# 14 mathematical puzzles (and their solutions)

The riddles are a playful way to pass the time, riddles that require the use of our intellectual capacity, our reasoning and our creativity in order to find their solution. And they can be based on a large number of concepts, including areas as complex as mathematics. That is why in this article we will see **a series of mathematical and logical puzzles, and their solutions** .

- Related article: "13 games and strategies to exercise the mind"

## A selection of mathematical puzzles

This is a dozen mathematical puzzles of different complexity, extracted from various documents such as the book Lewi's Carroll Games and Puzzles and different web portals (including the Youtube channel on mathematics "Derivando").

### 1. The Einstein riddle

Although it is attributed to Einstein, the truth is that the authorship of this riddle is not clear. The riddle, more logic than mathematics itself, reads as follows:

“**On a street there are five houses of different colors** , each one occupied by a person of a different nationality. The five owners have very different tastes: each of them drinks a kind of drink, smokes a certain brand of cigarette and each one has a different pet from the others. Keeping in mind the following clues: The Brit lives in the red house The Swedish has a dog as a pet The Danish drinks tea The Norwegian lives in the first house The German smokes Prince The green house is immediately to the left of the white The owner of the green house drinks coffee The owner who smokes Pall Mall raises birds The owner of the yellow house smokes Dunhill The man who lives in the house of the center drinks milk The neighbor who smokes Blends lives next to the one who has a cat The man who has a horse lives next to the one who smokes Dunhill The owner who smokes Bluemaster drinks beer The neighbor who smokes Blends lives next to the one who takes water The Norwegian lives next to the blue house

**Which neighbor lives with a fish as a pet at home?**

### 2. The four nines

Simple riddle, it tells us "How can we make four nines result in a hundred?"

### 3. The bear

This riddle requires knowing a bit of geography. "A bear walks 10 km to the south, 10 to the east and 10 to the north, returning to the point from which it started. What color is the bear?"

### 4. In the dark

"A man gets up at night and discovers that there is no light in his room. Open the glove box, in which **there are ten black gloves and ten blue** . How many should you take to make sure you get a pair of the same color? "

### 5. A simple operation

A riddle in simple appearance if you realize what it refers to. "At what time will operation 11 + 3 = 2 be correct?"

### 6. The problem of the twelve currencies

We have a dozen **visually identical coins** , of which all weigh the same except one. We do not know if it weighs more or less than the others. How will we find out what it is with the help of a balance in at most three opportunities?

### 7. The horse's path problem

In the game of chess, there are chips that have the possibility of going through all the squares of the board, like the king and the queen, and chips that do not have that possibility, like the bishop. But what about the horse? Can the horse move around the board **in such a way that it passes through each and every one of the squares of the board** ?

### 8. The paradox of the rabbit

It is a complex and ancient problem, proposed in the book "The Elements of Geometrie of the most auncient Philosopher Euclides of Megara". Assuming that the Earth is a sphere and that we pass a rope through the equator, in such a way that we surround it with it. If we lengthen the rope one meter, in such a way **that forms a circle around the Earth** Could a rabbit pass through the gap between Earth and the rope? This is one of the mathematical riddles that require good imagination.

### 9. The square window

The next mathematical puzzle **was proposed by Lewis Carroll as a challenge to Helen Fielden** in 1873, in one of the letters he sent him. In the original version we talked about feet and not meters, but the one we put to you is an adaptation of this. Say the following:

A nobleman had a room with a single window, square and 1m high by 1m wide. The nobleman had an eye problem, and the advantage allowed a lot of light to enter. He called a builder and asked him to alter the window so that only half of the light entered. But it had to remain square and with the same dimensions of 1x1 meters. Nor could I use curtains or people or colored glasses, or anything like that. How can the constructor solve the problem?

### 10. The riddle of the monkey

Another riddle proposed by Lewis Carroll.

"On a simple pulley without friction hangs a monkey on one side and a weight on the other that perfectly balances the monkey. Yes **the rope has neither weight nor friction** What happens if the monkey tries to climb the rope? "

### 11. Number chain

On this occasion we find ourselves with a series of equalities, of which we must resolve the last one. It is easier than it seems to be. 8806 = 6 7111 = 0 2172 = 0 6666 = 4 1111 = 0 7662 = 2 9312 = 1 0000 = 4 2222 = 0 3333 = 0 5555 = 0 8193 = 3 8096 = 5 7777 = 0 9999 = 4 7756 = 1 6855 = 3 9881 = 5 5531 = 0 2581 =?

### 12. Password

**The police are watching closely a den of a gang of thieves** , which have provided some type of password to enter. They watch as one of them reaches the door and knocks. From the inside it says 8 and the person answers 4, answer before which the door opens.

Another man arrives and they ask him for the number 14, to which he answers 7 and it also happens. One of the agents decides to try to infiltrate and approaches the door: from the inside they ask him for the number 6, to which he answers 3. However, he must retreat since not only do they not open the door but he starts to receive gunshots from the inside. What is the trick to guess the password and what error has the police committed?

### 13. What number does the series follow?

A riddle known to be used in a test of admission to a school in Hong Kong and there is a tendency that children tend to have better performance in solving it than adults. It is based on guessing **what number has the parking space occupied by a car park with six seats** . They follow the following order: 16, 06, 68, 88,? (the occupied square that we have to guess) and 98.

### 14. Operations

A problem with two possible solutions, both valid. It is about indicating what number is missing after seeing these operations. 1 + 4 = 5 2 + 5 = 12 3 + 6 = 21 8 + 11 =?

