[Riddle me this,]

@Trahern:

The first two solutions you just posted are among the few that most people managed to solve last. Interesting.

Good luck with the other 7, they should be easier.

[Riddle me this,]

@my maths puzzle:

I don't know if anyone's still working on it , but here's a hint:

All possible solutions to the solvable settings have a few things in common. If you can solve one or more of them, you can use a similar method to solve the rest.

My friends at the other forum manage to solve it already. One of them summed up the nature of the solutions quite nicely.

Exact quote:

assumptions

ใน operator สามอันที่ต้องเลือกมี / แน่นอน (ไม่งั้นคงคิดออกไปนานแล้ว)

operator ที่ทำอันสุดท้ายต้องเป็น *,/ (เพื่อให้เศษส่วนที่เกิดขึ้นระหว่างทางกลับเป็นจำนวนเต็ม)

อีก operator นึงเป็น +,- (สังเกตจากคำตอบของโจทย์ข้อที่คิดได้แล้ว)

Hmm.... what? What do you mean, you can't read it?

[Riddle me this,]

Hmm...

there are no E's?

@What am I?

A mosquito!

You got both right.

That paragraph is unusual because

it is lacking the most frequently used letter in English, E.

[Support Group for Simplyzero]

Yea I know that saco, but I never said watching a man go insane isn't fun to watch.

If you haven't noticed I'm kinda hoping for another rampant rpc (albeit he is weaker than the other rpc) as I mentioned him as a cool agent of destruction So I'm hoping for your options 2 or 3 .

I know it's cruel, but it's also awesome and the process and end result is b/funny.

Let's just say there's a reason you're joining team A and I'm joining team B.

[Riddle me this,]

Unusual paragraph:

The only article it contains is "an"; neither "a" nor "the" are used in the paragraph.

I don't know if that's what you had in mind, but it's odd enough for me to mention.

No, it's not what I have in mind, but it's very similar. Try something a bit more "

elemental

".

[Riddle me this,]

Phleg got mine right, and if you have the chance, any of you, do find the time to watch thunderbirds, tis awesome

@Saco, is it about

a book

?

I'm not sure which riddle you're referring to, but that answer doesn't fit any one of them.

If you're referring to the first riddle that is not crossed out.... again, no, because

it's a living thing.

Oh, and if you haven't figured out the crossed out "What am I?" riddle yet, shame on you. It's something very, very close to you at this very moment.

[Support Group for Simplyzero]

A crude way out:

If he suspects the mysterious double to be a shade in disguise, then dragging it into water or rolling the boat over would confirm things. I don't think it is that easy, but again, why give the 'shades can't go into water' hint?

A more subtle way out:

If the whole encounter is metaphorical like MB suggested, and that other simplyzero is really his split personality, then I don't think killing himself or his double would do him any good, that is assuming that he does want to rest now. If the original simplyzero now indeed desires peace and the double represents the urge for more aggression, then resorting to force would never end well for him - either he kills himself and the savage identity survives, or he kills the other half and confirms his violent tendency in the process. (In most fictions, killing half of yourself is never a good idea. In most cases, you either: 1. Die along with it; 2. That half really dies and you lose a sane chunk of your mind; 3. That half doesn't really die and you somehow embrace its values by 'killing' it.)

I'd go with Faraday's suggestion and look for a way to accept the other half and find a compromise.

@Lulu: Is that supposed to be japanese in the thread topic? It really looks like japanese, but I can't make any sense out of it.

[SERVER ISSUE]

Now that the super regen period has lasted this long, I hope it will last exactly 24 hours for fairness. It's painful to see those 1,000 hours worth of VE wasted away while I'm working.

EDIT: Oh, it changed back already. Nevermind then.

[Riddle me this,]

@Big Ben:

Either phlegmtheorem's answer or

the bell chimes once every 5 minutes.

(probably not )

Assuming you're already at the broken pattern gazebo and at the navigation screen, what is the minimum number of clicks required to solve the broken pattern gazebo puzzle?

... just kidding. DON'T answer it since it might be considered spoiler.

I have a lot of MP3 players but only one song. What am I?

... sorry, that's just for laughs, please read along.

