Puzzles of logic are one of the best ways to measure your intelligence, quick wits and ability to think outside the box. The following selection of riddles, brain-teasers and numeric sequences are designed to separate the deep thinkers from the dunces. They start off easy and get progressively harder — best grab a pen and paper!

The following puzzles range from a popular maths exam question that an estimated 90% of students fail to get right to the self-proclaimed hardest logic puzzle in the world. If you manage to ace all of these you're either a borderline genius or a puzzle addict with a very good memory. We'll be updating the article with complete solutions next week. In the meantime, take your best stab in the comments!

## #1 Pot-boiler

“A watched pot never boils unless it is watched.” Is this statement true or false?

## #2 Family Matters

A man was looking at a portrait when a passerby asked him, “Whose picture are you looking at?” The man replied: “Brothers and sisters have I none, but this man’s father is my father’s son.”

Whose picture was the man looking at?

Suppose, in the above situation, the man had instead answered: “Brothers and sisters have I none, but this man’s son is my father’s son.”

Now whose picture is the man looking at?

## #3 There's a Star Hiding in This Image. Can You Find It?

Some of you will spot it quickly. Others of you will not.

## #4 Six Matches, Four Triangles

Using six matchsticks of equal length, create four identical, equilateral triangles. (There's no need for snapping, burning, or otherwise altering the matchsticks.)

## #5 Blowin’ in the Wind

An airplane flies in a straight line from airport A to airport B, then back in a straight line from B to A. It travels with a constant engine speed and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from A to B?

## #6 The Monk And The Mountain

At precisely 7:00 a.m., a monk sets out to climb a tall mountain, so that he might visit a temple at its peak. The trail he walks is narrow and winding, but it is the only way to reach the summit. As he ascends the mountain, the monk walks the path at varying speeds. Though he stops occasionally to rest and eat, he never strays from the path, and he never walks backwards. At exactly 7:00 p.m., the monk reaches the temple at the summit, where he stays the night.

The following morning at 7:00 a.m. sharp, the monk departs the temple and begins his journey back to the bottom of the mountain. He descends by way of the same path, again walking slowly at times and quickly at others, stopping here and there to eat and drink and rest, but never deviating from the path and never going backwards. Twelve hours later, at 7:00 p.m. on the nose, the monk arrives back at the foot of the mountain.

Is there any point along the path that the monk occupied at precisely the same time on both days? How do you know?

## #7 Links in a Chain

The task: Join the four three-link chains on the left to form the circular chain on the right. To join two chains, you must cut, and then re-weld, a link. What is the minimum number of links you must cut and re-weld to complete the circle?

## #8 Car park numbers

Can you explain the numbering system in this parking lot? What number is the parked car obscuring?

## #9 Newton’s Trees

How can nine trees be arranged in ten rows, such that each row contains exactly three trees?

## #10 Complete the series

Hint: The answer to this riddle is not "6".

## #11 String Around the Rod

Twenty years ago, this puzzle appeared on a test administered to top-tier math students from 16 countries around the world. Apparently, only 10% of test takers got it right. Can you find the “simple” solution that so many intelligent students missed?

## #12 Just the two of us

You and a fellow traveler are caught trespassing through the kingdom of the fear-instilling Mad King. As a punishment, the king imprisons you both in separate cells and presents you with a riddle, which you must solve correctly in order to save your lives.

“From your separate cells, you can each see half the land of my kingdom,” says the king, “across which are distributed either 10 or 13 villages. Each day at 5 p.m., I will give each of you an opportunity to tell me the number of villages in my kingdom. If your answer is correct, you will both be freed. But if your answer is wrong, you will both be killed.”

On the fifth day, the two of you are freed. How many villages are there in the king’s kingdom, and how many villages did you each see? How do you know?

## #13 In Search of an Unusual Book

In a certain library, no two books contain the same number of words, and the total number of books is greater than number of words in the largest book.

How many words does one of the books contain, and what is the book about?

Need a hint? Here’s a conceptually similar puzzle, the answer to which can help put you in the right frame of mind for tackling this problem:

Why must there certainly be at least two people in the world with exactly the same number of hairs on their head?

## #14 Counting Coins

You are blindfolded and brought to a table, on which rest fifty coins. Sixteen of them, you are told, are heads up. Thirty-four of them are tails up. Your task is to sort the coins into two groups, such that each group contains the same number of coins that are heads up. With your blindfold on, you cannot see if a coin is heads up. Neither can you feel a coin to determine its orientation. You can, however, move the coins and flip any number of them over.

