If you have other brainteasers, please e-mail them to
Thanks - Chris
Which number has to be put into the first of the seven empty fields (greyed out) if you know that the sum of three consecutive numbers is always 19?
We are standing at the beginning of a very long row of light bulbs (let's say 20'000 of them, but could be infinitely many), each with a little chain hanging from it, which - when pulling on it - switches the respective light on if it was off, or off if it was on. In the beginning all of the lights are off.
Now the first person comes along and pulls every string, turning every light on. Then the second person comes and pulls every second string, turning every second light off. The third person will pull every third string, turning light 3 off, light 6 on again, etc. And so it continues with the nth person switching every nth light.
After the last person (e.g. after 20'000 or infinitely many, whatever you are more comfortable with) has gone through, which lights are on, and which are off?
Sorry, I could not think of a better title. Puzzler is quite mathematical (that is a warning I guess) and was invented by Tyler.
Given a 2x1 pool table, we shoot a (lonely) ball from one of the pockets in a line with slope m (or with some angle tan-1(m) ). Assume there is no friction (i.e. that it keeps going forever) until it perfectly hits a pocket. How many times will it bounce?
The puzzler being mathematical in nature (and to actually simplify things), we assume the ball as well as the pockets being points, i.e. infinitely small.
Imagine there is a (wooden) plank floating in the ocean. Let's say it is 100 [feet/meters/light years/whatever you prefer] long and one Lemming wide. Now we have Lemmings. Those little creatures just walk straight ahead (let's say with a velocity of one unit of length per unit of time) until they either fall off the plank at one of the two ends or bump into another Lemming. As they cannot cross they both instantaneously (let's assume) turn around and walk in the directions they just came from.
You are now free to place as many Lemmings onto that plank as you wish, wherever you want them and have them face whichever direction you prefer. Then let them go. After they started walking you cannot influence them anymore, i.e. you cannot add or turn around Lemmings.
What is the longest time you can have Lemmings on that plank? How do you arrange them? Why?
Look at the 'proof' in this pdf file (or here for a German version).
Let me know what went wrong.
I got this one from Tyler today, who got it from a friend
of his, Anthony Hodsdon, a couple days before.
See his original post of it,
I only put it here to not have to depend on his whim. That's what he wrote (i.e. I take
no responsability for the writing style):
An insane king sits alone on an infinite chessboard. He's probably crazy due to vast overwhelming loneliness of being the only chess piece on an infinite board, much as Euler probably felt while exploring the far reaches of mathematics he could hardly hope to explicate to the far-inferior second-best savants of the world.
In any case, the point is that our king is bound in a straightjacket. And which way must one travel in said garment? Why straightly, of course (being a king he moves stately as well). This means the king cannot move diagonally. Rather, he can move one square per turn either up, down, left, or right.
Enter the advisary. A demoness removes squares from the board one at a time. Once a square is removed, it may not be replaced, and the king may not move there. The king and demoness take turns, the king going first (albeit a moo point). The demoness wins if she can completely surround the king -- that is, if she can remove enough squares so that the king has finitely many places to move to. Implicitly, the king "wins" if he has a strategy which will indefinitely avoid such confinement.
The question is: who will win? Assume that both players have infinite mental capacity, id est, that they both play their own optimal strategy. Of course, I [Tyler] wouldn't have been so eager to post this if I did not have my own solution ready as well.
I heard this one today on CarTalk so you can also read it there. It's not too hard, but I had an idea for a nice (I think) extension of it.
You have seven stacks of 100 coins. Each regular coin weighs 10 grams. But some of the stacks of coins are fake, making them 11 grams each. Within a stack of 100 all the coins are identical. Several, all or none of the stacks can contain fake coins, so not only one of them.
With a scale that gives you absolute weights, how can you identify all the fake coin stacks with one weighing?
Fake coins may be heavier as well as lighter, i.e. 11 or 9 grams per coin. To make it a bit easier, let's say we have only 5 stacks of 100 coins.
Can you still identify all the fake coins as well as if they are 'light-fake' or 'heavy-fake' with one weighing?
You have nine identically looking balls, one of them is heavier than the others though.
How do you find the heavier ball with two weighings of a balance (one that doesn't give absolute weights but that lets you compare).
If that was too easy, do the one with twelve balls below.
You have 12 balls, all looking identical. One of them is lighter or heavier than the other ones (you don't know which). The difference is slight enough that you can't tell just by hand.
You have a balance, to compare two weights, i.e. two groups of balls.
How can you tell with three weightings which one is the fake?
Supposedly there are four ways of doing it, I only figured out one though.
Thirteen pirates found a big treasure, put it in a chest and want to lock it up with several locks. They want to distribute keys in a way, so that any seven (but not less) of them can open the chest.
How many locks do they need (at least) and which pirate gets how many (and which) keys to accomplish that?
