Can you draw 4 triangles in this 5 by 5 grid, covering all dots?

Your challenge is to draw 4 triangles in a regular 5 by 5 grid. Of course, there are some restrictions:

- all vertices of all triangles have to lie on the dots;
- all 25 dots have to be covered by the vertices/edges of the triangles; and
- no triangle may have a right angle (thus, the image above is a bad start: the triangle to the right has a right angle).

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

I originally found this puzzle in the Puzzling Stack Exchange website.

Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:

- David H., Taiwan;
- Michael W., USA;
- Pedro G., Portugal;
- B. Praveen R., India;
- Kees L., Netherlands;
- Jerry J., USA;
- Mihalis G., Greece;
- Alfredo E., Mexico;

Join the list of solvers by emailing me your solution!

The solution to this problem will be posted here after this problem has been live for 2 weeks.

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]]>Figure out the number I'm thinking of with a single question!

I'm thinking of a number amongst 1, 2, and 3. You are allowed to ask me a single question, to which I will truthfully answer with “yes”, “no”, or “I don't know”.

How can you find out the number I picked?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

(I originally found this puzzle on Reddit.)

Know how to solve this? Be the first to join the list of solvers by emailing me your solution!

Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:

- Carlos C., Portugal;
- Ventsislav K., Bulgaria;
- Reza K., Iran;
- Martin J., Czech Republic;
- Macdara M., Ireland;
- Luke J., UK;
- Panicz G., Poland;
- Santiago V., USA;
- David H., Taiwan;
- Pedro G., Portugal;
- Boro S., North Macedonia;
- Jerry J., USA;
- Lucas K., Brazil;
- Kees L., Netherlands;

Join the list of solvers by emailing me your solution!

**Lots** of people sent in their solutions and I was very pleased
with the amount of different solutions people came up with!
Hence, the solution I present here is not the *only one*,
but it's one I find very elegant.

The question you will ask me is

“If I pick one of the other two numbers, is yours larger?”

If I picked 3, then I know 3 is larger than any of the two other numbers (1 and 2). Therefore, I answer “Yes.”.

If I picked 1, then I know 1 is smaller than any of the two other numbers (2 and 3). Therefore, I answer “No.”.

If I picked 2, then I have no idea if you got 1 or 3, and that implies I need to reply “I don't know.”.

That's a pretty elegant solution, isn't it?

Feel free to share alternative solutions in the comments!

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]]>Can you solve this simple-looking arithmetic challenge?

The image above represents an arithmetic calculation:

SEND + MORE = MONEY

Each letter represents a unique digit, and each word represents a number with as many digits as letters (there are no leading 0s).

Can you figure out the numeric value of each word?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:

- Giorgio N., Italy;
- B. Praveen R., India;
- Nishant M., India;
- David H., Taiwan;
- Matthias W., Germany;

Join the list of solvers by emailing me your solution!

The solution to this problem will be posted here after this problem has been live for ~~2~~ 3 weeks.

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]]>If I scramble a Rubik's cube for long enough, will it solve itself?

A Rubik's cube is a toy like the one you can see in the picture above. It's a 3 by 3 by 3 cube, where each face has one of six colours. The cube can be scrambled, and at that point the colours of the faces no longer match.

The challenge I have for you involves proving something. I want you to prove the following statement:

“If you take a solved Rubik's cube and start scrambling it by following a fixed set of steps, you eventually end up with a solved Rubik's cube again.”

A silly example of how this is true is if you start turning the top face. You rotate it once. Twice. Thrice. A final turn, and it's back at the initial position!

But this also applies to more complicated sequences of turns!

If you have one, go grab a Rubik's cube and give it a go! (It's not the same thing, but you can also try this online simulator.)

For your convenience, here is a short GIF of me scrambling a cube by repeating the same set of steps.

First, here is an excerpt of the process, showing the set of steps I do repeatedly. If you look closely, you can see that I only do two movements:

Now, here is a sped up version from start to finish. (It took me around 3 minutes total to get back to the beginning through repeated scrambling!)

I started out with a solved cube and ended up with a solved cube.

Why?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

- David H., Taiwan;

Join the list of solvers by emailing me your solution!

