The Puzzling Thread

[tarnheld]

I hope i didn't shy anyone away with my comments, i guess i was too eager to post -- it's not an easy task to keep your mouth shut and not reveal too much in too short a time to not discourage others to start puzzling. I guess it will be better if puzzle posters refrain from answering and commenting every posted solution right away, so i will keep quiet about solutions posted for about a week after the first solution is posted (only adding general hints for all interested) and reveal the original solution and comments afterwards. Added these guidelines in post #1 for others who want to challenge us with their puzzles. 👍

So here is another one for you to enjoy:

Thief Parcel Service

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?

 

[GTsail]

Monks:

Spoiler

my Monks solution:
Like Imari, I'm assuming that each pair of monks look at each other
So the top two (looking at monk diagram provided by tarnheld) look at each other (I will label the monks: T1 and T2)
The bottom two also look at each other (I will label the monks: B1 and B2)
When the bell rings B1 and B2 walk towards each other and meet each other after 15 minutes and then stop
When the bell rings T1 and T2 walk towards each other and meet each other after 15 minutes, but don't stop and they continue to walk around the square and after walking for another 60 minutes, they meet up with B1 and B2 who have stopped at the bottom of the square.
Total elapsed time is 75 minutes and they are all together at the end
Monks B1 and B2 traveled 1/2 mile
Monks T1 and T2 traveled 2 1/2 miles

 

[tarnheld]

Interesting trivia: The monks mentioned in the Monks puzzle were later featured in Monty Pythons Quest for the Holy Grail.

 

[GTsail]

Moles question:

Spoiler

Can a mole eat another?
The first two moles eat the third and fourth moles and then the remaining 8 moles all dive into the 8 holes even if its a tight fit for the first two moles who are now fat and happy??

 

[GTsail]

Explosive Timing, preliminary thoughts:

Spoiler

My initial solution for the Explosive Timing:

diagram of sample fuses:
___ ___
|13| |13|
---- ----
|14| |14|
---- ----
|16| |16|
---- ----
|17| |17|
---- ----

60 sec's each


Lay the two fuses side-by-side as above in the section.
Cut the fuses in half
Take the top half of the first fuse and cut it in half
Take the bottom half of the second fuse and cut it in half
Burn the first fuse from the top (which should burn for about 13 seconds)
Burn the second fuse from the bottom (which should burn for about 17 seconds)
So you should now have measured 30 seconds of time


diagram of remaining half fuses:
___ ___
| 8| |6.5|
---- ----
| 8| |6.5|
---- ----
|8.5| | 7 |
---- ----
|8.5| | 7 |
---- ----

33sec's, 27sec's


Now take the bottom half of the first fuse that you cut in half and cut it in half and then half again
Then burn this fuse from the top (which should burn for about 8 seconds)
Now take the top half of the second fuse that you cut in half and cut it in half and then half again
Then burn this fuse from the bottom (which should burn for about 7 seconds)
So you should have measured an additional 15 seconds of time

And it doesn't seem to matter which end to these second two fuses you burn, as long as one is the top end, and the other is the bottom end.

For a total of 45 seconds!
What do you think? Does this work?

 

[IfAndOr]

Pondering Pesky Pyramids.

Spoiler: Solution?

The answer I believe is that 5 faces will result in the joining of the two different pyramids. Fixing two of the triangular faces together will result in the two adjacent triangular faces becoming coplaner.
Since I've not got any software to draw the resulting shape I've been trawling through internet to find something suitable. While not exactly what I wanted this image does show what I mean - if you remove the second square based pyramid from the final shape. (If I find a better image I'll update it)

Tarn the moles; is it the positioning of the holes that is the key to the solution?

 

[polysmut]

Thief Parcel Service

Spoiler: Solution

The sender puts a lock on the cargo.
The recipient adds their own lock then returns the cargo to the sender (like Elvis).
The original sender removes their lock leaving the intended recipients lock as the only remaining security device. Then sends the cargo again.
The ultimate recipient receives the cargo, removes their own lock & enjoys their new smartphone/7" single/painting.

 

[IfAndOr]

Here's a (probably) relatively easy one.

