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.
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Join the list of solvers by emailing me your solution!
The solution will be posted here 2 weeks after the problem.
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]]>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.
Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:
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:
(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.
Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:
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?
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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]]>Can you find a really large triangle that is also really tiny?
Does there exist a triangle that has an area larger than the Earth's surface (approximately 510 million km²) but whose heights sum up to less than one centimetre?
Give it some thought and send me your solution!
I saw this problem in the Facebook page of a group of maths students from a local university.
If you need any clarification whatsoever, feel free to ask in the comment section below.
Yes! There exists such a triangle.
Let \([ABC]\) be a triangle with sides \(a\), \(b\), \(c\), and let its respective heights be \(h_a\), \(h_b\), \(h_c\).
The area of this triangle is, therefore,
\[ \frac{ah_a}2 = \frac{bh_b}2 = \frac{ch_c}2 ~~~.\]
From the equalities above we conclude
\[ \begin{cases} ah_a = ch_c \iff h_a = \frac{ch_c}a \\ bh_b = ch_c \iff h_b = \frac{ch_c}b \end{cases} ~~~.\]
Because we are talking about a triangle, we know that \(c < a + b\), which means we can take a look at the equation above and remove \(c\) from there:
\[ h_a = \frac{ch_c}a \leq \frac{(a + b)h_c}{a} ~~~.\]
If, additionally, the triangle \([ABC]\) has two equal sides, \(a = b\), we get
\[ h_a \leq 2h_c ~~~,\]
and, similarly,
\[ h_b \leq 2h_c ~~~.\]
Now, all we have to do is sum the three heights of the triangle:
\[ h_a + h_b + h_c \leq 2h_c + 2h_c + h_c = 5h_c ~~~.\]
Therefore, in our isosceles triangle with equal sides \(a\), \(b\), we have that the sum of the three heights is less than or equal to \(5h_c\), where \(h_c\) is the height relative to the side that is unequal to the other two. With that in mind, we can make a really “long” triangle (that is, for which \(c\) is large) that is short (that is, for which \(h_c\) is small) so that it satisfies the restrictions in the problem statement. In fact, our reasoning shows that there are triangles with areas that can be arbitrarily large, while their three heights sum up to values that are arbitrarily small.
Makes sense?
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]]>This is an algorithmic puzzle where you just have to turn some coins.
You are inside a dark room, sitting at a table. The table in front of you is covered with coins, which you can feel, but you can't really see what their face up is. You are told, however, that exactly 20 of those coins have heads facing up and all the others have tails facing up.
Your task is to divide the coins in two groups, such that both piles have the exact same number of coins with their heads face up.
The only things you can really do are
Recall that the room is so dark, you can't really tell which face each coin has up, nor can you feel it with your touch.
Give it some thought and send me your solution!
If you need any clarification whatsoever, feel free to ask in the comment section below.
This problem was shared by @TodiLiju, who comments these “problem” posts from time to time.
Congratulations to the ones that solved this problem correctly and, in particular, to
for sending in a correct solution.
Because there are so many things that we don't know about the coins, when we start, and because of how little we control (we can't look at a coin and see which face is up) there are not many things that we can try to do. The only thing we know for sure is that there are \(20\) coins that have the “heads” face up.
This “\(20\)” is the only thing we have/know, so why not make a group of \(20\) coins, and leave the rest of the coins in their own group? Assume \(n\) is the total amount of coins on the table, and say that the group of \(20\) coins has \(k\) coins that have their “heads” face up.
Here is a summary of the situation:
Coins in group | Heads | Tails | ||
---|---|---|---|---|
\(20\) | \(k\) | \(20 - k\) | ||
\(n - 20\) | \(20 - k\) | \(n - 40 + k\) |
What we need, now, is to have the same number of coins with their “heads” face up in both groups. Currently, the group with \(20\) coins has \(k\) coins like that and the group with \(n - 20\) coins has \(20 - k\) coins like that. However, if we flip all the coins in the group of \(20\) coins, then all coins would swap their face up, which means we would get a total of \(20 - k\) coins with their “heads” face up, matching the other group.
In conclusion, what we need to do is:
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]]>Two doors, one gives you eternal happiness and the other eternal sadness. How can you pick the correct one?
Today's problem is a classical one. If you solve lots of logic puzzles and riddles then it is likely that you already know this one, but here it goes for the others that do not know it yet:
Two identical doors are in front of you. One of them is between you and eternal happiness, whereas the other one will give you eternal sadness if you cross it.