If you are comfortable with the basics of group theory, the proof can be stated succinctly. If you don't know what I'm talking about, that's fine, skip this!

Let \(M\) denote the permutation we are using to scramble the cube, as an element of the group of permutations of the Rubik's cube.

For \(m\) large enough, we have that in

\[ M^0, M^1, \cdots, M^m, ~ ,\]

two permutations are going to be the same, \(M^n = M^m\), with \(n < m\). (This follows from the pigeonhole principle and the fact that there's a finite number of permutations of the Rubik's cube.) If \(M^{-1}\) denotes the inverse permutation, then

\[ M^m = M^n \iff M^{m-n} = M^0 ~ .\]

This shows that \(M^{m - n}\) (where \(m - n \geq 0\)) is the solved cube.

Here is the generic overview of the solution we will produce together:

- we prove that if we keep scrambling the cube, we...

Can you solve this little minesweeper puzzle?

The xkcd website published its cartoon #2496, containing a minesweeper grid, and that is the problem for this post: to locate all of the mines in the minesweeper grid above.

For those of you who don't know the minesweeper game, here are the rules:

- some squares contain mines, others don't;
- squares that do not contain mines contain numbers instead, and those numbers indicate how many mines are neighbours to that numbered square (the neighbourhood of a square contains the squares that are directly adjacent to it, plus the squares that share a vertex in the diagonal).
- gray squares can contain mines or other numbers.

You must find all the mines by reasoning about the numbers you see.

You can play minesweeper online or you can play my minesweeper remake.

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

- David H., Taiwan;
- Gerard M., Ireland;
- Peter K., USA;
- Rodrigo H. M., Mexico;
- Rita F., Portugal;

Join the list of solvers by emailing me your solution!

Let's look for the mines together!

This is the mine field we have to work with:

Let's look at the lower-left corner, at the `3`

and the `1`

.

The `3`

has four neighbouring cells:

The two cells that have a double circle are next to the `1`

,
and therefore we can't have mines in both of those cells.
Therefore, the two top circles *must* contain mines.

Two mines have been found!

The `3`

that we were looking at now has two mines placed
next to it, so we only have to find the last one.

Regardless of where that mine is (one of the two double circles below),
it is always going to be next to the bottom-left `1`

.
This means that there can be *no* mines between the two `1`

s
on the bottom row.

By crossing out those two locations, we know that the bottom-right
`1`

has a mine right above it.

For the next step, we can look at the top-right `1`

.
The top-right `1`

already has a neighbouring mine
(to the left, and down) which means all other neighbouring
cells must be empty.

In crossing out those positions,
we realise that the `3`

on the second row, to the right,
only has one position left for its third mine.
Therefore, we can place a mine there.

As we place that mine, the puzzle is complete! Take a look for yourself:

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]]>It's night time and 4 friends need to cross a fragile bridge, but they only have one torch. What's the order in which they should cross?

4 friends are doing some trekking by night, when they hit a rope bridge that they need to cross.

They decide that it is best if only 2 cross at a time, because the bridge looks really fragile, and the bridge has some holes in it, so they better only cross if they have any way of lighting their way. Because the friends only have a single torch, they figure out that every time 2 people cross to the other side, someone has to come back with the torch, which will cost them some time...

The 4 friends want to cross that bridge as fast as possible. How can they do that if they each take 1, 2, 5, and 10 minutes to cross the bridge?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

- Jairo, Brazil;
- Martin J., Czech Republic;
- David H., Taiwan;
- Gerard M., Ireland;

Join the list of solvers by emailing me your solution!

If you follow what is called a greedy approach, you will get to the following answer: 19 minutes.

*However*, there is a way of arranging the crossings that is even faster than that and that only takes 17 minutes.
Can you figure it out?

I'll explain how, but in order to make it simpler for me, I'll refer to the people by the time they take to cross the canyon.

- 1 & 2 cross (
**+2**); - 1 comes back (
**+1**); - 5 & 10 cross (
**+10**); - 2 comes back (
**+2**); and - 1 & 2 cross again (
**+2**).

This takes up only 17 minutes! (There is another similar arrangement that also takes 17 minutes, and that's if 2 comes back alone first, and then 1.)

