### Education, Secondary education and schools

# How to solve the magic square (3rd class)? Benefits for schoolchildren

There are an unimaginable number of mathematical mysteries. Each of them is unique in its own way, but their charm lies in the fact that for the solution it is inevitable to come to formulas. Of course, you can try to solve them, as they say, by poking, but it will be very long and almost unsuccessful.

This article will talk about one of these mysteries, and to be precise - about the magic square. We will discuss in detail how to solve the magic square. 3 class general education program, of course, it goes, but maybe not everyone understood or does not remember at all.

## What is this riddle?

A magic square, 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 to have these figures in the sum along the vertical, horizontal and diagonal values equal.

In addition to the magic square, there is also a semi-magical one. It means that the sum of the numbers is the same only vertically and horizontally. A magic square is "normal" only if natural numbers from one were used for filling.

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 in relation to the center.

It is also important to know that squares can be of any magnitude other than 2 by 2. A square of 1 by 1 is also considered magical, since all conditions are fulfilled, although it consists of a single number.

So, we got acquainted with the definition, now let's talk about how to solve the magic square. The 3rd grade of the school program is unlikely to explain everything in detail as this article.

## What are the solutions?

Those people who know how to solve the magic square (the third grade knows exactly) will immediately say that there are only three solutions, and each of them is suitable for different squares, but it is still possible to avoid the fourth solution, namely "at random" . After all, to some extent, there is a possibility that an unknowing person can still solve this problem. But we will drop this method into a long box and proceed directly to the formulas and methods.

## The first way. When the square is odd

This method is only suitable for solving such a square, where the number of cells is odd, for example, 3 by 3 or 5 by 5.

So, in any case, you must first find a magic constant. This is the number that will be obtained when the sum of the digits is diagonal, vertical and horizontal. It is calculated using the formula:

In this example, we will consider a square three by three, so the formula will look like this (n is the number of columns):

So, in front of us is a square. The first thing to do is to enter the number one in the center of the first line from the top. All subsequent digits must be placed on the same cell of the right diagonally.

But then immediately the question arises, how to solve the magic square? Class 3 is unlikely to use this method, and most will have a problem, how can this be done in this way, if this cell does not exist? To do everything right, you need to include the imagination and draw a similar magic square from above and it will turn out so that the number 2 will be in it in the lower right cell. So, in our square we also put the deuce in the same place. This means that we need to write the numbers in such a way that they give a value of 15 in total.

The following figures fit in exactly the same way. That is, 3 will be in the center of the first column. But 4 on this principle can not be entered, because in its place already there is a unit. In this case, the number 4 is located at 3, and continue. The five is in the center of the square, 6 in the upper right corner, 7 at the 6, 8 in the upper left, and 9 in the center of the bottom line.

You now know how to solve the magic square. Demidov's third class passed, but this author had a slightly simpler task, however, knowing this method, it will be possible to solve any such problem. But this is if the number of columns is odd. And what if we have, for example, a 4 by 4 square? About this further in the text.

## The second way. For a square of double parity

A double-parity square is the one whose number of columns can be divided into 2 and 4. Now we consider the square 4 by 4.

So, how to solve the magic square (3 class, Demidov, Kozlov, Thin - the task in the textbook of mathematics), when the number of its columns is 4? It's very simple. Easier than in the example before.

First of all, we find the magic constant by the same formula that was quoted last time. In this example, the number is 34. Now we need to build the numbers so that the sum along the vertical, horizontal and diagonal lines is the same.

First of all you need to paint some cells, you can do it with a pencil or in the imagination. We paint all the corners, that is, the upper left cell and the upper right, lower left and lower right. If the square was 8 by 8, then it is necessary to paint not one cell in the corner, but four, 2 by 2 in size.

Now it is necessary to paint the center of this square, so that its corners touch the corners of the already painted cells. In this example, we will get a square in the center 2 to 2.

We proceed to filling 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 filled cells. It turns out that we enter 1 in the upper left corner and 4 in the right corner. Then the central one is filled 6, 7 and further 10, 11. Lower left 13 and right-16. We think the order of filling is clear.

The remaining cells are filled in exactly the same way, only in descending order. That is, since the last inscribed digit was 16, then at the top of the square we write 15. Next 14. Then 12, 9 and so on, as shown in the picture.

Now you know the second way how to solve the magic square. 3 class will agree that the square of double parity is much easier to solve than others. Well, we turn to the last method.

## The third way. For a square of single parity

A square of single parity is called a square whose number of columns can be divided into two, but not by four. In this case, this is a 6 by 6 square.

So, we calculate the magic constant. It is equal to 111.

Now we need to divide our square into four different squares 3 by 3. We get four small squares of size 3 by 3 in one big 6 by 6. The upper left is called A, the lower right is B, the upper right is C and the lower left is D.

Now you need to solve each small square, using the very first method that is 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, the work will start over A and D. It is necessary to select three cells in square A visually or using a pencil, namely the upper left, center and bottom left. It turns out that the selected digits are 8, 5 and 4. Similarly, we must select the square D (35, 33, 31). All that remains to be done is to swap the selected digits from D to A.

Now you know the last way how you can solve the magic square. 3rd class does not like the square of a single parity the most. And this is not surprising, of all the presented it is the most difficult.

## Conclusion

After reading this article, you learned how to solve the magic square. Grade 3 (Moro - the author of the textbook) offers similar tasks with only a few filled cells. There is no point in considering its examples, since knowing all three methods, you can easily solve all the proposed tasks.

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