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I have to make a 3x3 magic square using the numbers 0-10 without 1 and 9. I have tried various things but am not good at this. The sums of each row, column, and diagonal have to be equal; I added all the numbers up and divided by three so I believe each row should add up to 15?

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    $\begingroup$ Hi, Tyler, if f"'s helped you, make sure to consider checking that great big green checkmark next to his answer! As an added bonus it will give you a (albeit small) bit of rep! $\endgroup$ – warspyking Aug 29 '15 at 13:29
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 2 10  3
 6  5  4
 7  0  8

This is the only one, not counting reflections and rotations.

Proof: As you found, the sum of all the numbers is three times the sum of each row (or column), so each row must add up to 15. There are only two ways to make 15 with 0 and two numbers: 0+7+8 and 0+5+10. Similarly, 10 can only be used in 2+3+10 and 0+5+10.

Each corner is in one row, one column, and one diagonal, so it must have three ways to make 15. The center must have four ways to make 15. Therefore 0 and 10 must be in the middle of the sides, opposite each other, with 5 in the center. Then, if we fill in 2 and 3 next to the 10, all the rest of the numbers are forced.

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  • $\begingroup$ Damn... I was just about to post my answer :D $\endgroup$ – user14478 Aug 25 '15 at 22:53
  • $\begingroup$ Nice explanation - a lot more intuitive for me than the rigorous formulas from Joe Z. answer. $\endgroup$ – Falco Aug 26 '15 at 13:56
  • $\begingroup$ How would you be able to determine this if 1 and 9 weren't specifically excluded? $\endgroup$ – SirParselot Sep 16 '15 at 12:07
  • $\begingroup$ @SirParselot Can you clarify your question? Do you mean "Make a magic square using the numbers 0 to 10"? $\endgroup$ – LeppyR64 Sep 16 '15 at 14:08
  • $\begingroup$ @LeppyR64 Yes, make a magic square using the numbers 0 to 10 $\endgroup$ – SirParselot Sep 16 '15 at 14:15
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In general, to know whether this sort of problem is solvable (given a list of 9 numbers, form a magic square out of them), you need to look for three arithmetic sequences that are themselves arithmetically sequenced — namely, three sequences of the form:

\begin{matrix} a & a+c & a+2c \\ a+b & a+b+c & a+b+2c \\ a+2b & a+2b+c & a+2b+2c \end{matrix}

In this case, $a = 0$, $b = 2$, and $c = 3$, because the sequences are $(0, 2, 4)$, $(3, 5, 7)$, and $(6, 8, 10)$.

Then, you arrange them like this:

\begin{matrix} a+b & a+2b+2c & a+c \\ a+2c & a+b+c & a+2b \\ a+2b+c & a & a+b+2c \end{matrix}

to get your magic square. Note that every row, column, and diagonal sums up to $3a + 3b + 3c$. In your problem, this becomes:

\begin{matrix} 2 & 10 & 3 \\ 6 & 5 & 4 \\ 7 & 0 & 8 \\ \end{matrix}

We can prove that this solution is unique. There are exactly eight combinations of the nine terms above that add up to $3a+3b+3c$:

\begin{matrix} a & (a+b+c) & (a+2b+2c) & (1) \\ a & (a+b+2c) & (a+2b+c) & (2) \\ (a+b) & (a+c) & (a+2b+2c) & (3) \\ (a+b) & (a+2c) & (a+2b+c) & (4) \\ (a+b) & (a+b+c) & (a+b+2c) & (5) \\ (a+2b) & (a+c) & (a+b+2c) & (6) \\ (a+2b) & (a+2c) & (a+b+c) & (7) \\ (a+c) & (a+b+c) & (a+b+2c) & (8) \\ \end{matrix}

Out of these, the four terms that appear in the corner of the square appear exactly three times, the four terms that appear in the edges of the square appear twice, and the term that appears in the center $(a+b+c)$ appears four times. Therefore, if those terms are to appear in a magic square, they cannot appear anywhere but those places.

So, suppose you place $a+b+c$ in the center, the only place it can go. Then you choose an edge square to put $a$ in. Whichever square you choose to put it in will rotate the entire magic square, but let's choose the bottom one for now:

\begin{matrix} ? & ? & ? \\ ? & a+b+c & ? \\ ? & a & ? \end{matrix}

Then the entry in the top square must be $a + 2b + 2c$.

\begin{matrix} ? & a+2b+2c & ? \\ ? & a+b+c & ? \\ ? & a & ? \end{matrix}

Now, in order for the top row to sum up to $3a+3b+3c$, you still require one $b$ and one $c$. Therefore, only $a+b$ and $a+c$ can go in those squares, because you've already used $a$ and $a+b+c$, and the rest will make one of the coefficients go past 3. Whichever order you choose $a+b$ and $a+c$ to go in will reflect the magic square. After that, the rest of the magic square can be solved by summing.

Therefore, we've proven that only rotations and reflections of the $3 \times 3$ magic square are also magic squares.

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