Let $A$ be the set of all complex numbers $z$ such that $z^{24}=1$ and let $B$ be the set of all complex numbers $w$ with $w^{54}=1.$ That is \begin{align*} A&=\{z\;|\;z^{24}=1\}\\ B&=\{z\;|\;z^{54}=1\}\\ \end{align*} Finally let $C$ be the set of all numbers that can be formed by multiplying an element of $A$ by an element of $B:$
$C=\{zw\;|\;z\in A,w\in B\}.$
How many distinct elements are there in $C?$
Hi guys, I have been having some trouble with this problem. I have tried using the 24th roots of unity, but has gotten me nowhere so far. Can anyone help me?
The $24$th roots of unity are $e^{2\pi i x/24}$ and the $54$th roots are $e^{2\pi iy/54}$, where $x,y$ are integers. So the elements of $C$ are $$e^{2\pi i((x/24)+(y/54))}=e^{2\pi i(9x+4y)/216}\ .$$ So, we need to know how many different residues modulo $216$ are taken by the expression $9x+4y$. The answer is, all of them: since $9$ and $4$ are relatively prime, the equation $$9x+4y=m$$ has a solution for any integer $m$, and so it certainly has a solution for $m=0,1,\ldots,215$.
Thus, the number of elements in $C$ is $216$.