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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?"
They drive me nuts. Array and vecor logic. Fun
-B
I wouldn't say I have a favorite problem but often when I'm bored I'll pen down the Pythagorean theorem and solve it manually. 0 = ax*x + bx + c. I'll work it out until I get the solution that (I hope) everybody knows and loves! It helps to keep my math skills alive during boring meetings.
Is obviously 42
prove that a^n=b^n+c^n for any n.
What is the next line in the following sequence? 1 11 21 1211 111221 312211
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"
None.
You haul your ass to a bakery, shell out twenty bucks, and get a box or two full of cupcakes, then you go Cid Highwind on everyone.
"Siddown and eat your goddanm cupcakes!"
Striking fear in the authors of godawful fanfiction, I am here, appearing in darkness, Tuxedo Jack!
*** ZOT! ***
Tsunami -- You can't bring a good wave down!
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?
42
--
make install -not war
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).
None, just keep the sprinkles on the side as an option.
I'm told that it depends upon the price of fish in Tibet on Tuesdays.
Game Overdrive - Gaming News
Here is the little brain teaser I thought up-- see if you can solve
it...
In the following sequence:
1, 4, 8, 13, 21, 30, 36, 44...
What is the next number and why:
A. 48
B. 50
C. 53
D. 57
E. 61
F. There is no pattern
The problem, as stated, is incomplete. If it is being defined recursively, we need some starting conditions, like x(1) = 1. However, as the OP didn't actually ask a question, I'll state what I think he was trying for here:
... Does this always happen? The answer is, alas, unknown.
This is called the "Collatz Conjecture": given a positive integer a_1 = n, let a_i = a_{i-1}/2 if a_i is even, and a_i = 3a_{i-1}+1 if n is odd. Repeat. In other words, take a number, divide by two if it's even and take three times it plus one if it's odd, and repeat ad nauseum. Try a few integers, and you'll find that they eventually end up cycling: 1, 2, 4, 1,
This problem fascinated me through high school, and I eventually ended up going into mathematics partly because of the fun I had exploring its ins and outs.
There's a good write up of this on MathWorld.
At least 20, but as many as 34 cupcakes, right?
40 people = 20 teenagers (1/2) 10 adults (1/4) and 10 babies (remaining 1/4)
Half the babies (5 people) don't like cupcakes and one fifth of the babies left (1 person, 1/5 of the five babies left after the 5 that don't like cupcakes). This leaves 34 people who are still wanting cupcakes.
Chocolate cupcakes and sprinkled cupcakes are not exclusive of each other because you can have chocolate cupcakes with sprinkles, so you can disregard the whole thing about who likes cupcakes with sprinkles and who likes chocolate cupcakes as long as you make all the cupcakes chocolate and with sprinkles.
I met a man with seven wives. Every wife had seven sacks, and every sack had seven cats, and ever cat had seven kits. Kits, cats, sacks, and wives, how many were going to St. Ives?
Rank my idea: http://www.sinceslicedbread.com/node/531
How many cans can a canner can, if a canner can can cans?
If you don't want crime to pay, let the government run it.
http://www.websudoku.com/ is my sudoku fix of choice
In these days, bleeps and bloops mean something more
Epimenides was a Cretan who made one immortal statement: "All Cretans are liars."
"The Epimenides paradox is a problem in logic. This problem is named after the Cretan philosopher Epimenides of Knossos (flourished circa 600 BC), who stated , "Cretans, always liars". There is no single statement of the problem; a typical variation is given in the book Gödel, Escher, Bach (page 17), by Douglas R. Hofstadter.
"Academicians are more likely to share each other's toothbrush than each other's nomenclature."
Cohen
You find yourself before indistinguishable two doors, each with a statue. One door will lead to salvation, the other to death. The statue that guards the door to salvation always tells the truth, the statue to the door to death always lies. You may pose only one question to only one statue. What do you ask to determine which door is which?
