Puzzle with numbers in a square of different colors. Logic games
The day before yesterday I was 25. And next year I will turn 28.
What day is my birthday?
Simple deduction
The teacher said that he had two consecutive numbers in mind from 1 to 10. After that, he told one student one of these numbers, and the second - the other. The following conversation followed:
1st student: “I don’t know another number.”
2nd student: “I don’t know the other number either.”
1st student: “Now I know another number.”
Find all 4 possible combinations of two numbers.
The number known to the students cannot be 1 and cannot be 10, otherwise they would easily guess what number their friend knows.
The solution I propose involves counting from the beginning and from the end of the sequence from 1 to 10. The fact that the second student does not know the number told to the first student is a crucial point in the first student's reasoning. If the number told to the first student is 2, then he will expect that the number told to the second student must be either 1 or 3. Since the second student says that he does not know the number of the first student, then this number is definitely not 1. Therefore, the first possible the combination is 2 and 3.
If the first student's number is 3, then the second student's number must be 2 or 4. But if the first student's number is 2 (and the second student was aware that the first student's number is not 1), then he would know the first student's number. However, the second student also does not know the number of the first student (judging by his words), which means that his number is 4. Thus, the second possible combination is 3 and 4.
If we start counting from the other end of the sequence in a similar way, then the other two possible combinations will be 9 and 8, 8 and 7.
Complex deduction
This problem is one of the most difficult in this section.
The teacher said that he had planned two natural numbers greater than one. He told the first student the product of these numbers, and the second student their sum. The following conversation ensued:
1st student: “I don’t know the amount.”
2nd student: “I knew you didn’t know. The amount is less than 14.”
1st student: “Now I know these numbers.”
2nd student: “Me too.”
Find these two numbers.
The numbers guessed by the teacher were 2 and 9. Below is the entire logical chain of reasoning. (Note: If the solution below seems not entirely clear to you, then below you will find a more detailed analysis of the logorithm for solving the problem using the example of two number combinations.)
So, it is necessary to determine two natural numbers greater than 1 (one). The first student knows their product, and the second knows their sum. We know that the sum of the conceived numbers is less than 14, so consider the following options:
2 2 – NO – otherwise the first student would also know their sum...
2 3 – NO – otherwise the first student would also know their sum...
2 4 – NO – otherwise the first student would also know their sum...
2 5 – NO – otherwise the first student would also know their sum...
2 6
2 7 – NO – otherwise the first student would also know their sum...
2 8
2 9
2 10
2 11 – NO – otherwise the first student would also know their sum...
3 3 – NO – otherwise the first student would also know their sum...
3 4
3 5 - – NO – otherwise the first student would also know their sum...
3 6
3 7 – NO – otherwise the first student would also know their sum...
3 8 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14 (for example, 2+12).
3 9 – NO – otherwise the first student would also know their sum...
3 10 – NO – the product of these numbers does not give such options that all other possible factors that give the same product will add up to less than 14.
4 4
4 5
4 6 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
4 7 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
4 8 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
4 9 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
5 5 – NO – otherwise the first student would also know their sum...
5 6 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
5 7 – NO – otherwise the first student would also know their sum...
5 8 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
6 6 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
6 7 – NO – the product of these numbers does not give such options that all other possible factors that give the same product total less than 14.
So, the following possible combinations remain, which we will consider in more detail:
2 6 – NO – for the sum of these two numbers it is impossible to select other terms that give the same result (8), so that by multiplying these terms (for example, 4x4), you would get the product (16), the other possible factors of which add up to more than 14 (for example, 2+8= 10).
2 8
2 9
2 10
3 4 – NO – for the sum of these two numbers it is impossible to select other terms that give the same result, so that by multiplying these terms you would get a product whose other possible factors add up to more than 14.
3 6 – NO – for the sum of these two numbers it is impossible to select other terms that give the same result, so that by multiplying these terms you would get a product whose other possible factors add up to more than 14.
