Now that the new school and university years are well under way, many may feel a bit out of practise with their mental maths after several months off. Thanks to number theory, however, there are a few cool tricks that can make life a lot easier! Here are some interesting properties of the natural numbers and the explanation behind them:

1. Divisibility by 3

A commonly known fact: if the sum of the digits of any number is divisible by 3, then so is the original number.

For example 5631. 5+6+3+1 = 15, 15/3=5 and so using the rule we know that 5631 is also divisible by 5631. 5631/3 = 1877.

To understand the maths behind it we simply break down 5631:

5000 + 600 + 30 + 1 = 5 x 1000 + 6 x 100 + 3 X 10 + 1 = 5 x (999 + 1) + 6 x (99 + 1) + 3 x (9 + 1) +1

=( 5 x 999 + 6 x 99 + 3 x 9) + (5 + 6 + 3 + 1) = 9 x (5 x 111 + 6 x 11 + 3) + (5 + 6 + 3 + 1).

As the first part of the equation is multiplied by 9, then divisibility by 3 is dependent on the second, (5+6+3+1), the sum of the digits. This shows the same rule also stands for divisibility by 9. This breakdown could apply for any number and so the rule is true for all. To prove it you need to work through the same method but define your original number algebraically. Why not give it a go!

2. Divisibility by 11

Take some natural number n. If the sum of the even number digits subtracted from the sum of the odd digits is divisible by 11, then n is also divisible by 11.

Slightly more complex to visualise, take for example 94523. Then as (9 + 5 + 3) – (4+2) = 11. 11/11 = 1 and so 94532/11 must also be a natural number.

Using the mod function in number theory can provide a generalised proof but it’s possible to create an understanding of why this rule holds without it.

For each odd digit, for example 90000, we can instead think of this as

9 x 10000 = 9 x (9999 + 1) = 9 x 101 x 11 + 9 x 1. = 11(9 x 101) + 9.

For each even digit, for example 4000 can be written in a similar form to

4000 = 4 x 1000 = 4 x (1001 -1) = 4 x 11 x 91 – 4 x 1 = 11 x (4 x 91) – 4.

Expressing the number as a whole in this form, it can be seen that a factor of 11 can be taken out from all parts except for (9 – 4 + 5 – 3 + 2). And so, as the rule describes, if this value is divisible by 11 then the original integer must be as well!

3. Remembering Pi!

Not quite as algebraic as the previous two theorems, but here’s a handy way to remember the first few digits of Pi!

I wish I could remember Pi. "Eureka!" cried the great inventor. "Christmas pudding, Christmas pie is the problem's very centre."