A few slightly-modified-but-not-so-original riddles:

I am a distraction and a bringer of doom.

I might be behind you and lurking in your room.

Ignore me and my deeds, and you risk painful death.

Impede my and my works, and your blood will be spilled.

What am I?

This is an unusual paragraph. It is so unusual, but you might not find anything wrong with it. In fact, nothing is wrong with it, but it is still curiously unusual. Can you pinpoint what is so unusual about this paragraph?

[Riddle me this,]

Oh well, you know, not all weddings are

Christian

.

[Riddle me this,]

The groom's father?

[Hello!]

(˚д˚")

[Riddle me this,]

I think it's time I put this clunky self-made riddle to rest:

I cannot exist without people, yet people can exist without me.

I can work without people, yet people cannot work without me.

I cannot be seen, yet people see things better through me.

I cannot be heard, yet people understand me.

I cannot be touched or moved, yet people exploit me.

I am Her twin, yet people mark me as Her enemy.

What am I?

The answer (which I intended) is:

Science

Note that I play around with a few different definitions of the word.

Science is our knowledge and methodology. It is what we know about the world around us, and also what we learn to do to make the best out of each situation. We define Science, so it cannot exist without us, but we can abandon science altogether and still be able to survive with just instincts. Since Science is everything we know, we cannot work or accomplish anything without Science, but the laws and mechanisms of Science are always at work, even those unknown and undefined by us. In fact, if those laws weren't working, we wouldn't even exist.

Science is just a concept. It is intangible, so it cannot be seen, heard or touched, but it can indeed be understood and exploited, and it allows us to understand the world better.

Nature is the existence that encompass us. When we study our world, when we study out "Science", we are essentially studying Nature, thus these two concepts are twins in a sense.

But Science has become a powerful tool that enables us to disturb Nature's balance to a worrying degree. Some people blame others for the deed, some blame political or financial forces, some simply blame us humanity as a whole, and some put the blame on our tools: the existence of Technology and Science.

I'll try to be more thorough with the clues next time I compose another riddle.

On the other hand, I don't plan to spoil my other two unsolved puzzles (the 10 sets of numbers, and the 40 coins & scale) because there is no ambiguity in them. Feel free to send me PMs about those puzzles if you need help or want to declare your progress but I'm not going to publish the solutions, at least not yet

[Riddle me this,]

Yes, it's more complicated and not the same. Further more, both methods require the use of fixed-interval pauses, but while phlegmtheorem's method requires either "0 or 1" pause between the answers, mine requires "0 to N" pauses between the calls which may not practically work.

So unless someone can find another method without the need for waits, I'm convinced that phlegm's method is the best one, and my method is just a clever way to be stupid.

[Riddle me this,]

@Trahern:

Yes, I thought of the same assumption we were making too, but decided to keep it because if that assumption didn't hold true, we'd just have

a random 10-sided polygon with alternating convex and concave angles (please correct me if I'm using the wrong terminology again)

, and there would be no point in solving it...... or is there?

Example:

If the "arms" of the star were made of 5 equilateral triangles of the same size, the total angle of the "points" of the star would be 60*5 = 300, which is not 180.

If the assumption didn't hold, any answer between 0 and 540 would be possible.

@Metal Bunny:

Just when I thought we have a few good, new riddles, two of them had been solved so quickly.

By the way, just asking in advance, can I borrow some of your bunnies for some future riddles? Some riddles might be in need of some psychotic, almighty entities, and I think your bunny subjects could fit perfectly roleplay properly.

@canivorous bunny riddle:

Cheating method #1:

Starting from the backmost person, whisper the color of the hat of the person in front, then guess his or her own hat's color.

But they're sooo eaten by the bunny.

Cheating method #2:

Develop two accents to say "white" and "black". Each accent has a hidden meaning hinting at the color of the hat of the person in front of you.

But they're sooooo eaten by the bunny.

Real attempt:

During the given 24 preparation time, practice counting silently at the same speed. Also memorize each person's voice and their standing order just to be safe.

Let's number the people #1 to #20, #1 being the person in the front who can see no hats and #20 being the person in the rear who can see 19 hats.