4 4 – NO – for the sum of these two numbers it is impossible to select other terms that give the same result, so that by multiplying these terms you would get a product whose other possible factors add up to more than 14.
4 5 – NO – for the sum of these two numbers it is impossible to select other terms that give the same result, so that by multiplying these terms you would get a product whose other possible factors add up to more than 14.
The second student (who knew the sum of the hidden numbers) knew that the first student (who knew the product of the hidden numbers) did not know the sum of the numbers, and thought that the first student did not know that the sum of the numbers was less than 14.
There are only three possible combinations left:
2 8 – product =16, sum =10
2 9 – product=18, sum=11
2 10 – product=20, sum=12
Let's discard the sums that are formed by adding unique combinations of numbers - if such a product of numbers is known for which the sum is obvious (we could have stipulated this point much earlier, but then all the charm of the puzzle would have been lost) - because the second student knew that the known the sum is definitely not from this combination of numbers. So the sum cannot be equal to 10 (due to 7 and 3, where the product of 21 would clearly produce those numbers). The second student knows that the first student does not know the sum, but if the sum were equal to 10, then the first student would know the sum if the combination of numbers were 7 and 3. In a similar way, we discard the sum 12 (due to 5 and 7, when multiplying distinguishing themselves in a unique work 35).
And there is only one option left - the numbers 2 and 9. The problem is solved.
If the above solution does not seem entirely clear to you, now we will look in more detail at the main logarithm for solving the problem using the example of two number combinations.
Let's take the numbers 6 and 2 and see if this combination works.
This means that the first knows the product 12, and the second knows the sum 8.
First: “I don’t know the amount.”
The product I know is equal to 12, and you can get such a product like this: either 6x2 or 3x4. This means that the second person knows the sum equal to either 8 or 7.
The sum I know is 8, and you can get this sum by adding 6+2, 5+3 or 4+4. The first version of the terms will give the product 12, the second - 15, the third - 16.
A product equal to 15 can be immediately crossed out (that is, the option with numbers 5 and 3 can be discarded), because the number 15 is unique - it can be obtained exclusively through the natural numbers 5 and 3, so if it were just such a combination of numbers, the student would know both product and sum from the very beginning.
Consider the product 16. It can be obtained if the factors are either 4x4 or 8x2. In this case, the phrase that the sum of these factors would represent a number<14, другому студенту никак не поможет (4+4 и 8+2 <14).
Consider the product 12. In this case, the student will expect that the possible combinations of numbers are 4x3 or 6x2. But even in this case, the phrase that the sum of these factors would represent a number<14, другому студенту никак не поможет (4+3 и 6+2 <14).
Therefore, it is impossible to find a combination of numbers that add up to the number 8, where other terms that add up to the same amount, if multiplied, give a product whose other possible factors add up to more than 14. For example, if it is 4 and 4, then no such a sum from the possible other factors of the product 4x4, which in total would give a number greater than 14 (2+8=10).
I didn’t know whether it was 6x2 or 3x4, and the second student told me that the sum is less than 14. But it is absolutely obvious that he thought that from a sum equal to 8 or 7, one could find this variant of the terms, the product which will serve as a sum that must be greater than 14.
But his words did not help me at all, because 6+2 and 3+4 are in any case less than 14. Thus, the combination of numbers 6 and 2 is incorrect.
Now let's take the numbers 9 and 2 and see if this combination is suitable.
The first student knows the product, and the second student knows the sum of these numbers.
This means that the first knows the product 18, and the second knows the sum 11.
First: “I don’t know the amount.”
The product I know is 18, and you can get such a product like this: 9x2 or 6x3. This means that the second person knows the sum equal to either 11 or 9.
Second: “I knew you didn’t know. The amount is less than 14.”
The sum I know is 11, and you can get this sum by adding 9+2, 8+3, 7+4 or 6+5. The first version of the terms will give the product 18, the second - 24, the third - 28, the fourth - 30.
