### Instant Insanity

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

### Tree Graphs

A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree has exactly one more vertex than it has edges.

### Magic Caterpillars

Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.

# Magic W

##### Stage: 4 Challenge Level:

Bob has sent us one solution:

$6\quad\quad 4\quad\quad 9$
$\ 5\quad 7\quad 8\quad 3$
$\quad 2\quad \quad 1$
George used algebra to tackle the problem:

Call the numbers in the different dots $a, b, \ldots, i$, so the W looks like

$a\quad\quad e\quad\quad i$
$\ b\quad d\quad f\quad h$
$\quad c\quad \quad g$

Then $a+b+c=13$, $c+d+e$=13, $e+f+g=13$, and $g+h+i=13$. Adding these equations together, we get $a+b+2c+d+2e+f+2g+h+i=52$. But we know that the letters are just the numbers $1, 2, \ldots, 9$ in some order, so $a+b+c+d+e+f+g+h+i=45$, so $c+e+g=7$. Now $c$, $e$ and $g$ are all different, so the only possibility is that they are 1, 2 and 4 in some order. If $c$=1 and $e=2$, then we'd have $d=10$, and that's not possible. So of the three dots $c$, $e$ and $g$, the 1 and the 2 must be in ones that aren't next to each other. So we could have $c=1$, $e=4$ and $g=2$. Now we know that $d=8$ and $f=7$, and $a+b=12$ and $h+i=11$, and $a$, $b$, $h$ and $i$ are $3, 5, 6$ and $9$ in some order. So $a$ and $b$ are $3$ and $9$, and $h$ and $i$ are $5$ and $6$, but it could be either way round. So there are four different solutions, not counting reflections. One is shown below, and you get the others by switching the two numbers on one end or the other (or both!).

$3\quad\quad 4\quad\quad 5$
$\ 9\quad 8\quad 7\quad 6$
$\quad 1\quad \quad 2$

We can do something very similar if we want the sums to be $14$. This time, we get $c+e+g+45=4\times 14=56$, so $c+e+g=11$. There are several possibilities, which we'll consider separately.

Case $1$: ($1$,$2$,$8$). As above, we can't have $1$ and $2$ next to each other, so the only possibility (apart from reflection) is $c=1$, $e=8$, $g=2$. So $d=5$ and $f=4$, and $a+b=13$ and $h+i=12$, with $\{a,b,h,i\}=\{3,6,7,9\}$. So we get $\{a,b\}=\{6,7\}$ and $\{h,i\}=\{3,9\}$.

Case $2$: ($1$,$3$,$7$). This time, we can't have $1$ and $3$ next to each other either (as $14-1-3=10$), so the only possibility (again not counting reflection) is $c=1$, $d=6$, $e=7$, $f=4$, $g=3$, and we have $a+b=13$, $h+i=11$, with $\{a,b,h,i\}=\{2,5,8,9\}$. So $\{a,b\}=\{5,8\}$ and $\{h,i\}=\{2,9\}$, and there are four possibilities (by swapping ends), of which one is

$5\quad\quad 7\quad\quad 2$
$\ 8\quad 6\quad 4\quad 9$
$\quad 1\quad \quad 3$

Case $3$: ($1$,$4$,$6$). Now we can't have $4$ and $6$ next to each other, or we'd need another $4$ in the gap between them (to make $14$), so the only possibility is $c=4$, $d=9$, $e=1$, $f=7$, $g=6$, $a+b=10$, $h+i=8$, and so we must have $\{a,b\}=\{2,8\}$ and $\{h,i\}=\{3,5\}$.

Case $4$: ($2$,$3$,$6$). This time we can't have $2$ and $6$ next to each other, so the only possibilitiy is $c=2$, $d=9$, $e=3$, $f=5$, $g=6$, $a+b=12$, $h+i=8$, and so we get $\{a,b\}=\{4,8\}$ and $\{h,i\}=\{1,7\}$.

Case $5$: ($2$,$4$,$5$). Now we can't have $4$ and $5$ next to each other, so the only possibility is $c=5$, $d=7$, $e=2$, $f=8$, $g=4$, $a+b=9$, $h+i=10$, and we get the only solution as $\{a,b\}=\{3,6\}$, $\{h,i\}=\{1,9\}$.