(1) Let N be a running index starting from 20. Note that N will always be an even number.

(2) Have person #N count the number of white hats and black hats among the visible hats. There should be an odd number of hats in total.

(3.1) If he/she sees exactly 1 more white hats than black ones, call "white". The people in front of him/her will now know exactly how many white and black hats there are in total among them. Starting from person #N-1 to #1, count the hats in front of him/her and subtract from that total, then call out the remaining color, then the next person subtract that answer from the current total and repeat the same process. They will all be able to give the correct answers with absolute certainty. Go to step (4).

(3.2) If he/she sees exactly 1 more black hats than white ones, call "black". The people in front of him/her will be able to make the correct answer with the same method as in case (3.1). Go to step (4).

(3.3) Otherwise, he/she remains silent for a designated period of time (that's what the counting training was for), then subtract 2 from N and go back to step (2).

(4) If there are still people who haven't given an answer (i.e. if #20 couldn't make the answer yet), repeat the whole process from (1) with the remaining people. Otherwise, all's done. Persons #1 to #19 will surely survive, and #20 has a 1/2 chance of survival.

Some clarifications:

- During each run, one person will eventually be able to call "white" or "black", the last turn being the turn of the second person from the front -- Because there is only 1 visible hat, there is always 1 more hat of a color than another color. (1 vs 0)

- With the exception of N=20, when person #N calls "white" or "black", he/she can be sure that it's also the color of his/her hat, because if it weren't so ... that would mean there were equal number of black and white hats in #1 to #N ... which would mean that there were either exactly 1 more whites than blacks or exactly 1 more blacks then whites in #1 to #N+1 ... which would mean that person #N+2 would've already made an answer, which he/she hadn't.

EDIT: That was a very good puzzle. It took me hours and I just couldn't sleep until I got it. It's 4 a.m. and I'm still typing.

[Riddle me this,]

Because "similar" and "equivalent" are similar, but not equivalent...

Mine: The sum of the exterior angles of any polygon is 360.

Yours: The sum of the exterior angles of any ten-sided polygon is 180*(10+2) = 2160.

360 is not equal to 2160.

I'm not sure where your equation comes from, but it doesn't make much sense. It would claim that the sum of the exterior angles of a triangle would be 900, when here as well the sum is 360.

One point to keep in mind about my claim is that they're also directional angles, such that "left turns" would be negative, and "right turns" would be positive, or vice versa. I simply said "the difference" to make it sound more "human" rather than go through a description of sign notation.

For a visual representation of this, I like this example:

http://www.mathopenref.com/polygonexteriorangles.html

And for interior angles, this one is nice too:

http://www.mathopenref.com/polygoninteriorangles.html

Thank you. I get it all now after reading a bit through the links you've given. Basically, I didn't know the definition of "exterior angle", specifically the definition you were using. (That math article in the links also states that the "sum of exterior angles" could mean more than one thing.) I've never learned about that concept in school, and I didn't take my math classes in English either, so there's lots of room for misunderstanding.

By the way, in my last post, I just mistakenly presumed that the definition for "exterior angle" is the absolute angular span between a pair of intersecting line segments outside the enclosed area of a polygon which those line segments belong to.

Relax you two.

Math is one of my few ways of relaxing. No harm intended.

[Riddle me this,]

@5-point star problem, and

@Sacophilz for inferring that I don't speak human

The response Sacophilz posted seemed sound enough to me, but here's some human for ya...

The exterior of the star consists of ten angles.

The sum of the exterior angles of any polygon is 360.

The sum of the interior angles of a pentagon is 540.

Vertical angles are congruent.

The five of the ten exterior angles that are not the "points" of the star are the vertical angles of the interior pentagon, and thus their sum is also equal to 540.

The difference between 540 and 360 is 180... which is the sum of the remaining angles, the sum of the angle of the points of the star, and the answer to the problem.

You lost me at these two lines:

The sum of the exterior angles of any polygon is 360.

The difference between 540 and 360 is 180... which is the sum of the remaining angles, the sum of the angle of the points of the star, and the answer to the problem.

Umm.... how? Am I misunderstanding something?