## Solutions

If you have stayed with the intrigue of knowing what are the answers to these riddles, then you will find them.

### 1. The Einstein riddle

The answer to this problem can be obtained by making a table with the information we have and **going discarding from the tracks** . The neighbor with a pet fish would be the German.

### 2. The four nines

9/9+99=100

### 3. The bear

This riddle requires knowing a bit of geography. And it is that the only points in which carrying out this way we would arrive at the point of origin is **at the poles** . In this way, we would be facing a polar bear (white).

### 4. In the dark

Being pessimistic and foreseeing the worst case, the man should take half plus one to make sure he gets a pair of the same color. In this case, 11.

### 5. A simple operation

This riddle is solved with great ease if we take into account that we are talking about a moment. That is, time. **The statement is correct if we think about the hours** : if we add three hours at eleven, it will be two o'clock.

### 6. The problem of the twelve currencies

To solve this problem we must use all three occasions carefully, rotating the coins. First of all we will distribute the coins in three groups of four. One of them will go on each arm of the scale and a third on the table. If the balance shows a balance, it means that **the counterfeit coin with a different weight is not between them but between those of the table** . Otherwise, it will be in one of the arms.

In any case, on the second occasion we will rotate the coins in groups of three (leaving one of the originals fixed in each position and rotating the rest). If there is a change in the inclination of the balance, the different currency is among those that we have rotated.

If there is no difference, it is among those that we have not moved. We remove the coins on which there is no doubt that they are not false, so that in the third attempt we will have three coins. In this case it will be enough to weigh two coins, one in each arm of the balance and the other in the table. **If there is a balance, the fake will be the one on the table** , and otherwise and from the information extracted in the previous occasions, we can say what it is.

### 7. The horse's path problem

The answer is affirmative, as proposed by Euler. To do this, you should do the following path (the numbers represent the movement in which you would be in that position).

63 22 15 40 1 42 59 18 14 39 64 21 60 17 2 43 37 62 23 16 41 4 19 58 24 13 38 61 20 57 44 3 11 36 25 52 29 46 5 56 26 51 12 33 8 55 30 45 35 10 49 28 53 32 47 6 50 27 34 9 48 7 54 31.

### 8. The paradox of the rabbit

The answer to whether a rabbit would pass through the gap between the Earth and the rope lengthening a single meter the rope is affirmative. And it is something that we can calculate mathematically. Assuming that the earth is a sphere with a radius of around 6.3000 km, r = 63000 km, even though the rope that surrounds it completely must have a considerable length, extending it by a single meter would generate a gap of around 16 cm . This would generate **that a rabbit could pass comfortably through the gap between both elements** .

For this we have to think that the rope that surrounds it will measure 2πr cm in length originally. The length of the rope lengthening one meter will be If we lengthen this length by one meter, we will have to calculate the distance to be distanced from the rope, which will be 2π (r + extension needed to lengthen). So we have 1m = 2π (r + x) - 2πr.Doing the calculation and clearing the x, we obtain that the approximate result is 16 cm (15,915). That would be the gap between the Earth and the rope.

### 9. The square window

The solution to this riddle is **make the window a diamond** . Thus, we will continue to have a window of 1 * 1 square and without obstacles, but through which half of the light would enter.

### 10. The riddle of the monkey

The monkey would arrive at the pulley.

### 11. Number chain

8806=6 7111=0 2172=0 6666=4 1111=0 7662=2 9312=1 0000=4 2222=0 3333=0 5555=0 8193=3 8096=5 7777=0 9999=4 7756=1 6855=3 9881=5 5531=0 2581= ¿?

The answer to this question is simple. Only **we have to look for the number of 0 or circles that there are in each number** . For example, 8806 has six since we would count the zero and the circles that are part of the eights (two in each) and the six. Thus, the result of 2581 = 2.

### 12. Password

Looks are deceiving. Most people, and the policeman who appears in the problem, would think that the answer thieves ask for is half of the figure they ask. That is, 8/4 = 2 and 14/7 = 2, which would only need to divide the number that thieves gave.

That is why the agent answers 3 when they ask for the number 6. However, that is not the correct solution. And that is what thieves use as a password **it is not a numerical relation, but the number of letters of the number** . That is, eight has four letters and fourteen has seven. In this way, in order to enter it would have been necessary for the agent to say four, which are the letters that have the number six.

### 13. What number does the series follow?

This riddle, although it may seem a mathematical problem of difficult solution, really only requires observing the squares from the opposite perspective. And it is that in fact we are before an ordered row, that we are observing from a concrete perspective. So, the row of squares that we are observing would be 86, ¿?, 88, 89, 90, 91. In this way, **the occupied square is 87** .

### 14. Operations

To solve this problem we can find two possible solutions, being as we have said both valid. To be able to complete it, we must observe the existence of a relationship between the different operations of the riddle. Although there are different ways to solve this problem, we will look at two of them below.

One of the ways is to add the result of the previous row to the one we see in the row itself. So: 1 + 4 = 5 5 (that of the result above) + (2 + 5) = 12 12+ (3 + 6) = 21 21+ (8 + 11) =? In this case, the response to the last operation would be 40.

Another option is that instead of a sum with the figure immediately above, let's see a multiplication. In this case we would multiply the first number of the operation by the second and then we would do the sum. So: 1*4+1=5 2*5+2=12 3*6+3=21 8*11 + 8 =? In this case the result would be 96.