Paper Letters.

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?

 

[LeMansAid]

Only one pass (no double check), so I might have made a fool of myself.

Spoiler: Paper Letters

C, E, B, F, H, G, D, A

 

[IfAndOr]

Only one pass (no double check), so I might have made a fool of myself.

That didn't last for long.

 

[tarnheld]

Explosive Timing, preliminary thoughts:

Tarn the moles; is it the positioning of the holes that is the key to the solution?

The key to the moles is to not overthink it. More trivia: There is no word for overthinking in the german language.

Spoiler: Mole hint #2

Here are your eight holes: ◯◯◯◯◯◯◯◯
Now how are we supposed to get those ten moles inside?

 

[BobK]

Pondering Pesky Pyramids.

Spoiler: Solution?

The answer I believe is that 5 faces will result in the joining of the two different pyramids. Fixing two of the triangular faces together will result in the two adjacent triangular faces becoming coplaner.
Since I've not got any software to draw the resulting shape I've been trawling through internet to find something suitable. While not exactly what I wanted this image does show what I mean - if you remove the second square based pyramid from the final shape. (If I find a better image I'll update it)
View attachment 619887

While what you say is correct, however in the actual puzzle only one of the pyramids has a square base; the other has a triangular base. In other words, one has five sides and the other has four.

I keep meaning to construct some pyramids out of paper or some such but have yet to get around to it.

 

[LeMansAid]

While what you say is correct, however in the actual puzzle only one of the pyramids has a square base; the other has a triangular base. In other words, one has five sides and the other has four.

I keep meaning to construct some pyramids out of paper or some such but have yet to get around to it.

Just look at the second image and remove one of the square-based pyramids, leaving the other square-based pyramid and the "middle bit" (tetrahedron).

It's the resultant rhomboid faces that bring it down to five faces in total.

 

[BobK]

Just look at the second image and remove one of the square-based pyramids, leaving the other square-based pyramid and the "middle bit" (tetrahedron).

It's the resultant rhomboid faces that bring it down to five faces in total.

Ah, yes indeed. Now I see it. Thanks!

 

[tarnheld]

Explosive Timing, preliminary thoughts:

Ok, seriously: the fuses (manufactured by ACME) used in the explosive timing are only guaranteed to measure one minute when burned fully, and burn erratically in between: one fuse might burn down half the length in 3 seconds, then down to a quarter in 56 seconds and the last quarter in 1 second. The next fuse from the same pack might burn to half it's length in 5 seconds, then burn to quarter in 35 second and the last quarter takes 20 seconds. Just the right fuses for Coyotes hunting Road Runners.

 

[GTsail]

*** the fuses (manufactured by ACME) used in the explosive timing are only guaranteed to measure one minute when burned fully, and burn erratically in between.*** Just the right fuses for Coyotes hunting Road Runners.

Erratic fuses!!

Yeah, my solution assumed that each fuse was identical and burned down in a linear fashion, even if each inch didn't burn down at the same speed. Well, I guess its back to ACME to pick up a few more fuses so I can continue on with my explosive testing. I still have most of my body parts so I've got no excuses.

 

[TB]

I keep meaning to construct some pyramids out of paper or some such but have yet to get around to it.

It's the resultant rhomboid faces that bring it down to five faces in total.

AutoCAD didn't want to play nice this morning so please excuse the crudeness of the accuracy.

Spoiler: Pyramid Faces

Top View:



Side View:



The total number of faces is indeed 5.

 

[GTsail]

Explosive Timing: second attempt

Spoiler

Grab two new ACME fuses
Light both ends of the first fuse
When the fuse finishes burning and the two flames meet somewhere in middle of the fuse, 30 seconds will have expired
At the same time as you light both ends of the first fuse, you also light one end of the second fuse
As soon as the first fuse is fully burned (30 seconds) you also light the second end of the second fuse
The second fuse will burn down towards the middle and after 15 seconds it will be fully burned (since it essentially started out as only a 30 second fuse)
Total time elapsed should be 45 seconds
Does this work?