## #15 Potato Dryer

You have 100kg of potatoes, which are 99 percent water by weight. You let them dehydrate until they’re 98 percent water. How much do they weigh now?

## #16 Pigs in pens

How can you distribute all 21 of your pigs into four pigpens and still have an odd number of pigs in each pen? You may place the same number of pigs in any number of pens (for example, the first and third pens can both contain 3 pigs), but the number of pigs in each pen must be odd. How do you distribute your pigs?

## #17 The Perilous Bridge-Crossing

Four people fleeing a fire come to a river in the night. Spanning the river is a narrow bridge, which they must cross to safety before they are consumed by the fire.

The bridge is narrow and ill-kept, and can therefore support just two people at any one time. The four people share but one dim torch, which they must use to traverse the dilapidated bridge safely. Anyone who attempts a crossing without the torch is as good as dead.

All four people are injured in different ways, and so it takes them different amounts of time to cross the bridge. Person A can cross the bridge in one minute, person B in two minutes, person C in five minutes, and person D in ten minutes. When two people cross together, they must do so at the slower person’s pace.

With the fire growing closer by the minute, time is of the essence. What is the fastest time in which all four people can cross the bridge to safety?

## #18 Buttered toast

Consider the preparation of three slices of hot buttered toast. The toaster is the old-fashioned type, with hinged doors on its two sides. It holds two pieces of bread at once but toasts each of them on one side only. To toast both sides it is necessary to open the doors and reverse the slices.

It takes three seconds to put a slice of bread into the toaster, three seconds to take it out, and three seconds to reverse a slice without removing it. Both hands are required for each of these operations, which means that it is not possible to put in, take out, or turn two slices simultaneously. Nor is it possible to butter a slice while another slice is being put into the toaster, turned, or taken out. The toasting time for one side of a piece of bread is thirty seconds. It takes twelve seconds to butter a slice.

Each slice is buttered on one side only. No side may be buttered until it has been toasted. A slice toasted and buttered on one side may be returned to the toaster for toasting on its other side. The toaster is warmed up at the start. In how short a time can three slices of bread be toasted on both sides and buttered?

## #19 A Devious Selection Task

On the table before you are the above four cards. Your task is to turn over as few cards as possible to verify whether the following statement is true: **Every card with a vowel on one side has an even number on the other side.** You must decide in advance which cards you will examine.

## #20 Three Boxes, Two Lies

The Fair Maiden Rowena wishes to wed. And her father, the Evil King Berman, has devised a way to drive off suitors. He has a little quiz for them, and here it is. It's very simple:

Three boxes sit on a table. The first is made of gold, the second is made of silver, and the third is made of lead. Inside one of these boxes is a picture of the fair Rowena. It is the job of the White Knight to figure out – without opening them – which one has her picture.

Now, to assist him in this endeavor there is an inscription on each of the boxes. The gold box says, "Rowena's picture is in this box." The silver box says, "The picture is not in this box." The lead box says, "The picture is not in the gold box." Only one of the statements is true. Which box holds the picture?

## #21 100 green-eyed dragons

You visit a remote desert island inhabited by one hundred very friendly dragons, all of whom have green eyes. They haven't seen a human for many centuries and are very excited about your visit. They show you around their island and tell you all about their dragon way of life (dragons can talk, of course).

They seem to be quite normal, as far as dragons go, but then you find out something rather odd. They have a rule on the island which states that if a dragon ever finds out that he/she has green eyes, then at precisely midnight on the day of this discovery, he/she must relinquish all dragon powers and transform into a long-tailed sparrow. However, there are no mirrors on the island, and they never talk about eye color, so the dragons have been living in blissful ignorance throughout the ages.

Upon your departure, all the dragons get together to see you off, and in a tearful farewell you thank them for being such hospitable dragons. Then you decide to tell them something that they all already know (for each can see the colors of the eyes of the other dragons). You tell them all that at least one of them has green eyes. Then you leave, not thinking of the consequences (if any). Assuming that the dragons are (of course) infallibly logical, what happens?

If something interesting does happen, what exactly is the new information that you gave the dragons?

