Find a a cyclic number that start with 3 and after mutiplying it by 1.5 shift the 3 at the end of the sequence

Question

I have to find a number that start with 3 and if I multiply it for 1.5 (or multiply it by 3 and divide it by 2) shift the 3 a the end of the number sequence. I guess it must be a cyclic number but I can't find it!

Answer 1

Let the number be $3\cdot10^L+N$, where $N$ has $L$ digits. Then $$\frac32(3\cdot10^L+N)=10N+3$$ $$\frac92\cdot10^L-3=\frac{17}2N$$ $$9\cdot10^L-6=17N$$ Now we check which values of $L$ satisfy $17\mid9\cdot10^L-6$, starting from $0$. The smallest such $L$ turns out to be $15$, so the smallest number satisfying the question's requirement is $3$ prepended to $\frac{9\cdot10^{15}-6}{17}$, or $$3529411764705882$$