Brain Teasers For The Pub
Man alive, I’m burnin’ up on my brain.
She knows when I’m just teasin’
But it’s not likely in the season
To open up a passenger train.
 SILENT WEEKEND
Complete change of scenery today. I haven’t thought about puzzles for many years, but a recent post reminded me how much I like them. So I figured I’d dredge up a list I put together about 13 years ago and publish them here. You should be able to figure them all out over a beer without pen and paper. Seeing as I didn’t invent any of these myself, you’ll also probably be able to Google the answers if you’re the lazy, lowdown cheating type. I’m not going to include the Monty Hall Problem as it starts too many fights. Credit to all the people that created these. I can’t remember where I found them all.
Problem #1: The Dangling Cube
Take a hollow glass cube, exactly half filled with a coloured liquid. If you place the cube flat on a table, the surface of the liquid, seen from above, makes a square. What two dimensional shape does the surface make if you dangle the cube from a piece of string attached to one of the corners?
Problem #2: The Mutilated Chess Board
You have an 8×8 square (say a chessboard), and 32 dominoes each exactly the same size as two squares. It is easy to cover the chessboard (64 squares) totally using the 32 dominoes. Now, you take away one domino, and cut off two opposite corners of the chessboard (a1 and h8 for the chessplayers). Now you have 31 dominoes and 62 squares. Is it still possible to cover the board. If so, how? If not, why not?
Problem #3: Fly on the Windshield
You have two trucks, 200km apart. Both trucks are heading towards each other at a constant speed of 50km/hour. On the windshield of one truck is a fly. He flys at a constant speed on 70km/hour from the windshield of one to the windshield of the other. When he hits a windshield, he turns instantly without slowing down. He continues to do this until the 2 trucks collide, squashing the poor fly. The question is, how far does the fly travel before getting squashed.
Problem #4: The Annoying Piles
You have 10 piles of 10 coins each. One pile consists of 10 counterfeit coins, while the other 9 piles consist of 10 real coins. A counterfeit coin weighs 11g, while a real coin weighs 10g. You have a scale (a normal kitchen scale, not a balance scale). What is the minimum number of weighings needed to determine which pile is counterfeit? You are, of course, extremely unlucky.
Problem #5: A Bridge Too Far
Four people want to cross a bridge. It is a very rickety bridge, so at most 2 people can cross at the same time. Unfortunately, it is also very dark, so in order to cross the bridge, the group must be holding a flashlight. The have only one. So, clearly 2 people must cross, 1 must come back, 2 go across etc. The four people take 1, 2, 5 and 10 minutes to cross the bridge. If 2 cross together, it takes the time of the slower to cross (if 2 and 5 crossed together, it would take 5 minutes). What is the minimum time needed for them all to cross.
I got asked this in interviews with Microsoft
Problem #6: Casanova’s Conundrum
One night at a bar, you meet 3 beautiful girls (or guys). You really want to sleep with all 3, but you only have two condoms. Each of the four people involved suspects all the others may have some kind of infectious disease. The question is, how do you arrange to have sex with all three without anyone having any chance of catching a disease from any other? (I’m sure there is another less graphic way to state the problem, but I like this one).
Problem #7: Annoying Algebra
Simplify the following multiplication:
(xa)(xb)(xc) .... (xy)(xz).
That is, the are 26 terms in the multiplication. It does simplify considerably.
I got asked this in the South African Mathematics Olympiad where you had 30 mins per question.
Problem #8: The Mad Hatter
There are three black hats and two white hats. Three people, each with one hat, line up in a row like so: A B C. C can see both A and B, B can only see A, but A can’t see the other two. They can all hear each other. When person C is asked what colour hat he is wearing, he says he doesn’t know. When person B is asked what colour hat he is wearing, he doesn’t know either. When person A is asked, he knows and says it correctly. What colour hat is each person wearing?
Problem #9: The Mystery of the Missing Dollar
You and two friends decide to spend a night at a hotel. Rooms are 10 dollars per night per person. You each take 10 dollars (30 in total) and give it to the bellboy to go and order your rooms. When the bellboy gets to the reception,
he sees that there is a 3 for 25 dollars special. He pays with the 30 dollars, so gets 5 dollars change. As 5 dollars cannot easily be divied amoung 3 people, he give each person back 1 dollar, and keeps 2 for himself. Now, each person has paid 9 dollars (10 paid, 1 change), so they paid 27 dollars between them. The bellboy has 2 dollars. But 27+2=29, not 30. Where is the missing dollar?
