### Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

### Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

# Weekly Problem 5 - 2011

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

Suppose it is possible to make a list of all ten numbers.

The number $7$ must be at one end and must be next to $1$ since $7$ has no other factors or multiples under $10$. Without loss of generality we can assume $7$ is the first number, followed by $1$.

The number $5$ only has two possible adjacent numbers, $1$ and $10$. The same is true for $9$ which can only be next to $1$ or $3$. Hence either we must start with $7$, $1$, $5$, $10$ and end with $9$; or we start with $7$, $1$, $9$, $3$ and end in $5$. Either way this means that $1$ cannot be next to any other numbers.

The diagram below shows the only possible connections that can be used.

It is clearly impossible to link all ten numbers together without using $2$ twice. To see this imagine the sequence starts $7$, $1$, $5$, $10$, then the only possibility after $10$ is $2$ but the only possibility before $6$ is $2$ which means $2$ has to appear twice. Similarly if the sequence start $7$, $1$, $9$, $3$ we must also use $2$ twice.

However, the diagram suggests a possible list of nine numbers: $6$, $3$, $9$, $1$, $5$, $10$, $2$, $4$, $8$.

There are many other sequences of nine numbers that follow the rule. Can you find them all?

This problem is taken from the UKMT Mathematical Challenges.

View the previous week's solution
View the current weekly problem