I need some really difficult riddles, puzzles, and logic challenges.
November 3, 2011 9:58 AM Subscribe
Co-workers of mine insist on posting little brain teasers and such all around the office: riddles, mazes, jumbles, crosswords, etc. Help me find remarkably difficult (but possible) puzzles that I can use to troll them. Bonus points if they look, at first glance, relatively simple.
This one's been doing the rounds recently.
posted by Chocolate Pickle at 10:06 AM on November 3, 2011 [4 favorites]
posted by Chocolate Pickle at 10:06 AM on November 3, 2011 [4 favorites]
Check out Scam School. It's more of a bar scam thing so some stuff will be impossible, or at least really hard to pull off.
posted by theichibun at 10:15 AM on November 3, 2011
posted by theichibun at 10:15 AM on November 3, 2011
Response by poster: kagredon: "The Self-Referential Aptitude Test (as linked in this MeFi thread.)"
This is so exactly what I meant. Printed and posted. More like this!
posted by Plutor at 10:39 AM on November 3, 2011
This is so exactly what I meant. Printed and posted. More like this!
posted by Plutor at 10:39 AM on November 3, 2011
How about one that's actually pretty simple, but seems like it's hard and/or just busywork? Most of the people I've told it to have given up after a few minutes.
You have a string of 100 christmas lights with a toggle switch. To start with, the lights are all off. On the first switch, every first light (which is to say, every light) changes state, so they all turn on. On the second toggle, every second light changes state (in this case, they turns off). On the third toggle, every third light changes state. And so on, until you've toggled this string of lights 100 times.
After 100 toggles, how many (and which) lights are left on?
Answer: (hover for spoiler)
posted by phunniemee at 10:41 AM on November 3, 2011
You have a string of 100 christmas lights with a toggle switch. To start with, the lights are all off. On the first switch, every first light (which is to say, every light) changes state, so they all turn on. On the second toggle, every second light changes state (in this case, they turns off). On the third toggle, every third light changes state. And so on, until you've toggled this string of lights 100 times.
After 100 toggles, how many (and which) lights are left on?
Answer: (hover for spoiler)
posted by phunniemee at 10:41 AM on November 3, 2011
Take a 1 mile by one mile square. Create fences around/through this square, without going outside the bounds of the square, such that there is no place where one can draw a straight line from a point outside the square to another point outside the square, crossing the borders of the square but not the fence. Minimize the length of the fence.
posted by novalis_dt at 10:49 AM on November 3, 2011
posted by novalis_dt at 10:49 AM on November 3, 2011
Other paradoxes are also fun. I've always had fun with Russell's Paradox:
Some sets are members of themselves, and some are not. For example, the set of all squares is not a member of itself, because it is a set, not a square. While the set of all non-squares is a member of itself because it is a set rather than a square. Is the set of all sets which are not members of themselves a member of itself?
Or:
Suppose there is a town with just one barber. In this town, men keep themselves clean-shaved by one of two methods: (a) they shave themselves; or (b) they visit the barber. Or, stated another way, the barber shaves each and only those men who do not shave themselves. Who shaves the barber?
posted by Hylas at 10:54 AM on November 3, 2011
Some sets are members of themselves, and some are not. For example, the set of all squares is not a member of itself, because it is a set, not a square. While the set of all non-squares is a member of itself because it is a set rather than a square. Is the set of all sets which are not members of themselves a member of itself?
Or:
Suppose there is a town with just one barber. In this town, men keep themselves clean-shaved by one of two methods: (a) they shave themselves; or (b) they visit the barber. Or, stated another way, the barber shaves each and only those men who do not shave themselves. Who shaves the barber?
posted by Hylas at 10:54 AM on November 3, 2011
I also heard a couple of good tree-related ones:
(1) Where do trees get their mass? That is, where does the matter that makes up a tree mostly come from?
(2) If you look at successive cross-sections of a branch as it leaves the trunk of a tree, will the branch get smaller or larger?
Both of these can be figured out using only what a child knows about trees, but
posted by novalis_dt at 11:25 AM on November 3, 2011
(1) Where do trees get their mass? That is, where does the matter that makes up a tree mostly come from?
(2) If you look at successive cross-sections of a branch as it leaves the trunk of a tree, will the branch get smaller or larger?
Both of these can be figured out using only what a child knows about trees, but
posted by novalis_dt at 11:25 AM on November 3, 2011
The Monty Hall Problem seems simple, but a lot of people inevitably get it wrong, and can't be convinced that they're wrong. At least I couldn't until I saw a complete diagram of all the possible outcomes.
A man goes bear hunting. Facing due south, he spots a bear running across his field of view. Still facing due south, he fires, and shoots the bear. What color is the bear?