If the first student knows the product is 18, then he will consider the possible combinations: 9x2 and 6x3, so if I tell him that the sum must be less than 14, this will tell him that I have another probability in which the sum will be greater or equal to 14. So it is (see the next three paragraphs): 12+2, 14+2 and 15+2.
If the first student knows the product equal to 24, then he will consider the combination options 6x4, 8x3 and 12x2, but 12+2 is already 14, so if the product known to the first student were 24, then he could not be absolutely I'm sure the amount will be less than 14.
If the first student knew the product to be 28, then he would consider combinations of 7x4 or 14x2, but 14+2=16, so if the product known to the first student was 28, then he could not be absolutely sure that the sum will be less than 14.
If the first student knew the product to be 30, he would consider combinations of 5x6, 10x3, and 15x2, but 15+2=17, so if the product known to the first student was 30, he couldn't be absolutely sure. that the amount will be less than 14.
First: “Now I know these numbers.”
I didn't know if it was 9x2 or 6x3, and the second student tells me that the sum is less than 14. He must have had options with a sum ≥14, but this is not possible for a sum of 9 obtained with a combination of 6 and 3. Therefore, the sum known to him is 11, and it was obtained by adding 9 and 2.
How old are the children?
Two friends talking:
- Peter, how old are your children?
- You know, Thomas, I have three of them. And if you multiply their ages, you get 36.
- This is not enough...
- The sum of their ages is equal to the number of bottles of beer that we drank today.
- This is still not enough.
- Fine. The last thing I can say is that the eldest son wears a green cap.
How old are Peter's children?
Let's start with the product of three factors - 36. Write on paper all the options for the three factors that give the product equal to 36. Since we cannot be sure about the sum of bottles of beer, we will write only those two options that are possible with three factors (1-6-6 and 2-2-9), which add up to the same number. We also know that the eldest son likes to wear some kind of headdress from time to time. Therefore, option 1-6-6 is eliminated, since we need an option where there is only one older child.
Math sign
What mathematical sign can be placed between the numbers 5 and 9 to make a number greater than 5 and less than 9?
Fraction
Place all 9 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 9 in the numerator and denominator of the fraction, using each digit once and only once, so that the resulting fraction is equal to 1/3.
Five digit number
If you assign the number 1 in front of a certain 5-digit number, you will get a number that is 3 times smaller than if you add the number 1 at the end of the same number. Find this number.
Cipher
Find the number if:
- This number consists of 6 different digits.
- Even and odd digits alternate (zero can also alternate and will be considered an even number).
- Every two adjacent digits differ by more than 1.
- The number consisting of the first two digits, as well as the number consisting of the middle two digits, are divided without a remainder by the number made up of the last two digits.
There is more than one solution to this problem.
The last two digits in the number can be the following: 03, 05, 07, 09, 14, 16, 18, 25, 27, 29 and 30. Multiple (divisible without a remainder) two-digit numbers (and at the same time consisting of even and odd alternating digits) for 03, 07, 09 and 18 will be as follows: 03 – 27, 63, 69, 81 07 – 49, 63 09 – 27, 63, 81 18 – 36, 72, 90. There are 5 six-digit numbers that satisfy the conditions of the task , which can be made from these two-digit numbers: 692703, 816903, 496307, 816309 and 903618.
(Provided that the number 903618 satisfies the conditions of the task despite the reverse order of even and odd digits.)
Make a table of three numbers arranged vertically and three numbers horizontally, as shown in the example below. Numbers can only be taken from the list provided. You can use the same number several times. Having compiled a table, calculate the sum of all the numbers in it. What is the maximum amount that can be received?
Table List of numbers 
Example using each of the numbers: 40067 04802 78215 twice 
The amount in this example is: 73. But, of course, this result can be improved.
Mysterious number
Find the number indicated by asterisks if you know the following:
- All 4 digits of the unknown number are different.
- None of the numbers are zero.
- Below are auxiliary 4-digit numbers, where each “0” to the right of the number means that there is a digit in this number that coincides with one of the digits of the desired number, but is in a different position.
