ZA Furries

I sometimes require links, sometimes have links between myself and my partner. I come off with improvisations, and I can either restrain or adorn. What am I?

Cuffs

Bingo. Your turn.

Cape, this is just a reminder that it is still your turn to post a puzzle or riddle. If you do not have one, you may give your turn as riddler to another player.

You are in a pharmacy.
You notice a man walking into the pharmacy.
He whispers into the ear of pharmacist behind the counter.
The pharmacist draws a hand gun from the counter and points it at the man.
The man smiles and walks away.

What did he whisper?

He needed a laxative.

He wanted his hiccups cured?

... Personally, I like Rakuen's answer more.

Rakuen is on a mission to be gross lately.

Do with your puzzle privilege what you want Galahad.

How about a poetic riddle of my own?

Having me, I would likely be your best friend
Following you, twelve years, right to the end
Teach me, feed me, but please do not ever forsake me
For without your ownership, I would be lost in this city
Doomed to spend the rest of my days jailed in a cage
Until another claims me, or I fall to the grip of old age

What am I?

I guess if you factor in poetic licence it could be a dog.

What? Did I get something wrong?

Correct! It is indeed a dog.

Well dog life spans differ greatly, they don't all last exactly 12 years. They don't all live in the city. Sometimes they get put down if they are abandoned. But yeah... poetic licence.

That is indeed true, I suppose. Alas, as you said: poetic license. Your turn, Rakuen.

In light of several people not posting riddles/puzzles when it is their turn (due to understandable reasons), I believe it might be necessary to amend the original rules:

When someone answers a riddle or puzzle correctly, s/he reserves the first right to post his/her own. S/he may also delegate this right to post a puzzle to one of the people who attempted to answer the previous riddle (or, if s/he is the only one to have answered, s/he my choose anyone who attempted to answer the puzzle before that, excluding the person who got it right first and posted the latest riddle). However, if s/he gives no indication of answering or picking someone else after 24 hours, anyone except the previous riddler can post their own riddle/puzzle on a first-come-first-serve basis.

In other words: If you're the new riddler and don't say anything at all after one day, your right is forfeited and anyone (except the previous riddler, whose puzzle the inactive riddler answered) can post their riddle or puzzle - the first person to do so becomes the substitute riddler.

Right. Time for another puzzle!

There are TEN children in a class, and each has either a RED or a BLUE cap on his head. The teacher decides to play a game: every child closes his eyes, and the teacher then swaps around all the caps so that each child has a different one. The children then open their eyes.

Each child can see the colours of the caps of everyone else except his own.

"If you see FOUR OR MORE red caps, you get a red balloon," says the teacher, "Otherwise, you get a blue balloon."

That day, SOME children left with red balloons, and SOME children left with blue balloons.

How many RED caps were there?

The first hint, out of three, will be revealed tomorrow afternoon (in a manner so that you do not have to read it if you do not want to.)

The puzzle:

There are TEN children in a class, and each has either a RED or a BLUE cap on his head. The teacher decides to play a game: every child closes his eyes, and the teacher then swaps around all the caps so that each child has a different one. The children then open their eyes.

Each child can see the colours of the caps of everyone else except his own.

"If you see FOUR OR MORE red caps, you get a red balloon," says the teacher, "Otherwise, you get a blue balloon."

That day, SOME children left with red balloons, and SOME children left with blue balloons.

How many RED caps were there?

Here is the first hint: [CLICK]

Hint 2/3: [CLICK]

There must be four red caps. Then the six blue caps each get a red balloon because they see four red but all the red wearers can only see three red caps and get blue balloons. If there are more than four red caps then everyone will get a red balloon so it can't be that and if there are fewer than four red caps then no one will get a red balloon. Neither of those situations fit the scenario.

That is correct. Well done, Rakuen.

Hm... The participation is dwindling, and I wonder if people are still interested in playing.

I posted a puzzle, but then removed it so I could rewrite it in a more solveable form. My apologies for the inconvenience.

