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breannabryan1017
01.07.2019 •
Mathematics
Atrue mathematician would know this problem. so if you know the magic square(you can look it up)i need each of a 3 by 3 square to have one of these each number in it: 5,6,7,8,9,10,11,12, and 13. you can only use one number once.but each row,column and from diagonal must all equal up to 27.
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Ответ:
Attached is one version.
_____
The key is that 9 must go in the middle.
Ответ:
If you know how to work a 3 by 3 magic square using the digits 1 to 9, you can make any 3 by 3 magic square starting with any number.
Starting Square
8 1 6
3 5 7
4 9 2
This magic square adds up to 15. That is 3 * the center number.
That is not the magic square you want, but it can be made into yours.
Just add 4 to every number in the base magic square.
12 5 10
7 9 11
8 13 5
And the sum is 3 * 9 = 27 in all the rows columns and diagonals, just what you requested. What happens when you have 0 in the center?
That means you take away 5 from every number in the base square.
3 - 4 1
-2 0 2
-1 4 -3
It still works. Every row column and diagonal add to 0.
Ответ:
2) 109/12 cups of sugar
note: i am unsure that i did this right so rework the problem if you think i did it wrong.