More Adventures with Modular Arithmetic

Why do this problem?

This problem follows on from Clock Arithmetic. It gives students the opportunity to consider the properties of numbers when we add or multiply numbers using modular arithmetic, and to then move on to equations with modulo arithmetic.

At the end students are challenged to discover that in general $ax \equiv b \text{ mod }n$ has solutions if, and only if, the highest common factor of $a$ and $n$ divides $b$.  This means that if $a$ and $n$ have no common factors greater than 1 then $ax \equiv b \text{ mod }n$ has a solution for all values of $b$ (where $0 \le b \le n-1$).

Possible approach

This problem featured in an NRICH Secondary webinar in April 2021.

Start by asking students to think of two numbers, and find what they are modulo $7$.  Ask them to add their original numbers and find out what this sum is modulo $7$.  What do they notice?

It might be helpful to work through an example:

Then give the students time to try a few examples of their own and write down what they notice. Encourage them to try some modulos other than $7$.

Bring the students together and display the proof sorter to work on as a class.  

Work on multiplication in the same way as addition, but this time students can have a go at writing their own proof before comparing it to the proof on the problem page.

For the final part of this problem, students could start by playing with the Stars interactivity and seeing which combinations of number of dots and step size ensures that every dot is visited.  This can be linked to the equation $ax \equiv b \text{ mod }n$ by taking number of dots $=n$ and step size $=a$.

Key questions

Possible support

This problem follow on from Clock Arithmetic.  Students might like to work on Stars before trying to link this to   modular equations.

Possible extension