To moriarty: You have a point, so I hope the clarification below would help.

1. All the letter I in Lithium are crossed out.
2. The word "hEEL" is intentional and has no error. For those who used the client, it spelled like this: a small letter h, two capital letter E, and a capital letter L.
3. The four-digit code is not from the numbers found in the table of elements or the position of letters in the english alphabet.
4. The code could already be seen in the room if put in the right angle.

The answer will be revealed after 12 hours! I will post another riddle, maybe, a math riddle next! More guesses please!

@imprisonedindarkness hmm, it's just a guess, but could it be as simple as 1334?

thanks everyone for these riddles, i'll post some more when we are done with these

We didn't have a riddle for a while, so here is one from me.
A blind guy finds himself on a deserted island. The rescue will come in three days, but meanwhile, in the two spare-ones, he needs to take pills to stay alive.
In a bag, he has two white and two red pills. They're all of identical shape and texture, indistinguishable by touch.
Each day, he needs to take one white and one red pill. Taking a wrong combination could result in death.
What should he do?

Giving me a study in scarlet vibes.
Anyway, what about smell, or tasting one, then the other, and matching it. don't take it, just pop it into your mouth briefly or taste it with the tip of your tongue.

Below is the answer of my escape riddle above.

To moriarty: Here's my guess. For the first pill, the blind dude can just fish a pill from inside the bag. At this moment, it doesn't matter whether the pill is white or red. The second pill is the tricky part. There's a 33 percent that he may take out the wrong pill, but it beats dying, right? So as long as he gets the right pill (66 percent probability), he'll live, but he'll die otherwise. By the way, is he alone in that island? If so, how did he know the colors of the pills at the first place? The riddle didn't say.

Hello everyone! Sorry for not posting the answer these few days. Got busy, but here's the answer.
The four-digit code is ... 7334.

Here are the explanation for the clues.
For the clue pertaining to the list of three elements. It's actually half misleading and half pointing you to the right answer. Tungsten, aluminum, and lithium when converted to their symbols turns into W, Al, and Li. Since all the I's in the lithium are crossed out, you have to do the same with its symbol, thus, L. When you read the symbols altogether, you'll get the word: wall. It means you have to solve the clues on the wall and nothing else.
For the clues on the wall, Moriarty got it. Since we want to go out, we have to turn the arrow pointing to the bomb 180 to the direction of the door. The "hEEL" will turn upside down due to that. So the upside down upper case L looks like the number 7, the two upper case E looks like the number 33, and the lowercase h looks like the number 4, thus, 7334.

@imprisonedindarkness nice-one! Wall honestly didn't come to my mind, I considered the symbols, but somehow missed it.
@MisterKaaminari & @imprisonedindarkness
Interestingly, there is a way to get 100% certainty without the need to smell or taste the pills.
I'm not sure how did the guy managed to get there, perhaps he fell of from a boat. But yes, he's alone, and the only thing he knows is he got two white and two red pills, and there is no way to tell them apart.

To MisterKaaminari: What's the answer to your riddle above? Is it a merlock?

To moriarty: I can't see how to get the hundred percent certainty in this. Can you give more clues? Haha, my brain hurts.

@moriarty I believe the only way is to break all 4 pills in half and then he can take one half from each pill. is this allowed, or are the pills not to be tampered with. if he breaks the pills in half, he wil take two halfs of the white pill, making it one white pill and two halfs of red pill making it 1 red pill.

You are paid thirty cents by your best friend, using just two coins. One is not a nickel, what are the two coins that total thirty cents?

this is quite popular. one is not, but the other one is

Yes it's murlock hahah.

To sky360: Nice job! This is a good answer. By halving them, you will get two halves of the white ones and two halves of the red ones, which totals into one white and red pills, which answers the riddle perfectly!

To MisterKaaminari: Haha! I lucky-guessed it! When I read the riddle, it made me thought of the game Hearthstone. That's why I answered that. Haha!

To GodSaveTheQueen: I don't know your currency. Can you give the convertion, like from the smallest coin to the largest ones and how many units of the smallest coin when converted to the next higher one?