- Each “+” to the right of a number means that there is a matching digit in that number in the same position as the digit of the desired number.
1996
Using the numbers: “1”, “9”, “9” and “6” and the signs of arithmetic operations: “+”, “-”, “x”, “:”, the root sign and parentheses, get the following results:
29, 32, 35, 38, 70, 73, 76, 77, 100 and 1000.
All four digits must be used only in the specified order, each digit only once, and the digits must not be turned upside down.
100
Using four sevens (7) and one one (1), you get the number 100. In addition to 5 digits, you can use the usual arithmetic operations: “+”, “-”, “x”, “:”, root sign and parentheses .
Equation
Rearrange only one digit so that you get an equality:
101 – 102 = 1
Sequences
There is an infinite number of formulas (functions) that will satisfy a given finite sequence of numbers. Try to find the simplest formulas for the following sequences.
- 8723, 3872, 2387, ?
- 1, 4, 9, 18, 35, ?
- 23, 45, 89, 177, ?
- 7, 5, 8, 4, 9, 3, ?
- 11, 19, 14, 22, 17, 25, ?
- 3, 8, 15, 24, 35, ?
- 2, 4, 5, 10, 12, 24, 27, ?
- 1, 3, 4, 7, 11, 18, ?
- 99, 92, 86, 81, 77, ?
- 0, 4, 2, 6, 4, 8, ?
- 1, 2, 2, 4, 8, 11, 33, ?
- 1, 2, 6, 24, 120, ?
- 1, 2, 3, 6, 11, 20, 37, ?
- 5, 7, 12, 19, 31, 50, ?
- 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ?
- 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ?
- 4, 7, 15, 29, 59, 117, ?
- 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ?
- 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ?
Motto
Science is not and will never be a finished book.Albert Einstein
How to solve magic squares?
A puzzle like Sudoku is commonly called a magic square. This is a square whose cells are filled with numbers so that the sum at the end of any row, column and diagonal is the same. In magic square puzzles, some numbers are missing, and you need to arrange them in such a way as to satisfy the equal sum condition described above. How to solve magic squares?
Methods for solving magic squares
In order for the solution of magic squares to be correct, you need to know the very magic sum that should be obtained when adding numbers in rows, columns and diagonals. After this, placing the missing numbers becomes much easier. How to find this amount?
Method 1
The simplest version of a magic square is when one of the rows, one of the columns or one of the diagonals is completely filled with numbers. In this case, all that remains is to calculate the sum of these numbers and select solutions.
Method 2
The sum of the numbers at the ends of rows, columns and diagonals can be calculated using special formulas. In this case, the formula for squares with an even number of cells in one row will differ from squares with an odd number of cells.
So, for even squares the following formula is suitable:
- n + ((n+1) * n * (n-1) / 2) , where n is the number of cells in one row.
For odd squares the formula is:
- n * (n 2 +1) / 2, where n is also the number of cells in one row.
Example solution
Let's consider solutions to a magic square of nine cells with numbers from 1 to 9. First, let's calculate the sum that should be obtained at the ends. We have 3 cells in one line, that is, n = 3. Substitute the value into the formula:
- 3 * (3 2 +1) / 2 = 3 * 10 / 2 = 15
Now we select the numbers so that the sum is 15.
Next, the algorithm will require a little spatial imagination. Place the number 1 in the middle of the top line. We place each next number on the right diagonally upward. We try to put 2. But there are no cells there, if we substitute another identical imaginary square above our square, then the number 2 will appear in the lower right corner of this 
new square. We transfer it to our square and place it in the lower right corner. We also put the number 3 on the right diagonally upward - and again there is no cell there, using an imaginary square we find out that its place is in the middle of the left column. We put the number 4 according to the same principle, but this cell is occupied by one - in this case we put it directly under the number 3. The number 5 diagonally up and to the right of 4 is in the very center, and the number 6 is in the upper right corner. The number 7, with the help of imagination, should have ended up in the lower left corner. But there is already a 4 there, so we put it directly under the number 6. The number 8 appears with the help of an imaginary square in the upper left corner, and the number 9 in the remaining cell in the middle of the right column. The general algorithm is as follows: put the next number at the top right diagonally, if there is no space, use an imaginary square, and if the cell is occupied, then put the number directly below the previous one.
