Riddle me this,

[Metal Bunny]

Hu-uh....

Well I'm just posting quickly now, because it's post nr 100 (I feel like those people that post: 'w00t f1rst p0st!')

Anyway, I'm going to give a hint and answer other riddles later, need some more time to think

The noise obviously upsets him, but there are many sounds, so even if you remove the knife, or remove the kung fu kick sound, there are still other sounds, so it's not very valid. In this case, as with other provided solutions, the problem is that you're focusing to much on the noise. Remove the noise, the walls and think outside the box

[Grido]

sorry to disapoint you MB but you did the 101st post, not the 100th

[Metal Bunny]

100th Reply then... detaildevil.. grmbl..

[tobajas]

A few riddles.

Riddle1

I'm a riddle in nine syllables,

An elephant, a ponderous house,

A melon strolling on two tendrils

O red fruit, Ivory, fine timber!

The loaf's big with it's yeasty rising

Money's new minted in this fat purse.

I'm a means, a stage, a cow in calf.

I've eaten a bag of green apples

Boarded the train there's no getting off.

(this is also a famous poem)

Riddle2

Black I am and much admired, men seek me until they're tired.

When they find me, they break my head, and take from me my resting bed.

What am I?

Riddle3

Whoever makes it, tells it not.

Whoever takes it, knows it not.

And whoever knows it wants it not.

[thrall]

THE 3 RIDDLE IS DEFINETLY

THE COFFIN, BECOUSE YOU DONT KNOW YOU HAVE IT IF YOU R DEAD AND NOBADY WANTS A COFFIN

THAT IS FOR THE LAST 2 LINES

BUT I DONT KNOW THE FIRST 2

[Grido]

hehe i cheated slightly on the first riddle so i wont say what it is, but i like the answer

i still dont know what the ''O red fruit, Ivory, fine timber!'' line refers too still though, will have to ponder slightly

*thinks about next riddle*

[phlegmtheorem]

ANSWER TO THIS RIDDLE

A guy gets home at 5pm and sees his wife cooking, he then takes a nap and when he wakes up (at 8pm) a superb dinner was ready to be eaten. At 11 pm they go to bed. In the morning at 6am, he killed her. Why? What happened here?

So my clue was to take each part individually, and not to assume that it was meant to be sequential. If you read it with that in mind, then you can see that... The man killed his wife at 6am. He put her in the oven. When he got home from work, she was still cooking at 5pm. When he got up from his nap at 8pm, she was done cooking, and ready to be eaten (so her ate his dead wife). His girlfriend and he went to bed at 11pm. Soooo, the man killed his wife in the morning, then ate her for dinner before bed.

Guesses for tobajas's riddles:

#1:

It's a poem about being pregnant, i'm not sure what the exact answer to the riddle is though.

#2:

The break my head part as me stumped, but I think it's about mining diamonds (they are dark until they get cut, and they are found in a bed of rock)

#3:

The last one could be a coffin like thrall said, but it doesn't make as much sense as forged docuements or fake money. (you don't want people to know it's fake, if you accept it you don't know it's fake, and if you know it's fake then you wont accept it).

[Metal Bunny]

Mmmhh, never thought about the

cannabalism perspective

Riddle 1,

She's pregnant , that's the only answer

But phlegm was first.

Riddle 2,

It's coal. The stuff you burn

Riddle 3,

Same answer as phlegm and thrall, lol, he was first

[phlegmtheorem]

Nice MB, that is a much better answer for #2.

[Trahern]

My response to Sacosphilz's crown puzzle...

Prisoner 1 responds White when Prisoner 3's crown is white.

Prisoner 2 responds White when the crowns of 1 and 3 are not the same color.

Prisoner 3 responds Black when Prisoner 1's crown is Black.

If I did this right, they should have a 5 in 8 chance, which might not be optimal, but it still better than half.

... my apologies if I missed a previous post with the answer while skimming for it.

[tobajas]

The answers to my riddles are.

Riddle 1 . Yes phlegmtheorem she's pregnant but the answer is pregnancy i will give you correct for pregnant tho.

Riddle 2. Correct Bunny it's coal and not diamonds.

Riddle 3. No it's not a coffin but phlegmtheorem you wrote the answer sort of it is fake money or rather to be precise the answer i counterfeit money.

[Sacosphilz]

My response to Sacosphilz's crown puzzle...

Prisoner 1 responds White when Prisoner 3's crown is white.

Prisoner 2 responds White when the crowns of 1 and 3 are not the same color.

Prisoner 3 responds Black when Prisoner 1's crown is Black.

If I did this right, they should have a 5 in 8 chance, which might not be optimal, but it still better than half.

... my apologies if I missed a previous post with the answer while skimming for it.

Wow

The "optimal" method has already been given in one of the posts. It is probably simpler to put into words, and has a bit better chance than your method.

