Making a 3x3 magic square that adds up to 1 I got this problem from my sister in 6th grade. I spent 3 hours on this and couldn't figure it out for the life of me.
You're given an empty 3x3 square. You have to fill it with numbers from -9 to 9. Columns and rows and diagonals add up to 1. You cannot repeat any number.
I can't seem to find any combination that would give sums of 1 for all of them. Is there some algorithm for this?
The teacher claimed that none of her students had been able to solve this problem the past 14 years she's given it to them.
 A: Suppose the sum of
each row, column, and
diagonal is s.
Then the sum of
all the digits is 3s.
If z is the center value,
then,
adding all the lines
through the center,
since the center is
counted 4 times
and all the other cells once,
4s-3z = 3s,
or
s = 3z.
Therefore
the sum of all the digits
is 3s or 9z.
Therefore
the sum of all the digits
is divisible by 9.
But if s = 1,
then the sum of all the digits
is 3,
which is not divisible by 9.
This is a contradiction,
so no such square can exist.
A: Take a standard magic square and divide all of the numbers inside by $15$:
$$\begin{matrix}
  8/15 & 1/15 & 6/15 \\
  3/15 & 5/15 & 7/15 \\
  4/15 & 9/15 & 2/15
 \end{matrix}$$
Each entry is a (rational!) number between $-9$ and $9$.
A: First row: $3,-4,1$
Second row: $-2,0,2$
Third row: $-1,4,-3$
The sum of each row/column/diagonal adds up to $0$. In order to add up to one, the center number ought to be $1/3$ So with integers, I do not think that it can be done. No wonder nobody found it in 14 years...
A: I don't think it can be done. I made sort of a program to see all the possible answers and the closest you can get is rows and columns to add 1, but only one of the diagonals. The other is going to be a multiple of 3, ranging from -27 to 27.