There are various techniques for constructing squares of single parity and double parity.
Calculate the magic constant. This can be done using the simple mathematical formula /2, where n is the number of rows or columns in the square. For example, in a square 6x6 n=6, and its magic constant is:
- Magic constant = / 2
- Magic constant = / 2
- Magic constant = (6 * 37) / 2
- Magic constant = 222/2
- The magic constant for a 6x6 square is 111.
- The sum of the numbers in any row, column and diagonal must be equal to the magic constant.
Divide the magic square into four equally sized quadrants. Label the quadrants A (top left), C (top right), D (bottom left), and B (bottom right). To find out the size of each quadrant, divide n by 2.
- Thus, in a 6x6 square, the size of each quadrant is 3x3.
In quadrant A, write the fourth part of all numbers; in quadrant B, write the next fourth of all numbers; in quadrant C, write the next fourth of all numbers; in quadrant D, write the final quarter of all numbers.
- In our example of a 6x6 square, in quadrant A, write the numbers 1-9; in quadrant B - numbers 10-18; in quadrant C - numbers 19-27; in quadrant D - numbers 28-36.
Write down the numbers in each quadrant as you would for an odd square. In our example, start filling quadrant A with numbers starting from 1, and quadrants C, B, D - starting with 10, 19, 28, respectively.
- Always write the number from which you begin filling in each quadrant in the center cell of the top row of a particular quadrant.
- Fill in each quadrant with numbers as if it were a separate magic square. If an empty cell from another quadrant is available when filling a quadrant, ignore this fact and use the exceptions to the rule for filling odd squares.
Highlight specific numbers in quadrants A and D. At this stage, the sum of the numbers in columns, rows and diagonally will not be equal to the magic constant. Therefore, you must swap the numbers in certain cells of the upper left and lower left quadrants.
- Starting from the first cell of the top row of quadrant A, select a number of cells equal to the median number of cells in the entire row. Thus, in a 6x6 square, select only the first cell of the top row of quadrant A (the number 8 is written in this cell); in a 10x10 square you need to select the first two cells of the top row of quadrant A (the numbers 17 and 24 are written in these cells).
- Form an intermediate square from the selected cells. Since you have selected only one cell in a 6x6 square, the intermediate square will consist of one cell. Let's call this intermediate square A-1.
- In a 10x10 square, you selected the two cells in the top row, so you need to select the first two cells in the second row to form an intermediate 2x2 square of four cells.
- On the next line, skip the number in the first cell, and then highlight as many numbers as you highlighted in the intervening square A-1. Let's call the resulting intermediate square A-2.
- Obtaining intermediate square A-3 is similar to obtaining intermediate square A-1.
- Intermediate squares A-1, A-2, A-3 form the selected area A.
- Repeat the process described in quadrant D: create intermediate squares that form the selected area D.
I love games where you have to think. Therefore, our series of “top 10” articles smoothly flows into puzzles. Today I will talk about ten number puzzles. When I rushed to compile this rating, I was faced with the problem of finding ten good games, despite the fact that there are tons of digital puzzles in the App Store! The bad thing is that there are a lot of clones, repetitions and low-quality crafts... But when the top was compiled, I realized that everyone would find something new in it! Even I got to know three great games. Let's go!
Threes!
There are numbers on the playing field. The player can move all the numbers in any of the 4 directions. Moreover, if the movement of any row or column is impeded by a wall and there are:
a) identical numbers greater than or equal to 3
b) 1 and 2
then they add up and instead of two numbers, a third one appears - the sum. The goal is to score as many points as possible. The game is endless, but it is very difficult to score a lot of points.
After the release of Threes! the App Store became inundated with clones under the name “2048”.
