Beta

Slashdot: News for Nerds

×

Welcome to the Slashdot Beta site -- learn more here. Use the link in the footer or click here to return to the Classic version of Slashdot.

Thank you!

Before you choose to head back to the Classic look of the site, we'd appreciate it if you share your thoughts on the Beta; your feedback is what drives our ongoing development.

Beta is different and we value you taking the time to try it out. Please take a look at the changes we've made in Beta and  learn more about it. Thanks for reading, and for making the site better!

Your Favorite Math/Logic Riddles?

Cliff posted more than 8 years ago | from the top-10-mind-twisters dept.

Math 1965

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?"

cancel ×

1965 comments

Soduku (3, Interesting)

beacher (82033) | more than 8 years ago | (#13800847)

They drive me nuts. Array and vecor logic. Fun

-B

Link to online version (1)

beacher (82033) | more than 8 years ago | (#13800895)

Kind of annoying background music - link to an online Soduku game is here [uclick.com]

Math and science are obsolete (-1, Offtopic)

shanen (462549) | more than 8 years ago | (#13800956)

Soduku drives you nuts? (Whatever "soduku" is?) Anyway, the puzzles of this modern age are things like how we (or BushCo claiming to be acting on our behalf) can cut taxes for the rich while spending billions on wars while ignoring and eliminating infrastructure spending like flood prevention and education without going bankrupt. And don't waste your time looking for the linked post with the answer. There ain't no answer on this side (the outside) of the funny farm.

Welcome to 2005 going on 1984 but with our final destination sometime around 1840...

Re:Math and science are obsolete (0)

Anonymous Coward | more than 8 years ago | (#13800969)

Get over it.

Re:Math and science are obsolete (-1, Flamebait)

Anonymous Coward | more than 8 years ago | (#13800980)

You're an idiot. Just thought I'd let you know.

Have a nice day!
-Jacob S.

Another online version (5, Interesting)

Enti (726249) | more than 8 years ago | (#13800972)

http://www.websudoku.com/ [websudoku.com] is my sudoku fix of choice

FP (-1, Offtopic)

Anonymous Coward | more than 8 years ago | (#13800848)

FP!

Riddle (1)

connah0047 (850585) | more than 8 years ago | (#13800849)

You are making cupcakes for a party at which there will be 40 people. Half of them will be teenagers, a quarter of them will be adults, and the rest will be babies. Half of the babies don't like cupcakes, and one fifth of the babies left are too young to eat cupcakes. Half of the adults and three-quarters of the teenagers like chocolate cupcakes, and the rest of the people like cupcakes with sprinkles on them.

How many sprinkle cupcakes should you make?

Re:Riddle (3, Funny)

Tuxedo Jack (648130) | more than 8 years ago | (#13800874)

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!"

Re:Riddle (1)

ZeldorBlat (107799) | more than 8 years ago | (#13800879)

>How many sprinkle cupcakes should you make?

14

Re:Riddle (1)

Alien54 (180860) | more than 8 years ago | (#13800984)

Sorry to say, you made a mistake in your mental math. The answer is not 14

You are making cupcakes for a party at which there will be 40 people. Half of them will be teenagers, a quarter of them will be adults, and the rest will be babies. Half of the babies don't like cupcakes, and one fifth of the babies left are too young to eat cupcakes. Half of the adults and three-quarters of the teenagers like chocolate cupcakes, and the rest of the people like cupcakes with sprinkles on them.

This gives us

20 Teeagers

10 Adults

10 Babies

5 babies do not like cupcakes, 2 are too young, so 3 babies eat cupcakes

5 adults and 15 teenagers do chocolate cupcakes

This leaves us with 5 adults, 5 teens, and 3 babies to eat cupcakes with sprinkles.

Trivia note: Fibonacci, who is known for the fibonacci number series, is the person who introduced and first promoted the use of the arabic number system to Medieval Europe

Re:Riddle (2, Insightful)

st0rmshad0w (412661) | more than 8 years ago | (#13800908)

None, just keep the sprinkles on the side as an option.

