### Nine Colours

You have 27 small cubes, 3 each of nine colours. Use the small cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of every colour.

### Muggles Magic

You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.

### Cogs

A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the cogwheel A as the wheels rotate.

# Conway's Chequerboard Army

##### Stage: 3 Challenge Level:

Duncan from Warden Park School, Zach from Rosemellin School and Stephen all got the minimal method for achieving both targets $C$. Stephen explains his method for $C$:

Position five counters in the row under $A$ (row $1$), with the middle counter directly under $A$. Then place three counters in the row below that (row $2$), with the left most directly under $A$.
Two under $A$ jumps one under $A$, so there is a counter on $A$. Then the counter two to the right of $A$ jumps to be directly under $A$ in row $1$. Then this counter jumps the one on $A$ so it sites on $B$.
The counter in row $1$ two to the left of $A$ jumps to directly under $A$, and the counter in row $2$ two the the right of $A$ jumps to be directly under $A$ in row $2$. Then this counter jumps the one above to land on $A$ then jumps the next to land on $C$.

Although all of you also found a method for finding $d$, you did not find the minimal method, so using the least possible counters and moves. Andy from the Garden International School explains the minimal way of achieving the target $D$.

$6c -> 4c$
$5e -> 5c -> 3c$
$5a -> 5c$
$6a -> 6c -> 4c -> 2c$
$6e -> 6c$
$7c -> 5c$
$7a -> 7c$
$8c -> 6c -> 4c$
$7e -> 7c$
$8e -> 8c -> 6c$
$8a -> 8c$
$9c -> 7c -> 5c -> 3c -> 1c$

Andy goes on to explain why you cannot achieve further than $D$ on the grid provided.

With the counters above needed to reach $D$, then we are left with:

Which you cannot reach beyond $D$ as you would need to get counters on $D$, $B$, and below $A$. Though there is a counter on $D$ which isn't in my picture. This cannot be made.

Congratulations to Andy, can you think of another way of reaching $D$?