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# Plum Tree

This is a long and detailed hint which provides ideas, not just for this particular problem, but also for other magic graph problems where similar reasoning can be used.

Vertex Magic Investigation

There are 8 vertices and 7 edges. Let's suppose the magic constant is $h$ and this is the same at each vertex. The total of all the numbers $1 + 2 + ... + 15 = 120$ but the numbers on the edges, say $a,b,c,d,e,f,g$ are each counted twice so

$$(a+b+c+d+e+f+g)+120=8h.$$

So $a+b+c+d+e+f+g$ is a multiple of 8 and it is at least $1+2+...+7=28$ so it must be $32$ or more. Hence $8h\geq 152$ and $h\geq 19$.

It is easy to check, in a similar way, for the largest value that $h$ can take. Now it is a matter of systematically checking the possibilities for magic labellings with $h=19$ etc.

Edge Magic Investigation

How do we discover what edge magic graphs there are? Here the three order-3 vertices $a, b,c$ are counted three times and the sum of all the numbers is again $120$.

$$2(a+b+c)+120=7k.$$

Hence $k$ is even and since $a+b+c\geq 6$ we have $7k\geq 120+12=132$ and so $k\geq 20$.

If $k=20$ then $a+b+c=10$ so this triple is $1+2+7$ or $1+3+6$ or $1+4+5$ or $2+3+5$.

The possible edge sums for $k=20$ are:

$1+4+15$, $1+5+14$, $1+6+13$, $1+7+12$, $1+8+11$, $1+9+10$, $2+3+15$, $2+4+14$, $2+5+13$, $2+6+12$, $2+7+11$, $2+8+10$, $3+4+13$, $3+5+12$, $3+6+11$, $3+7+10$, $3+8+9$, $4+5+11$, $4+6+10$, $4+7+9$, $5+6+9$, $5+7+8$.

It is easy to find the edge magic labelling for $k=20$.

Moreover, substituting $16-P$ for each of the P-labels giving a magic total of $k$ will give the same numbers $1$ to $15$ and a magic total of $48-k$. So for all the magic graphs you find with a magic total of $20$ there are corresponding graphs with a magic total of $28$.

It is left for you to find the edge magic labels for $k=20$ and for $k=22$ (and correspondingly $k=26$) andfor $k=24$. Remember $k$ must be even and at least $20$ so these are the only possibilities.

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This is a long and detailed hint which provides ideas, not just for this particular problem, but also for other magic graph problems where similar reasoning can be used.

Vertex Magic Investigation

There are 8 vertices and 7 edges. Let's suppose the magic constant is $h$ and this is the same at each vertex. The total of all the numbers $1 + 2 + ... + 15 = 120$ but the numbers on the edges, say $a,b,c,d,e,f,g$ are each counted twice so

$$(a+b+c+d+e+f+g)+120=8h.$$

So $a+b+c+d+e+f+g$ is a multiple of 8 and it is at least $1+2+...+7=28$ so it must be $32$ or more. Hence $8h\geq 152$ and $h\geq 19$.

It is easy to check, in a similar way, for the largest value that $h$ can take. Now it is a matter of systematically checking the possibilities for magic labellings with $h=19$ etc.

The techniques for finding the labellings are the same as used
for the
Caterpillar problem.

Edge Magic Investigation

How do we discover what edge magic graphs there are? Here the three order-3 vertices $a, b,c$ are counted three times and the sum of all the numbers is again $120$.

$$2(a+b+c)+120=7k.$$

Hence $k$ is even and since $a+b+c\geq 6$ we have $7k\geq 120+12=132$ and so $k\geq 20$.

If $k=20$ then $a+b+c=10$ so this triple is $1+2+7$ or $1+3+6$ or $1+4+5$ or $2+3+5$.

The possible edge sums for $k=20$ are:

$1+4+15$, $1+5+14$, $1+6+13$, $1+7+12$, $1+8+11$, $1+9+10$, $2+3+15$, $2+4+14$, $2+5+13$, $2+6+12$, $2+7+11$, $2+8+10$, $3+4+13$, $3+5+12$, $3+6+11$, $3+7+10$, $3+8+9$, $4+5+11$, $4+6+10$, $4+7+9$, $5+6+9$, $5+7+8$.

It is easy to find the edge magic labelling for $k=20$.

Moreover, substituting $16-P$ for each of the P-labels giving a magic total of $k$ will give the same numbers $1$ to $15$ and a magic total of $48-k$. So for all the magic graphs you find with a magic total of $20$ there are corresponding graphs with a magic total of $28$.

It is left for you to find the edge magic labels for $k=20$ and for $k=22$ (and correspondingly $k=26$) andfor $k=24$. Remember $k$ must be even and at least $20$ so these are the only possibilities.