Re:Riddle (2, Insightful)

iamdrscience (541136) | more than 8 years ago | (#13800959)

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.

Here's a political riddle. (0, Troll)

Anonymous Coward | more than 8 years ago | (#13800850)

What brain benders tickle your fancy?

Yeah, how did Bush get elected? Even more mind numbing is how he got re-elected. How'd that happen? Seriously, we have a man who is not the least bit curious, managed to squeak through school, managed to run two companies into the ground and yet still get elected as President of the United States where he is proceeding to run the country into the ground. Bush is cutting basic science research of all sorts (unless it is defense or energy related) and is pushing a religious agenda to the detriment of science and science education.

Re:Here's a political riddle. (0)

Anonymous Coward | more than 8 years ago | (#13800918)

You lost, hippie, get over it.

Actually, please don't. Please keep fighting against Bush and make sure for the next three years he's the focus of all you do and say, because you don't want him re-elected again.

Please, please, PLEASE keep fighting a battle you've already lost.

Re:Here's a political riddle. (1)

rah1420 (234198) | more than 8 years ago | (#13800933)

because you don't want him re-elected again.

Absent a constitutional convention, that'll be difficult to do.

Infinity (0)

Anonymous Coward | more than 8 years ago | (#13800852)

I have always been intrigued by the concept of infinity.

Re:Infinity (4, Interesting)

Frequency Domain (601421) | more than 8 years ago | (#13800894)

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?

Solution (3, Interesting)

Frequency Domain (601421) | more than 8 years ago | (#13800906)

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).

2+ (1)

Nikkodemus (763778) | more than 8 years ago | (#13800853)

2=?

Re:2+ (1)

Vorondil28 (864578) | more than 8 years ago | (#13800929)

Oh crap, I know this one...

Keeping my skills fresh (3, Interesting)

Paladine97 (467512) | more than 8 years ago | (#13800854)

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.

Re:Keeping my skills fresh (1)

mjc_w (192427) | more than 8 years ago | (#13800941)

Uh, that's the quadratic formula.

The Pythagorean theorem is about the sides of a right angled triangle (a^2 + b^2 = c^2).

Re:Keeping my skills fresh (1)

avxo (861854) | more than 8 years ago | (#13800974)

You realize that what you wrote is not the Pythagorean theorem. That is: a + b = c

Re:Keeping my skills fresh (1)

avxo (861854) | more than 8 years ago | (#13800989)

Silly browser... a^2 + b^2 = c^2

Re:Keeping my skills fresh (1, Informative)

LeonGeeste (917243) | more than 8 years ago | (#13800992)

That's only the Theorem if you're expressing it in terms of quadrances (squares of distances), like that weirdo's book that got posted a short while ago on /. Ah, here it is:

http://science.slashdot.org/article.pl?sid=05/09/1 7/1313249&tid=228&tid=14 [slashdot.org]

In normal trig, it's a^2 + b^2 = c^2.

Thinking back, wouldn't it be cool to have a quadrance-ruler? It would be marked off in units of inches squared, and the reading on it would be the square of the number of inches in distance. (Of course, if you were a communist, that would be centimeters.)

thrice-plus-one-or-half (1)

Diego and Aline (914966) | more than 8 years ago | (#13800856)

x(n) = x(n-1)/2 if n is even
                = 3*x(n-1)+1 if n is odd

Re:thrice-plus-one-or-half (2, Interesting)

dcclark (846336) | more than 8 years ago | (#13800930)

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:

This is called the "Collatz Conjecture [wikipedia.org] ": 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, ... Does this always happen? The answer is, alas, unknown.

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.

Re:thrice-plus-one-or-half (1)

dcclark (846336) | more than 8 years ago | (#13800949)

A few corrections:

We divide by two if a_{i-1} is even, not if a_i is even. Likewise for a_{i-1} being odd.