Answer(ROT13): Nfx nal dhrfgvba gb juvpu lbh nyernql xabj gur nafjre. Gb qrgrezvar juvpu qbbe vf juvpu lbh arrq gb xabj gur eryngvbafuvc bs gur nafjre lbh ner tvira gb gur gehgu. Gur guvat V yvxr nobhg guvf evqqyr vf vg sbeprf lbh gb pbafvqre gur bcrengbe va gur ybtvpny fgngrzrag gb or gur inevnoyr. Nqqvgvbanyyl crbcyr nera'g hfrq gb nfxvat dhrfgvbaf jura gurl nyernql xabj gur nafjre fb gurl graq abg gb or noyr gb guvax bs n fbyhgvba evtug njnl. Gur jubyr guvat orpbzrf boivbhf jura lbh cbfr n dhrfgvba fhpu nf "Ner gurer gjb fgnghrf urer?"
- 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.
What is the degree of the angle between the hour hand and the minute hand when it is 2:15?
:14, = 67.5. The difference is 22.5
22.5 degrees.
Yes, you can do it iteratively until inifinity, but the minute hand is at 90 degrees off 12, and the hour hand is at 60 for 2, plus 30/4 for the
-- Is "Sig" copyrighted by www.sig.com?
H = first 3 digits
T = last 4 digits
[250*(80*H + 1) + 2*T - 250]/2
[20000*H + 2*T]/2
10000*H + T
What the fuck is that in response to?
Rank my idea: http://www.sinceslicedbread.com/node/531
This is one of my favorites - it has stumped many self-professed math geeks, yet high school freshmen have spotted the solution immediately.
x=y
x^2=xy
x^2-y^2=xy-y^2
(x+y)(x-y)=y(x-y)
x+y=y
2y=y
2=1
Every step uses perfectly valid algebra, yet something is obviously very wrong somewhere.
Enjoy...
An object at rest cannot be stopped!
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
Generalized problem: find a similar method (to the division by 3 or 6 rules) for any integer. Not all are fast ways, but there is a relatively simple solution for any integer. A fun one to try is 11. Hint: write the number in base 11 and use modular arithmetic.
Not really hard to prove, but it's cute.
Take any six-digit number that's of the form ABCABC where A,B,C are any integers (yes, they can be the same, yes they can be zero, although that might make it less than six-digits if A, or A and B, are zero), and that number is guaranteed to be divisible by 13.
1) No object can violate conservation of angular momentum.
2) To rotate an object one needs to give it angular velocity, hence angular momentum.
3) To have finite angular momentum, an object needs torque applied to it (or a force applied away from the center of moment).
4) Gravity acts on the center of moment and does not result in torque on any free falling object.
5) Cats dropped feet up manage to land on their feet.
6) Does this mean cats violate conservation of angular momentum; no wonder Egyptians worshiped them.
What is wrong with this discussion; no math involved from my Classical dynamics class.
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.
Is ..:: Riddles ::... In has (amogst others) the famous "prison with a lamp" problem:
100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.
Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?
Right here.
Incidentally, that page has been slashdotted in the past.
There is a king and there are his n prisoners. The king has a dungeon in his castle that is shaped like a circle, and has n cell doors around the perimeter, each leading to a separate, utterly sound proof room. When within the cells, the prisoners have absolutely no means of communicating with each other.
The king sits in his central room and the n prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners' rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers "yes" but is wrong, they all will be beheaded. If a prisoner answers "yes," however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
Here's one last monkey wrench to toss in the gears, though. The king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice. The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.
(written by a former roomate)
let's see... the problem goes roughly like this:
you have five hats (two red, three black) and three people. you queue the people up in order of height and have them face the same way (this way the tallest person can see the two people in front of him/her, the middle person can see the shortest person, and the shortest person can't see anyone). you put a hat on each person's head and instruct them that they are not allowed to take the hat off or turn around. you then ask them to tell you what color their hat is. after a while, the person at the front of the line correctly announces the color of his/her hat. how did the person at the front of the line know and what were the other hat colors?
a = first 3 digits
b = last 4 digits
((a*80+1)*250 + b+b -250)/2
(a*20000 + 250 + b*2 - 250)/2
a*10000 + 125 + b - 125
a*10000 + b
It's only amazing if you don't know algebra, and no, a calculator is not required. Then again, if the point is to encourage people to eventually put down their calculator and instead try understanding why something works, then I'm all for it.
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.
If you balance all of the quarters on their edge you can choose any two arbitrary groupings - no coins would be heads up.