*These puzzles were originally compiled by Robbie Gonzalez for iO9. *

## Comments

Ah yes, once upon a time I could solve these riddle, I was diagnosed with an IQ of 138.

Now at age 70 I am endowed with a little wisdom. I could not be fussed about such stuff,

There is only one truth.

We are only here to have children

sounds like your IQ has increased, by having the wisdom to not waste the time you have left on this mudball doing silly puzzles... :-P

Yeah Why do pleasure ships have the helmsman on the front, and industrial ships have the helmsman on the back? Don't answer. Me being silly again, but asking questions.

If you put all our accumulated wisdom together we can only get one answer.

We are here to keep our species alive. If you approach any question no matter how we ask, you get the same answer... to live, and ensure our children live

We, the Human Race are on the cusp of outliving any disaster that can befall us, cosmic or internal.

We need to take the next step and explore Mars. That is the next place for us to be.

Would you hold back Columbus?

Let's go.

1. True

2. His son, his father

3. Just to the upper right of the center.

4. Tetrahedron

5. -

6. Yes. Take two people, one travelling up, one travelling down. No matter their pace, they'll meet somewhere.

7. Three. Cut all the links on one chain and use them to connect the rest.

8. The nubers are upside down, thus the obscured number is 87

9. Can't do this one with ASCII art, what a shame.

10. ? = 5

1/2, 3/4. 5/5 (adding 1/4 each time)

11. 20cm (imaging the rod is actually a cardboard tube, then unravel it. The string is now the hypotenuse of four triangles of height 4 and base of 12/4, or three. Pythagorean triads says that the hypotenuse is 5)

12. This takes way to long to explain.

13. One book has no words and is about nothing.

14. -

15. 50kg

16. Three of the pens contain seven pigs each, the final pen encompasses the other three and thus contains twenty-one pigs.

17. -

18. -

19. Three cards. A, B and 7. A and seven to check if the statement is true, B to check if the statement is false.

20. Silver box.

21. Knowing that at least one of them has green eyes, the dragons sit around waiting until midnight to watch their neighbours (who they can see have green eyes) to turn into sparrows. When none of them do (because they believe they're the only one's with not green eyes), they come to a realization that they themselves have green eyes, so all one hundred of them turn into sparrows.

Last edited 29/04/16 7:03 pm6. The problem with this answer is that in this case it is not two people on the same day, but the same person on different days. It is not necessarily the case that there is one spot where the monk will be at the same time on each day. example - imagine a large rock about half way ... on the first day the monk rests just downhill, until 12pm, and then proceeds. On the second day the monk rests just uphill from the rock until 12pm, and then proceeds downhill. I think you provided the expected answer, but the problem was poorly configured.

a clock that is wrong, is still correct twice a day!

I believe that's only a clock that has stopped working. a clock that is set early or late will never be correct assuming that it progresses at the correct rate

Last edited 30/04/16 8:30 pmYes, it's true.

But even if you graphed location against time for the monks two trips, they have to cross at some point, even if they're on different days.

Using the two people anology just makes it easier to understand.

Last edited 30/04/16 11:52 amedit; ah okay, I see what you mean.

Last edited 30/04/16 8:22 pmOne monk. Read the question. There's 4 places. The ONLY times he's ever in the same place guaranteed Is the 'start' and 'finish' of each journey - regardless of where the 'starts' and 'finishes' are.. Start 7am finish 7pm

1. false. my logic is rusty, but i'll have a crack: if (P -> (P and ~Q)) is false

let P = a watched pot

let Q = boils or negate it ~Q = doesn't boil

The statement “A watched kettle never boils unless it is watched” can be reworded to say “If a watched kettle boils, then it is watched.”

Which is true because a watched kettle is watched, the whether it boils or not is irrelevent.

disagree: the statement should read, "if it a pot is watched, then the watched pot never boils"

eg:

P = a watched pot

~Q = not boiled (i.e. not boiled, ~ notates a negative)

P -> (P ^~ Q)

see here: https://en.wikipedia.org/wiki/Propositional_calculus

edit: to be proper, it really should read: "if and only if t a pot is watched, then the watched pot never boils"

which still fails

Last edited 30/04/16 10:12 pmedit 2: there's an assumption that the pot is actually on the stove!

in logic, to negate an argument, find a counterexample: i watched my pot boil on the stove, therefore it boiled. === boiled pot watched

Last edited 30/04/16 10:31 pm4. make a square with a cross inside, matches can overlap

But there's an issue with 21. A dragon must find out that they have green eyes to transform. So at midnight the fact that none transform does not tell any individual dragon that they have green eyes, just that the other dragons are unaware that they do. The new information does not rule out one dragon having not green eyes, so every dragon is free to assume they are potentially that dragon, which they must logically already be doing as they can already see eye colours. Your solution works if there are two dragons, not if there are three or more.

