# Why 24?

*This problem is in two parts. The first part provides some building blocks which will help you to solve the final challenge. These can be attempted in any order. Of course, you are welcome to go straight to the Final Challenge without looking at the building blocks!*

Click to reveal any of the questions below to get started.

**Question A**

What happens when you multiply the numbers either side of it?

For example, if you choose $7$, work out $6 \times 8$. Repeat several times.

Notice anything interesting? Convince yourself it always happens.

**Question B**

**Question C**

**coprime**(they have a highest common factor of $1$).

Multiply them together and record the result. Repeat several times.

Notice anything about your results?

Start with numbers other than $120$. Does the same thing always happen? Convince yourself.

**Question D**

Multiply them together and record the result. Repeat several times.

Notice anything interesting? Convince yourself it always happens.

**FINAL CHALLENGE**

Notice anything interesting? Convince yourself it always happens.

Take any prime number greater than 3, square it and subtract one. Repeat several times and record your results.

What is the highest common factor of the numbers you get? How is this related to the observations you made when working on the building blocks?

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.

### Why do this problem?

This problem involves a significant 'final challenge' which can either be tackled on its own or after working on a set of related 'building blocks' designed to lead students to helpful insights. It is well suited for students who are working on the difference of two squares.

Initially working on the building blocks then gives students the opportunity to work on harder mathematical challenges than they might otherwise attempt.

The problem is structured in a way that makes it ideal for students to work on in small groups.

### Possible approach

This task might ideally be completed in groups of three or four.

Each student, or pair of students, could be given their own building block to work on. After they have had an opportunity to make progress on their question, encourage them to share their findings with each other and work together on each other's tasks.

Alternatively, the whole group could work together on all the building blocks, ensuring that the group doesn't move on until everyone understands.

When everyone in the group is satisfied that they have explored in detail the challenges in the building blocks, hand out the final challenge.

The teacher's role is to challenge groups to explain and justify their mathematical thinking, so that all members of the group are in a position to contribute to the solution of the challenge.

It is important to set aside some time at the end for students to share and compare their findings and explanations, whether through discussion or by providing a written record of what they did.

### Key questions

What important mathematical insights does my building block give me?

### Possible support

Encourage groups not to move on until everyone in the group understands. The building blocks could be distributed within groups in a way that plays to the strengths of particular students.

### Possible extension

Students could be offered the Final Challenge without seeing any of the building blocks.