The cycle is 4, 2, 1, 4, ..., not 1, 2 ,4 ,1, ...

Someone remind me not to write math at 1:00 am again... especially when it comes to my thesis!

The Answer.... (5, Funny)

Omnieiunium (872399) | more than 8 years ago | (#13800857)

Is obviously 42

easy one (5, Funny)

zanderredux (564003) | more than 8 years ago | (#13800859)

prove that a^n=b^n+c^n for any n.

Re:easy one (5, Interesting)

calvin1981 (922478) | more than 8 years ago | (#13800899)

That one's really easy. Set a=42, b=0 and c=42, for any n :)

Re:easy one (4, Informative)

NitsujTPU (19263) | more than 8 years ago | (#13800988)

Uhmm, if n = 0, that is not true.

a^0 = 1
b^0 = 1
c^0 = 1

1 != 2

So, I would submit that that might be true for all nonzero values of n.

Re:easy one (1, Funny)

Anonymous Coward | more than 8 years ago | (#13800905)

I have discovered a truly remarkable proof of this theorem that the length of this comment is too small to contain.

Re:easy one (0)

Anonymous Coward | more than 8 years ago | (#13800925)

If all of the factors of a^n are prime (or irregular), than the square root of the left side of the equation is going to be equal to b(a^n)^c for all negative integers.

So obviously, b^n != c^n for any positive number. Think about it. If b^n isn't the same as c^n, how can b^c possibly be the same as the square root of a^n+b^n????

Maybe I'm missing something obvious here.

Sequence (2, Interesting)

blystovski (525004) | more than 8 years ago | (#13800860)

What is the next line in the following sequence? 1 11 21 1211 111221 312211

Re:Sequence (1)

blystovski (525004) | more than 8 years ago | (#13800876)

Correct formatting...sorry!

1
11
21
1211
111221
312211

Re:Sequence (0)

Anonymous Coward | more than 8 years ago | (#13800897)

13112221

Re:Sequence (1)

Doc Ruby (173196) | more than 8 years ago | (#13800928)

Better yet, what's the preceeding line?

Re:Sequence (1)

Hawthorne01 (575586) | more than 8 years ago | (#13800962)

ummn, "Correct formatting...sorry! " ? :-)

Re:Sequence (0)

Anonymous Coward | more than 8 years ago | (#13800886)

13112221 ?

Re:Sequence (1)

Noksagt (69097) | more than 8 years ago | (#13800891)

13112221

Re:Sequence (1)

eu_neke (415715) | more than 8 years ago | (#13800898)

13112221. Followed by 1113213211, then 31131211131221, then 13211311123113112211. I'm getting bored now ;)

Look and Say (5, Informative)

Noksagt (69097) | more than 8 years ago | (#13800953)

There's a good write up [wolfram.com] of this on MathWorld.

Re:Sequence (1)

DeuceTre (785433) | more than 8 years ago | (#13800968)

132231 232221 134211 14131231

Oldie but goodie... (1)

Moofie (22272) | more than 8 years ago | (#13800864)

How much wood would a woodchuck chuck, if a woodchuck could chuck wood?

Re:Oldie but goodie... (0)

Anonymous Coward | more than 8 years ago | (#13800888)

If a woodchuck could cut wood, would he chuck cut wood?

Re:Oldie but goodie... (2, Funny)

markov_chain (202465) | more than 8 years ago | (#13800889)

*** ZOT! ***

Re:Oldie but goodie... (2, Funny)

Dorothy 86 (677356) | more than 8 years ago | (#13800920)

I'm told that it depends upon the price of fish in Tibet on Tuesdays.

Re:Oldie but goodie... (0)

Anonymous Coward | more than 8 years ago | (#13800932)

He would chuck what a woodchuck could chuck, if a woodchuck could chuck wood.

