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Your Favorite Math/Logic Riddles?
shma asks: "Whether you're involved in the Sciences, Mathematics, or Engineering, you undoubtedly enjoy finding simple solutions to seemingly difficult problems. I'm sure you all have a favorite mind-bender, and who better to share it with than the Slashdot community? Post your own problems and try to solve others. Just one request: If you have figured out the solution, link to it in a post, rather than write it out where anyone can see it." What brain benders tickle your fancy?
"Here's a sample to consider: You're in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it's too dark to tell what you're moving or flipping (no, you can't figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?"
Is obviously 42
prove that a^n=b^n+c^n for any n.
Turn a light on.
Slashdot - where whining about luck is the new way to make the world you want.
Bill Gates is said to have solved the problem by memorizing the combinations first, the brute force approach.
It ones of those that requires a knack for seeing the simple things
"It is a greater offense to steal men's labor, than their clothes"
Then you may like this one: X to the X to the X to the... = 2. What is X if the left hand side is an infinite sequence of powers?
There's a good write up of this on MathWorld.
http://www.websudoku.com/ is my sudoku fix of choice
In these days, bleeps and bloops mean something more
- One room has three switches, labeled A, B, and C.
- Another room has three light bulbs, labeled 1, 2, and 3.
- Each switch is connected to one bulb, but you do not know which is connected to which.
- When inside either room, you cannot see the other room.
- You begin in the room with the switches and may turn the switches on and off in any way you choose.
- Once you leave the room with the switches, you may not reenter it. You may, however, go to the room with the light bulbs.
How can you determine which switch is connected to which light? Here is a hint and solution.I like this problem because people are ordinarily good at logic have so much trouble with it. I once had the pleasure of meeting Donald Knuth and stumped him with this puzzle.
Wow! It's my age! How did you do that?!?
Correct Horse Battery Staple: 72 bits of entropy. Enter "Correct H" into google. When it generates the phrase, that's
Let's say I have a stack of sticks: all identical, inflexible, unbreakable. Sticks can touch only at their ends, not in between.
If I give you 3 sticks, you can make one triangle. If I give you 2 more sticks (5), you can make 2 triangles. If I give you another stick (6), how can you make 4 triangles?
--
make install -not war
Simply place any 18 coins into the second group and flip those over.
If you flip a coin over that was heads, it is now tails and is eliminated from consideration. If you flip a coin over that was tails, it marks with heads a coin selected that was not heads. Therefore after 18 coins are flipped, the number of heads in the second pile is equal to the number of heads that are left in the first pile.
Dividing by zero is not "perfectly valid algebra". Division is not closed on the set of real numbers. Its not really a riddle if you lie in the problem description. Otherwise the solution to the sample problem could be "Pull out 9 of the quarters into a seperate group. I was lying when I said you couldn't see any of them."
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
I guess you could say 0^0 = 1, for sufficiently large values of 0. :)
Since it's an infinite sequence, you can separate the left-most X and rest still equals 2. Thus X^2 = 2, so X = sqrt(2).
... = n. ...) = n. ... equals 4, not 2, so it can't be the solution to the original problem.
;-)
Disprufe(TM) by contradiction:
1. Suppose sqrt(2) ^ sqrt(2) ^ sqrt(2) ^
2. Then, sqrt(2) ^ (sqrt(2) ^ sqrt(2) ^
3. Hence, sqrt(2) ^ n = n.
4. Therefore, n obviously equals 4, because sqrt(2) ^ 4 = 4.
5. Hence, sqrt(2) ^ sqrt(2) ^ sqrt(2) ^
What's wrong with this logic?
Weeks of coding saves hours of planning.
Oh, woe is me. I have a perfect logic puzzle, but was unlucky enough to be otherwise engaged when this story was posted. (By the way: a soft couch, a carefully selected DVD, half a bottle of rum, and a girl. Guess which element to this excellent scenario was fucking ruined by copy protection? I'll give you a hint: I may have just switched sides in this movie piracy debate. Fuck the RIAA. It was a perfectly legal store-bought DVD. Fuck them all.)
But anyway, logic puzzles. This logic puzzle is excellent. I've had it up on my site (http://www.xkcd.com/blue_eyes.html), and after I got boingboing'ed I got a lot of email about it, so I've been able to tweak the wording to get rid of most of the confusing stuff, leaving only the logic. It's extremely subtle; I've never seen anything like it.
Here's the puzzle:
A group of people 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. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight.
On this island live 100 blue-eyed people, 100 brown-eyed people, and the Guru. The Guru has green eyes, and does not know her own eye color either. Everyone on the island knows the rules and is constantly aware of everyone else's eye color, and keeps a constant count of the total number of each (excluding themselves). However, they cannot otherwise communicate. So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes, but that does not tell them their own eye color; it could be 101 brown and 99 blue. Or 100 brown, 99 blue, and the one could have red eyes.
The Guru speaks only 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 with 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, 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."
xkcd.com - a webcomic of mathematics, love, and language.
I had a conversation with Brian0918 on AIM this morning, in which he revealed he's really trolling when I pointed out to him there is no solution (see my other posts) on this topic. Here's a little tidbit: ". i usually just post the problem to get people into big disputes, which so far has worked 2 out of 2 times". If you want the full conversation, email me at sbartaNOSPAM_at_MAPSONgmail.com.
Rank my idea: http://www.sinceslicedbread.com/node/531