But I prefer your solution.
This is a non trivial problem. ...
...
Once upon a time, in a land far far away, there lived a knight who had
just rescued his first damsel in distress. The knight was called before
the king to receive a reward. The king told the knight that he had
written an amount of gold on a piece of paper and twice the amount of
gold on another piece of paper. He placed the two pieces of paper face
down in front of the knight, and told him he could chose either one.
The king would give the knight the amount of gold on the paper as a
reward. Or, the knight could opt to get the amount of gold on the other
paper instead.
This was the knight's first reward, so he had no idea what he was likely
to get. But the knight reasoned that no matter what amount he saw on the
paper he chose, he would take the other one because he had more to gain
than lose. For example, if the paper he chose had 16. He might win
another 16, and at worst he only loses 8!
The damsel points out that if the knight is going to end up with the amount
on the other piece of paper anyway, why not just choose it first and not
switch. His reward will be the same.
Is the damsel correct? Or is the knight's plan sound?
Refute the argument you disagree with. (Refuting the incorrect argument
is the challenge.)
The following two problems appeared in IMOs 1993 and 1994 (you can find the answers using Google, but I won't give a direct link).
A solitaire game is played on an infinite square grid. Initially, there are n^2 pieces in an n*n square formation. On each move, the player moves a piece either horizontally or vertically over an immediately adjacent piece into the square beyond, which must be unoccupied, and removes the piece that was jumped over. The objective is to end up with only one piece on the board. For which values of n is this possible?
Show that there exists a set A of positive integers with the following property: given any infinite set S of prime numbers, there are positive integers k>=2, n and m such that both n and m are the product of k primes in S, n is in the set A and m is not in the set A.
Here's a riddle.
It's Saturday night and some geek (sorry Cliff) is desperate and dateless so he spends 3 hours composing a submission, and YES!, it gets accepted by Slashdot just after midnight.
The riddle? How sad is that?
Assuming the earth is a perfect sphere, describe the solution set of points where you can go 1 mi south, 1 mile east, and 1 mile north and return to your starting point. Hint: the cardinality of the set is R cross Z + 1 (and yes, I know that's equal to R, but expanding it makes it a more effective hint). Feel free to email me for more hints.
U.S. War Crimes blog. Email for free Mandriva support.
You are a tourist, visiting a desert island just off the coast of South America. There's only one reason that you would be visiting this one-acre island, and that is that there is a tiny plateau reaching up a mile into the air, with the ruins of an Aztec temple on it.
As you walk along a path, you come to a fork. In the fork are two men, of which you know little, except that they must have come from one of the villages on the other islands nearby.
There are three villages--the Marqetteres always lie, the panguons always tell the truth, and the Shie'ep always do what everybody else is doing.
You may ask one question to one of the men. What do you do?
Answer(ROT-13):
Fvzcyr. Lbh vtaber obgu zra, naq jnyx fgenvtug gb gur zvyr uvtu cyngrnh, juvpu jbhyq or ivfvoyr sebz nal cbvag ba n bar-nper qrfreg vfynaq.
abcabc/13=(abc+1000*abc)/13=abc*1001/13=abc*77 nice!
About 5 seconds after the moving end of the rope comes loose.
Hmm...if the 'teenager' is 18 or 19, does that also count as an adult?
Ok, here is a progression of questions which require no special training. Make sure you only ROT13 one answer at a time if you're trying these yourself:
Assume Earth is a perfect sphere.
Q1) Where can you stand such that if you go 1km North, then 1km East, then 1km South, you're back where you started?
A1 rot13'ed) gur fbhgu cbyr. pregnvayl abg gur abegu cbyr, nf lbh pna'g tb abegu sebz gurer. naq vs lbh fnvq 1xz fbhgu bs gur abegu cbyr v'q fnl ab gbb, nf lbh pna'g tb rnfg sebz gur abegu cbyr, bayl fbhgu.
Q2) OK smarty. Where ELSE can you do it from, on the Earth's surface? No tricks are involved either, just a bit of thinking.