(The bear is white. The only place where you are always facing due south is the North Pole).
posted by Gilbert at 11:34 AM on November 3, 2011
A man goes bear hunting. Facing due south, he spots a bear running across his field of view. Still facing due south, he fires, and shoots the bear. What color is the bear?
(The bear is white. The only place where you are always facing due south is the North Pole).
posted by Gilbert at 11:34 AM on November 3, 2011
Oh I have a few...here are two that's almost naughty:
What am I?
You can hold me and when you need to use me, you open my legs. When you're done with me or don't need me, you toss me aside.
Answer: scissors
What am I?
I am dirtier and dirtier as you soap and wash.
Answer: water
I have another, but it requires a diagram. It's very simple and cute. How would I share that with you?
posted by Yellow at 12:43 PM on November 3, 2011
What am I?
You can hold me and when you need to use me, you open my legs. When you're done with me or don't need me, you toss me aside.
Answer: scissors
What am I?
I am dirtier and dirtier as you soap and wash.
Answer: water
I have another, but it requires a diagram. It's very simple and cute. How would I share that with you?
posted by Yellow at 12:43 PM on November 3, 2011
I've always enjoyed the Birthday Paradox.
How many people need to be in a room, for there to be a 50-50 chance that two will share a birthday?
posted by LeanGreen at 12:50 PM on November 3, 2011
How many people need to be in a room, for there to be a 50-50 chance that two will share a birthday?
posted by LeanGreen at 12:50 PM on November 3, 2011
This one is fun, and it can be presented as a written question without the interactive aspect.
Also, one of my all-time favourites is the twelve ball weighing. You have twelve apparently identical balls and a simple balance. You know that one of the balls is a different weight to the others, but you do not know whether it is lighter or heavier. You have precisely three weighings to determine not only the odd ball out but also whether it is heavier or lighter than the rest.
Memail if you can't figure it out. :-)
posted by Decani at 2:09 PM on November 3, 2011
Also, one of my all-time favourites is the twelve ball weighing. You have twelve apparently identical balls and a simple balance. You know that one of the balls is a different weight to the others, but you do not know whether it is lighter or heavier. You have precisely three weighings to determine not only the odd ball out but also whether it is heavier or lighter than the rest.
Memail if you can't figure it out. :-)
posted by Decani at 2:09 PM on November 3, 2011
Draw a rectangle, wider than it is tall. Draw a line inside this rectangle, bisecting it horizontally. Bisect the top half vertically, and draw lines in the bottom half to cut it in thirds. There should now be five inner sections of your rectangle. Now tell them to draw a line such that it goes through the center of each line segment exactly once. The twist? It can't be done.
posted by Night_owl at 11:55 PM on November 3, 2011
posted by Night_owl at 11:55 PM on November 3, 2011
You and a bunch of your friends are taken prisoner. There are n of you.
You are given some time to plan a strategy with your friends, after which you will be placed in individual cells so you can't communicate.
There is a room with two switches in it, each with two positions.
Every so often (who knows how long) the guards will pull one prisoner into a room. This prisoner will be required to flip one, and only one, of the switches. You can't write on the walls, scuffle the dust, or do anything else-- the only thing a prisoner can do is switch one switch, and then leave.
You are told that at various points in time, everyone will have been pulled in the same number of times-- but who knows how many. In other words, prisoner A could be pulled in 13 times before prisoner B ever comes in once, and there's no known order so the prisoners could be pulled in in any ordering. But eventually, if you wait long enough, every prisoner will have been in the same number of times (you just don't know what number of times this will occur at).
If at any point one of the prisoners can say, with certainty, that every single prisoner has been pulled into the room at least once, then everyone goes free. (If the prisoner is wrong, you all die, so don't guess.)
So, what's your strategy? What do you agree to do in order to survive and be set free as soon as possible?
posted by nat at 12:34 PM on November 4, 2011
You are given some time to plan a strategy with your friends, after which you will be placed in individual cells so you can't communicate.
There is a room with two switches in it, each with two positions.
Every so often (who knows how long) the guards will pull one prisoner into a room. This prisoner will be required to flip one, and only one, of the switches. You can't write on the walls, scuffle the dust, or do anything else-- the only thing a prisoner can do is switch one switch, and then leave.
You are told that at various points in time, everyone will have been pulled in the same number of times-- but who knows how many. In other words, prisoner A could be pulled in 13 times before prisoner B ever comes in once, and there's no known order so the prisoners could be pulled in in any ordering. But eventually, if you wait long enough, every prisoner will have been in the same number of times (you just don't know what number of times this will occur at).
If at any point one of the prisoners can say, with certainty, that every single prisoner has been pulled into the room at least once, then everyone goes free. (If the prisoner is wrong, you all die, so don't guess.)
So, what's your strategy? What do you agree to do in order to survive and be set free as soon as possible?
posted by nat at 12:34 PM on November 4, 2011
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posted by kagredon at 10:02 AM on November 3, 2011 [4 favorites]