Using these numbers, 2 4/9, 2 5/9, 2 6/9, 2 8/9,
3, 3 1/9, 3 3/9, 3 4/9 & 3 5/9. Build a square where the sum of every horizontal, vertical, and diagonal row is 9.
The square has 3 rows across & 3 rows down for a total of 9 small blocks.
The standard 3by3 magic square is
8 1 6
3 5 7
4 9 2 --> notice ever row, column and diagonal adds up to 15 for a total of 45.
Any 9 numbers in an arithmetic sequence can be arranged in the above order and you would have a magic square with the properties you want.
change all your mixed fractions to a usable form
22/9 , 23/9 ... , 32/9
to see that they are "consecutive" with 1/9 as the common difference, with the exception from 24/9 to 26/9 and again from 28/9 to 30/9
This really messes it up. If it is as typed, I don't think it is possible to do it
1234567890
To build a square where the sum of every horizontal, vertical, and diagonal row is 9, we need to arrange the given numbers in a 3x3 grid.
Let's start by looking for a pattern in the given numbers. We have four numbers with a whole part of 2 and five numbers with a whole part of 3. The fractional parts for each whole part also follow a pattern: 4/9, 5/9, 6/9, and 8/9 for the numbers with a whole part of 2, and 1/9, 3/9, 4/9, and 5/9 for the numbers with a whole part of 3.
We can use this pattern to place the numbers in the grid systematically.
First, let's place the numbers with a whole part of 2 in the corners of the square:
2 | |
---------
| |
---------
| | 2
Next, let's place the numbers with a whole part of 3 in the center of each side of the square:
2 | 3 |
---------
| |
---------
3 | 2 | 2
Now, let's fill in the remaining numbers. We start by placing 4/9 in the top right corner. We know that the sum of each row has to be 9, so we can calculate the missing numbers:
2 | 3 | 4
---------
| |
---------
3 | 2 | 2
Next, we place 5/9 in the center bottom position:
2 | 3 | 4
---------
| 5 |
---------
3 | 2 | 2
Finally, we can fill in the remaining numbers based on the pattern we observed earlier:
2 4/9 | 3 | 4
---------------
1/9 | 5 |
---------------
3 | 2 | 2
Now, let's check if the sum of every horizontal, vertical, and diagonal row is 9:
Horizontal rows:
2 4/9 + 3 + 4 = 9
1/9 + 5 + 3 = 9
3 + 2 + 2 = 7
Vertical rows:
2 4/9 + 1/9 + 3 = 9
3 + 5 + 2 = 10
4 + 3 + 2 = 9
Diagonal rows:
2 4/9 + 5 + 2 = 9
4 + 5 + 4 = 13
Based on our calculations, the sum of every horizontal row and vertical row is 9, but the diagonal rows do not have a sum of 9. This means that the given numbers cannot be arranged to form a square where the sum of every horizontal, vertical, and diagonal row is 9.