### Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

### 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.

# Why 24?

##### Stage: 4 Challenge Level:

Weida, from Collaton St Mary School, noticed something about question A:
$5 \times 5 = 25, 25 - 1 = 24$
$7 \times 7 = 49, 49 - 1 = 48$
$11 \times 11 = 121, 121 - 1 = 120$
I noticed that the result for the second calculation in each example was equal to the product of the numbers on either side of the prime number that I started with.

Masood, Akeel and Dolapo from Wilson's School, and Rajeev from Fair Field Junior School used algebra to explain:

$(x - 1)(x + 1) = x^2 - x +x -1 = x^2-1$

Aidan, Elliott, Mark, Mark, Michael and Miriam, from Curtin Primary school in Canberra, sent in their explanations for each part:
A) If you multiply the numbers on either side of x by each other you get one less than the square of x. This can be represented as :$( x + 1 ) \times ( x - 1 ) = x^2 - 1$

B) Because there is only two numbers between two consecutive multiples of 3, every third number is a multiple of 3.

C) If you multiply coprime factors of x you get another factor of x or x itself.

D) They are all divisible by 4 because even numbers are every second number. Multiples of 4 are every fourth number therefore, every second even number will be a multiple of 4. That means the one of our numbers will be a multiple of 4.

Final Challenge They are all divisible by 24. Why? What if we put the numbers in the formula for A? Since we are squaring the prime $x$ and subtracting one we have $(x+1)\times(x-1)$

$x$ cannot be divisible by three so one of the numbers either side must be - $(x+1)$ or $(x-1)$
This means that $x^2-1$ is a multiple of 3.
Every second even number is a multiple of 4 (see D) therefore every other even number is a multiple of 2. Since $x$ cannot be divisible by 2, then $x+1$ is divisible by 2 & $x-1$ is divisible by 4 (it could be the other way around).

We are multiplying a multiple of 2 by a multiple of 4 so $x^2-1$ is a multiple of 8. That means that $x^2-1$ is a multiple of 8 & 3. We can't multiply normal factors but 3 & 8 are coprime and multiplying coprime factors will result in another factor so 3 x 8 is 24. $x^2-1$, the final number is divisible by 24.
Isabella and Ian, from Cashmere High School in Christchurch, and Richard from Comberton Village College, sent us similar algebraic arguments.

Hannah, from Munich International School, Preveina, from Crest Girls' Academy tried squaring some prime numbers and subtracting 1, and verified that they were all divisible by 24.

Finally, well done to Patrick from Woodbridge School, and Andrew who didn't give his school name, who both used modular arithmetic. Here is Andrew's explanation:

The challenge asked to pick a prime number greater than 3, so let p be such a prime.
Squaring p then subtracting one is the difference of two squares: $p^2 - 1 =(p-1)(p+1)$
So we can look at the properties of $(p-1)$ and $(p+1)$ rather than $p^2 -1$.

Since p is a prime, p is not divisible by 3. Hence it either has the remainder 1 or the remainder -1 when divided by 3.
Case 1: $p = 1$ (mod 3) $\Rightarrow p - 1= 0$ (mod 3)
Therefore $p - 1$ is divisible by 3.

Case 2: $p = -1$ (mod 3) $\Rightarrow p + 1 = 0$ (mod 3)
Therefore $p + 1$ is divisible by 3.
We have concluded regardless of the remainder when p is divided by 3, either $p-1$ or $p+1$ is divisible by 3.

Now consider p modulo 4. p cannot equal 0 (mod 4) otherwise it would not be prime. p also cannot equal 2 (mod 4) because it would then be divisible by 2, and p is a prime greater than 2. Therefore $p = 1$ or $p = 3 = -1$ (mod 4).

Case 1: $p = 1$ (mod 4) $\Rightarrow p - 1 = 1 - 1 = 0$ (mod 4) and $p + 1 = 1 + 1 = 2$ (mod 4)
Therefore $p-1$ is divisible by 4 and $p+1$ is divisible by 2.

Case 2: $p = -1$ (mod 4) $\Rightarrow p - 1 = -1 - 1= -2 = 2$ (mod 4) and $p+1= -1 + 1 = 0$ (mod 4)
Therefore $p-1$ is divisible by 2 and $p+1$ is divisible by 4.
Therefore since either $p-1$ or $p+1$ is divisible by 4 and the other 2, their product is divisible by 8. But it was also established previously that $(p-1)(p+1)$ is divisible by 3. Therefore $(p-1)(p+1)$ is divisible by 8 and 3, and therefore 24.

This means for any prime $p > 3$, $p^2 - 1 = (p-1)(p+1)$ is divisible by 24.