Problem #10: The Gender Mind Bender
There is a town in a remote part of the world with a large population, and some strange habits. They all believe that one boy per family is enough. So, each couple continues to have children until they have their first boy. After this, they have no more children. Assuming that there is a 50 percent chance of each child being a boy, what will the ratio of boys to girls settle at after 20 generations? Why?
You could put the answers up here for the really lazy people and in the interests of saving the planet by avoiding google searches
Yeah, I will in a few days. Too much to post, too little time.
I hate you.
1. hexagon
2. I hate you.
3. too far for his own good
4. log2(10), rounded up == 4.
5. take a helicopter, to hell with the bridge
6. I hate condoms, screw ‘em.
7. If you’re saying a == 26 && z == 1, and x > 26, then (x1)!/(x26)! will work, I think, but I’m sure that’s not what you had in mind. In which case: I hate you.
8. Polka dot.
9. The total amt of money hasn’t changed. $25 went to the hotel, $2 to the bellboy, and each customer has $1. It all adds up to $30. Where’s the problem?
10. 50%. Lots and lots of babies will be born in this scenario. It doesn’t matter how many. By odds, half will be boys, half will be lapdogs of Satan.
1. Hexagon
2. No. Each domino covers 1 black and 1 white square, but 2 black squares have been removed > impossible.
3. 140 km.
4. 1 try (scale 1 coin from pile 1, 2 coins from pile 2, etc.)
5. 19 min.
6. ?
7. 0 (xx = 0).
8. ?
9. Nothing is missing. 25 to the hotel, 2 to the bellboy, 3 back in the pocket of the customers.
10. 1 to 2.
8. Black
[The fact that C didn't know his hat eliminates the possibility that C saw two white hats. So, C saw at least one black hat. If B had seen a white hat, B would have known that he was wearing the black hat seen by C. B didn't know, therefore B was seeing a black hat on A's head.]
I like no.8
A, B, C are rowed up: b is looking at a’s back, c is looking at a & b’s back, a is looking at the wall
c can see a and b, b sees a, a sees no one
c must see 1 black hat and 1 white as he’s not able to make a guess. He is therefore be wearing black
b hears nothing from c but can see a is wearing black. He is in fact wearing white
a knows he’s wearing black and states this.
When we finally get to meet, I’m going to NOT buy you a pint!
Here’s one for you:
2 doors — one leads to life, the other, death. Problem: they are not marked *life* or *death*.
Both doors have a guard. You may ask one guard one question to make your selection.
Further problem: one guard tells the truth, the other lies {no doubt the manager!}. There is no discernible difference between the guards.
What one question could ask either guard to ensure you make the right choice?
Grrrrrrrrr! What a way to start Friday!!!!!!!
you see, this is why projects go wrong, you are all reinventing the answers I shall wait for the answers and appreciate the cleverness.
Oh and I think the answer to 3 is no where. When the train and the fly collide they are not moving. If you assume squashed to me the collision then no where is the answer. Anyway, I shall not try and reinvent anything.
I got #6 while running yesterday! You start with 2 condoms! You do A and carefully remove the outtermost condom. Reserve.
Now you do B with the innermost condom.
Finally, you put the reserved condom back on, inside out!
And do C. Done.
Ok, here’s my attempt:
1. Triangle
2. yes,if arranged properly – just a guess
3. i’ll go with stefane’s 140 since i am too lazy to calculate it.
4. 1 try – stefane
5. 17 mins (1,2 go together, 1 goes back, sends 5,10 together, 2 comes back and picks 1 up)
6. 3sum/inside out – but i didn’t think of Mo’s soln.
7. x^26 – x^25(a+b+c+…+z) – x^24(az+bz+cz+..+yz) + x^23(acz+bcz+…+xyz)+… – abc…z
8. A – black, B – white, C – ?
9. no missing dollar – just $25 and a $2 tip
10. wild guess – 1:2^20
waiting (impatiently) for the answers.
My solutions (from 13 years ago!) below. I think at least one person got the correct answer for all the questions.
Solution #1: The shape made is a regular hexagon.
Solution #2: It is impossible. Consider the colours of the squares of the chessboard. There are 32 white squares, and 30 black squares. Each domino will cover one black and one white square. No matter how you do things, the last domino wil have to cover two white squares, which is impossible.