A2) n ovg bire bar xz fbhgu bs gur abegu cbyr: nsgre jnyxvat gur 1xz abegu, n 1xz jnyx rnfg pbzcyrgryl pvepyrf gur abegu cbyr, zrnavat lbh'ir qbar n ebhaq gevc. 1xz fbhgu gura ergheaf lbh gb gur vavgvny cbfvgvba. n srj crbcyr pbzcynva nobhg guvf bar, nf lbh nera'g jnyxvat va n fgenvtug yvar, rira gubhtu lbh'er nyjnlf urnqvat rnfg. lbh pna erzvaq gurz gung gurl jrera'g tbvat va n fgenvtug yvar va n1 rvgure. naq nfx gurz gb qrsvar 'rnfg' vs gurl fgvyy nera'g unccl.
Q3) You really think you're good don't you? OK, I want to know where ELSE!
(read this when you think you have it, before you read the real answer: gur nafjre vf abg nabgure cbfvgvba ba gur rnegu'f fhesnpr qhr rnfg (be jrfg) bs gur nafjre gb d2. jryy vg vf, ohg vg'f abg tbbq rabhtu, gurer'f fbzrjurer ryfr.)
A3) guvf nafjre vf nyzbfg gur fnzr nf gur ynfg, ohg vafgrnq bs cynpvat lbhefrys fb gung gur bar xz rnfgreyl jnyx vf n pbzcyrgr ybbc, lbh'er rira pybfre gb gur abegu cbyr, naq znantr gjb ybbcf! be, sbe gung znggre, lbh pna zbir rira pybfre, naq nf lbh nccebnpu gur '1xz fbhgu' cbvag sebz gur abegu cbyr lbh jvyy svaq zber naq zber fbyhgvbaf.
Enjoy.
If you do the population curve (takes a couple minutes, and somewhat tedious) it teaches an interesting lesson. The more money we donate to starving countries, like we do to various countries in Africe (both government and public donations), the more they reproduce. Thus they require more money, eventually dying out anyway. If we don't donate money, they drop to incredibly low numbers, and stabilize, but not as low as the other alternative. Thus the dilema. Either you donate and save some lives, causing more death in the end. Or you sit idle by, letting people die.
Weak ;)
(20000a + 250 + 2b - 250) / 2
10000a + b
John von Neumann (1903-1957) [Hungarian/US mathematician and scientist] The following problem can be solved either the easy way or the hard way.
Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour. A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour. It does this until the trains collide and crush the fly to death. What is the total distance the fly has flown?
The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows:
Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it.
When this problem was posed to John von Neumann, he immediately replied, "150 miles."
"It is very strange," said the poser, "but nearly everyone tries to sum the infinite series."
"What do you mean, strange?" asked Von Neumann. "That's how I did it!"
Seven.
Oh, yes, the 'Laffer curve' is real to the extent that 0% of any number is 0, and 100% of 0 is also 0. However, arguments based on it presume that we are ABOVE the ideal tax level - a claim for which there is no supporting evidence. http://en.wikipedia.org/wiki/Laffer_curve
You're actually making a better point than your toxic attitude would otherwise suggest. The rich can much more easily avoid taxes, legally or otherwise, than the average person, so raising their taxes really won't help revenue. (Also, shoving the tax load onto them removes any popular support for restraining government, but that's another issue.) If the rich see that one nation's attitude toward them has changed, they all just revise all future plans about investment. In the extreme case, since the rich, being rich, really don't need to work, if their taxes are too high, they just "consume" more leisure. You could, of course, just seize all their assets, but don't expect any more geese to be laying golden eggs where you live anytime soon.
I know you hate the rich and all, I'm just talking about the practical consequences of trying to milk more revenues out of them.
Plus, the premises of the argument behind the Laffer curve really are true. Again, you might not be past a Laffer point, but make no mistake there is a Laffer point. Denying this just makes wise people revise their estimation of the merit of your statements downward.
Rank my idea: http://www.sinceslicedbread.com/node/531
I like my no-frills version of the game:
Multiply the first three digits of your SSN by 1,000,000
Multiply the middle two by 10,000
Add those to the last four digits
Holy shit, how did you do that!?
Bush doesn't initiate tax changes. They come from Congress and have to start in the House of Reps as set forth in Article I, section 7 of the US Constitution. You might wanna read Article II aswell, the limited powers of the presidency, U.S. Constitution.