Shikaku
A simple and non-pop puzzle from the creators of Sudoku. The goal in this game is to divide the field with numbers into rectangles so that the area of the rectangles is equal to the number inside it. There is only one implementation of this game for iPad.
Numtris: A game of logic and numbers
This is an original adventure game. Tetris with numbers. Numbers fall from above and you need to either collect them according to the Threes principle (1 and 2 will give 3), or remove them by collecting several identical ones (for example, four identical fours). Numtris has a full campaign with many missions. The missions are varied: from holding out for 40 seconds to killing a monster... You can compete with friends both online and on the same iPad.
The game is very stylish with nice graphics. I recommend trying it, since it's free.
Download Numtris for free (in-app purchases available)
GREG — A mathematical puzzle game
An interesting game for speed and the ability to quickly add numbers. There are numbers on a 4 by 4 field. It is necessary to type the sum from these numbers so that you get the number in the circle on top. As soon as the number is collected, it changes and you need to select the numbers again. The less you use some numbers on the field, the more they heat up... After 5 such “heatings” the game may end. Reset occurs after each level. At the end the game rewards you with some title. Can you knock out "Math Genius"?
Few people loved mathematics in childhood, but mathematical puzzles on the Internet always become hits, because solving them usually does not require in-depth knowledge, but requires ingenuity and innovative thinking. We invite you to test yourself on the five main logic puzzles of this year.
Task No. 1
Kumar Ankit invited Facebook users to count how many triangles are shown in his drawing. Almost none of the users coped with the seemingly simple task of counting figures. Many are close to the correct answer, but most lack a little care.
Answer:
There are 24 triangles inside the large triangle, it’s not difficult to count, but most users did not pay attention to another triangle hidden in the author’s signature. Thus, there are 25 triangles in total in the picture.
Task No. 2
An unusual problem with two solutions was offered to Internet users by the creators of the site gotumble.com. According to them, one solution to the puzzle is simpler, about 10% of people are able to find it, but only one person in a thousand can reach the second solution. Try it yourself.
Answer:
First solution consists of adding to each subsequent example the result of the previous one. So, adding 5 to the sum of 2 and 5, we get 12. Adding 12 to the sum of 3 and 6, we get 21. And so on. In this case, the correct answer to the puzzle will be 40.
But second solution, which only one person in a thousand understands, consists of adding the first digit of the example with the product of two digits:
2 + 2*5 = 12, 3 + 3*6 = 21, 8 + 8*11 = 96.
Task No. 3
We have a triangle consisting of four parts, but if we rearrange the parts, it appears as an empty square. How can this be?
Answer:
This is not an optical illusion at all. It's all about different angles of inclination of the hypotenuse of the red and turquoise triangle - hence the different sizes of the figures.
Task No. 4
Guardian columnist Alex Bellos invited readers to solve a problem that is part of the final math exam in some countries. According to statistics, only one person out of 10 solves it.
We have a cylinder around which a thread is wound symmetrically four times. The circumference of the cylinder is 4 cm and its length is 12 cm. You need to find the length of the thread.
Answer:
The task seems too complicated to most schoolchildren, but in fact, you just need to understand that by turning the cylinder onto a plane, we get an ordinary rectangle with sides of 4 and 12 cm, which can be divided into four smaller rectangles with sides of 4 and 3 cm. Thread in this case, it will be the hypotenuse of a right triangle and its length in each of the four figures can be calculated using a simple school formula, it is equal to 5 cm. As a result, the total length of the thread is 20 centimeters.
Problem #5
And finally, the latest mathematical puzzle that blew up social networks. According to the author of the post, it depicts a riddle that is given as a bonus question to students in Singapore. The compilers of the riddle suggest studying the number sequence and filling in the four empty windows with the missing numbers.
Answer:
Netizens puzzled over this problem for a long time, but even serious mathematicians could not cope with it. And the Ministry of Education of Singapore disowned this task, saying that it had nothing to do with it. So most likely the puzzle was just someone's cruel joke.