Chidoric

5. FIF

6. Cyberspace

7. IDNE

8. scsw

9. Grayven

10. Y-san

11. CB

12. allherdanger

13. woon

14. Marth

15. Kay

16. plasmid

Backup:

1. LIS

Too bad I'm gonna have to watch all the giggling without being able to see the pics though.

Is this the answer:

Hit the two moles at a time in the following order: 2 & 4, 1 & 3, 3 & 4. This combination can go in an endless loop ensuring that regardless where the mole comes up, it will be under one of the mallets. To begin the round, if the mole is coming up 1 or 3, hit 1& 3 together. If the mole is coming up 2 or 4, hit 2& 4 together, then continue the sequence. If the mole is coming up 5, then hit 5 by itself, then the next turn hit 3 & 4 together.

You got it! That sure was quick, we haven't even had any wiseguys post some non-logical answers yet. But puzzle II in the series is still out there waiting to be solved.

Continuing from How would you cross puzzle land? and How would you cross puzzle land? Part II, (which by the way still has a solution that hasn't been found).

You have crossed the river of lava with your cash and stroll into the city, and you find that the carnival has come to town, complete with carnival games. But before you get too engrossed in them, you notice a particularly cute person by the name of Sam (I'll let the reader choose whether that's short for Samuel or Samantha) observing one of the games. You walk up and ask "Have you tried this game out?" Sam replies "No, I'm a little suspicious of this one. I think I'll see how some other people do first." You're faced with a carny running a shell game. The carny hides a marble under one of three shells and shuffles them around at breathtaking speed. His challenge: "Care to guess which shell holds the marble? I'll even sweeten the deal: I'll turn over an empty shell after you pick, and you can change your choice if you like." You give him a short stare through squinted eyes as if you're sizing him up, and say "All right, but I'm gonna call forth some supernatural spirits to help me with this one." You then close your eyes and start chanting "Hum, baya baya baya", and start a strange dance that could easily be confused with a neurological disorder. After flailing and spinning around a bit, you say "sorry, the spirits couldn't help me out" and walk off without playing the game.

Sam seems a little put off, "What's the deal with all that 'channeling spirits' mumbo-jumbo?" You reply, "cover, for me to bump the table with my hips while I was doing that stupid looking dance. When I bumped it, I didn't hear a marble rattling under any of the shells, so I walked away." Important point: for those of you who haven't seen the first two in this series, these problems can often be attacked by both logic and, um, less sophisticated wit. Even more important point: your explanation drew a smile from Sam. "I saw some other games that seemed more legit earlier on if you want to try them out" Sam offers. "Sure thing, lead the way."

At the next stand, Sam notices a piece of onyx on the shelf of prizes and asks the carny "What would we need to do to win that?" "For that," the carny replies, "you'll need to get a perfect score on the whack-a-mole game." You take a look at the fair-goers who are currently taking a whack at it, and you're immediately struck by how fast the mole is. Unless you're a drummer, you won't be able to win this game with reflexes alone; you'll have to anticipate where the mole will be. "Let's win that game!" Sam urges, and you sure don't want to disappoint.

PUZZLE TIME: The machine has five holes that the mole can emerge from, numbered 1-5, and two mallets that you and Sam can use for mole whacking. There will only be one mole emerging from one hole at a time. From watching the previous players, you notice that the game starts with some music and a pseudo-random flashing light display that will actually give away the first hole the mole will emerge from. From that point on, the mole will appear from subsequent holes based on which holes were whacked with mallets on the previous turn.

If only one hole gets whacked on a turn, the machine will calculate 1) the number of the hole that was whacked minus one, and 2) the number of the hole that was whacked minus two. If either of those numbers is <1, it will add 5 to that number to bring it in the range of 1-5. The next mole with then jump out from either of those two holes.

If two holes are whacked at the same time, the machine will calculate both 1) the sum and 2) the product of the holes that are whacked. If either of those numbers is >5, it will keep subtracting 5 from that number until it falls in the range of 1-5. The next mole with then jump out from either of those two holes.

If the same set of holes is whacked twice in a row, the game will forego the above rules and the mole will jump out wherever it pleases.

Finally, since a carny is watching you play the game, you're not likely to win a prize if you resort to any obvious shenanigans. Each mallet can only take care of one mole-hole at a time. Remember that you need a perfect score, so no missing any moles. Have at it!

You don't need any surgical gloves at all. Non-sterile exam gloves will do just fine for that exam.

Total number of gloves used is n+1.