Re:Oldie but goodie... (2, Interesting)

toddbu (748790) | more than 8 years ago | (#13800966)

Alternatively:

How many cans can a canner can, if a canner can can cans?

Re:Oldie but goodie... (1)

spacecowboy420 (450426) | more than 8 years ago | (#13800993)

As much as he could.

One possible solution: (5, Funny)

heinousjay (683506) | more than 8 years ago | (#13800866)

Turn a light on.

Fork in the road (1, Interesting)

Anonymous Coward | more than 8 years ago | (#13800867)

You are stranded on an island, on a path which splits in two directions. One direction takes you to "The Village of Death", the other path takes you to "The Village of Life." There are two tribes of people living on the island, one which ALWAYS TELLS THE TRUTH and one which ALWAYS LIES. A person is standing at the fork in the road. What is the ONE QUESTION (micro-variants don't count) you can ask this person which will ALWAYS get you to the Village of Life. Remember that you don't know which tribe the person is from.

Re:Fork in the road (1)

Xavic (826170) | more than 8 years ago | (#13800913)

which way is your village...

Re:Fork in the road (1)

rah1420 (234198) | more than 8 years ago | (#13800917)

Assuming that the truth-teller comes from the village of life (you didn't specify) then simply point down one of the forks and ask the person "Is this the road to your village?"

If the person answers "Yes" then take the path.

Reasoning is that the truth teller will always tell the truth and that's the right way. Since the liar will always lie, he'll say "yes" as well and that will be a falsehood.

Re:Fork in the road (1)

BJH (11355) | more than 8 years ago | (#13800990)

No.

IF truthteller AND path_to_own_village THEN YES
IF liar AND path_to_own_village THEN NO
IF truthteller AND NOT path_to_own_village THEN NO
IF liar AND NOT path_to_own_village THEN YES

That still leaves you with no way to know which one is the correct path.

Re:Fork in the road (1)

75th Trombone (581309) | more than 8 years ago | (#13800934)

Answer being:

'If I ask someone from the other tribe which path leads to the Village of Life, what will he tell me?'

Re:Fork in the road (1)

Robber Baron (112304) | more than 8 years ago | (#13800944)

You are stranded on an island, on a path which splits in two directions. One direction takes you to "The Village of Death", the other path takes you to "The Village of Life." There are two tribes of people living on the island, one which ALWAYS TELLS THE TRUTH and one which ALWAYS LIES. A person is standing at the fork in the road. What is the ONE QUESTION (micro-variants don't count) you can ask this person which will ALWAYS get you to the Village of Life. Remember that you don't know which tribe the person is from.

That's easy. You pick either guy (it doesn't matter which one) and ask him: "If you were the other guy, which way would you tell me to go to get to the village of life?", and then choose the opposite path to the one he tells you.

Re:Fork in the road (1)

JohnnyBigodes (609498) | more than 8 years ago | (#13800946)

That's an easy one. You just ask him where he's from. You will always get to go to the Village of Life.

Re:Fork in the road (1)

iamdrscience (541136) | more than 8 years ago | (#13800998)

You can ask them a question about anything that you already know, i.e. hold up 3 fingers and ask how many fingers you're holding up. You can ask about their tribe or villages because you don't know where they're from so you won't know the truth from a lie.

Petals of the Rose (4, Interesting)

Alien54 (180860) | more than 8 years ago | (#13800871)

I personally like the petals of the rose [google.com]

Bill Gates is said to have solved the problem by memorizing the combinations first [borrett.id.au] , the brute force approach.

It ones of those that requires a knack for seeing the simple things

Re:Petals of the Rose (1)

two_stripe (584918) | more than 8 years ago | (#13800963)

Hehe. Thats quite a good one. I figured it out after two rolls, though i can see how frustrating it would be if you didnt pick up on this quickly :)

What angle forms when it is 2:15? (1)

imstanny (722685) | more than 8 years ago | (#13800875)

What is the degree of the angle between the hour hand and the minute hand when it is 2:15?

