The Puzzling Thread

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This is a place for training that muscle between your ears! Here you can post puzzles, riddles and brain teasers, try to solve them and discuss your quandaries and all things puzzling.

If you want to post a puzzle for others to solve please follow these guidelines:
  • You should give a short title to your puzzle, like The SuperHard Riddle, to distinuish it from others and for easy reference.
  • You should have an answer to the puzzle. (No unsolved problems please)
  • The possible anwer(s) should be derivable from the description of the puzzle alone or combining publicly available information and should not involve personal secrets. (Like "What is currently under my pillow?")
  • Keep in mind that most puzzles and brain teasers are easily googlable: try to change the wording or invent a different story altogether.
  • You may provide hints and after a sufficient time with no progress you can post the answer but please use the Spoiler feature always when giving away hints or the final answer.
  • You should refrain from commenting every solution as it is posted, better wait a fair amount of time and post general hints not targeted to a particular solution approach
For people trying to solve the riddles the last point applies too:

When giving away hints or answers to any puzzles in this thread, plase use the Insert...Spoiler feature to hide it and not spoil the fun of others trying to solve it. Use the Puzzle Title as Spoiler Title and add "hint" or "solution".

I'll try to keep a record of all puzzles and hints and answers posted so far in post #1. Here is my preliminary system of rating riddling and solving success: each riddler gets a point for each day his puzzle can't be solved, with a capping of points to be determied. Each solver gets a point for solving a puzzle -- barring obvious rephrasing of earlier answers. Lets see if this results in a fair rating system to determine a leaderboard of puzzle solvers and riddle posters.

Puzzles
Can you put ten moles into eight holes, so that each is in one and only one hole?
Clarification: a hole has only room for one.
The young craftsman is in deep thought: he has been assigned to link the five pieces of chain to a single chain. If you count the opening of a chain link and the closing of a chain link as single operations his foreman told him that it is done in eight operations. Can he do better?

IMG_1168.JPG


Clarification: The opening of a link counts as one operation, the closing of a link is a spearate operation, opening and closing of a link are therefore two operations. A chain has each link attached to not more than two other links.
Somewhere in the east there is a monastery on a square shaped property with an area of a square mile. Each day at dusk four monks go to the lookouts at the edges of the fence surrounding the monastery to guard the property. Each monk looks along the side of the fence to another monk, no monk is looked at by more than one monk. At night the bell of the church at the center of the monastery tolls and it's time to change shifts. It is so dark that each monk sees only the lantern of the other monk along the fence he is looking at. So they all start to go towards the lantern of the other monk they see at a speed of two miles an hour.
  1. Will they ever meet again and where will that be?
  2. What distance have they traveled?
  3. What time did the trip take?
Clarification: The guarding monks are arranged like this:
Code:
M--->M
^    |
|    |
|    v
M<---M
You are moving into your new house with your new spouse after your marriage in Vegas (you don't remember much about this part of the story...). Going around the place you notice three switches and remember that the estate agent told you that one of them switches a single light fixture in the basement and the others do nothing (the agent said it would be an opportunity to put your own stamp on...). You ask your spouse to go downstairs and help you find out which switch is the right one, but turning around you can only see a moving shadow and a yell from the distance: "Honey, you know you have to only go down once to find the switch... I'm off to the pool!"

So how can you find the right switch by going downstairs to the basement only once?
I propose to play a game of 66 coins of various denominations. I will arrange these 66 coins in one row on the table in an order of my liking. We then take turns to take one of the coins from either end of the row until they are all gone, with you beginning to take the first coin. The one of us with the highest total value of their coins will win them all, if we have the same total value at the end you win. Can you win this game?
A group of people from the Privacy and Equality Party go on a drinking bout. At the end when it is time to pay they have a special way to split the bill. They calculate the average value of their montly incomes and determine the part they have to pay based on the proportion of their income to the average of the group. So if Jim earns twice as much as the average of a group of eight persons he has to pay a part twice as much as the average part of 1/8, i.e. 1/4 of the bill. But to live up to their privacy promise they have found a way to determine the average value of their incomes without any person in the group having to disclose their individual income to other persons and only by communicating with persons in the group. How could they have done that?
You find a box with fuses (lengths of string to be used for timing explosives) and the instructions on the box say that they each burn for exactly one minute, but not uniformly along their length, i.e. they are not half as long as they were when you let them burn for 30 seconds for example. There are two fuses left in the box. Can you measure 45 seconds using them?
You have two pyramids, one with a square base and one with a triangular face, both having all edges of length one. You glue one face of the triangular pyramid on a triangular face of the other pyramid. How many faces does the resulting solid have?
In days of yore there was a country that was inhabited by thiefs. One of them wanted to quit stealing and had the idea to start a parcel delivery service. But all employees he could find were tempted to steal the contents if not properly secured, so his company seemed to be set to fail right from the start. As everyone in this country needed locks to secure everything he could assume that every customer had many padlocks available to secure the goods to be delivered themselves. But of course the key to each lock was unique so at first it seemed like the receivers weren't able to open the padlocked parcels. But he found a way to make the business work and became a wealthy man. How does delivery work with his service?
8 pieces of paper have been used to create the image below. All pieces are of the same size (A).
In which order have they been placed down?