But how can we be sure that this is the *fastest* way to cross the canyon?
We can try *all* the valid canyon crossings,
and check all of them take 17 minutes or more.
Or we can try to reason about the problem, and conclude that any way
we cross the canyon is going to take 17 minutes or more.

Let's go for the second approach, where we reason about the problem.

Because the torch has to be present in all the crossings,
we see the canyon needs to be crossed a total of 5 times.
Let `o`

represent a person, `x`

the torch, and `|`

the canyon:

- two people go forward, and we are at
`o o | o o x`

; - one person comes back with the torch:
`o o o x | o`

; - two people go forward:
`o |...`

Three friends are given three different numbers that add up to a dozen. Can you figure out everyone's numbers?

Three friends, Alice, Bob, and Charlie, are assigned three different positive whole numbers by their fourth friend, Diane. Furthermore, Diane told them that their three numbers add up to 12 and that Charlie's is the largest one.

Diane then asks the three of them if they know everyone's numbers, to which Bob replies “I do!”, whereas Alice and Charlie remain silent. After Bob's revelation, Alice and Charlie think for a couple of seconds and confirm that now they also know everyone's numbers.

What were Alice's, Bob's, and Charlie's numbers?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

- Ashok M., India;
- David H., Taiwan;
- Attila K., Hungary;
- Jason P., US;
- “Starsmer”, US;

Join the list of solvers by emailing me your solution!

For Bob to know Alice's and Charlie's numbers, Bob's number must be big enough so that Charlie's number doesn't have much wiggle room.

If Bob had the number 6, then Charlie would have at least 7 and their two numbers would add up to 13, which is already above 12. However, if Bob's number is 5, then Bob knows that Charlie can only have a 6 (if Charlie had 7 or more, then Bob and Charlie alone would add up to 12 or more) and hence Alice has to have a 1.

Thus, we conclude that Bob can guess everyone's numbers if Bob is given the 5.

From Alice's point of view, holding a 1 doesn't give her enough information to deduce what Bob and Charlie have, because they could have 2 and 9, for example, or 3 and 8.

From Charlie's point of view, holding a 6 doesn't give him enough information to deduce what Alice and Bob have, because they could have 1 and 5, or 2 and 4, for example, not to mention that Charlie wouldn't know who has the largest number.

After Bob announces he knows everyone's numbers, the other two can reverse engineer this reasoning and discover the missing numbers as well.

An alternative approach would be to list all possible number assignments and then look for the assignment that attributes a unique number to Bob. In other words, we can go through the numbers 1 to 12 and ask: “If Bob had this number, in how many different ways could I attribute numbers to Alice and Charlie?”.

This problem was taken from this Reddit post, and shared with permission.

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]]>You have two magical ropes that you can set on fire and you need to count 45 minutes. How do you do it?

You are given two magic ropes and a lighter. The ropes are magic because you are told they burn in a weird way: each rope takes exactly 1 hour to burn from end to end, but they don't burn at a constant rate. (What that means is that the time elapsed doesn't have to be proportional to the length of burnt rope. For example, it may happen that the first half of the rope takes 35 min to burn, then a huge portion of the remaining rope burns in 10 min, and then the final tip of the rope takes 15 min to burn.)

Given two magic ropes like this, how do you use them to measure 45 minutes?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

Congratulations to the ones that solved this problem correctly and, in particular, to the ones who already sent me their correct solutions:

- Christ van W., The Netherlands;
- Attila K., Hungary;
- Ashok M., India;
- David D., US;
- Greg K., US;
- Shivam T., US;
- Marco M., Italy;
- David H., Taiwan;
- Cody B., US;

(The list is in no particular order.)

Email me your solution to get your name (or an alias) featured in here!

I find this problem to be really interesting! The fact is that it looks like there is not much that we can do, because the first thing that pops into our minds is to cut the rope into portions, however, the problem statement tells us that the time that a piece of rope takes to burn will not be proportional to its length. Therefore, cutting the rope isn't an option.

Our only other option is to actually light the rope(s) on fire,
but that can't be *just* it, because a rope takes 60 minutes to burn
and we want to time 45 minutes.

