### Doodles

A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not 'triple points'. Number the vertex points in any order. Starting at any point on the doodle, trace it until you get back to where you started. Write down the numbers of the vertices as you pass through them. So you have a [not necessarily unique] list of numbers for each doodle. Prove that 1)each vertex number in a list occurs twice. [easy!] 2)between each pair of vertex numbers in a list there are an even number of other numbers [hard!]

### Russian Cubes

How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted in order 'blue then red then blue then red then blue then red'. Having finished one cube, they begin to paint the next one. Prove that the girl can choose the faces she paints so as to make the second cube the same as the first.

### N000ughty Thoughts

Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!

# Weekly Problem 27 - 2008

##### Stage: 3 and 4 Challenge Level:

Firstly, there are $12$ unit squares which contain an even number. Every $2\times2$ square in the diagram has entries which consist of two odd numbers and two even numbers and hence have an even total. There are $16$ of these. Each $3\times3$ square in the diagram, however, has entries which consist of five odd numbers and four even numbers (giving an odd total), or four odd numbers and five even numbers (giving an even total). There are $4$ of the latter: those with $8$, $12$, $14$ or $18$ in the centre. Every $4\times4$ square in the diagram has entries which consist of eight odd numbers and eight even numbers and hence have an even total. There are $4$ of these. Finally, the full $5\times5$ square contains $13$ odd numbers and $12$ even numbers, giving an odd total. So the required number is $12+6+4+4$, that is $36$.

This problem is taken from the UKMT Mathematical Challenges.

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