paper-gif.619895
You sit in a imensely boring presentation of your boss about the financal situation of the company. After not so much time has passed your mind starts to wander and you start to doodle away on your sheet of paper, some lines here, some lines there and all of a sudden your sheet ends up like this:
fullsizerender-jpg.620225

In horror you realize that both rectangles labeled A are the same size, and B and C too. That leaves 4 square units in the left upper part but only 3 in the lower right, and the triangles next to the diagonal of same size. So the upper left part is one unit square bigger than the lower right part! Did you just destroy math? What is going one here?
After you have quickly erased the devilish doodle but still ages away from being finished with watching the presentation you want to find solace in honeycombs and try to draw regular hexagons. But they won't come out as regular hexagons if the corner points are put on the intersections of the square unit grid:
fullsizerender-jpg.620227

Either the sides are not all equal, or they don't lie all on the same circle. So can you find a way to put the corners of a regular hexagon on the corners of a square grid, or is it impossible?
On most days, Camile and Melissa (sisters) walk a short distance towards the city (for the exercise) down a narrow road in front of their house as they wait for the bus (which has doors on both sides) to come by and pick them up for the trip into the city where they work.

On one occasion, as they are walking down the road, they spot two cars on the side of the road. On the left side of the road is a Chevy Camaro which is sitting just over 40 meters from the road. On the right side of the road is a Ford Mustang which is sitting just over 41 meters from the road. The two sisters are huge car enthusiasts, so Camile walks over to the Chevy Camaro and Melissa walks over to the Ford Mustang for a look.

Suddenly, they look up and see that their bus is coming along down the road. When they look up, the bus is 70 meters further back up the road from where the sisters left the road to view the two cars. The bus is traveling slowly because the road is only a wide as the bus, and because it normally expects to pick up the two sisters, but it won't stop since the bus driver doesn't like to stop unless there is someone actually waiting on the road.

The sisters immediately begin to run towards the bus at an angle towards where they expect to meet/catch the bus once they reach the road. Since the bus is traveling slowly towards the city, the two sisters are able to run at half the speed that the bus is traveling.

Can the sisters catch the bus?
How far will the bus have traveled from where they first noticed it?
The common sudoku solver is delighted when one puzzle is finally solved and all numbers are entered and the sudoku conditions are met. But in fact not just one is solved: imagine rotating the sudoku grid by 90 degree and you have another assignment of numbers for all grid cells, that makes two sudokus. Are there more ways to rearrange a sudoku? When you solve one sudoku instance, just how many sudokus have you solved actually?
In a small county not far away the 65 citizens revolted and disempowered the king, now each citizen including the king gets an hourly wage of 1 dollar. The king can no longer vote himself, but he retained the power to suggest wage changes. The wage of each citizen must be a whole number of dollars, and the sum of all wages must be the number of citizens.
The suggestions are voted on and carried if there are more "yes" votes than "no" votes. Each citizen can be counted on to vote "yes" if his wage increases, "no" if it decreases and not to bother voting if it doesn't change at all.

The king is of course selfish and clever. What is the maximum salary he can get for himself and how long does it take to get it?
Fill in the following cross-number square:
_ _ _ _ _
_ _ x x _
_ _ _ _ x
x _ _ _ _
_ x _ x _

All horizontal and vertical numbers must be either squares or cubes of an integer unless the integer is the squareroot of 121.