The way these values are determined is by equating a function like f(taxation) to total GNP or GDP and graphing the results then finding the highest point on a curve where GNP+f(taxation) is a max. Historically large taxes tend to reduce the GDP and small taxes are insuffient for the government to perform its duties as required by the constitution.
The US currently collects about 3 trillion dollars in taxes every year.
How they do it and what their reasoning is is all freely available online
Maybe you can figure out where they're going wrong and run for office.
I guess you could say 0^0 = 1, for sufficiently large values of 0. :)
Yes, it is undefined and you don't need a calculator. Actually, you shouldn't. But the bottom line is that it approaches 1, in the limit: lim(x->0) x^0 = 1, and for x^x this holds as well; you can try both on a calculator with very small values for x, if you want. In some "practical" problems, such as fitting a polynomial on a set of points, you really need 0^0 to be 1.
However, if you're going to try the other limit, lim(x->0) 0^x, the answer would seem to be 0 (but since this function is not continuous, there is no need to make it equal 0).
The theoretical answer? it depends on the function you're looking at.
The pratical answer? Try 1.
One wonders if it is in fact you who is dismissive of the poor, your attitude toward them in your posts here certainly is.
...RIGHT?
*sigh*
I'm trying to explain to you the practical consequences of what you're proposing. I was just describing aspects of reality. You'll notice I never said anything was good or bad, just what will or won't happen. If you can find the place where I was dismissive of the poor, I'd really like to know where it is.
Now, I already explained enough so you could understand the difference between cracking down on the rich vs. the poor. The rich can easily scurry away and/or stop producing. The poor can neither easily scurry away nor stop producing. Again, this is not to say anything is "good" or "bad", just that it "will" or "won't" raise tax revenues. Contrary to your staunch refusal to dispassionately analyze the topic, there really are relevant practical considerations in raising taxes.
In fact, I'd like nothing more than to test out your ideas. Check out the link in my sig. I submitted an idea to a policy site. The idea is that basically, in one state, we do what you propose: high taxes on the rich, high minimum wage, good workers protections and workplace safety requirements, etc. In the other state, do the oppose: no min. wage, low taxes on the rich, no safety requirements, etc. If you're really serious about your views, you'd leap at the chance to do this and see who's right based upon which state people flock to.
You do think you're right, right?
Rank my idea: http://www.sinceslicedbread.com/node/531
The BART is the SF Bay area's excuse for a subway / mass transit. To ride, you buy a ticket at a kiosk full of $x worth of "BARTness". When you get board the train, you stick your ticket into a turnstile, and it hands it back to you. When you reach your desgination and get off the train, you again stick your ticket into a turnstile. It deducts the cost of your trip (based on how far you traveled) and gives your ticket back. You keep using the ticket until $x is used up. BUT: Suppose you walk toward the train, put your ticket in the entrance turnstile. You pick up your ticket, then you change your mind and leave, putting your ticket in the exit turnstile to get out. The cosmic BART megamachine will charge you the maximum possible fare, even though you haven't gone anywhere. For a good reason. What's the reason?
I think the answer has to deal with the math being misleading. The value of $9 is provided to stump the reader, as it really doesn't pertain to the original cost of the room in the way suggested (9+9+9+2 != 30). Instead, 9+9+9 resembles what the men end up paying for the room (27), but the total is expected to be 25. The real question then becomes, "Who pocketed the extra two dollars?" (9+9+9-2 = 25). In the end, it seems to be just a matter of the logic problem playing off of misplaced signs (+/-).
In these days, bleeps and bloops mean something more
Since that's basically a probability question, I thought I'd follow up with my own favorite probability mind bender:
Your married-with-two-kids co-worker invites you over to dinner. When you arrive a son of the coworker answers the door. What is the probability that the other child is a girl?
Followup:
The co-workers oldest child, a son, answers the door. What is the probability that the other child is a girl?
Most who have gone through a formal stats class have seen this one before, but it is always fun to try and wrap your head around it the first time.
Education is a better safeguard of liberty than a standing army.
