Devinettes app has about 3 or 4 riddles that I haven't been able to crack in 8 years.
https://f-droid.org/en/packages/com.workingagenda.devinettes/
It gives no answers or hints, just reports 'correct' if you type the right answer in.
Cool app! I tried it and I have 7 riddles to go. Which ones have you not cracked yet?
Edit: a little break helped, 4 to go!
I'd say, your 4 are in this 6:
Name your heroine after me.
Name me the skin
Without this common currency
Without you I am incomplete 1
Without you I am incomplete 2
This: Brought by man
- PS, I am not looking for any answers here at all. (Spoiler-tags do not work in several clients)
I'm fine solving these 15yrs later or not at all. The unsolved ones are good for campfires, or travelling through other countries.
"failed to install due to unknown error" :(
Do you feel like posting the riddles here?
I've always been fond of:
The person who makes me, doesn't want me
The person who buys me, doesn't need me
The person who uses me, doesn't even know it
What am I?
Maybe not the hardest, but it's a good one, and most people I ask, don't get it very quickly.
I'm not sure how to spoilers the answer, so I'll just give you time to guess the answer. It's more fun that way.
https://join-lemmy.org/docs/users/02-media.html
The table doesn't highlight how necessary the literal word 'spoiler' is needed after the 3 colons. Very confusing imo
Agreed, the first time I used it I think I edited my post 3 or 4 times before getting it right.
Light as a feather, but no man can hold me for long.
Answer
A breath
I think I've seen this one on a morhedel wordlock in Betrayal at Krondor :o
The version I heard was. “What’s lighter than a a feather, lighter than a flower, the strongest man cannot hold me, for a quarter of an hour”
What rock group has four guys who don't sing?
Click for answer
Mount Rushmore.
Poor people have it.
Rich people need it.
If you eat it, you die.
What is it?
Tap for spoiler
Nothing
Oh that's clever, I'm keeping this one in my back pocket
Shit?
I was going to say “your own heart”
Some android apps don't honour spoiler tags (Boost) :( Feel free to link to your answer somewhere instead of spoiler tagging it if you feel extra nice :)
I'm using boost. I've just been scrolling very slowly.
Eternity app does not seem to handle the spoiler tags too.
What, no Riddle of the Sphinx yet?
Be the change in the world you wish to see.
It was posted by rhythmisaprancer.
'What is my hobby ?'
Look at these equations:
1^3 = 1^2
1^3 + 2^3 = (1+2)^2
1^3 + 2^3 +3^3 = (1+2+3)^2
1^3 + 2^3 +3^3 +4^3 = (1+2+3+4)^2
Question:
Can it go on like this forever, is it always a true equation? If yes, why? If no, why?
Proof by induction?
1±2±3±...±n =(1+n)*n/2plugging that into the right side of the equation to transform it:
((1+n)*n/2)^2 = (1+n)^2*n^2/4=n^2(n^2+2n+1)/4 = (n^4 + 2n^3 +n^2)/4If this holds for n:
1^3 + 2^3 +3^3 + ... + n^3 = (n^4 + 2n^3 +n^2)/4Then for n+1:
(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (1+n + 1)^2*(n+1)^2/4(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^2+4n + 4)(n^2 +2n + 1)/4(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^4 + 4n^3 + 4n^2 + 2n^3 + 8n^2 + 8n + n^2 + 4n + 4)/4(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^4 + 2n^3 + n^2)/4 + (4n^3 + 12n^2 + 12n + 4)/4(n+1)(n^2 +2n + 1) =? n^3 + 3n^2 + 3n + 1n^3 + 2n^2 + n + n^2 + 2n + 1 =? n^3 + 3n^2 + 3n + 1n^3 + 3n^2 + 3n + 1 =? n^3 + 3n^2 + 3n + 1Which is obviously true.
So yes, it holds forever.
This is the way.
Corresponding Wikipedia page with a graphical proof, among others
The graphical proof is really nice :)
Your math teacher might not approve of this proof
The given examples suffice to prove the general identity. Both sides are obviously degree 4 polynomials, so if they agree at 5 points (include the degenerate case 0^3 = 0^2), then they agree everywhere.
You are right LOL: I do not approve. But somehow I like the lazy approach :)
Everyone knows the hardest riddle of all time:
"What's in my pocket?"
You even get to keep a gold ring if you use that one!
What walks on four legs at dawn, two legs at noon, and three legs at dusk?
spoiler-title
Humans in their three stages of life: crawling at birth, walking upright in main life, and walking assisted at the end
Famously the hardest logic puzzle in the world: Blue Eyes
https://xkcd.com/blue_eyes.html
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
The answer
The answer is >!the 100 people with blue eyes leave on the 100th night.!< Double spoiler tagged so it hopefully works on every Lemmy client.
