Math, asked by saraswatidandin2, 8 months ago

construct 3 magic square by using number from 1to9
please help me​

Answers

Answered by soujanyamazumder
2

Answer:

Step: 1

Write nine consecutive numbers  in three rows and columns as shown along side.nine consecutive numbers  in three rows and columns as shown along side. Say for example, we are using natural numbers from 1 to 9.(Normally we use natural numbers only) Here numbers from 1 to 3 is written in the first row followed by 4 to 6 in the second  and 7 to 9  in the third rows..

Step: 2

Now draw lines diagonally under the numbers from both sides as shown in the figure. Once it's done you shall get a  3 x 3 square grid as high lighted by the red line in the figure. But here 4 cells are empty and 4 numbers are lying outside the  square grid. All you will  have to do is to use these numbers to fill the empty cells in the opposite side.

Final answer

There after draw a neat  3 x 3 grid and rewrite the numbers as completed in STEP 2. 

Now see the magic. the sum across the rows, the columns and diagonals are 15.

  Rows                   Columns              Diagonals

 4 + 9 + 2 = 15       4 + 3 + 8 = 15      4 + 5 + 6 = 15

 3 + 5 + 7 = 15       9 + 5 + 1 = 15      2 + 5 + 8 = 15

 8 + 1 + 6 = 15       2 + 7 + 6 = 15

Formula 

Of course we have formula for finding the numbers (Arithmetic Progression)  used for filling the Magic Square for a given sum.

For any Magic Square of the order 3 x 3; the first term of the progression will be

      F = S/ 3 - 4D

Here S denotes  the Magic Sum, F the first number of the sequence used for filling and D the common difference between the numbers  in the sequence.

To understand the concept follow the second example. See how to form a MS 3 X 3 with a magic sum of 132 and a sequence of common difference 5.  Divide 132 with 3,we get 44. 44 minus four times 5- (44 - 20) we get 24. There fore the sequence is 24, 29, 34, 39, 44, 49, 54, 59, 64.

Answered by gumapathi9865
3

Answer:

sorry I can't able to answer your recent question...

So Iam answering here.

Answer is in the above attachment.....

Hope this will help you.

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