Judy sent us the following solution:

Suppose that we have a labelling with magic total

T

that looks like this:

A E I

B D F H

C G

Then we can replace each entry

x

10 - x

and get another labelling (because if

x

is one of 1, 2, ..., 9, then

10 - x

will also be). Also, this label will have magic total

30 - T

, because instead of adding up

A + B + C

, for example, we'll add up

10 - A + 10 - B + 10 - C = 30 - T

, and all of the sums will be the same. So if we have a labelling with magic total

T

, then we certainly have one with magic total

30 - T

Using the solution to the Magic W problem, we know that there is 1 magic labelling for $T = 13$ , and there are 5 for $T = 14$ . So (using what we worked out above) there's also one for $T = 17$ and 5 for $T = 16$ . (There can't be any more for $T = 17$ , for example, because any labelling of $T = 17$ is also one of $T = 13$ .)

When do magic labellings exist? Well, again using the ideas from the solution to the Magic W problem, we must have $C + E + G + 45 = 4 T$ . But $C + E + G \geq 6$ , so $4 T \geq 45 + 6 = 51$ , so $T \geq 13$ . Also, $C + E + G \leq 24$ , so $4 T \leq 45 + 24 = 69$ so $T \leq 17$ . So we only need to check whether there are any magic labellings for $T = 15$ . Suppose that there is one. Then we have $C + D + E = 15$ , and also $C + E + G = 4 \times 15 - 45 = 15$ , so $D = G$ . But that's not allowed, so there are no labellings of $T = 15$ .

To summarise, there are magic labellings for $T = 13$ , 14, 16 and 17 (and no others), and there are $1 + 5 + 5 + 1 = 12$ magic labellings.