This shouldn't be too hard.
I have found two answers so far.....
Good luck!
:cheers:
On your quest for the Holy Grail you encounter the Knights who say Ni!. They refuse to allow passage through their woods unless you bring them a shrubbery, one that looks nice, is not too expensive and is arranged in the form of the light gray part of this diagram:

After you harrassed the local villagers to fullfill their wish, you find them no longer being the Knights of Ni, but the Knights who say "Ekke Ekke Ekke Ekke Ptang Zoo Boing!". Also they now demand a rectangular shubbery. Can you fullfill their wish by cutting the original shrubbery into pieces (using a herring provided by the Knights who until recently said Ni!) and rearranging them into a rectangle and not wasting any piece of it?


Happy Puzzling! :gtpflag:
 
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The Thoughtful Craftsman

The young craftsman is in deep thought: he has been assigned to link the five pieces of chain to a single chain. If you count the opening of a chain link and the closing of a chain link as single operations his foreman told him that it is done in eight operations. Can he do better?

View attachment 618432

The Thoughtful Craftsman
6 operations.
Open all three links 3 of one of the pieces.
Use the open links to join the remaining 4 pieces & close them in position.
Alternatively, could you also cut four end links, assemble and close the four you cut?
20170106_130434.jpg
 
Alternatively, could you also cut four end links, assemble and close the four you cut?View attachment 618503

And that are how much steps, if you count the opening of a link and the closing of a link as separate steps? :) Remember, eight steps is the number to beat.

Ok, here is another one especially for the math-haters:

The Basement Light

You are moving into your new house with your new spouse after your marriage in Vegas (you don't remember much about this part of the story...). Going around the place you notice three switches and remember that the estate agent told you that one of them switches a single light fixture in the basement and the others do nothing (the agent said it would be an opportunity to put your own stamp on...). You ask your spouse to go downstairs and help you find out which switch is the right one, but turning around you can only see a moving shadow and a yell from the distance: "Honey, you know you have to only go down once to find the switch... I'm off to the pool!"

So how can you find the right switch by going downstairs to the basement only once?
 
And that are how much steps, if you count the opening of a link and the closing of a link as separate steps? :) Remember, eight steps is the number to beat.
That would be eight. But...
The young craftsman is in deep thought: he has been assigned to link the five pieces of chain to a single chain. If you count the opening of a chain link and the closing of a chain link as single operations his foreman told him that it is done in eight operations. Can he do better?

View attachment 618432
So technically I did it in four... right?
 
Sorry, i should have worded that better. Hopefully the others are as clearly stated as possible, given their riddle nature. :nervous: If you have any questions this is the place to ask!

And here is a hint:

If you look from the right perspective, this could be a hole: ◯

Looking forward to other riddle posters, keep 'em coming! :)
 
Moles

Put them in holes however you like. Each mole is only in one hole. Just stop the moles from leaning over and grabbing beers out of the hole beside them. I suggest whacking them. Moles hate that.

Craftsman

Is a chain by definition a long thing? Because if all you have to do is link them then break one link, thread one end of the other four through it and close again. All five pieces are now connected.

Two operations, and a crappy chain.

Monks

There's two ways this works. Either each monk is looking at the next, clockwise or counterclockwise. In which case they spiral into the centre, and I'm too lazy to do the math for that.

Or the two monks at the southern corners are looking at each other, and the two monks at the northern corners are looking at each other. Each pair walks together, they meet in the middle of the north and south sides, a mile away from where they started and it took them an hour.
 
Craftsman

Is a chain by definition a long thing? Because if all you have to do is link them then break one link, thread one end of the other four through it and close again. All five pieces are now connected.

Two operations, and a crappy chain.

That was the first solution I thought of, but thought to myself, "That probably does not count for what they are looking for :lol:"
 
That was the first solution I thought of, but thought to myself, "That probably does not count for what they are looking for :lol:"

I suspect the answer is what @S_Bridge posted, because that's actually clever.