@sky360 spot on!
Yes, all he needs to do is to break the pills in halves, either manually, or using knife/whatever sharp object is available, then eat the one half of the result and save the other for the next day.
Personally, I'm always fascinated thinking about this riddle, even when I already know the solution. You're implicitly dealing with a problem, distinguishing the pills, which is unsolvable, but do something smart that suddenly makes it irrelevant and gets you to the goal.

@moriarty hmmm. I feel you, some times you will have to think out of your limits to find reasonable solutions. this answer was not so hard to come by for me how ever, because here in India we get higher doze pills for low cost, so we are used to breaking pills in half and taking them. so I thought what will happen if he breaks the pills in half, then it was just putting 2 and 2 together. btw I have a riddle too!
A bat and ball combo costs one dollar and 10 cents, the cost of the bat is a dollar more than the ball. what is the individual cost of the bat and the ball. looking forward to your answers. cheers, sky

My new riddle is written below.
To sky360: The bat is a dollar and five while the ball is five scents?
Before we start with the riddle, I just want to warn you that it's a visual one, so I'll just describe the scenario and you can do it at home. Here's another riddle from me.

The Riddle:
You have a pen and a paper. On the paper, there are nine dots in a three-by-three arrangement, which means that on the first row, there are three dots from left to right, another three dots directly below the first row, and another three dots just directly below the second row. Thus, they create a square formation. By using four connected lines, you have to connect all the nine dots without leaving a single one. How do you do this?

To make it easier to explain the answer, let's assign each dot a corresponding number, so it should look like this:

1st Row: 1 2 3
2nd Row: 4 5 6
3rd Row: 7 8 9

Example Answer: First line starts from 1, down to 4, down to 7. Second line starts from 7, right to 8, right to 9. Third line starts from 9, up to 6, up to 3. Four line starts from 3, left to 2, left to 1. Since dot 5 isn't included, this answer is wrong. Or you can explain your answer in another way as long as each line and dot are clearly stated.
Note: For screen reader user that doesn't understand my English above, view this in the web. I'm not sure if I explain the whole thing correctly, so better to explore it bit by bit using screen reader navigation.
Have fun everyone!

@imprisonedindarkness hmm, interesting. Is there a definition for what "line" means, and perhaps what exactly is meant under "connected"?
For example, if connected referred to any kind of connection between lines, we could draw something in style 1-2-3, 4-5-6, 7-8-9, 1-4-7. But since this is pretty obvious, I assume it's not the case.
I guess it refers to connected by ends, where each line has to start from the ending of the previous-one i.e. you're supposed to connect everything in one drag.
Here, the question is, what exactly is a line. After a bit of thinking, I have doubts it's possible to connect everything using four perfectly straight lines, i.e. lines where you're not allowed to turn a single degree, every turn is counted as a new line.
Though, if at least 45-degree turns were allowed, the solution could be as follows:
1-5-8, 8-6-3, 3-2-4-7, 7-8-9.
No line in this solution turns more than 45 degrees, so the turns can be viewed as curves, and there is 4 of them, possible to draw in one go.

Below are the definition of the terms, 'line' and 'connected', to my last riddle above.
To Moriarty: There's a perfect answer actually. My mom gave me this riddle way back in elementary, and unfortunately, I didn't get the answer after a week of headache. Hahaha! She got it from a parent seminar and told it to me since it was interesting. And of course, since I like such puzzles.

Clarification of Terms:

1. A line in this riddle means the straight one, no curviture allowed from point A to point B.
2. Connected means you have to do it in one drag as Moriarty aptly stated, so first line is point A to point B, and the second line is from point B to point C, and so on and so forth.

If there's more question about the riddle above, just ask away and don't be shy!

Hey, I think I got it!
First line, 2-3-3', where 3' is a fake point located on the place where 4 would be, if the grid size was 4x4.
Second line, 3'-6-8-7', where 7' is a fake point located at where 13 would be if if grid size was 4x4.
Third line, 7'-7-4-1.
Fourth line, 1-5-9.
;)

Note: A chess Riddle is written at the end of this post.

To Moriarty: Hahaha! You got it right! Congrats!
In the original text, the answer is expressed like this:
First line: 1-2-3-A, where A is a point next to 3;
Second Line: A-6-8-B, where B is a point below 7;
Third Line: B-7-4-1, and the last;
Fourth Line: 1-5-9.

