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Olivia from Roedean in the UK tried different values of x to solve the equations 3x \equiv 1 \text{ mod } 7 and 3x \equiv b \text{ mod } 7:
Mahdi from Mahatma Gandhi International School in India used the rules proved earlier in the problem:
For the equation 6x \equiv b \text{ mod } 7, Olivia wrote:
Mahdi wrote:
Since 7 is prime, for every value of a and b, there is a unique solution for x in the equation ax\equiv b \text{ mod 7}
For the equations 4x \equiv b \text{ mod } 10 and ax \equiv b \text{ mod } 10, Olivia wrote:
To find numbers that solve 4x \equiv b \text{ mod } 10, we make a table again
Mahdi wrote:
The problem that Mahdi mentions is Stars. Olivia also used the interactivity to explain how to choose possible values of a:
Considering the proof that, when A \equiv a \text{ mod }n and B \equiv b \text{ mod }n it is always true that AB \equiv ab \text{ mod }n, Amy asked whether this proof could be adapted to find which values of a and b give solutions to the equation ax \equiv b \text{ mod }n.
Since the proof shows that the first rule is true for all a and b, but later we want to find the specific values of a and b which give solutions, one cannot be adapted easily to give the other. However, a similar argument can be used to demonstrate which values of a and b will work for each n.
This is linked to Mahdi's first method. Consider applying Mahdi's first method to 4x\equiv b \text{ mod 10}:
4x \equiv 2\text{ mod 10}
\Rightarrow 4x \equiv 12\text{ mod 10}
\Rightarrow x \equiv 3\text{ mod 10}
However
4x \equiv 1\text{ mod 10}
\Rightarrow 4x \equiv 11\text{ mod 10}\equiv 21,31, 41, 51, ...\text{ mod 10}
And none of those numbers are divisible by 4.
Can you generalise this?