The next step is realising that 45 minutes is \(3/4\) of an hour, and \(3/4 = 1/2 + 1/4\). What is more, \(1/4 = (1/2)\times(1/2)\), i.e., three quarters of an hour is half an hour plus half of another half hour. The recurring theme here is halves. Therefore, it might be a good idea if we reframe the problem. Instead of trying to measure a specific amount of time with the burning of the ropes, can you do something to the rope so that it measures exactly half of the total time that the rope could burn for?

In other words, if a rope takes a full \(x\) minutes to burn, how could you work with that rope in order to measure \(x/2\) minutes?

Give it some thought.

If you light up the rope on both ends, then it will burn for exactly half of the time! So, if...

]]>You are on vacation and must find the most efficient way to cross all bridges. How will you do that?

If you like solving riddles and puzzles, it is likely that you have already encountered this puzzle. But even if you have, it is always good to go back and think about the classics. On top of that, I will formulate the problem in a slightly different way, so that you can be entertained for a bit even if you already know the more classic version.

Take a look at this satellite view from Kaliningrad, Russia, where I have highlighted seven bridges:

Your task is to figure out what route to take if what you want to do is cross all of the highlighted bridges at least once but, at the same time, keep the total number of crossed bridges as low as possible.

Having said that, what is the best route you can come up with?

(Just to be clear, I don't care about the length of the route – the number of miles/kilometres you would walk/drive – I only care about the number of bridges you cross.)

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

In case you are wondering, the classic version of this puzzle is dubbed “the seven bridges of Königsberg” because that is what this place was called when a famous mathematician first dwelled on this problem.

- Attila K., (Hungary)

There are seven distinct bridges that we want to traverse, so we know the shortest path has to go over seven bridges, minimum. What we will show is that, actually, we need to go over eight bridges in total in order to visit all seven bridges.

In order to show that is the case, consider the following figure:

In the figure above we can see that I numbered the four pieces of land to which the bridges are connected.

How many bridges does each piece of land connect to?

- connects to 3 bridges;
- connects to 3 bridges (as well);
- connects to 3 bridges (as well); and
- connects to 5 bridges.

Now we will use this information to show that it is impossible to create a path that visits all bridges exactly once.

Let us think about a hypothetical path we would do,
in order to traverse all the bridges exactly once.
More specifically, let us think about what happens
in the middle of our walk.
If we are in the middle of our path,
when we enter some piece of land through a bridge,
we have to leave that piece of land through another bridge.
In other words, for each time we *arrive* at a piece of land,
we also...

Alice and Bob sit across each other, ready for their game of coins. Who will emerge victorious?

Alice and Bob sit across each other at a circular table. They will now play the “game of coins”!

The “game of coins” is simple:
they have access to a *huge* pile of circular coins of the same size and,
in turns, they place a coin on the table.
That coin must not overlap with any other coins already placed; it must rest completely on the table.
The first player to not have enough room to place a coin, loses.

Alice will play first.

Can either of the two develop a winning strategy? And what strategy would that be?

Give it some thought!

If you need any clarification whatsoever, feel free to ask in the comment section below.

Congratulations to the ones that solved this problem correctly. Because no one sent me their solution in time, I cannot list any winners for this time.

Alice can win the game, because she is the first player. Her winning strategy is easy: she just has to play the first coin in the very centre of the table, and then play diametrically opposite to Bob.

How does this work? It is actually quite interesting! Alice starts by placing a coin in the very centre of the table so that the table starts to exhibit a very useful property (for her): given any valid coin position, the diametrically opposite location is also valid.

See it this way: when Alice places the first coin in the very centre of the table, think of it as her drilling a hole in the centre of the table, so that no coin can go there any more.

Now, it is Bob's turn to play, and he has to play a coin in a place with no holes, so he just places his coin somewhere:

What Alice does is to figure out what is the play that is diametrically opposite to Bob's:

After she replies, you can think of it as if their coins now also drilled holes in the table:

Now, at this point, there are three holes in the table, instead of the single hole that Alice “drilled” in the beginning, but one thing remains true: for each point in that table, if a coin would fit there, then it would also fit in the same location, but diametrically opposite to the centre.

So, Bob might play some coin somewhere:

Then, what Alice does is the same, she finds the diametrically opposite play:

And they do this back and forth, until Bob runs out of space to place a new coin. It's never Alice who runs out of space because she is always replying to Bob's moves.

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