How to make 11 19 horizontal and vertical oblique add up to equal

This is the same as 1-9, that is, all 10 is added, and it teaches you an odd-order magic square arrangement.

1. Write 1,2 in the middle of the first line, and then write 2,3 in order to the top right......(The right side is written to the left, and the top side is written to the bottom).

3. When encountering integer multiples of the side length, write down (such as 5-order (fill in 1-25), write 6 below 5, write 11 below 10, write 16 below 15......）

For example, the third-order :

Fifth Order: Seventh Order.

Support the upstairs one, I'll add one more.

There are three types of construction of plane magic squares: n is an odd number, n is a multiple of 4, and n is other even number (the form of 4n+2).

n is the easiest when it is an odd number.

1) Place 1 in the middle column of the first row;

2) The numbers from 2 to n n are stored according to the following rules:

Walk in a 45° direction, such as to the upper right.

Each number is stored in rows minus 1 from the number of rows of the previous number, and the number of columns is added by 1

3) If the range of rows and columns is outside the range of the matrix, it is winded.

For example, if 1 is in the first row, then 2 should be placed in the bottom row, and the number of columns should also be added to 1;

n is a multiple of 4.

The symmetrical element exchange method is adopted.

First, fill in the matrix from 1 to n n in order from top to bottom and from left to right.

Then the number of the two diagonal positions in all 4 4 sub-squares of the phalanx is paired with respect to the center of the square.

It is called the exchange, that is, a(i,j) and a(n-1-i,n-1-j) are exchanged, and the number in all other positions does not change.

Or the diagonal line is unchanged, and other positions can be symmetrically swapped).

n is an even number.

When n is an even number (i.e., 4n+2) that is not a multiple of 4: first, the large square matrix is decomposed into 4 odd (2m+1 order) sub-squares.

According to the above odd-order magic squares, the corresponding values of the four sub-squares of the decomposition are assigned.

The upper left sub-array is the smallest (i), the lower right sub-array is small (i+v), the lower left sub-array is the largest (i+3v), and the upper right sub-array is large (i+2v).

That is, the 4 sub-squares correspond to the element difference v, where v n*n 4

The four submatrices are arranged from smallest to largest

Then the corresponding elements are exchanged: a(i,j) and a(i+u,j) are exchanged in the same column (jn-t+2), a(t-1,0) and a(t+u-1,0); a(t-1,t-1) and a(t+u-1,t-1) are exchanged for two pairs of elements.

where u=n 2, t=(n+2) 4 The above exchange equalizes the sum of the elements in each row and column and two diagonals.

The law of horizontal and vertical oblique addition and equality is the famous nine-square grid problem, and the law to solve such a problem is generally to use the method of matching the size and the number of intermediate data. An array of natural numbers is called a magic square of order n, and the sum of the numbers in each row, column, and two diagonals is equal.

Here's how:

Put 1 (or the smallest number) in the middle of the first line. Arrange the remaining n2-1 numbers according to the following rules:

1. Each number is placed in the upper right box of the previous number.

2. If the grid to be placed in this number has exceeded the top row, then put it in the bottom row, and still put it in the right column.

3. If the number to be placed in the grid has exceeded the rightmost column, then put it in the leftmost column, still in the previous row.

4. If the grid to be placed in this number has exceeded the top row and beyond the rightmost column, then put it in the same column of the next row of the previous number.

Split Method:Finding some identical numbers in two rows of three squares of the three squares side by side, and then using the nine-squares to derive the position of the numbers in the other row, is very suitable for intermediate and advanced Sudoku.

Formula Method:The way to do this is to bring the mathematical formulas into it. With the middle number as the center, the diagonal is 'n-1,n,n,n=1', so that the sum of the number of each row is 3n.

Summary. Hello, I am glad to answer for you the sum of the sum of the three numbers of horizontal and vertical is to find the sum of the upper and lower numbers equal to the sum of the left and right numbers.

Hello, I am glad to answer for you the sum of the sum of the three numbers of horizontal and vertical is to find the sum of the upper and lower numbers equal to the sum of the left and right numbers.

The third-order magic square is a simple magic square, also called the nine-square grid. The basic form is to fill in the matrix of 3 3 with nine numbers of 1 9 jujube, so that the sum of the three numbers horizontally and vertically is the same. Preparedness.

How to do this.

Can you take it right?

Hello, excuse me.

Analysis: Since the average of these 5 numbers is between 22 and 23, and the addition of horizontal and vertical requires 69, which is 3 times that of 23, it is considered that the horizontal and vertical numbers are added with 3 numbers. Since a total of 6 numbers need to be added, it is considered that one of the numbers is used as the intersection of horizontal and vertical to participate in both horizontal and vertical calculations.

Solution: Horizontal addition: 21 + 24 + 24 = 69

Vertical addition: 22 + 23 + 24 = 69

As shown in the figure below:

<> analyze the numbers you provide, either , or stupid, with wild dust, if it is the first set of numbers, then 14 (maximum) has been filled in the upper right corner; If it is the second set of number ridge holes, then 7 (minimum) is filled in the upper left corner. Therefore, there is no solution to this problem.

Hello. This is called a third-order magic square.

The nine numbers must be a continuous series, and the middle number is in **, because the sum of the three numbers is 60, so the middle number must be one-third of its sum, which is 20

The sum of horizontal, vertical, and oblique numbers is 60

There are certain rules for playing magic squares.

Good luck and goodbye.

Summary. Kiss, send me the question.

Kiss, send me the question.

Both horizontally and vertically should be equal to 50

Kiss. In.

That's not the right question.

All of them must be equal to 50

The question is wrong, there is no original question, this is written by a child.

There's definitely something wrong with the title.

I know there's a problem with the problem and I can't solve it.

The question is set incorrectly.

Kiss. I read it wrong.

There are still problems, though.

There is still a problem with the topic.

The question should be equal to 50 horizontally and vertically

All questions should be equal to 50

There's definitely something wrong with the title.

The question is set incorrectly.

Kiss me wrong.

There are still problems, though.

There is still a problem with the topic.