How to solve magic squares? Numbers in a square puzzle title.
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 so as to meet the condition described above equal amount. 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 option magic square- 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.
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 reasoning of the first student. 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 you start counting from the other end of the sequence in a similar way, then the other two possible combinations are 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 total 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 specified this point much earlier, but then all the charm of the puzzle would have been lost) - because the second student knew that the known to him the sum is definitely not from this combination of numbers. Thus, the sum cannot be equal to 10 (due to 7 and 3, where the product of 21 will clearly produce these 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 seems not 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 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 version 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 the 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 combinations of 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 given 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
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.
There are an unimaginable number of mathematical riddles. Each of them is unique in its own way, but their beauty lies in the fact that to solve it you inevitably need to come to formulas. Of course, you can try to solve them, as they say, but it will be very long and practically unsuccessful.
This article will talk about one of these riddles, and to be more precise, about the magic square. We will look in detail at how to solve the magic square. 3rd grade of the general education program, of course, this goes through, but perhaps not everyone understood or does not remember at all.
What is this mystery?
Or, as it is also called, magic, is a table in which the number of columns and rows is the same, and they are all filled with different numbers. The main task is that these numbers add up vertically, horizontally and diagonally to the same value.
In addition to the magic square, there is also a semi-magic square. It implies that the sum of numbers is the same only vertically and horizontally. A magic square is “normal” only if one was used to fill it.
There is also such a thing as a symmetrical magic square - this is when the value of the sum of two digits is equal, while they are located symmetrically with respect to the center.
It is also important to know that squares can be of any size other than 2 by 2. A 1 by 1 square is also considered magical, since all conditions are met, although it consists of one single number.
So, we have familiarized ourselves with the definition, now let’s talk about how to solve a magic square. The 3rd grade school curriculum is unlikely to explain everything in as much detail as this article.
What are the solutions?
Those people who know how to solve a magic square (grade 3 knows for sure) will immediately say that there are only three solutions, and each of them is suitable for different squares, but still one cannot ignore the fourth solution, namely “at random” . After all, to some extent there is a possibility that an ignorant person will still be able to solve this problem. But we will throw this method into the long box and move directly to formulas and methods.
First way. When the square is odd
This method is only suitable for solving a square that has an odd number of cells, for example, 3 by 3 or 5 by 5.
So, in any case, it is initially necessary to find the magic constant. This is the number that is obtained by summing the numbers diagonally, vertically and horizontally. It is calculated using the formula:
In this example, we will consider a three by three square, so the formula will look like this (n is the number of columns):
So, we have a square in front of us. The first thing to do is to enter the number one in the center of the first line from the top. All subsequent numbers must be placed one square to the right diagonally.
But here the question immediately arises: how to solve the magic square? 3rd grade is unlikely to use this method, and most will have a problem, how to do it this way if this cell does not exist? To do everything right, you need to turn on your imagination and draw a similar magic square on top and it will turn out that the number 2 will be in it in the lower right cell. This means that in our square we enter the two in the same place. This means that we need to enter the numbers so that they add up to 15.
Subsequent numbers are entered in exactly the same way. That is, 3 will be in the center of the first column. But it will not be possible to enter 4 using this principle, since there is already a unit in its place. In this case, place the number 4 under 3 and continue. The five is in the center of the square, the 6 is in the upper right corner, the 7 is under the 6, the 8 is in the upper left, and the 9 is in the center of the bottom line.
You now know how to solve the magic square. I passed Demidov's 3rd grade, but this author had slightly simpler tasks, however, knowing this method, you will be able to solve any similar problem. But this is if the number of columns is odd. But what should we do if, for example, we have a 4 by 4 square? More on this later in the text.
Second way. For a double parity square
A double parity square is one whose number of columns can be divided by both 2 and 4. Now we will consider a 4 by 4 square.
So, how to solve a magic square (3rd grade, Demidov, Kozlov, Tonkikh - a task in a mathematics textbook) when the number of its columns is 4? It's very simple. Easier than the example before.
First of all, we find the magic constant using the same formula that was given last time. In this example, the number is 34. Now you need to arrange the numbers so that the sum vertically, horizontally and diagonally is the same.
First of all, you need to paint over some cells, you can do this with a pencil or in your imagination. We paint over all the corners, that is, the upper left cell and the upper right, the lower left and the lower right. If the square were 8 by 8, then you need to paint over not one square in the corner, but four, measuring 2 by 2.
Now you need to paint the center of this square, so that its corners touch the corners of the already shaded cells. In this example, we will get a 2 by 2 square in the center.
Let's start filling it out. We will fill in from left to right, in the order in which the cells are located, only we will enter the value in the shaded cells. It turns out that we enter 1 in the upper left corner, 4 in the right corner. Then we fill in the central one with 6, 7 and then 10, 11. The lower left 13 and 16 in the right. We think the order of filling is clear.
We fill out the remaining cells in the same way, only in descending order. That is, since the last number entered was 16, then at the top of the square we write 15. Next is 14. Then 12, 9 and so on, as shown in the picture.
Now you know the second way to solve the magic square. Year 3 will agree that the double parity square is much easier to solve than the others. Well, we move on to the last method.
Third way. For a square of single parity
A square of single parity is a square whose number of columns can be divided by two, but not by four. In this case it is a 6 by 6 square.
So, let's calculate the magic constant. It is equal to 111.
Now we need to visually divide our square into four different 3 by 3 squares. You will get four small squares measuring 3 by 3 in one large 6 by 6. Let's call the upper left one A, the lower right one - B, the upper right one - C and the lower left one - D.
Now you need to solve each small square using the very first method given in this article. It turns out that in square A there will be numbers from 1 to 9, in B - from 10 to 18, in C - from 19 to 27 and D - from 28 to 36.
Once you have solved all four squares, work will begin on A and D. It is necessary to highlight three cells in square A visually or using a pencil, namely: the upper left, central and lower left. It turns out that the highlighted numbers are 8, 5 and 4. In the same way, you need to select square D (35, 33, 31). All that remains to be done is to swap the selected numbers from square D to A.
Now you know the last way to solve the magic square. Grade 3 doesn't like the square of single parity the most. And this is not surprising, of all those presented it is the most complex.
Conclusion
After reading this article, you learned how to solve a magic square. Grade 3 (Moro is the author of the textbook) offers similar problems with only a few filled cells. There is no point in considering his examples, since knowing all three methods, you can easily solve all the proposed problems.
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 will 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 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"?