 

[GTsail]

Moles: second attempt

Spoiler

create a circle with the eight holes
put all the moles inside the circle
Done!!!
All the moles are inside "one" "hole"

Moles: third attempt

Spoiler

create a circle of four mole holes
Put a mole in the center of the created circle and into each of the four mole holes

create a second circle of four mole holes
Put a mole into the center of this second circle and into each of these four mole holes

The two extra moles are inside circles, so there is only one mole in each of these created circles, and only one mole in each of the other eight holes. Can I define a "circle" as a hole?

 

[tarnheld]

To all Pyramid Faces solvers here are the juicy background details:

Spoiler: Pyramid Faces background

The problem was actually a part of the US SAT test, but the creators didn't recognize the coplanarity and incorrectly marked 7 faces as the right answer. The 17-year old Daniel Lowen noticed the coplanarity but his correct answer of 5 faces was marked as wrong. He then sucessfully challenged the results. For details see here or here.

Explosive Timing: second attempt

Spoiler

Moles: second attempt

Moles: third attempt

Spoiler

 

[GTsail]

I have a non-whispering extension to @Imari 's Fair Split solution:

Spoiler

To keep my explanation easier, lets say there are four people in the drinking party.

A has income of $10k, B has income of $20k, C has income of $30k, D has income of $40k
As Imari mentions, each of the four people divide their income into four non-equal parts, so for example:
A >>> 1/2/3/4 = 10
B >>> 3/4/6/7 = 20
C >>> 6/7/8/9 = 30
D >>> 8/9/11/12 = 40

The first person randomly writes down on a piece of paper, one part of their income, and then passes this paper along to the next person. The next person, at random, adds one part of their income to the number that they were handed, writes this down on a new piece of paper, and passes this paper along to the next person. And so on, and then the paper comes around to the first person, and the first person randomly adds a second part of their income, and then passes the paper along. And then the second person adds their second part of their income and passes the paper along. And so on......

Since each person only sees the total, and doesn't know which random component was added each time the paper circulated around the table, none of the drinking buddies should be able to determine anyone else's actual income.

Eventually, the total will grow until it finally totals everyone's income. At that point divide the total by four and you have the average.

Using the example above without randomizing the order of the components, here are the numbers that would be shown on each piece of paper as they were passed around the table:
1,4,10,18,20,24,31,40,43,49,57,68,72,79,88,100 So total is 100k and average is 25k

The only problem I see is that once everyone knows the average, can't they watch what portion a person chips into the "kitty" to pay their portion of the drinking bill, and wouldn't they be able to determine what everyone else's income was at that point even if no one said what their income was as the average was being calculated?

 

[polysmut]

By the way; thank you @tarnheld for this thread. It's a great idea & something that I think quite a few people on GTP will enjoy. 👍

I'm not into it enough to have a go at every puzzle but mostly that's just because I'm busy with work at the moment.
If I see a solution straight away I'll aim to write it out, time permitting. For the puzzles I don't grasp immediately, I'll enjoy reading everyone else's efforts.

Maybe when I have more time later in the year I'll find some puzzles to add into the mix.

 

[tarnheld]

By the way; thank you @tarnheld for this thread. It's a great idea & something that I think quite a few people on GTP will enjoy. 👍

Thanks, i hope it's as much fun for you all to embark on the journey of solving as it is for me to search for and present the riddles. Mostly i take them out of books like those of Martin Gardner and transform them into a different story so that they can't be found easily by googling -- the rewrite is fun in ifself . I solved most of them myself before and to corner and cut the possiblities until it makes click and you stand before the solution takes a certain stubbornness but it's rewarding in many ways that must be experienced and cannot be really explained.

Maybe when I have more time later in the year I'll find some puzzles to add into the mix.

That would be great, i'd love to try solving a few of the riddles that puzzled you all before.

Here are some more for you (I admit they look innocent on the outset but one of them ventures (maybe too) deep into the realm of math):

Doodles

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:

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?

Regular Hexagon Doodles

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:

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?

 

[polysmut]

Doodles

Spoiler: Solution

There's no reason to assume that the two different arrangements of A, B & C would leave the same remainder when drawing the two different rectangles around them in the top left & bottom right.
So the answer is no, you didn't just destroy maths.