For the logic, imagine if it were actually just three people: the guru, one brown eyes, and one blue eyes. For the sake of clarity, I'll speak through the perspective of a blue eyes in these examples. Guru says "I see someone with blue eyes." Brown eyes also sees someone with blue eyes. But I see no one with blue eyes. I deduce that I must have blue eyes, and leave that same night.
Now >!imagine there are two blue eyes, two brown eyes, and the guru. Guru says "I see someone with blue eyes." Brown eyes both see two sets of blue eyes, but I only see one set of blue eyes. I figure if the other person sees no other blue eyes, they'll leave the first night. They don't, which can only mean that they also saw someone with blue eyes, which must be me. We both leave the second night.!<
You >!can expand this logic all the way out to 100, so on the 100th night, all the blue eyes leave.!<
I can't see how this expands from your last case of 2 blue eyes to any more blue eyes?
When there are two blue eyed people (and you can see one of them) then the guru saying they see a blue eyed person has value because you can wait and see what the only person with blue eyes does. If they do nothing it's because the gurus statement hasn't added anything to them (they already see someone with blue eyes). And this in turn tells you something about what they must see - namely that you have blue eyes.
But how does this work when there are 3 people with blue eyes?
There isn't anyone who might see no blue eyes. And you know this, because you see at least 2 sets of blue eyes. When no-one leaves on day 1 it's because they're still not sure, because there's no circumstance where the gurus statement helped anyone determine anything. So nothing happening on day 1 doesn't add any useful info to day 2.
So the gurus statement doesn't seem to set anything in motion?
Same way it expands to two: When there are three blue eyes, then each of them guesses they might have brown or something and there could be only two blue on the island, in which case as described those two would have left on the second night.
But they didn't... So there must be three total. Same with 4, the 3 you can see would have left on night 3 if each of them saw the other two not leave on night 2...
Leaving or not is the only communication, and what the guru really did was start a timer. It has to start at 1 even though everyone can see that there's more than one simply due to the constraints of the riddle - if the guru were allowed to say 'I see at least 50 blue eyed people' then it would start at 50 because there's no other fixed reference available. Everyone knows there's either 99 or 100, but they don't know which of those it is, so need a way to count to there. They also think everyone else can see anything from 98 to 101 depending, so it's not as straightforward as thinking the count could start at 99.
spoiler
I think the problem is underspecified: it doesnt define what happens if they get it wrong. If they get it wrong are they barred from exiting? Or can they make a guess every night? Whats stopping them from all saying blue and the boat leaving with 100 blue eyed people on night 1? Or do they only get one guess?
Also it would have saved more people if the guru had simply stated the counts they daw because everyone could count the other people and then see what color was missing (their own eye color) and they could have all left night 1.
The everybody is a perfect logician part means no one gets it wrong and no one make a guess.
They also don't say the color of their eyes when they leave. They can just use logic to figure it out.
There are some other problems with this versions wording though. Usually you need to specify that the speaker doesn't lie.
this is a great opportunity to bring in riddles from the Kids Write Jokes tumblr. not gonna spoiler the answers for... obvious reasons.
some of my favourites:
Knock Knock
Who’s there?
Doctor Dentist
Doctor Dentist who
How can you be a Doctor and a Dentist there is not enough time
what do dogs call Italy?
dogtaly
How many sharks does it take to make a grandpa shark?
1,280,000 sharks.
what does NASA stand for.
not another spaceship aaaaaaaaaaaaaahhhhhhhhhhhhhhhhhhh
and my all time fave
can you fart?
do it now
What's at the start of an era, twice in a generation, nowhere in an instant, but awaits you at the end of time?
spoiler
the letter e
What has four wheels and flies?
spoiler
Supposed to be a dump truck, but they have way more than four wheels.
spoiler
A carriage with horses
Blaine is a pain.
If you've never played Betrayal at Krondor it's full of these treasure chests that you have to solve riddles to unlock. Here's a few good ones:
Whoever has it is angry
Whoever loses it is angrier
Whoever wins it has it no more
!A Dispute!<
Brought to the table
Cut and served
Never eaten
!Cards!<
We love it more than life
We fear it more than death
The wealthy want for it
The poor have it in plenty
!Nothing!<
What has 11 heads, 2 wings and 22 legs?
Answer
A football team.
Not a particularly hard one, but I always enjoyed it.
A dolphin is neither a horse nor a boat.
A boat is neither a horse nor a dolphin.
A horse is neither a dolphin nor a boat.
But what is the one common thing that would happen if you were to bite any of these three things?
These will make you angry But they might make you laugh
These have been around quite long Yet you'd never expect them
Tap for spoiler
Deez nuts lmao gottem
ligma
Tap for spoiler
balls
