14 mathematical puzzles (and their solutions)

Several challenges to exercise mathematical and logical skills.

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..

A selection of mathematical puzzles

This is a dozen mathematical puzzles of varying 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 "Deriving").

1. Einstein's riddle

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

“In a street there are five houses of different colorseach occupied by a person of a different nationality. The five owners have very different tastes: each of them drinks a certain type of beverage, smokes a certain brand of cigarette, and each has a pet different from the others. Considering the following clues: The Brit lives in the red house. The Swede has a dog as a pet The Dane drinks tea The Norwegian lives in the first house The German smokes Prince The green house is immediately to the left of the white house The owner of the green house drinks coffee The owner who smokes Pall Mall breeds birds The owner of the yellow house smokes Dunhill The man who lives in the center house drinks milk The neighbor who smokes Blends lives next door to the one who owns a cat Man who owns a horse lives next door to the man who smokes Dunhill The homeowner who smokes Bluemaster drinks beer Neighbor who smokes Blends lives next door to the one who drinks water The Norwegian lives next door to the blue house

Which neighbor lives with a pet fish 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 a little knowledge of geography. "A bear walks 10 km south, 10 km east, and 10 km north, returning to the point from which he 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. He opens the glove drawer, in which there are ten black and ten blue gloves. there are ten black gloves and ten blue gloves.. how many must he take to make sure he gets a pair of the same color?"

5. A simple operation

A seemingly simple riddle if you realize what it refers to. "At what point will the operation 11+3=2 be correct?"

6. The problem of the twelve coins

We have a dozen visually identical visually identical coinsof which all weigh the same except one. We do not know whether it weighs more or less than the others. How can we find out which one it is with the help of a balance at most three times?

7. The problem of the horse's path

In the game of chess, there are pieces that have the possibility of passing through all the squares of the board, such as the king and queen, and pieces that do not have that possibility, such as the bishop. But what about the knight? Can the knight move around the board in such a way that it passes through all the squares of the board in such a way that it passes through each and every square of the board.?

8. The rabbit paradox

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

9. The square window

The following mathematical puzzle was proposed by Lewis Carroll as a challenge to Helen Fielden in 1873, in a in 1873, in one of the letters he sent her. In the original version it was about feet and not meters, but the one we present here is an adaptation of this one. It reads as follows:

A nobleman had a drawing room with a single window, square and 1m high by 1m wide. The nobleman had an eye problem, and the advantage let in too much light. He called a builder and asked him to alter the window so that only half the light would enter. But it had to remain square and with the same dimensions of 1x1 meters. Nor could he use curtains or people or colored glass or anything like that. How could the builder solve the problem?

10. The riddle of the monkey

Another riddle proposed by Lewis Carroll.

"On a simple frictionless pulley a monkey is hung on one side and on the other a weight that perfectly balances the monkey. If the rope has neither weight nor friction, what happens if the monkey tries to climb up the rope?"

11. Chain of numbers

This time we are faced with a series of equalities, of which we have to solve the last one. It is simpler than it seems. 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 keeping a close watch on a den of a gang of thieves, who have set up some kind of password.The police are keeping a close watch on a den of a gang of thieves, who have set up some kind of password to gain entry. They watch as one of them comes to the door and knocks. From inside they say 8 and the person answers 4, to which the door opens.

Another arrives and is asked for number 14, to which he answers 7 and also passes. One of the agents decides to try to infiltrate and approaches the door: from inside they ask him for the number 6, to which he answers 3. What is the trick to guess the password and what mistake has the policeman made?

13. What is the next number in the series?

A riddle known for being used in a Hong Kong school entrance exam and for the tendency that children tend to perform better at solving it than adults. It is based on guessing the number of the occupied parking space in a parking lot with six parking spaces.. They follow the following order: 16, 06, 68, 88, ? (the occupied parking space we have to guess) and 98.

14. Operations

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

Solutions

If you've been wondering what the answers to these riddles are, you'll find them below.

1. Einstein's riddle

The answer to this problem can be found by making a table with the information that we have and discarding from the clues. The neighbor with a pet fish would be the German.

2. The four nines

9/9+99=100

3. The bear

This riddle requires a little knowledge of geography. And the only points where we could reach the point of origin by following this path are at the poles. Thus, we would be in front of a polar bear (white).

4. In the dark

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

5. A simple operation

This riddle is very easily solved if we take into account that we are talking about a moment. That is, time. The statement is correct if we think about the hoursIf we add three hours to eleven o'clock, it will be two o'clock.

6. The problem of the twelve coins

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

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

If there is no difference, it is among those we have not moved. We remove the coins that we have no doubt are not the false one, so in the third attempt we will be left with three coins. In this case it will be enough to weigh two coins, one on each arm of the balance and the other on the table. If there is balance, the counterfeit coin will be the one on the table.If not, and based on the information obtained on previous occasions, we will be able to tell which one it is.

7. The problem of the horse's path

The answer is yes, as Euler proposed. To do so, it should make the following path (the numbers represent the motion in which it 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 rabbit paradox

The answer to whether a rabbit would pass through the gap between the Earth and the rope by lengthening the rope by one meter is yes. And it is something we can calculate mathematically. Assuming that the earth is a sphere with a radius of about 6,3000 km, r=63000 km, even though the string that completely surrounds it has to be of considerable length, extending it by a single meter would generate a gap of about 16 cm . This would generate that a rabbit could comfortably pass through the gap between the two elements..

For this we have to think that the rope that surrounds it is going to measure 2πr cm in length originally. The length of the rope lengthening one meter will be If we lengthen this length one meter, we will have to calculate the distance that the rope has to be distanced, which will be 2π (r+extension necessary for it to lengthen). Then we have that 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 string.

9. The square window

The solution to this puzzle is to make the window a rhombus. Thus, we will still have a 1*1 square window with no obstacles, but half as much light would enter through it.

10. The riddle of the monkey

The monkey would reach the pulley.

11. Chain of numbers

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 only have to find the number of 0's or circles 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

Appearances can be deceiving. Most people, and the policeman in the problem, would think that the answer the robbers are asking for is half the number they are asking for. That is, 8/4=2 and 14/7=2, so all that would be needed is to divide the number the robbers give.

That is why the agent answers 3 when asked about the number 6. However, this is not the correct solution. Because what the thieves use as the password is not a numerical ratio, but the number of letters in the number. That is, eight has four letters and fourteen has seven. Thus, in order to enter, the agent would have had to say four, which are the letters in the number six.

13. Which number follows the series?

This puzzle, although it may seem a difficult mathematical problem to solve, actually only requires observing the squares from the opposite perspective. And the fact is that we are actually looking at an ordered row, which we are observing from a specific perspective. Thus, the row of places that we are observing would be 86, ?, 88, 89, 90, 91, 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 puzzle. Although there are different ways to solve this problem, we will see 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. Thus: 1+4=5 5 (the one in the result above)+(2+5)=12 12+(3+6)=21 21+(8+11)=¿? In this case, the answer to the last operation would be 40.

Another option is that instead of an addition with the immediately preceding digit, we see a multiplication. In this case we would multiply the first digit of the operation by the second and then do the addition. Thus: 14+1=5 25+2=12 36+3=21 811+8=¿? In this case the result would be 96.