Edward Everett (1794 - 1865)
In the first case where you don't know if the boy is the first child or not there are four possibilites BB, BG, GB and GG. We can eliminate GG since that's now impossible. From the remaining 3 cases only BB would allow the second child to be a boy so the chance
that the other child is a girl is 2/3.
In the second case where we know the child is the oldest boy there are only 2 possibilites BB and BG. We can't eliminate any possibilites so the chance the second child is a girl is 1/2.
again, with breaks in it:
1
12
1112
3112
132112
1113122112
311311222112
what's the next line?
Fleur de Sel
When you arrive a son of the coworker answers the door. What is the probability that the other child is a girl?
<answer type="weasel">Zero—the co-worker must have two sons, otherwise the question would say "the son" rather than "a son" ;-)
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Never let logic and reality get in the way of a good, liberal hate-on for Republicans.
(-1: Post disagrees with my already-settled worldview) is not a valid mod option.
While it is true people are greedy and will continue to work hard, even as more and more gets filched by the government, at some point revenue will decrease because people will stop working so hard for diminishing returns.
It's very cynical to tout as good a government that lays like a 2000 pound pig on top of the population, and then shouting joy about it as the population barely moves it around, said movement, if nothing else, caused by the economic activity necessity to stay alive.
But see, if we take taxes, i.e. confiscate money that is not ours, to use for projects that we approve of then it doesn't feel like stealing to us.
Which I find as a disgusting attitude, and why I have little long-term hope for humanity. They preform all the age-old tricks of a dictator, but laundered because it's at the behest of power hungry, charismatic individuals who launder it via "see, the voteres elected me and want it!"
(-1: Post disagrees with my already-settled worldview) is not a valid mod option.
Yes. There are only 3 possibilities, and you've enumerated them all...
I18N == Intergalacticization
They each pay $10 when the room is $30. The room's price drops to $25, so they're each given $1 back, bringing their total to $27. The bellhop keeps the other $2 that would have brought their total down to $25. The $5 is still there, as $1 each for 3 people + $2 for the bellhop = $5. The total amount does not need to equal $30 after the price changes; it needs to equal $25.
These sorts of tricks are magical only to those who never took 7th-grade algebra (or won't apply it):
Let a be the three-digit prefix and b be the four digit number.
(((80a+1)250+2b)-250)/2
(20000a + 250 + 2b - 250)/2
(20000a + 2b)/2
10000a + b
Which is obviously your phone number. No, I didn't try it.
More interesting tricks (not puzzles) of this sort rely on less-obvious (unless you know them) arithmetic facts, like relationships between numbers and the sum of their digits.
Note to ACs: I usually delete AC replies without reading them. If you want to talk to me, log in.
what's the next line?
5.
No finite sequence determines the subsequent.
As such, "math puzzles" of the "what is the next number?" kind are not math puzzles at all--they are psychology and common-knowledge puzzles. They should be stated, "I'm thinking of a number. To me, the number is the next in the following sequence: (...). Your job is to guess, based on what you know of me (or people like me), of mathematics, and of common knowledge, which of the infinite number of mathematical relationships betweeen the numbers in that sequence is the one that is important to me."
People who work in numerical methods are only too aware of how little information finite sequences contain beyond their own bounds. Interpolation is hard enough. Extrapolation is virtually impossible. Even simple sequences like "1,2,3,4..." can have literally anything as the next value--it is trivial to come up with generating functions that give integers for the first few integer arguments and wildly varying irrational values after that. Unless you know what the generating function is, the finite sequence tells you nothing. Guessing the generating function from a finite sequence is all about guessing what the questioner knows and what kind of generating function a person with their knowlege (or common knowledge) is likely to choose that would produce the given sequence.
A modicum of mathematical knowledge is still required, but far more psychology is necessary.
Blasphemy is a human right. Blasphemophobia kills.
I am reminded of a short bit in the Journal of Irreproducible Results, where they had proof of -1 being the largest integer. Rough rephrasal:
The defining property of the largest integer is that there is no larger integer. But if you take any integer and add one, you get a larger integer. So the largest integer must fulfill x = x + 1, and taking it one step farther for good measure, x = x + 2.
x = x + 2
Square both sides.
x^2 = x^2 + 4x + 4
Subtract x^2.