But cheers to the hobos breaking the system! :cheers::cheers:
 
That's a clever solution to the Craftsman problem @Imari! The book i took it from was very explicit in the description of putting together the single pieces and contained one fully worked out example using eight steps. Now i know why they went such great lengths... ;)👍

Regarding the extraneous monk arrangement (two groups each looking at each other): This arrangement came up soon after the foundation of the monastery, but was quickly deemed a sacrilege by the eldest of the monks. ;)
The Paths Of The Monks is a recasted story of a problem that would be too easy to find directly, and i was too lazy to draw
a map of the monastery. :embarrassed: So here it is (not to scale):

Code:
M--->M
^    |
|    |
|    v
M<---M

Thanks for catching these "bugs", seems i was too lazy in the wording of these riddles. :ouch: There's a tension between being precise and giving away too much hints about how to approach the problem. But it's a lot of fun transforming these stories.:D

I'm too lazy to do the math for that

I'm sure you mean the boring math, like endless calculations or simplifying big equations, where you can be certain you will make a mistake every 4th step and can only hope they cancel out in the end. ;) I can assure you that i hate these too and i will favour puzzles where you can get to the answer by sidestepping these calculations with some clever argument. For the Monks you don't need calculus to get to the answer, otoh if you know your "limits" no one holds you back of solving the problem using these. :mischievous:

For the Basement Light problem any math approach will fall short and the Moles also can only be tackled by thinking out of the box. :)
 
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For the Basement Light problem any math approach will fall short and the Moles also can only be tackled by thinking out of the box. :)

Yeah, the light problem I already looked up because it was mathematically impossible to solve three bits of uncertainty with one bit of information. So I won't go into that.

As far as the moles, either I'm reading it wrong or it's really easy.

Can you put ten moles into eight holes, so that each (mole) is in one and only one hole?

Put 10 moles in the first hole? Each mole is in one and only one hole. Distribute them so that there are two holes with two moles, and six holes with one mole? Each mole is still in one and only one hole. Any combination of moles in holes? Each mole is in one and only one hole unless you start chopping moles in half or hanging them over two holes.

As written it doesn't seem like it requires anything beyond the concept of object permanence to solve. If I put a plate in the sink, that plate will be in one and only one sink unless I do something funny like put the sink in another sink or break the plate in half and put a piece in two sinks. Am I reading it wrong?
 
Hehe, poor wording from me, it's not that easy. :) Clarification: Only one mole fits in a hole.

So let's see if we can word this clearly. Because otherwise it's an exercise in mind-reading, not in thinking.

Can you put ten moles into eight holes, given that a hole may only contain one mole?
 
So let's see if we can word this clearly. Because otherwise it's an exercise in mind-reading, not in thinking.

Can you put ten moles into eight holes, given that a hole may only contain one mole?

Yes, that's better, but somehow now it sounds like it can't be done. :boggled::mischievous:
Somehow i sense your razorsharp mind might cut the Moles brain teaser into pieces and it might be better to distract you and add new and more suitable puzzles for you to ponder. How about a

Game of 66 Coins

I propose to play a game of 66 coins of various denominations. I will arrange these 66 coins in one row on the table in an order of my liking. We then take turns to take one of the coins from either end of the row until they are all gone, with you beginning to take the first coin. The one of us with the highest total value of their coins will win them all. Can you win this game?

EDIT: in case that we both have the same total value you win too.

And here is a really difficult puzzle whose solution can be put to good use:

Fair Split

A group of people from the Privacy and Equality Party go on a drinking bout. At the end when it is time to pay they have a special way to split the bill. They calculate the average value of their montly incomes and determine the part they have to pay based on the proportion of their income to the average of the group. So if Jim earns twice as much as the average of a group of eight persons he has to pay a part twice as much as the average part of 1/8, i.e. 1/4 of the bill. But to live up to their privacy promise they have found a way to determine the average value of their incomes without any person in the group having to disclose their individual income to other persons and only by communicating with persons in the group. How could they have done that?
 
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And here is a really difficult puzzle whose solution can be put to good use:

Fair Split

A group of people from the Privacy and Equality Party go on a drinking bout. At the end when it is time to pay they have a special way to split the bill. They calculate the average value of their montly incomes and determine the part they have to pay based on the proportion of their income to the average of the group. So if Jim earns twice as much as the average of a group of eight persons he has to pay a part twice as much as the average part of 1/8, i.e. 1/4 of the bill. But to live up to their privacy promise they have found a way to determine the average value of their incomes without any person in the group having to disclose their individual income to other persons and only by communicating with persons in the group. How could they have done that?
That's easy. Any reasonably competent programmer could provide you with a routine to accept input anonymously and display the average without breaking a sweat.