Using a similar explanation, as I see it:

The exterior of the star consists of ten angles.

The sum of the exterior angles of any ten-sided polygon is 180*(10+2) = 2160.

The sum of the interior angles of a pentagon is 180*(5-2) = 540.

Vertical angles are congruent.

The five of the ten exterior angles that are not the "points" of the star are the vertical angles of the interior pentagon, and thus their sum is also equal to 540.

The difference between 2160 and 540 is 1620, which is the sum of the exterior angles of the points of the star.

The sum of the interior angles of the points of the star, which is the answer to the problem, would then be 5*360 - 1620 = 180

[Riddle me this,]

That puzzle is one of my favorite as well. How nostalgic.

I agree that there's no point in spoiling the answer or the method. What's important is if you can solve it and how long it takes you.

As for the method, I think you can also use paper and it won't be any harder.

[New combat system]

As a new MP4 player, my best course of action now would be:

- Ditch the hope for fight balance.

- Farm as many losses as I can. If I'm willing to lose all my creatures' vitality and all my VE, then it is very easy to keep attacking with the same damageless ritual and gaining losses from whoever is available to attacks without disturbing them, as often as the game will allow me.

- After I'm satisfied with my losses-wins difference, maybe a few hundred or a few thousands, start trying to win fights.

If everyone does the same thing, then the following will happen:

- Forget about fight balance. It won't be possible after MP3 anymore. Even if you try to raise your win counts, the 20% VE loss condition makes it so that your creatures won't last many battles until you run out of gas, but if you try to lose you can keep losing forever.

- Loss farming will be different from before in that it will rarely give unwanted wins to other players, but it will still disturb the game's balance as a whole.

- Players who manage to farm more losses will benefit more in the long run. They get attacked less and they can choose who to attack without having to worry about honor. Ironically, stronger players will have more losses since they've bean around longer than weaker players.

I haven't resort to that method on a full scale yet because it doesn't feel right, but I'm ready to do it if it will prove a point, whatever that point is.

EDIT: I'm not calling the new system good or bad, and I definitely can't compare it to the old system because I don't know enough of the old one, I'm just speculating what would happen.

[Riddle me this,]

Doh, once you said

1. H+(40-H) + T+(60-T) = 100

, I understood it...just like the coin problem and the geometrical solution to the stars, although that one was easier My mind is kind of not geared for mathematics yet, I have that exam on next week wednesday and I was going to start studying on thursday. So I haven't done anything math-like in 5 weeks...

Lol :lol: let me give it a try.

Here's how it works:

First, let's treat the original 100 coins as the first "group". Any coin we take from it will be put into the second "group".

The first group starts with 40 heads, while the second group starts empty with 0 heads. (The number of tails is not important.)

Each time you take a coin from the first group, flip it, and put it into the second group, one of these two things happen:

1. You pick up a head and flip it into a tail. The first group will end up with 1 less head, while the second group's number of heads will stay the same.

2. You pick up a tail and flip it into a head. The first group's number of heads will stay the same, while the second group will end up with 1 more head.

In other words, each time you flip a coin from the first group into the second group, the difference between the two groups' number of heads will get closer by 1.

Since the first group starts with 40 more heads than the second group (40 vs 0), if you repeat this process 40 times, their numbers of heads will become equal, although that number may be anything from 0 to 40.

I hope that reads like a human language.

["Test" Announcement 05/19/2008]

In-game player name: Sacosphilz

Preferably group B, but either group is fine.

Just a new and weak MP4.

[Riddle me this,]

@Metal Bunny:

Looking at the rate we're solving them, I do hope someone will spam more riddles in here before we run out of challenges, unless those riddles are more train bogeys from hell.

@Trahern:

You solution for scale & 13 coins is correct, although it might be a bit hard to read for those who don't already know the answer.

Judging by a certain part of your method (there is more than one solution), you might learn something more by solving the 12-coin version, and it might help you solve the 40-coin version, if you're up to the challenge.

Your method for the coin dividing problem is also right, but please do me a favor and explain to MB why it works.

[Riddle me this,]

Thanks, MB.

