# Your Favorite Math/Logic Riddles?

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

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?"*

## Soduku (3, Interesting)

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

-B

## Link to online version (1)

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

## Math and science are obsolete (-1, Offtopic)

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

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)

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

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

Have a nice day!

-Jacob S.

## Another online version (5, Interesting)

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

## FP (-1, Offtopic)

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

## Riddle (1)

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

How many sprinkle cupcakes should you make?

## Re:Riddle (3, Funny)

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

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)

14

## Re:Riddle (1)

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

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)

## Re:Riddle (2, Insightful)

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

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)

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)

## Re:Infinity (4, Interesting)

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

## Solution (3, Interesting)

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

## 2+ (1)

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

## Re:2+ (1)

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

## Keeping my skills fresh (3, Interesting)

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

## Re:Keeping my skills fresh (1)

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

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)

## Re:Keeping my skills fresh (1)

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

## Re:Keeping my skills fresh (1, Informative)

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

http://science.slashdot.org/article.pl?sid=05/09/

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)

= 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)

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,

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)

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,

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)

## easy one (5, Funny)

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

## Re:easy one (5, Interesting)

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

## Re:easy one (4, Informative)

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

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)

## Re:easy one (0)

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

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)

## Re:Sequence (1)

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

1

11

21

1211

111221

312211

## Re:Sequence (0)

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

## Re:Sequence (1)

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

preceedingline?## Re:Sequence (1)

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

## Re:Sequence (0)

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

## Re:Sequence (1)

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

## Re:Sequence (1)

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

## Look and Say (5, Informative)

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

## Re:Sequence (1)

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

## Oldie but goodie... (1)

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

## Re:Oldie but goodie... (0)

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

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

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

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

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

## Re:Oldie but goodie... (0)

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

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

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

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)

## One possible solution: (5, Funny)

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

## Fork in the road (1, Interesting)

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

## Re:Fork in the road (1)

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

## Re:Fork in the road (1)

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

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)

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)

'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)

## Re:Fork in the road (1)

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

## Petals of the Rose (4, Interesting)

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

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)

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

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

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)

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

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

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)

## Sample Question Answer (0)

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

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)

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)

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)

## Spoiler (1)

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

"Log cabin plus sea".

## How about... (1)

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

## I + I = (0)

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

## why.. (0)

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

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

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

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

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

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

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

## My favorite logic riddle... (0)

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

## Phone Numbers (1)

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

## Re:Phone Numbers (0)

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

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)

## Re:Answer to the Sample Question (0)

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

## Re:Answer to the Sample Question (0)

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

## An original brain teaser (2, Interesting)

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

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)

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)

## Birthday Paradox (0)

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

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

## .999=1 (0)

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

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

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

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)

## Fuuuck... (0)

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

And I thought my head was hurting

before...## Cheat and find the answer (1)

## Dr.Gumby (923209) | more than 8 years ago | (#13800975)

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

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

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)

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)

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):

## Truth vs. Lies (3, Insightful)

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

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)

Other things I like are not necessarily problems, but things that just inspire awe, such as proving that

## $10 to Anyone Who Can Solve This (1)

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

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)

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.