That said, I'm amazed that you could come up with a completely different method that actually works and has better chance than 1/2. If this was some kind of test you'd get points for the uniqueness. :lol:

[phlegmtheorem]

Trahern, that's an awesome answer!

Here is a tricky math one. (wish I knew how to post pictures).

Draw a 5 pointed star on a piece of paper. Now make a dot in each of the points of the star. What is the sum of all 5 angles that have dots in them? What method did you use, and why?

Hint 1:

it doesn't matter if the star is lopsided or perfect, the angles will have the same summation

Hint 2:

Remember your algebra rules... the sum of the interior angles equals...

[Sacosphilz]

@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.

[Sacosphilz]

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?

[Metal Bunny]

Yeah, umh, this topic is getting a bit hectic with all the new riddles being introduced here and there. Not that there is anything wrong with it, it's just that it could get confusing. So I propose a suggestion.

Nobody is going to post a riddle when there are more than 4-ish riddles, so that each riddle gets the attention it deserves. This is just out of courtesy. As soon as there are 1-2 riddles left, people can post a bunch of riddles, for like a day or something, untill we have enough again. Rinse, reload and repeat

Right now these riddles are left.

1,2,8,9 -> 76

1,4,5,6 -> 24

1,4,7,8 -> 58

1,4,8,9 -> 74

1,6,7,8 -> 24

1,7,8,9 -> 36

2,4,6,8 -> 51

3,3,8,8 -> 24

3,4,7,8 -> 10

3,6,6,9 -> 22

That nasty train riddle from hell.

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?

Coin & scale riddles

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?

Really, it's just to get more structure into it and so that we don't overload ourselves. It's also out of courtesy, because every riddles deserves some attention untill it is solved.

People, it's just a suggestion and I'm no moderator, and you still have free will... (insert puppy eyes, which I sadly can't post, because I don't know how...)

[Sacosphilz]

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.

[Metal Bunny]

I could do that, you know, edit some of my posts (as I have the first post , w00t f1rst p0st ), but it will have to wait till weekend, doing this post as a quickie before I'm getting back to my studies again

[Trahern]

13 coins.

Let us call them ABCDEFGHIJKLM.

Weigh ABCD against EFGH

If ABCD is the same as EFGH,

Weigh IJ against KA.

If IJ is the same as KA, weigh L against A.

If L and A are the same, it's M.

If they're different, it's L

If IJ is different from KA, note the direction of IJ.

Weigh I against J, the fake is whichever goes in the same direction as IJ in the last weigh.

If ABCD is different from EFGH, note the direction of ABCD

Weigh BCDEF against IJKLM

If they're different, note the direction of BCDEF

If BCDEF moved the same way as ABCD, it's in BCD

Weigh B against C, whichever moves in the same direction as the past two is the fake.

If B and C are the same then D is the fake.

If BCDEF moved in the same direction as EFGH, then it's E or F

Weigh E against F, whichever moves in the same direction as EFGH is the fake.

If BCDEF is the same as IJKLM, then it's in AGH.

Weigh AG against LM.

If AG is the same as LM, then it's H.

If AG moves the same way as ABCD (in the first weigh), then it's A.

If AG moves the same way as EFGH (in the first weigh), then it's G.

It looks a lot prettier in my grid format... hopefully it makes sense.

And thank you guys for the kind words. I found the post with the optimal solution for the crown puzzle; that does make a lot of sense when you think of it from that angle.

[Trahern]

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?

Count out 40 coins.

Flip them over.

[Metal Bunny]

That's... I don't get it, how does that stop the 'you are not able to recognize the coins' problem?

Also I meant Wednesday, it's the day off

And of course there should be no limitation to the amount of riddles posted if you look it at that way, namely we get stuck with 5 hard riddles . But I do believe we know when an insane amount has been reached and we have to put a temporary stop, let's hope nobody spams riddles into the topic .

[Sacosphilz]

@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.

[Trahern]

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

I'd be glad to.

I started by expressing the coins as:

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

So that when you split them into two groups you have:

2. H + T = x and

3. (40-H) + (60-T) = 100 - x

From equation 2:

4. T = x - H

5. H = x - T

Given the form equation 3, I chose x=40

(x represents the number of coins in the group represented by equation 2. You could choose 60 as well I think, but since the other group is represented by 100-x, anything over 50 is semantic.)

So the revised equations are:

4. T = (40-H)

5. H = (40-T)

Which is to say the number of tails in the group of 40 is now equal to the number of heads in the group of 60.

Flipping over all the coins in the 40 group makes the tails heads and the heads tails, meeting the condition that the number of heads is 40-H in each group - H being the number of heads you initially moved into the group of 40, and can be anywhere between 0 and 40.

Hopefully that helps. If not, let me know and I'll make another try.

[Metal Bunny]

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...

[Sacosphilz]

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.