0 = 4x + 4
Subtract 4x.
-4x = 4
Divide by -4.
x = -1
Checking this result, -1 is indeed an integer and nothing is one more than -1. QED.
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.
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
3 1 2 2 1 1
1 3 1 1 2 2 2 1
1 1 1 3 2 1 3 2 1 1
3 1 1 3 1 2 1 1 1 3 1 2 2 1
continue the series!
Ah, arrogance and stupidity, all in the same package. How efficient of you. -- Londo Mollari
All very true, but if the question were phrased: "What's the simplest generating rule for this sequence?" There would be a lot fewer answers and quite likely one unique answer. The emphisis should be on compressing the available information and not on predicting future information.
Prior to entering the room (I am assuming it's a room where you can somehow touch the light bulb, the light works and is connected to one of the switches etc), flip one of the switches e.g. switch number 1 (we'll call them 1,2 and 3) and leave it on for 30 seconds. Flip that switch back to off and flip switch number 2 to on. Now enter the room. If the bulb is lit it is switch number 2 that controls the light. If the bulb is not lit but warm (feel by touching it) it is switch number 1 that controls the light. If the bulb is unlit and cold it is switch number 3 that controls the light.
Here's another nice one, courtesy of Raymond Smullyan.
Suppose there are three people, called A, B and C. Each of these is a "perfect logician"; that is, given some information, they all are able to immediately draw any and all conclusions that can possibly be drawn from this information. Furthermore, suppose there are four red and four green stamps.
Now, all three of them close their eyes, and two stamps are glued to their foreheads, each; the remaining two stamps are put away. Now, they all open their eyes again.
Then, the first, A, is asked whether he knows the colours of the stamps on his forehead. He says he doesn't. Then B is asked the same thing, and also says he doesn't, and afterwards, C is asked and says he doesn't, too. Now, A is asked a second time, and he still says he doesn't know. But then, when B is asked a second time, he now says he does know.
The question is: how?
quidquid latine dictum sit altum videtur.
If you really get one of 'those' meetings or classes, you can try this. It is so boring, you have already made another Tic Tac Toe crossed set of lines. Take all 10 numeral digits and put them in the Tic Tac Toe so that all horizontal, all vertical and all diagonal sums each add up to ... 15
I give no hints.
Ah I got it. Took 16 rolls (written down) and almost hour. The fact that I am ignorant about roses didn't help :)
:)
Anyway I read that story and it didn't appear to me that he was trying to solve it by memorization, but rather that after an hour, seeing hundreds of rolls, he remembered many of them, which isn't all that surprising. What I got out of the story, is that he persistently kept at the problem trying many different ideas until he finally got it, even after everyone else in the group had solved it or quit.
I don't think that how quickly someone solves any one particular problem is much of an indicator of how smart they are. We were doing brain teasers on an ACM trip my freshman year of college, and I was one of the first to get most of them. Then there was this one that everyone got right away and I couldn't get it. When I finally did a day later I was kicking myself because it was so obvious - I just wasn't looking at it the right way.
This story showed my that Bill Gates is a very persistent and determined person which is probably a big reason that he was so successful. That and he needed to get a girlfriend at that point, as do I apparently
man, you solved it the hard way. it says a windowless room. therefore, all I have to do is get down on the floor after flipping each switch. if I see light coming out from under the door, ergo, I have my solution. there is no other light source, doors are required to have those gaps underneath in order to allow for building ventilation, and whatever M$ doesn't tell me about the conditions of the room, I can modify to my advantage.
an even easier solution would be to get a fireaxe, chop a hole in the door, and flip switches while I looked thru the hole. haven't opened the door and I've solved it in even less time than your method.
when dealing with idiotic interview questions like this, use the unexpected solution.
I'm good with numbers -
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
if that's the answer to your riddle, you are an idiot. it's not within the constraints of the problem.
So what you're saying is that you wrote the problem ambiguously, and to resolve ambiguities we ought to choose the interpretation that yields a solution?
Your description of the problem does not say when k is fixed. A perfectly valid reading of the problem is to suppose that the prisoners are told that the king will decide k when the game begins.
You ought to admit that the problem was unclear, instead of insulting everyone who interprets your ambiguity the wrong way.