# Finding remainders from an encoding function

Let's consider the encoding function: $$y=ax \bmod 5$$ where $$x$$ is the number that corresponds to a certain letter in the alphabet. I'm trying to understand why for every number $$n$$, with $$n>4$$, I can find a number a such as $$ax \equiv nx \bmod 5$$.

First of all, I'm trying to understand why, given a certain $$x$$, if a ranges between $$[0,\ldots,4]$$ I cover all possible remainders (that is the remainders from $$0$$ to $$4$$). For example, if I take $$x=3$$, then:

$$0=0*3 \rightarrow$$ remainder is $$0$$

$$3=1*3 \rightarrow$$ remainder is $$3$$

$$6=2*3 \rightarrow$$ remainder is $$1$$

and so on...

But i don't know how to prove it... Can you help me?

As stated, your assertion is fals: if $$x=5$$ (or any multiple of $$5$$), then $$ax\equiv 0\pmod 5$$ regardless of what $$a$$ is.
But given $$x$$ not a multiple of $$5$$: if you want $$ax\equiv r\pmod 5$$ for some particular $$r\in\{0,1,2,3,4\}$$, then you can do so by choosing $$a\equiv rx^{-1}\pmod 5$$, where $$x^{-1}$$ is the modular inverse of $$x$$.
For a general modulus $$q$$ in place of $$5$$, this will be possiblie whenever $$\gcd(r,q)=1$$.
• Why do we know at prior that, in this case, $x^{-1} \in \left \{ 0,1,2,3,4 \right \}$ ? May 4, 2019 at 16:08