14 math puzzles (and their solutions)
Riddles are a fun 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 your mind"
A selection of math 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 the following:
“In a street there are five houses of different colors, each occupied by a person of a different nationality. The five owners have very different tastes: each of them drinks a type of drink, smokes a certain brand of cigarette and each has a different pet from the others. Considering the following clues: The Brit lives in the red house. The Swede has a pet dog. 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 one. 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 in the center drinks milk. The neighbor who smokes Blends lives next door to the one with a cat. The man who owns a horse lives next door to the one who smokes Dunhill. The owner who smokes Bluemaster drinks beer. The neighbor who smokes Blends lives next door to the one who drinks water. The Norwegian lives next to the blue house
What neighbor lives with a pet fish at home?
2. The four nines
Simple riddle, he tells us "How can we make four nines equal one hundred?"
3. Bear
This puzzle requires knowing a little 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 wakes up at night and discovers that there is no light in his room. He opens the glove drawer, in which there are ten black gloves and ten blue. How many should he catch to make sure he gets a pair of the same color? "
5. A simple operation
A seemingly simple riddle if you realize what he is referring to. "At what point will the operation 11 + 3 = 2 be correct?"
6. The Twelve Coin Problem
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 scale in 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 horse? Can the knight move across the board in such a way that it passes through each and every one of the squares on the board?
8. The rabbit paradox
It is a complex and ancient problem, proposed in the book "The Elements of Geometry of the most still scientist Philosopher Euclides of Megara". Supposing 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 make a circle around the Earth Could a rabbit pass through the gap between the Earth and the rope? This is one of the math puzzles that requires good imagination skills.
9. The square window
The following math puzzle was proposed by Lewis Carroll as a challenge to Helen Fielden in 1873, in one of the letters he sent her. In the original version they talked about feet and not meters, but the one that we put you is an adaptation of this. Pray 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 let in a lot of light. He called in a builder and asked him to alter the window so that only half the light would come in. But it had to remain square and with the same dimensions of 1x1 meters. He couldn't use curtains or people or colored glass or anything like that. How can the builder solve the problem?
10. The riddle of the monkey
Another riddle proposed by Lewis Carroll.
“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 frictionWhat happens if the monkey tries to climb the rope? "
11. String of numbers
This time we find a series of equalities, of which we have to solve 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
Police are closely monitoring a den of a gang of thieves, which have provided some kind of password to enter. They watch as one of them comes to the door and knocks. From the inside, 8 is said and the person answers 4, answer to which the door opens.
Another arrives and they ask him for the number 14, to which he answers 7 and also passes. One of the agents decides to try to infiltrate and approaches the door: from the inside they ask him for number 6, to which he answers 3. However, he must retreat since not only do they not open the door but he begins to receive shots from the inside. What is the trick to guess the password and what mistake has the policeman made?
13. What number does the series follow?
A riddle known for being used in an entrance exam to a Hong Kong school and for there is a tendency that children tend to perform better in solving it than adults. It is based on guessing what number is the occupied parking space of a car park with six spaces. 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 been left with the intrigue of knowing what the answers to these riddles are, then you will find them.
1. Einstein's 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. Bear
This puzzle requires knowing a little geography. And it is that the only points in which by following this path we would reach 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 anticipating the worst case scenario, the man should take half plus one to ensure he gets a pair of the same color. In this case, 11.
5. A simple operation
This puzzle is easily solved if we consider 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 to eleven o'clock, it will be two.
6. The Twelve Coin Problem
To solve this problem we must use the three occasions carefully, rotating the coins. First we will distribute the coins into 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 equilibrium, it means that the counterfeit coin with a different weight is not among them but among those on 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 tilt of the balance, the different coin 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 the false one, so that in the third attempt we will have three coins left. In this case, it will be enough to weigh two coins, one on each arm of the scale and the other on the table. If there is balance, the false one will be the one on the table, and otherwise and from the information extracted in the previous occasions, we will be able to say what it is.
7. The problem of the horse's path
The answer is yes, as Euler proposed. To do this, it should do the following path (the numbers represent the movement 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 one meter is yes. And it is something that we can calculate mathematically. Assuming that the earth is a sphere with a radius of about 6,3000 km, r = 63,000 km, despite the fact that the chord that completely surrounds it has to have a considerable length, extending it a single meter would generate a gap of around 16 cm. This would generate that a rabbit could comfortably pass through the gap between both 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 said length one meter, we will have to calculate the distance that the rope has to distance, which will be 2π (r + extension necessary for lengthen). So we have that 1m = 2π (r + x) - 2πr. Doing the calculation and solving 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 puzzle is make the window a rhombus. Thus, we will continue to have a 1 * 1 square window without obstacles, but through which half the light would enter.
10. The riddle of the monkey
The monkey would reach the pulley.
11. String 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 have to find the number of 0 or circles that 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 six. Thus, the result of 2581 = 2.
12. Password
Looks are deceiving. Most people, and the police officer who appears in the problem, would think that the answer that the robbers ask for is half the number they ask for. That is, 8/4 = 2 and 14/7 = 2, so it would only be necessary to divide the number that the thieves gave.
That is why the agent answers 3 when asked for the number 6. However, that is not the correct solution. And is that what thieves use as a password It is not a number relationship, but the number of letters in the number. That is, eight has four letters and fourteen has seven. Thus, in order to enter, it would have been necessary for the agent to say four, which are the letters that number six has.
13. What number does the series follow?
This riddle, although it may seem like a difficult mathematical problem to solve, actually only requires looking at the squares from the opposite perspective. And it is that in reality we are facing an orderly row, that we are observing from a specific perspective. Thus, 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, both being valid as we have said. In order to complete it, it is necessary to 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 from 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 a sum with the immediately previous figure, we see a multiplication. In this case we would multiply the first figure of the operation by the second and then we would do the sum. Thus: 14+1=5. 25+2=12. 36+3=21. 811+8=¿? In this case the result would be 96.
