My Favourite Problem No.3 Solution
The maximum number that can be reached is 23.
To prove this you need to show that 23 is possible but 24 isn’t.
Showing 23 is possible
To do this you need to draw out a solution for 23. One solution is given below.
Showing 24 is not possible
24 would need to connect to 12, 8, 6, 4, 3 and 2. This is not possible without crossing over one of
the existing lines.
One method of showing 24 is not possible is based on the solution to a problem you might be
familiar with. The problem is to join 3 houses to each of a gas, water and electricity supply without
any of the lines crossing. This is not possible on a sheet of paper (in 2D) as the attempt below
suggests. The final connection between E and H1 must cross over another connection.
These diagrams are called graphs, the points are called nodes or vertices and the connections
are called arcs or edges. If a graph can be drawn such that no arc crosses over another arc then it
is said to be a planar graph.
For more detail about this see: http://en.wikipedia.org/wiki/Planar_graph
We can rephrase the solution to the problem involving the gas, electricity and water supplies
mathematically as the complete bipartite graph on two sets of three vertices is not planar.
We can apply this to the factor problem for 24. By just considering the two set of numbers 2, 3, 6
and 12, 18, 24 you would need a complete bipartite graph, i.e. a diagram where each number in
one set is joined to one from the other set.
Therefore 24 is not possible.
Another method of showing 24 is not possible is based on the fact that it is not possible to draw 5
points and join each one to the other 4 without the lines crossing.
Or to express this mathematically: the complete graph with 5 vertices is not planar.
Each vertex is connected to the other 4, but the connection between C and D must cross over
another arc.
By just considering 24 and 4 of its factors, the numbers 2, 3, 6, and 12 it is possible to join up the
direct factors:
There is no connection needed between 2 and 3. However, there are indirect paths joining 2 to 3.
2 – 14 – 7 – 21 – 3 and 2 – 10 – 5 – 15 – 3. It is not possible to add either path to the diagram
without it crossing one of the other lines.
Further Investigation
You might like to investigate this topic further by visiting:
http://www.cimt.plymouth.ac.uk/projects/mepres/alevel/discrete_ch6.pdf
or play a game involving planar graphs http://planarity.net/#