Somehow, though, I don't think that's the kind of solution you had in mind.
 
66 coins

I actually think this one is harder than the fair split. Harder to prove anyway. But yes, I can always win if a draw counts as me winning.

Let's assume a simplified version of the game; we can only choose from one end. Number the coins from that end 1 to 66. So in this version of the game, I end up choosing all the odd numbered coins and you end up choosing all the even numbered coins.

Clearly you win this game, as you can simply stack the highest value coins into the even slots.

However, when I can choose from both ends the best you can hope for is a draw. Why? Because I can choose whether I want coin 1 or coin 66. At every turn I can choose whether I want the even coin or the odd coin, and I can force you to choose one from the other set. Therefore, either the odd and even sets are equal, or I can choose the one which is larger. Either way, I win.

That's easy. Any reasonably competent programmer could provide you with a routine to accept input anonymously and display the average without breaking a sweat.

Somehow, though, I don't think that's the kind of solution you had in mind.

I had a similar idea. Have a hat. Each person puts X indistinguishable beads in according to their income, without showing the other people. Shake it up. Count beads. Divide by 8. Average income.

Like you said, probably not what was in mind but it's by far the simplest way.

However, if they want to do it with no external apparatus at all then there is a method.

I divide my income into 8 non-equal parts, which may be zero, negative, or anything else. For the sake of kindness to my friends I probably try and make the numbers not too crazy, but hey, they wanted privacy.

I whisper a different one of these numbers to each of the other members, and keep one to myself. They all do the same. We each total up the 7 numbers that were whispered to us and the number that we kept for ourselves. We call this the Private Sum.

We can then publically share our Private Sums and add them together to come to a Public Sum. This can then be divided by 8, giving an average income for the group.
 
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That's easy. Any reasonably competent programmer could provide you with a routine to accept input anonymously and display the average without breaking a sweat.

That's what i told the people from the PEP but they are really stubborn and think entering their private data into a computer program could lead to the data getting leaked even if the programmer acted in good faith. (Call me oldschool but somehow i can relate :)) So using a computer program has to be considered "communicating with persons outside the group".

I had a similar idea. Have a hat. Each person puts X indistinguishable beads in according to their income, without showing the other people. Shake it up. Count beads. Divide by 8. Average income.

That's a very good method! 👍 I would even go as far as saying this solves the problem, but first it's a little impractical when it comes to big numbers, and you need a central secret store, the hat. The PEP wants to at least use mail to decentralize the procedure. They trust the mail provider to not intercept the contents or else will use encryption only known to people in the group.
 
That's a very good method! 👍 I would even go as far as saying this solves the problem, but first it's a little impractical when it comes to big numbers, and you need a central secret store, the hat. The PEP wants to at least use mail to decentralize the procedure. They trust the mail provider to not intercept the contents or else will use encryption only known to people in the group.

I suspected as much. Second method it is. Replace whispering with mail if necessary.
 
I suspected as much. Second method it is. Replace whispering with mail if necessary.

I'm currently pondering an attack on your method, but it looks good so far! Well done, that and the 66 coins solution cuts right to the chase of the matter! :bowdown: Can i persuade you to post a nice riddle yourself, Mr. Holmes? :)

EDIT: BTW, i have a different method for Fair Split that is a bit simpler than yours. Will post my solution after a few more days.
Anybody who wants to give the Fair Split a try: this problem has many possible solutions.
 
I'm currently pondering an attack on your method, but it looks good so far! Well done, that and the 66 coins solution cuts right to the chase of the matter! :bowdown: Can i persuade you to post a nice riddle yourself, Mr. Holmes? :)

EDIT: BTW, i have a different method for Fair Split that is a bit simpler than yours. Will post my solution after a few more days.
Anybody who wants to give the Fair Split a try: this problem has many possible solutions.

Is it...

Alice adds a random number to her salary and passes it to Bob. Bob adds his salary and another random number to the number given to him by Alice. Repeat around the circle.

When it gets back to Alice, she subtracts her number and passes to Bob. Bob then subtracts his number and so on. Once it gets to Harold, the last person, he divides by 8 and has the average.

