Find the smallest positive integer $n$ such that $ n[5]= [0]$ in the groups $i.\mathbb {Z}_{15}$ ,$ii.\mathbb {Z}_{20}$.

Question

Find the smallest positive integer $n$ such that $ n[5]= [0]$ in the groups $i.$$\mathbb {Z}_{15}$ , $ii.$$\mathbb {Z}_{20}$.

My solution goes like this:

If $n=3$ then $3[5]=[15]=[0]$, is true as $a\in [5]$ if $a\equiv 5\pmod{15}$ and hence $3a\equiv 15\equiv 0\pmod{15}$. So, if $a\in [5]$, then $3a\in [15]=[0]$, so $3.[5]=[0]$ and $n=3$. Similarly, for $\mathbb {Z}_{20}$ , $n=4$.

Is the above solution valid? Does the above proof holds true? If not, then where is the problem occurring?...

Answer 1

are u are looking for the smallest positive integer n satisfying similtaneously both conditions? if So then since 15 divides $5n$ iff 3 divides $n$ ,n must be of the form $n=3k$ for some positive integer $k$.in order to satisfy the second condition,$15k$ must be a multiple of 20,so(since 3 and 4 are coprime) $k$ must be a multiple of 4,the minimal value of $k$ is 4,so $n=12$.