If you have a piece of paper, and you draw any quadrilateral of any size (rhombus, rectangle, or square) on that piece of paper. How can u divide that piece of paper in half so that it also evenly divides teh quadrilateral?

If you have a 5 gallon jug and a 3 gallon jug of water, and a hose so u can refill any as u please. What are the steps to get exactly 4 gallons of water?

Jugs (2, Interesting)

Noksagt (69097) | more than 8 years ago | (#13800921)

If you have a 5 gallon jug and a 3 gallon jug of water, and a hose so u can refill any as u please. What are the steps to get exactly 4 gallons of water?
Fill the 5G jug. Pour it into the 3G jug, so you have 2G in the 5G jug. Empty the 3G jug & pour the 2G from the 5G jug into the 3G jug. Refill the 5G jug & finish filling the 3G jug. It will only take 1G, so you will have 4G in the 5G jug.

Re:What angle forms when it is 2:15? (1)

st0rmshad0w (412661) | more than 8 years ago | (#13800977)

Jugs-

Fill the 3G and pour it into the empty 5G. Refill the 3G and pour into the 5G until full, this leaves 1 gallon in the 3G. Empty the 5G and pour the contents of the 3G (1 gallon) into the 5G. Now fill the 3G and pour completely into the 5G, making 4 gallons total in the 5G.

SEND + MORE = MONEY (1)

Radio Shack Robot (640478) | more than 8 years ago | (#13800877)

SEND + MORE = MONEY What number does each word represent? The letters represent a single decimal digit. There is only 1 solution to this problem.

Sample Question Answer (0)

Anonymous Coward | more than 8 years ago | (#13800880)

I think the answer is either:

1. Turn the lights on, put 9 heads in each group
2. Cut all the coins in half and seperate...close enough to being split up

both may be technically correct, the best kind of correct...

cnidarian

Sleeping Beauty! (0)

Anonymous Coward | more than 8 years ago | (#13800883)

The case of Sleeping Beauty:

We plan to put Beauty to sleep by chemical means, and then we'll flip a (fair) coin. If the coin lands Heads, we will awaken Beauty on Monday afternoon and interview her. If it lands Tails, we will awaken her Monday afternoon, interview her, put her back to sleep, and then awaken her again on Tuesday afternoon and interview her again.
The (each?) interview is to consist of the one question: what is your credence now for the proposition that our coin landed Heads?

When awakened (and during the interview) Beauty will not be able to tell which day it is, nor will she remember whether she has been awakened before.

She knows the above details of our experiment.

What credence should she state in answer to our question?

-------
Some people, "halfers", consider that the answer to this question is obviously "1/2". Others, "thirders", consider that it is obviously "1/3".

Easy Solution using Advanced Physics (0)

Anonymous Coward | more than 8 years ago | (#13800885)

Fork universe as needed.

flip arbitrary number of quarters, seperate into two even groups.

If groups do not contain an equal number of heads up quarters, destroy universe.

The universe that remains will have the right number of quarters.

How do you turn a cabin into a beach hut`? (1)

Hektor_Troy (262592) | more than 8 years ago | (#13800887)

You integrate it.

Spoiler (1)

Hektor_Troy (262592) | more than 8 years ago | (#13800902)

Integrate cabin = log( cabin ) + c.

"Log cabin plus sea".

How about... (1)

mnemonic_ (164550) | more than 8 years ago | (#13800890)

a closed form solution to the Navier-Stokes equations? Quite a riddle I'd say.

I + I = (0)

Anonymous Coward | more than 8 years ago | (#13800892)

a window |+|

why.. (0)

Anonymous Coward | more than 8 years ago | (#13800896)

Why do they call them slashdot editors?