If there were three pirates, they would need three locks. The first pirate would get keys for lock 1 and 2, the second one for lock 2 and 3 and the third one keys for lock 1 and 3. Any two of them are now able to open the chest.
Now, hush hush, do the same for the thirteen guys, that captured you and make your very survival dependent on your fast answer, yuck, they look grim indeed, and... well, better hurry up.
You have two fuses that burn down in 30 seconds each. Unfortunately, they burn very irregularily, i.e. sometimes faster and sometimes slower, i.e. after half the time, much more or less than half of the length of the fuse might be burnt, you only know that the whole fuse takes 30 seconds to burn down completely. Also, the two fuses are not identical.
First Question: How can you measure 45 seconds?
Second Question: How can you measure 22.5 seconds?
Thanks to G for the solutions.
(a) Move two matches to make four squares of equal size. No "lonely" toothpicks allowed, of course.
_ _ _ _|_|_|_| |_|_|
(b) Move three matches to get three squares. (A good start might be to do it with four matches.)
_|_|_ _|_|_ | |
You have a field of four by five squares and five pieces as follows:
_ _ _ _ _ _ _ _ _ _ _ _ |_|_| |_|_|_ |_|_|_| _|_|_ |_|_|_|_| |_|_| |_|_| |_| |_|_|_|
Either fill the 4x5 field with those five pieces or show why it is not possible (there's a elegant proof).
What size staircase can I build if I have 100 toothpicks?
Look at the picture to see what is meant by "staircase".
The puzzle was for a friend's friend math education class. I have a solution.
An Italian mafia family of 30 in Alcatraz is about to be executed. But there is a tradition that all prisoners are placed in a line such that the 30th will see 29 in front of him, the 29th will see 28, and so forth until the first who doesn't see anyone.
Then all have a black or a white hat placed on their head whose color they do not know. Starting with the 30th (who sees 29 hats) a prisoner must try to guess the color of his hat. He says it out loud so everyone hears his answer.
If he is right, he is set free, otherwise he is shot.
Since the prisoners know about this, they can devise a plan beforehand. An obvious one is that every even person names the color of the hat on the person in front of him, and every odd person names the color he just heard (so 30th tells 29th his color, and 29th says his own, and so on). This way 15 people will definitely go free.
Can you find a better solution? I have one...
Thanks to Nikita for this teaser!
Four coins lie on a table and form a square, something like that:
o o o o
Now you are allowed to jump with a coin over another one, in any direction, i.e. also diagonally or however. You always have to jump all the way to the other side, so that the distance between 'jumper coin' and 'jumped over coin' is the same before and after the jump.
Is it possible to form a larger square? If yes, how? If no, why not?
I have a solution, ask for it if you'd like to get it.
A certain fast food restaurant sells chicken nuggets in packages of 6, 9 or 20. What is the largest amount of nuggests that you cannot buy purchasing combinations of 6, 9, or 20.
Thanks to G. for telling me! Ask for my solution by email.
I heard that one on CarTalk on NPR, so please read it there (or here if the link should be broken).
G. suggested this solution. CarTalk's official solution. Pretty much the same as G.'s, but with some pictures added.
Here is a picture explaining it all. I still couldn't solve it, even though it really looks trivial.
NK's solution, which I have not read yet (one has some pride). It is confirmed numerically by a CAD program on which NK's friend constructed the trianlge.
Aug 5, '03: CW also has a solution. (And he doesn't even seem to have to draw lines!)
Also as a graphic.
Easier. I have a solution.
There is a group of related puzzlers here. (Link can also be found in the link section below.
There was a monastery where the monks living in it spent most of their time in solitude in their cells, reading, writing or whatever they do. Only once a day they would meet for lunch. Even then, there is no communication whatsoever (no talking, no gesturing etc.) among them but they are able to see each other. In the whole monastery there are no mirrors or other ways of seeing oneself, only the abbot knows what he looks like.
One day, the abbot puts up a sign that reads:
Some of you have a sickness that expresses itself in coloring your ears green. The sickness is not transmitted among you but nevertheless, as soon as you know that you have the sickness, leave the monastary immediately. As soon as all the infected people are gone, I will remove this sign.
After seven days, all the sick monks had left and the abbot removed the sign.
How many monks had been sick?
More to come. Sooner or later. (Just have to type it up. Or YOU send me more!)
A group of Puzzles related to the Fibonacci numbers. Particularly, there is an extension to my Second Triangle Puzzle above.
Argentine Puzzle Contest. Final Test, July 5, 2000. That was part of the 9th world puzzle championship (who knew those existed?).
Two very nice puzzles by Chris Coyne.
Hoping and waiting for more to come.
I have solutions to both of them. Ask if you're interested.
Cartalk on NPR offers a weekly new puzzler. Of interst is mainly the "Really Hard Puzzlers" section.
The rec.puzzles archive