Solution #3: 140 km. It takes 2 hours for the trucks to collide. So the fly is flying for two hours at 70 km/h. He flies 140 km. In a test, university students took longer to answer this than 12 year olds.
Solution #4: 1 weighing. Take one coin from the first pile, two from the second and 10 from the last pile. You have 55 coins on the scale, which would weigh 550g if all coins were real. If it weighs 1g more, then the 1st pile is counterfeit, 2g more the second etc.
Solution #5: 17 minutes: 1 and 2 go across – 2 mins. 1 comes back – 3 mins. 5 and 10 go across – 13 mins. 2 comes back – 15 mine. 1 and 2 go across – 17 minutes
Solution #6: First, put on both condoms together, and ‘do’ A. Then, take off the top condom, and do B. Finally, turn the removed condom insideout, put it back on over the first, and do C. This is the only way to do it.
Solution #7: 0. The 24th term in the expansion is (xx) = 0. Anything multiplied by zero is zero. Sorry.
Solution #8: A has a black hat. As far as I can see, it is not possible to determine the hats of the other two. When C says he doesn’t know, this must mean that it is impossible for both A and B to have white hats. Otherwise, C would know he had a black hat. Thus, when B say he doesn’t know, A must have a black hat. If B saw that A had a white hat, he would know he had a black hat. So, A can determine that he is wearing a black hat.
Solution #9: The question is again completely misleading. Only twentyseven dollars were paid by the three men. 25 went to the hotel and 2 to the receptionist. Trying to add to get back to 30 is simply nonsensical.
Solution #10: 1:1, as in any normal town. Every birth is natural, and each child still has a 50/50 chance of being a boy or a girl. No babies are being killed or anything. It wouldn’t matter if they stopped after having a child with blue eyes, a child with 7 toes etc. Alternatively, you could sum the infinite series of probabilities. Each family has one boy. There is a 50% chance of no girls, and 25% chance of 1 girl, a 12.5% chance of 2 girl etc.
Wow.. I can’t believe I totally missed the (xx).
Problem #6 is a little trickier if you are a girl trying to “do” 3 boys. But a similar reasoning applies: Boy A puts 2 condoms on and do you. Then boy A gives the outer condom to boy B. Boy B does you. Finally, boy A gives the innermost condom to boy C, who wears it insideout. On top of it, boy C also wears the condom from boy B. Properly doubly protected, boy C does you.
#10. Your answer seems written up wrong. Does not 50% chance of no girls equal 50% chance of a girl first up. Then 25% chance of 2 girls (or a boy and a girl). Followed by a 12.5% chance of a girl ,girl, boy or girl, girl, girl combination?
I am totally confused, although I guess the probability numbers do add up equally for both sexes. My impression was that as only people with girls and no boys continue breeding, there would be more girls. Then I threw a coin 100 times and was dismayed to get the equivalent of 97 boys and 86 girls where heads symbolized a boy and tails a girl.
Not quite:
50% chance of 0 girls, 1 boy
25% chance of 1 girl, 1 boy
12.5% chance of 2 girls, 1 boy
6.25% chance of 3 girls, 1 boy
There will always a be 1 boy in there. But it makes more sense to think about it the other way. Every birth is just a normal 50%50%. Doesn’t matter when any particular family decides to stop.
Hardly for the pub are they !! Unless you bring dominoes, a chess board, a glass cube etc. Id say your local’s a barrel of laughs.
Solution#6: Is this one of the right solution for problem#6? Use first condom and do A. Use second condom and do B. Use the first condom insideout, and the second condom reversed on top of that, to do C. Is this right?
No, not quite. Your solution has a pretty messy reversed second condom, so C would slap you in the face and walk out the room. See my solution higher up in the comments.
That’s a great point of view, yet seriously isn’t make any sence by any means talking about which mather. Any means with thanks plus i’d try to reveal your own posting into delicius but it seems to be issues using your blogs is it possible to make sure you recheck the item. thanks once again.
Wearing two condoms puts you at risk of either or both of them tearing. So there’s no safe way to do it.
Hi, how’s it going? Just shared this post with a colleague, we had a good laugh.
I think one big problem with many SEO companies (both big and small) is a lack of transparency to the client.
Just randomly stumbled across these brain teasers and my colleagues and I decided to have a go at them!
That’s a smart answer to a diclifuft question.
Kewl you should come up with that. Exltnlece!
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