What do you get if you multiply 6 by 9? (2, Interesting)

Doc Ruby (173196) | more than 8 years ago | (#13800903)

42

Re:What do you get if you multiply 6 by 9? (1)

fcrick (465682) | more than 8 years ago | (#13800954)

if 42 is actually a number is base 13, this works out...4 13s and 2 is 54, which is 6 times 9

Re:What do you get if you multiply 6 by 9? (3, Funny)

Anonymous Coward | more than 8 years ago | (#13800967)

Nobody writes jokes in base 13.

My favorite logic riddle... (0)

Anonymous Coward | more than 8 years ago | (#13800910)

How many Slashdot editors does it take to change a light bulb?

Phone Numbers (1)

futurekill (745161) | more than 8 years ago | (#13800922)

1. Grab a calculator (You won't be able to do this one in your head) 2. Key in the first three digits of your phone number (NOT the area code) 3. Multiply by 80 4. Add 1 5. Multiply by 250 6. Add the last 4 digits of your phone number 7. Add the last 4 digits of your phone number again 8. Subtract 250 9. Divide number by 2 Do you recognize the answer?

Re:Phone Numbers (0)

Anonymous Coward | more than 8 years ago | (#13800982)

First 3 digits = x, last 4 digits = y

x
80x
80x + 1
2000x + 250
2000x + 250 + y
2000x + 250 + 2y
2000x + 2y
1000x + y

Do you recognize the answer?
Why, I certainly do.

Answer to the Sample Question (1)

FreemanPatrickHenry (317847) | more than 8 years ago | (#13800923)

I haven't thought about a more efficient way, but it seems that the surest bet would be to split the coins into two equal groups, and then flip all the coins in each group. You're most likely to end up with 25 heads up coins in each group (being the maximum entropy state).

Re:Answer to the Sample Question (0)

Anonymous Coward | more than 8 years ago | (#13800947)

I'd say take 25 quarters in each hand and hold them all on edge.

Re:Answer to the Sample Question (0)

Anonymous Coward | more than 8 years ago | (#13800999)

flipping all the coins is the same as flipping none of the coins... unless they are already an equal number of heads, you haven't accomplished anything

An original brain teaser (2, Interesting)

bratwiz (635601) | more than 8 years ago | (#13800927)

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 nth root of n (1)

ChipMonk (711367) | more than 8 years ago | (#13800937)

f(x)=x^(1/x)

It is not defined for negative values or 0. It is defined only for x>0. At 1, f(x)=1, then it peaks somwehere in 2.71
What I'm really interested in is the first derivative. Where f'(x) is 0, is the maximum of f(x). Just one catch: no limits in the formula. I don't want something that I need to calculate forever; I want a formula giving me a value I can calculate to an arbitrary precision.

The problem comes into play with the (1/x) in the exponent. All attempts to derive this result in a formula which contradicts itself, due to 0 being in either the base or the denominator of the exponent in the original f(x).

It is defeating all attempts of my college-intro-to-calculus. Searches at wolfram.mathematica.com don't help, either (or I'm just doing the wrong searches). Who can help?

Re:the nth root of n (1)

croto (909381) | more than 8 years ago | (#13801000)

I don't exactly understand what you you're interested in... is it just the first derivative? Ok it goes like that: log f=(1/x) log(x) ---> f'/f=((1/x)log(x))'=(1-log(x))/x^2 ---> f'(x)=x^(1/x) (1-log(x))/x^2 if you want to know where it peaks, well, f'(x)=0 when log(x)=1, that is, when x=e I just calculated it on a paper napkin, so it might have some mistake... but the idea should be clear

Birthday Paradox (0)

Anonymous Coward | more than 8 years ago | (#13800942)

If there are 23 people in a room, what are the chances that at least two of them will have the same birthday?

see http://en.wikipedia.org/wiki/Birthday_paradox [wikipedia.org] for more info.

.999=1 (0)

Anonymous Coward | more than 8 years ago | (#13800945)

'Nuff Said

For the sake of the rest of the discussion (1)

Y-Crate (540566) | more than 8 years ago | (#13800951)

Just to get an easy, universally-agreed-upon one out of the way.... .9999 =! 1

Now we can move on to questions that can generate some real debate.