The logic works for any n-amount of dragons. None of them initially transform at midnight because they can see that the other n-1 dragons have green eyes, and thus assume that they solely don't have green eyes. But the fact that none of them transform shows that they're all using this logic. Every single one of them see's n-1 pairs of green eyes, which means they all have green eyes.

19: No need to look at 4 - if it has a vowel the statement is true, if it doesn't have a vowel the statement is still true.

19. Only two cards, A and 7, are required to check the statement is true. It doesn't matter what is on the other side of a consonant.

18. 115 seconds?

17. 19 minutes?

It does matter.

It states that every card with a vowel on one side has an even number on the other side.

It also doesn't state that there's a number on one side and a letter on the other. The other side of the B could be a vowel and thus render the statement false.

The only card you don't have to check is the 4. This is because whether the other side is a number, consonant, vowel etc. it still satisfies the statement.

I started doing these and writing down the answers but I'm not sure if they're getting annoying (like number 10, which is by far the most ridiculous "puzzle" ever created) or if it's because it's late and I'm getting more tired and more grumpy.

#5 faster if both airports are at opposite sides of the earth and the pilot takes the opposite side on the way back (avoiding the subsequent headwind).

#9 Plant 7 of the trees first in a 3-1-3 configuration. This alone gives you 5 "rows" (or basically, lines as the question didn't stipulate the rows had to be all in one continuous line) - two horizontal, two diagonal and one vertical. Basically, a line picture of an hourglass with a line down the middle.

Next put the two remaining trees right next to the lone tree in the middle. this gives another 5 - one from the three trees together and 4 from the left and right trees on the top and bottom making a line to the middle tree on the other side (sort of like two X shapes)

#13 - the books can only have no words. Since no two books can have the same number of words and there are more books than there are number of words, it follows that normally there would need to be at least two that have the same number of words (otherwise we have an infinity + 1 scenario). However if all books have no words, they technically don't contain the same number of words.

Similarly, there must be at least two people in the world that are complete bald and have no hair

#14 "move" the coins so they are all standing on the side and divide into two equal groups. Both have the same amount of heads up coins (that is to say, none)

#19 all four cards.

There is no stipulation that in each card one side has a letter and the other has a number. Assuming that the rule to be verified applies to both the side that is face up and the side that is face down it could be possible that all the cards have vowels on the other side, including the A and B cards.

#20 silver box with the lead box being right

if the gold box is right, there is a contradiction with the silver box

If the silver box is right, there is a contradiction between the gold and lead box.

only the lead box can be correct

#21 they'd still remain dragons.

They can see that other dragons have green eyes, confirming the information of the traveller that at least one of them has green eyes. However it is still logically consistent for each individual dragon to not come to the conclusion that they have green eyes even if nobody else comes to that realisation.

#19 You don't need to check the 4. The rule holds regardless of what is on the other side of that card.

#21 On the 100th night they would all turn to sparrows. Say there was only 1 dragon with green eyes. That dragon would obviously turn into a sparrow on the first night. If there were 2 dragons with green eyes, they would both turn on the second night (since they both would've expected the other to turn on the first night if there was only one green-eyed dragon - so there must be 2 green-eyed dragons). And so on for any

ndragons. The new information imparted by the sailor is that *all of the other dragons* know that there is at least one green-eyed dragon.#14 Counting Coins

Put all the coins in one pile, then move 16 to a second pile while flipping every single one. This results in the same number of heads in both piles.

Try it with a smaller set and you'll see. Imagine you have 4 coins and 2 of them are heads, if you move 2 to a second pile flipping both of them then you have the following possibilities:

- You manage to pick both head coins, now both piles contain 0 heads.

- You pick one head and one tail, now both piles contain 1 head.

- You manage to pick both tails coins, now both piles contain 2 heads.

Last edited 01/05/16 1:12 amso what happened to posting the answers to these? I've been waiting!

Was waiting for answers too.

I've also been waiting....

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