The reason I like it less is that it is less secure. If Carl knows the numbers that Alice passes to Bob, he can know Bob's salary. It only takes two untrustworthy parties to potentially break the system. On the other hand, through my system Carl would have to know the other six numbers that Bob passed to others AND Bob's private number in order to find his salary.

Assuming that Bob is never going to tell his own private number, Carl then needs all the other numbers that were given to him as well. So Carl needs Bob's six numbers that he told the others, and the six numbers that the others gave to Bob. That's a lot of information, and it requires Carl to either monitor all the others or that they all betray Bob. Even if a couple of people are colluding to gain information, it's still secure.

Both systems are secret, but one is more robust again potential failures. Although it's true that you can make it even harder by combining the two methods, have the final sums added up by passing from person to person secretly instead of simply sharing them publically as I suggested.

It's yet another barrier to attack, any attacker would need to have all final sums before the person they were attacking.

I kind of like crypto systems. They're fun. :) PM me the attack you're thinking of and I'll look into it as well if you like. I'm still not 100% sure that what I suggested is secure, but I think it is.
 
For the Basement Light problem any math approach will fall short
Hmm, so a practical approach maybe?

I think this would work.

Turn on two of the switches for a few minutes. Then switch one of them off again and go into the basement. If the light is on you found the correct switch straight away. If the bulb is off but warm to touch it's the switch you turned off. If it's off and cold it's the switch you didn't try. Of course with today's LEDs it wouldn't be so easy.
 
Even simpler... :P

The reason I like it less is that it is less secure. If Carl knows the numbers that Alice passes to Bob, he can know Bob's salary. It only takes two untrustworthy parties to potentially break the system. On the other hand, through my system Carl would have to know the other six numbers that Bob passed to others AND Bob's private number in order to find his salary.
Your first system is very ingenious in solving this shortcoming of your second method, my method depends also on all people being trustworty. But you are making a lot more information public than the second method and i have a feeling that somehow it could be used to infer part of the private information, especially if the initial partition of the private information is chosen poorly. But that's just a guess, i haven't found a real attack. The details of cryptosystems are notoriously difficult and your method is a brilliant solution to the problem anyways. 👍

I kind of like crypto systems. They're fun. :) PM me the attack you're thinking of and I'll look into it as well if you like. I'm still not 100% sure that what I suggested is secure, but I think it is.
I really like this problem as it shows that private information doesn't have to be shared but you can still do useful computation with it -- can you hear me Google, Facebook? :) And you don't have to know advanced number theory or handle 4096 digit numbers to use it -- that makes it even more intriguing. :dopey:

Hmm, so a practical approach maybe?

I guess it's called Life Hack nowadays... :sly:👍


So, with all that bright minds around here it's time for more riddle posters! As we await their arrival, here are some more from me:

Explosive Timing

You find a box with fuses (lengths of string to be used for timing explosives) and the instructions on the box say that they each burn for exactly one minute, but not uniformly along their length, i.e. they are not half as long as they were when you let them burn for 30 seconds for example. There are two fuses left in the box. Can you measure 45 seconds using them?

Pyramid Faces

You have two pyramids, one with a square base and one with a triangular face, both having all edges of length one. You glue one face of the triangular pyramid on a triangular face of the other pyramid. How many faces does the resulting solid have?
 
Pyramid Faces
You have two pyramids, one with a square base and one with a triangular face, both having all edges of length one. You glue one face of the triangular pyramid on a triangular face of the other pyramid. How many faces does the resulting solid have?
Sounds easy -- what's the catch?
The triangular-based pyramid has four faces and the square-based one has five faces for a total of nine. Gluing two triangular faces together effectively removes those two faces from the exterior of the solid, so there would be seven faces. Since all triangles are guaranteed to be congruent since each triangle has a length of one, there is no overhang. However if the square base is glued to one of the faces of the tetrahedron, then not all of the square is covered so there would be eight total faces.

We can get still more faces if we deliberately misalign faces when gluing them together; eg gluing one triangle "right side up" against another triangle "upside down" would yield a total of thirteen faces.
 
There is always a catch in the puzzling thread! :mischievous: https://en.wikipedia.org/wiki/File:Evillaugh.ogg

First a clarification: No misalignment allowed, only face onto face. :) Now the hint: If you glue together two cubes with all edges of size one, the resulting solid has how many faces?
In that case I stand by the answer I gave in the first sentence of my response. Your hint is in fact a bit of misdirection in this case.
 