As I was walking to St. Ives... (3, Funny)

LeonGeeste (917243) | more than 8 years ago | (#13800964)

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?

Fuuuck... (0)

Anonymous Coward | more than 8 years ago | (#13800973)

I picked the wrong night to log on to Slashdot after some heavy drinking.

And I thought my head was hurting before...

A True/False Oldie but Goodie (2, Interesting)

Quirk (36086) | more than 8 years ago | (#13800976)

Since I tend to muck about in philosophy, history and epistemology I'll go with perhaps the most ancient riddle.

Epimenides was a Cretan who made one immortal statement: "All Cretans are liars."

"The Epimenides paradox [wikipedia.org] 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.

Solution!! (0)

Anonymous Coward | more than 8 years ago | (#13800981)

My girlfriend figured it out, I guess that makes her hot.

Seperate 18 of the coins. The amount of heads remaining in the original pile will be equal to the amount of tails in the new pile. Flip over all the coins in the new pile to get the piles to have an equivilent number of heads.

Example
New Pile -- 5 heads, 13 tails
Old Pile -- 13 heads, 29 tails

flip new pile and get:

New Pile -- 13 heads, 5 tails
Old Pile -- 13 heads, 29 tails

Divisible by 3 or 6? (1)

jkevin99 (801278) | more than 8 years ago | (#13800983)

I learned this trick in grade school many years ago; I am often surprised at how many people haven't heard of it:

How can you *quickly* determine if an integer is divisible by 3? By 6?


Answer (with words reversed and spaces removed so you can think about it without seeing the answer too easily):


.wellas6bydivisibleisitthen2byand3bydivisibleevenl yissumtheifAnd.well as3bydivisibleisnumberoriginalthethen,3bydivisible evenlyisnumbertheindigitstheof allofSUMtheIf

Truth vs. Lies (3, Insightful)

sheetsda (230887) | more than 8 years ago | (#13800987)

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?"

a few of my favs (1)

lawpoop (604919) | more than 8 years ago | (#13800994)

First up, does Goedel's incompleteness theorem imply that computers will never be able to have human-like intelligence [wikipedia.org] ?

Other things I like are not necessarily problems, but things that just inspire awe, such as proving that .99... = 1, or that formula that shows pi*log^-1 = 0 or whatever it is... is always struck me as the Grand Unification Theory of algebra and geometry. It's so simple and shows that these numbers, which are so hard for me to work with, combine in some fashion to show some property that is strikingly simple, like finding a beatiful crystal of clarity in a quarry of grey, difficult mathematics.

$10 to Anyone Who Can Solve This (1)

Comatose51 (687974) | more than 8 years ago | (#13800995)

Very simple math "riddle" here [claymath.org] . If you solve it, just send me the answer and I'll PayPal you the $10 for your work...*

1.Find problem on Slashdot
2.Solve problem
3.Email to generous stranger
4.Get $10 via PayPal
5.Profit!
See how easy that is! No tricks here...

* - by accepting the $10, you transfer all claims to any further monetary compensation from any party to Comatose51, including but not limited to $10 million from the Clay Institute

Lightbulb problem (4, Interesting)

Ellen Spertus (31819) | more than 8 years ago | (#13800996)

Given:
  • 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 [nicemice.net] and solution [nicemice.net] .

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.

Load More Comments
Slashdot Account

Need an Account?

Forgot your password?

Don't worry, we never post anything without your permission.

Submission Text Formatting Tips

We support a small subset of HTML, namely these tags:

  • b
  • i
  • p
  • br
  • a
  • ol
  • ul
  • li
  • dl
  • dt
  • dd
  • em
  • strong
  • tt
  • blockquote
  • div
  • quote
  • ecode

"ecode" can be used for code snippets, for example:

<ecode>    while(1) { do_something(); } </ecode>
Create a Slashdot Account

Loading...