Since we have finished with this riddle, let's have another one. My next riddle is a chess riddle, so here it goes!

The Riddle:
You are in the endgame of chess. You have four pieces left on the board,: a white knight at a2, a king at e1, a rook at h1, and another rook at h2; while your opponent only has his king at b1. You move your white knight to c3 and check the king at b1. Your opponent moved his king to c1 as a response to your check. Now, you have to checkmate your opponent's king in just one move. How would you do it?

The board should look like this:

-- a3, -- b3, WN c3, -- d3, -- e3, -- f3, -- g3, -- h3.
-- a2, -- b2, -- c2, -- d2, -- e2, -- f2, -- g2, WR h2.
-- a1, -- b1, BK c1, -- d1, WK e1, -- f1, -- g1, WR h1.

The letter-number combination is the coordinates of the square.
The two dashes are the empty squares.
The letter W before another letter is the white, while the letter B is black.
N is knight.
K is king.
R is rook.

And that's it. Good luck to all!
Hahaha! I'm about to run out of riddles in this rate. Hahaha!

@imprisonedindarkness thanks! Interestingly, the most difficult part for me was getting beyond the idea of connecting the dots using just the dots themselves, even though there was indeed no such limitation in the task.
I think it can be shown it's impossible, because the straight line restriction means each line can be at most three points long. With the condition of having them connected, it means in four lines, three can cover each at max two unique points, what implies you can reach 9 if and only if you use three dots for every line you draw and don't touch any already covered point, if you use a two-point line or cross something you've already covered, you're not getting to 9 in 4 lines.
Now, considering the symmetries of the square, there are just about 4 unique variations you can draw using only three-dot lines, which are easy to manually review and see it's not possible.
Someone familiar with the graph theory would probably laugh at my proof haha, I'm not sure if we do have this subject somewhere along the study, I need to check. But I think it's valid.
After having this certainty and getting myself to believe I didn't make a mistake anywhere, it was not that hard to find the turn points you can use outside of the formation and form the pattern, that took me just few minutes. It's very interesting how easy are we, or at least me, to swap one problem for another, and struggle with it for days, while there is an easy direct solution discoverable in minutes.

Regarding the chess puzzle, hmm. Don't move the knight to c3. Do Kd1 instead, black will respond Ka1, then white plays a tempo with rook (Rg2), black is forced to Kb1 and you can Nc3+, Ka1 Ra2#.

In the given position, I can think of a pat XD, but a checkmate...
Though, it would apparently be no problem for ChatGPT, according to some matches I saw, it would play something along the lines of Rd1#, or Kxc1. XD

Edit: Wait, that's it! You can't play Rd1, but you could castle!
It's not typical for this stage of the game for the king and rook to be on their original places during the whole match, but it is theoretically possible.
This is genius! I love it!

Here is one riddle I encountered back on the elementary school. Kids are apparently much better in solving it than adults:

6683 = 4
537213 = 0
78 = 2
01234567 = 2
98640 = ?

To Moriarty: I agree to that. I'm one too, haha! But most people are like that. We overcomplicate things even though the answer is already staring us in the face. When you're inside the problem, just like that box formed by nine dots, you restrict your thinking even though everybody outside can see the answer clearly. Actually, that's the point of that riddle as I understand. It's from a parental coach as I said above. You have to think out of the box.

For your answer to my chess riddle above, that's right! You have to do a castling to achieve a checkmate on the next move! Hahaha! Although it's infinitely difficult to achieve that in real games, it is theoretically possible.

For your riddle, my answer is five. This riddle counts the circles in the number rather than doing any mathy thing on any of them. Hahaha! When first seeing the numbers, you would think it's a math riddle even though in essence, it's a visual riddle. Whoever made this likes to prank people for sure. Hahaha!

For how I answered it, actually, it stumped me for a bit, until I remembered a similar riddle where the number of strokes while writing the numbers is the answer, so I put two and two together.

Since you've given a number riddle, I will do the same. Haha! But only a simple one though.

The Riddle:
When you check the time, there's only five seconds before it hits three. When you check it again, it's now five seconds before it hits six. When you check it for the third time, it's five seconds again until it hits nine. How many seconds are there left before it hits twelve?