Here is the revised version:

Person A and person B have ordered french fries from a fast food outlet with a peculiar way of ordering: you can tell them how large or small, or thick and thin, to cut your french fries! However, they use the same amount of potato for each order.

Person A orders THIN french fries, whereas Person B orders THICK french fries - the two orders of french fries both have the same length, but different cross-sectional areas. The cooks use the same amount of potato for both orders.

The two then go to collect some tomato sauce for their french fries. If both have the habit of dipping each french fry into the tomato sauce to the same depth (i.e the same length of each french fry is covered in tomato sauce)...

Who will need more tomato sauce in the end? Person A, Person B, or will they use the same amount of tomato sauce?

Person A, because a smaller object has a greater total surface area to volume ratio, and he has a larger number of fries.

I am somewhat disappointed more did not participate - although perhaps it was because they knew Lee posted the correct answer. And correct it is! Well done, Lee, and your reasoning is sound. Would you like to post a puzzle or riddle, or would you prefer me to create another one?

I'll leave it to you, not in brain mode right now.

Very well.

The last one was rather easy. Let us step it up just a bit from easy to easy-medium:

You have a computerphile friend who loves sending you text messages. However, this rather strange friend has sent an equally strange message:

<W = Q
I = >U
O = L'
G = T.

<DD'<R
>TL'U
IN
<V>KZ'W.W.

As it turns out, it is a message that has been encrypted with a cypher. With the above information, you have enough information to decipher it.

What does the message say?

see
you
in
class

That is correct. Well done, Rakuen.

The key to solving this puzzle lies in looking at your keyboard. From the definitions in the beginning, you will notice that a < or > indicates "to the left of" or "to the right of" the letter is is next to, respectively. For example, <O will be I, because I is to the left of O.

Similarly, ' or . indicates "above" or "below" the letter it is next to.

Here is another to tide any puzzle-lovers over, while I think of a more unique and original one. (Though I nonetheless created this one as well, based on another puzzle similar to it that I saw.)

You are assigned to decrypt a three-digit passcode from a system of six clues below. Each of these clues consists of three digits - the same length as the passcode - followed by how many digits contained in that sequence are correct (part of the actual passcode), and, if there are correct digits, how many of these are in the wrong position. (Left, center or right)

4 1 5 [One correct, one in the wrong place]
3 7 4 [None correct]
5 6 1 [Two correct, one in the wrong place]
6 0 2 [Two correct, two in wrong places]
1 6 8 [One correct, none in a wrong place]
0 5 6 [Two correct, two in wrong places]

Deducing from these clues, what is the three-digit passcode?

Are you sure you didn't make a mistake? I can't get this.
If 374 has none correct and 415 has one correct then either 1 or 5 has to be correct. No matter which it is, 561 has two correct so the second correct number must be 6. Either 5 and 6 are the two correct ones or 1 and 6 are the two correct ones. But 168 has only one correct so it cannot be 1 and 6. That means based on the first three codes, 5 and 6 have to be correct. Then if 602 and 056 both have two correct then 2 must be the third correct digit. There are only six possible arrangements of these digits but every single one is ruled out.
562 - No because 561 has one in the wrong place.
526 - No because 056 has two in the wrong place.
256 - No because 056 has two in the wrong place.
265 - No because 415 has one in the wrong place.
652 - No because 602 has two in the wrong place.
625 - No because 415 has one in the wrong place.

Unless maybe I missed something somewhere but this looks impossible.

Whoops! Thank you for the notification, Rakuen.

To correct:
"5 6 1 [Two correct, none in the wrong place]"
"6 0 2 [Two correct, one in the wrong place]"

That still doesn't make sense. 562 will still be the only numbers that can be in the code but now 602 having everything in the wrong place breaks it while all the previous objections still stand... o.0

Edit: Nvm... with the second correction I definitely say 562. And it's a lot harder when the clues are wrong. >.>

Does anyone else have an answer, too?