If you know that: \begin{eqnarray*} a^2 = a \\ a + a = b \\ y^2 =y \\ y + y = y \\ z^2 = bb \\ z + z = aa \end{eqnarray*} How can the decimal number $99$ be written in this number system?
** What methods do I need to do in order to find any result? Just give me a direction **
I tried to substitute everything but the result was very weird.
$a^2=a$ so $a=0$ or $a=1$. $a+a=b$ and $b \neq a $ so $a=1$ and $b=2$.
The third & fourth equations give $y=0$.
Now let $B$ be the base, so $2B+2=z^2$ and $2z=B+1$, giving $B=7$.
$99_{10}=\color{red}{201_7}$ which is $\color{blue}{bya}$ in their language.