@imprisonedindarkness,
The answer can be 185.
since, you only ask how many seconds, let's assume you checking time in a minutes.
so as you check for last time, we assume, time is 8 minutes, 55 seconds.
so if time want to become 12 minutes, it should spend 185 seconds, to reach 12 minutes.
here is, a next question, specially for math lovers!
I am a three-digit number. My second digit is equal to the sum of my first and third digits. My first digit is one less than my third digit. What number am I?

To Abhishek: That's not the right answer. But you can try it again though.

Here's my answer to your riddle above. If your digits are arranged from left to right, it should be 132. If not, it's 231.

also, 253, 374 and 495 can be the answer

@Zelon
your answers are correct!

@imprisonedindarkness
I have attempted various methods to solve your riddle, but was only able to arrive at the answer that I provided earlier.
about my riddle:
I would like to inform you that answer 132 is correct, while answer 231 is incorrect as it does not satisfy the given conditions. The conditions state that the first digit plus the third digit must equal the second digit, and the third digit minus 1 must equal the first digit.
There are two approaches to solving this riddle: logical and equational. Using the logical approach, we can determine that the first and third digits are consecutive, and the third digit cannot be greater than 5 since it is the answer. Therefore, we can form pairs of consecutive digits and add them up to find the middle digit. The pairs are 1:2, 2:3, 3:4, and 4:5, which yield the answers 132, 253, 374, and 495, respectively.
Using the equational approach, we can use the given conditions to form equations in terms of x, y, and z, where x is the first digit, y is the second digit, and z is the third digit. We can then solve for z and obtain the values of x and y. The solution that satisfies the conditions is 132.

@imprisonedindarkness correct!

Btw, speaking of chess, what would you say if we played a chess match? The puzzle of all puzzles!

Hmm, for the clock puzzle, my first guess would be 10805. Because 3 hours times 360 plus 5.

I have another riddle below.

To Moriarty: That's right! It's 10805! For your chess invitation, I have to decline though. When playing chess, I often overthink, leading to my turns getting longer and longer. You would have become impatient I know. Haha. That's a weakness of mine. Haha.

To Abhishek: Actually, the second answer is just a precaution. When you put the word 'first', you didn't say if it's from the left or from the right. I have become paranoid because of that riddle about the house, pets, nationalities, etc., above, that one about who owns the horse and drinks water. Yes, that's right. I'm blaming you, Moriarty, for that. Hahaha!

The Riddle:

There are three people on a small boat, a family of one woman and two men. When it sinks, every single person dies. How many survives and what is their relationship to the deceased?

BTW, sorry for not replying to the messages. I was busy these few days.

@imprisonedindarkness well, it's upto you, though note I'm used to three hours matches on tournaments having two games per day, i.e. usually about 6 hours of playing per day, so time is not really a problem for me haha.

Anyway, to your riddle.
"There are three people on a small boat, a family of one woman and two men. When it sinks, every single person dies. How many survives and what is their relationship to the deceased?"

Hmm. This depends on the interpretation. If every single person refers to all the people on the boat, given the condition that everyone dies, I'm pretty confident the survivors will be an empty set.
Since the empty set has cardinality equal to zero, we can deduce there will be zero survivors.
Also, what's their relation to the deceased? Well, in line with the set theory, it can be shown the empty set is a subset of any set, thus we can say the survivors are a subset of the deceased.

There is also an alternative interpretation, where every single person refers to a set of people from the boat who have zero mates.
There are four solutions in terms of survivor count - 0, 1, 2 and 3.
When it comes to the relation between survivors and the deceased, let's mark their count s and d, respectively, the number of possible relationships is |W|^(s*d) for s>0 & d>0, where W is the set of labels you use for marking genealogic relationship.
While getting its exact value can be exhausting, we can get some insights into its form by thinking about its finity, i.e. whether the set is finite or infinite, as the number of relationships will behave in the same way.

In special cases of s=0 or d=0, we can say survivors are a subset of the deceased, or their superset, respectively.

So much for what we can assume from the given conditions. (smile)

To Moriarty: Here's my honest reaction after I read that: "..." and "?". Hahaha!
Alright, jokes aside, here's the answer to that riddle. The number of surviver is two. Their relationship to the deceased is his parents. The "single" there meant the marital status of the passengers.