Furthermore, there are two people close by, and they know which door is which. You are allowed to ask one of those persons a single question. However, one of the persons always lies and the other person always tells the truth.
What question should you ask in order to determine the correct door?
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:
(The list is in no particular order.)
I am yet to meet someone who can walk me through their thought process while they solve this for the first time, so I will have a hard time trying to show you how you could have gotten to the solution by yourself if you haven't yet. This is one of those type of problems where it is just easier to tell the solution and then verify that it works.
Because you know one person always lies but the other person always tells the truth, you have to figure out a way to have your question ran by both of them at the same time, even though you can only direct your question at one of them. Having said that, you can ask a hypothetical question to person A, saying something along the lines “If I had asked person B ...”. This is the key.
What you should ask to either person is
“If I had asked which door would bring me eternal happiness to the other person, what door would that person have pointed to?”
The person you directed your question at will point to a door, and you should then use the other one. Let me break this down for you.
For the sake of the explanation, suppose the door on the left brings eternal happiness and the person on the left is the liar.
If you direct your question to the liar, their train of thought will be along the lines of “the truth-teller will definitely point to the door on the left, so I should point to the one on the right”, therefore the liar lies about the honest person's hypothetical answer and points to the door on the right.
If you direct your question to the honest person, their train of thought...
Syncro is a beautiful game where you have to unite all the petals in a single flower. In how many moves can you do it?
Look at the image above. There are four “flowers”, each one with four “petals”. Notice that each of the flowers has one coloured petal. There are also arrows going from one flower to the other: this means that the corresponding shape sends the coloured petals in that direction.
For example, the square makes all coloured petals rotate once in the clockwise direction, or if you take the image above and do “circle + square” then the coloured petals end up like this:
Your objective is to find a sequence of circles and squares that put all coloured petals in a single flower, in the smallest number of steps possible.
Give it some thought...
If you need any clarification whatsoever, feel free to ask in the comment section below.
Syncro is a desktop/mobile game developed by some friends of mine, and the objective of the game is the same as that of this problem, except the game itself has several levels.
This game can be played online and there is also an Android app.
If you complete the game, you can even end up in the hall of fame!
Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:
(The list is in no particular order.)
This problem is really hard to solve without a visual representation of what is going on, so let me remind you of what the problem looks like:
We want to figure out what is the shortest sequence of circles and squares sends the four white triangles in the four petals into a single flower.
We see that the square rotates the whole thing while the circle swaps the top corners and joins the bottom corners on the bottom left.
Making a couple of smart remarks will make it easier to solve the problem:
This shows that a hard minimum amount of moves we need is 5,
because that is the sequence ○⎕○⎕○
, which is the absolute
minimum we could need, as we need at least 3 circles and
we cannot have consecutive circles.
This clearly doesn't work, so the solution is at least 6 steps long.
After that, and playing around a bit, you can figure out that the shortest solution has eight steps and it is
○⎕⎕○⎕⎕⎕○
There is, obviously, a more rigorous solution that proves that this is the shortest solution. Doing so in paper involves keeping track of some branching possibilities, so I will do my best to...
]]>A waiter at a restaurant gets a group's order completely wrong. Can you turn the table to get two or more orders right?
An arbitrary number of people are people sitting down, around a round table, waiting for their orders. When the waiter brings their orders, they mix everything up and no single person is served what they ordered.
Can you always turn the table so that at least two orders are correctly aligned? You can start by trying with the example in the figure above, assuming you are not colour-blind (sorry if you are :/).
Give it some thought...
Congratulations to the ones that solved this problem correctly and, in particular, to the ones that submitted correct solutions:
(The list is in no particular order.)
I received some really interesting solutions and I'll be sharing a paraphrasing of the one I think is the most elegant.
Yes, it is always possible to turn the table some number of places in order to fix the dishes of at least two people at the same time. Let's see how to do it.
Each person sitting at the table will look to its left and count the number of plates until its own plate.
For example, if we consider this placement:
Then the corresponding numbers would be the following:
If there are \(n\) people sitting at the table, that number is an integer between \(1\) and \(n - 1\), inclusive. It is not \(0\) because that would mean that person got the correct plate and it is not \(n\) because that is a whole turn around the table, also meaning that person would've gotten its own plate.
We have seen that each person has a number inside the set
\[ \{ 1, 2, \cdots, n-1 \} ~~~,\]
and there are \(n\) people, but that set only contains \(n - 1\) distinct numbers. Therefore, the pigeonhole principle tells you that at least two persons have the same number \(d\), which means you can turn the table \(d\) times in the counter-clockwise direction to deliver the correct plate to them.
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