Regular Hexagon Doodles

Spoiler: Solution

It's impossible, because the cosine of 30 degrees is irrational therefore y and z can't both be multiples of the same unitof length.

The angles were obtained from calculating that the interior angles of a hexagon add up to 720 degrees. In a regular hexagon that's 120 each.

 

[tarnheld]

Spoiler: about S_Bridge's Doodles

There's no reason to assume that the two different arrangements of A, B & C would leave the same remainder when drawing the two different rectangles around them in the top left & bottom right.

But would you agree that if you have a triangle and you cut it into pieces and reassemble them forming a triangle with the same lengths, angles and thus the same area, there shouldn't be any holes left? Wouldn't the hole in the devilish doodle contradict the concept of area, i.e. that area of a shape made up of disjoint shapes is the sum of the area of the pieces?

Spoiler: about S_Bridge's Hexagon Doodles

It's impossible

Aw, that's sad but true. 👍 That square grid seems rubbish for helping to draw anything regular ... maybe it might work out with regular Octagons?

I have a non-whispering extension to @Imari 's Fair Split solution:

Spoiler: discussion, Imari might want to take a look too

That's neat! 👍 But if two of the group collude, they can work out the income of another one in the group: For example if A and B conspire to keep all numbers they get in the rounds, they can use it to work out B's income: they just subtract the received numbers of each round and add up. But there's a simple cure to make your procedure as safe as @Imari's original solution, can you guess how? It can also be used to cure @Imari's second procedure.

 

[polysmut]

Spoiler: Doodles cont'd.

But would you agree that if you have a triangle and you cut it into pieces and reassemble them forming a triangle with the same lengths, angles and thus the same area, there shouldn't be any holes left? Wouldn't the hole in the devilish doodle contradict the concept of area, i.e. that area of a shape made up of disjoint shapes is the sum of the area of the pieces?

There's obviously something in the way the riddle is presented that is supposed to trick the reader into a false assumption?
I just can't see what that something is.

Nothing about the doodle looks wrong to me & nothing about the description makes me think the areas described should be equal.

I stand by the answer of: No, you didn't break maths.

Edit:

Spoiler: Regular Hexagon Doodles cont'd.

Aw, that's sad but true. 👍 That square grid seems rubbish for helping to draw anything regular ... maybe it might work out with regular Octagons?

Octagons, no. The same proof can be adapted easily to show this.

You could do it with a regular quadrilateral.

 

[Imari]

Spoiler: about S_Bridge's Hexagon Doodles

Aw, that's sad but true. 👍 That square grid seems rubbish for helping to draw anything regular ... maybe it might work out with regular Octagons?

Spoiler

Is it? I feel like there might be some wiggle room if there was no side that was parallel to the grid, but that's just intuition. I'll see if I get bored enough at work tomorrow to put some time into proving or disproving it.

Spoiler: Doodles cont'd.

There's obviously something in the way the riddle is presented that is supposed to trick the reader into a false assumption?
I just can't see what that something is.

Nothing about the doodle looks wrong to me & nothing about the description makes me think the areas described should be equal.

I stand by the answer of: No, you didn't break maths.

Spoiler

It's not as hard as it looks. 8x8 =/= 5x13

 

[Imari]

Argh, double post.

 

[polysmut]

Spoiler: Regular Hexagon Doodles cont'd

Is it? I feel like there might be some wiggle room if there was no side that was parallel to the grid, but that's just intuition. I'll see if I get bored enough at work tomorrow to put some time into proving or disproving it.

Of course, yeah. My proof isn't actually a proof as we can rotate the shape, not just expand it.
I'm not going back to it though, I'm no longer just the right amount of drunk to put pen to paper.

Edit:
I see what I'd missed.

Spoiler: Regular Hexagon Doodles

We're supposed to wrongly assume the long diagonal is straight? Which would give both sides of that long diagonal an equal total area?

It's not straight though, if it passes through the corners of the smaller shapes. Or it is straight but it doesn't pass exactly through the points of the smaller shapes.
Otherwise this would be zero.

 

[GTsail]

Cars and Buses:

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?

Good luck!