Take one glove and designate it the "bum glove". Take 2n of the doctors and put them in pairs that will each share a glove - the first one wears the pair's glove with the bum glove on top for the exam, the second doc takes the pair's glove and turns it inside out so his hand is facing the unused surface before putting the bum glove over it and examining. After the 2n docs have gone, the +1 doc can do his exam with the (still unused) inner surface of the bum glove.

You could use a similar approach to cover the subset of case 3 where there are an even number of both doctors and patients. Pair them off (pairs of doctors and pairs of patients) that will each share a glove. Each doctor uses his/her surface of the exam glove, and covers it with the appropriate surface of the patient's glove for the exam that's about to be performed. Total number of gloves used would be:

patients/2 + doctors/2

If both the number of docs and the number of patients are odd, then remove one doctor and one patient from the group and handle the remaining even number of doctors and patients as above. The odd ones out also get to split a glove between the two of them: the doctor uses the inside surface of the odd glove out and covers it with the other patients' gloves for all of their exams. The odd patient out uses the other surface for all of his exams with the other doctors wearing their gloves on the inside. The odd pair out can do their exam with just the odd glove out and nothing in between. Total gloves is still:

(patients + doctors) / 2

That still leaves the question open: if there are an odd number of patients and even doctors (or vice versa), can you do it with fewer than (patients + doctors + 1) / 2 gloves? And also, did you have something different in mind, 'cause the OP makes the p <= d stipulation for case 3 that wouldn't seem to make any difference with this set of solutions.

[Pirates of the Caribbean - the sequel]

This "solution" is a little more elaborate but I think it still works. I hope my crummy diagram helps you visualise.

Consider a chest that is held closed by 4 rods. The placement of the 4 rods is such that two run from one side of the chest to the other (parallel to one another) and the other two rods are positioned diagonally between the end points of the first 2 rods. The end of each rod is secured by one of the locks and each lock secures the end of 2 rods.

(1) (2)
|\ /|
| \ / |
| \ / |
| \/ |
| /\ |
| / \ |
| / \ |
|/ \|
(3) (4)

Each rod is secured in place by 2 locks. Unlocking any 3 locks enables the pirates to remove all of the rods and thus open the chest. For example. Opening locks 1 and 3 enables the lefthand rod to be removed. Opening lock 2 allows the diagonal to be removed. Lock 4 is still locked to the remaining rods but since these rods have one free end (since the other locks have been opened) these rods can simply be slipped out, still attached to lock 4. This works regardless of which pirate is missing.

I may need to clarify this some more.

I agree with Romulus' assessment, but that's an interesting approach that inspired another solution. Such creative people here at brainden!

The treasure chest would need to have a "lip" sticking out from the bottom of the lid and the top of the base. If you drill holes in the lip and thread a chain through, you could put the locks on the chain's links as shown in this picture, and the chain could only be removed and the chest opened when 3/4 pirates remove their locks.

BOW TO MY MAD PAINT SKILLZ

[Pirates of the Caribbean - the sequel]

Maybe the chest is just the top, placed on the bottom part. Put 1 lock at each side and when 3 people unlock it, the fourth lock with just act as sort of like a pivot point to open it. There doesn't have to be any "master key" or anything.

Got it, Avalanche!

I was also trying to think of a way to use all 12 edges of a cube in a problem like this, but that just got too complicated for me.

[Pirates of the Caribbean - the sequel]

Carification: by "one key for each lock" I mean that there are four keys which will each open one lock, not that there's a single master key that opens all of the locks.

[Pirates of the Caribbean - the sequel]

As I was pondering Pirates of the Carribean, I thought of some approaches that were probably off topic but nonetheless interesting. Suppose you have four pirates and want to make a treasure chest that will open when any three decide to open it. You only have four locks, and one key for each lock. How would you do it?

[How would you cross puzzle land? Part II]

Maybe this hint will open some more possibilities.

Although I'm looking for a solution that's mostly logic-based, that doesn't mean that it's completely without tactics that are less than scholarly.

[Pirates of the Caribbean]

Just put one lock on it and give me the key. I'll make sure the treasure is nice and safe.

Oh, wait, this isn't one of my puzzles, is it.

1 - Fallout (from nuclear tests)

2 - Traffic

3 - Collaboration

4 - Negligence

5 - Eno (Brian)

6 - [A fragmented plate that says Australia,] (apparently not a tectonic plate: Indo-Australian doesn't start with O, Okinawa & Okhotsk are too far away)

7 - [They might just cry when they see a Monet, (art athletes?)]

8 - Serf (surf, dude)