I agree with having some more structure in general, but I don't think there is a need to restrict the number of "active" riddles. The riddles/puzzles are of different natures and are not of equal difficulty to begin with. Limiting the number would just make us end up with a bunch of nigh-unsolvable puzzles that cannot please all of us. For instance, I know my math puzzle is hard as hell, I didn't expect anyone to be able to solve it in a short time, and it's not exactly friendly for everyone either, so I won't mind if the puzzle doesn't get a lot of attention. (Just a little sad inside ) I think it's healthier for this thread if we just post as many riddles as they come up, with some organization, that is.

As for the structure, here are a few things that could be done. These are separate suggestions and not all are necessary.

1. Number the riddles. Just simple running numbers would do, but some related ones might be tied together by their numbers. (Such as #13.1, #13.2, #13.3 for the coins&scales puzzle) This is what my friends are doing in another forum, and it helps in referencing the riddles, whether you're attempting an answer, giving hints, discussing or whatnot. The posts would then look like this:

#3

That's a classic one indeed. I think I've seen another version from a DnD game for PC.

The answer is

Nothing

With some good system for reference, you'd know where to look up the riddle itself.

2. Put a summary into the first post of this thread. It may include links to all posted riddles and our current progress.

3. Put a summary on each page or every few pages. A new page is coming in 4 posts after this one.

[Riddle me this,]

A thread similar to this one was started in another forum I frequent. I'll be translating and exchanging puzzles between these threads as long as both are active.

This is one of them:

100 identical coins are randomly scattered on a table without overlapping one another. 40 coins are showing up heads and 60 coins are showing up tails.

With your eyes closed, divide the coins into two groups so that both groups have the same number of coins that show up heads.

Assuming that you cannot memorize the coin faces before closing your eyes, and that you cannot tell heads and tails apart just by touching and "feeling it", how do you do it?

[Riddle me this,]

@5-point star problem:

(No picture either. Some imagination is required. )

There are 10 points of line intersections in a 5-point star. I'm naming them this way:

The outer points (the "tips" of the star) are A, B, C, D and E.

The inner points are V, W, X, Y and Z.

Let them be laid out in a way that the star consists of a pentagon VWXYZ in the middle and five triangular "arms": AVW, BWX, CXY, DYZ and EZV.

For the sake of sanity, I'm also going to assign the following variables to represent the value of each relevant angles.

The size of angle...

VAW = a1; AVW = a2; AWV = a3

WBX = b1; BWX = b2; BXW = b3

XCY = c1; CXY = c2; CYX = c3

YDZ = d1; DYZ = d2; DZY = d3

ZEV = e1; EZV = e2; EVZ = e3

ZVW = v; VWX = w; WXY = x; XYZ = y; YZV = z

The answer to the question would be equal to the sum of a1+b1+c1+d1+e1, which is what I'm going to find.

Let's examine the triangle AVW first.

Here are the relevant information we know about AVW triangle and the middle pentagon:

- The sum of the angles a1, a2, a3 = 180 degrees; because they are three internal angles in a single triangle.

--> a1+a2+a3 = 180

- The sum of the angles a2, v = 180 degrees; because they are on the same line.

--> a2 + v = 180

- The sum of the angles a3, w = 180 degrees; because they are on the same line.

--> a3 + w = 180

- The sum of the angles v, w, x, y, z = 540 degrees; because they are five internal angles in a single pentagon.

--> v+w+x+y+z = 540

From the above information, we can imply that:

--> v + w = (180 - a2) + (180 - a3)

----> v + w = 360 - (a2 + a3)

------> v + w = 360 - (180 - a1)

--------> v + w = 180 + a1

----------> a1 = v + w - 180

By using the above method as the model and examining the other four triangles in the same manner, we can imply that:

> a1 = v + w - 180 (from above)

> b1 = w + x - 180

> c1 = x + y - 180

> d1 = y + z - 180

> e1 = z + v - 180

and then:

> (a1+b1+c1+d1+e1) = v + w + w + x + x + y + y + z + z + v - 900

--> (a1+b1+c1+d1+e1) = 2(v+w+x+y+z) - 900

----> (a1+b1+c1+d1+e1) = 2(540) - 900 = 180

Answer: 180 degrees.