In that case I stand by the answer I gave in the first sentence of my response. Your hint is in fact a bit of misdirection in this case.
The misdirection is intended of course, can't give everything away... ;)
Would you agree that the solid formed by glueing the cubes has 6 faces, just as each of the two cubes before (If two faces are joined that are coplanar they become one face of the new solid)? Doesn't that mean that if you glue two solids at faces identical on both solids the equality

(faces of glued solid) = (faces of solid 1) + (faces of solid 2) - 2

doesn't hold in all cases? What about the pyramids?
 
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Hidden in case it happens to be the answer, or part of it.

The way that it's worded, I can't see how the pyramid one can be answered conclusively. There's not enough information. The two pyramids may both be square based versions. They may be a configuration of one square, one octagonal. Or a something different again. Without knowing the bases of both, we can't know the number of faces of both.
 
The misdirection is intended of course, can't give everything away... ;)
Would you agree that the solid formed by glueing the cubes has 6 faces, just as each of the two cubes before (If two faces are joined that are coplanar they become one face of the new solid)? Doesn't that mean that if you glue two solids at faces identical on both solids the equality

(faces of glued solid) = (faces of solid 1) + (faces of solid 2) - 2

doesn't hold in all cases? What about the pyramids?
Okay then...
I'd certainly agree that joining two cubes will result in a solid with six faces. The two cubes have a total of twelve faces, minus the two which are joined, minus the four coplanar surfaces, However, when joining our square pyramid and tetrahedron, none of the resultant sides are coplanar. Hence the new object has seven surfaces (4 + 5 - 2).

Edit:
@LeMansAid, the two pyramids were specified to have a triangular base (a tetrahedron) and a square base (a square pyramid).
 
@LeMansAid, the two pyramids were specified to have a triangular base (a tetrahedron) and a square base (a square pyramid).
Are you sure?
You have two pyramids, one with a square base and one with a triangular face, both having all edges of length one. You glue one face of the triangular pyramid on a triangular face of the other pyramid. How many faces does the resulting solid have?
I wouldn't expect to be using "base" and "face" as interchangeable terms. Each pyramid is referred to twice, but only once out of the four mentions is a base characteristic specifically mentioned.

If there genuinely is one square based, and one triangular based, my answer would be 5 faces as the end result, but that would be weird since the trick to coming up with the right answer would be compromised in the question itself.

I don't know exactly what to make of "the triangular pyramid" though. Could all pyramids be considered triangular? If that's where the second pyramid is meant to gain it's definition as being distinct from the first pyramid (square one), then I could see that maybe using face instead of base initially, could happen without the riddle confounding itself. My answer would still be 5 in that case.
 
The two pyramids may both be square based versions.
Sorry, i had a typo in the statement of the problem: the second pyramid has all faces triangular (a tetraheadron), or a triangular base (not just one triangular face as originally and misleadingly stated). A pyramid with a triangular base is a tetrahedron -- it has only triangular faces.

@BobK look here:
Okay then...
I'd certainly agree that joining two cubes will result in a solid with six faces. The two cubes have a total of twelve faces, minus the two which are joined, minus the four coplanar surfaces, However, when joining our square pyramid and tetrahedron, none of the resultant sides are coplanar. Hence the new object has seven surfaces (4 + 5 - 2).
Are you sure?
Draw two square pyramids, touching along one edge of their bases. Look at the line between their apexes, how long is it? And what about the space in between them?

@LeMansAid look here:
If there genuinely is one square based, and one triangular based, my answer would be 5 faces as the end result, but that would be weird since the trick to coming up with the right answer would be compromised in the question itself.

I don't know exactly what to make of "the triangular pyramid" though. Could all pyramids be considered triangular? If that's where the second pyramid is meant to gain it's definition as being distinct from the first pyramid (square one), then I could see that maybe using face instead of base initially, could happen without the riddle confounding itself. My answer would still be 5 in that case.
Sorry about the typo. If your answer is "The resulting solid has 5 faces", can you show how you have come to this conclusion? The triangular pyramid, or tetrahedron has 4 faces, the square pyramid has 5 faces, how can you get rid of 4 faces by gluing together 2 faces to arrive at your answer?
 
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