Question

For any natural number $n>1,$ write the infinite decimal expansion of $\frac{1}{n}$ (for example, we write $\frac{1}{2}=0.49$ as its infinite decimal expansion, not $0.5$). Determine the length of the non-periodic part of the (infinite) decimal expansion of $\frac{1}{n}.$

Answer 1

For any prime p, let $v_p\left(n\right)$ be the maximum power of p dividing n; i.e $p^{1/2}p\left(n\right)$ divides n but not higher power. Let r be the length of the non-periodic part of the infinite decimal expansion of 1/n.
Write
$\frac{1}{n}=0.a_1{a}_2.....a_r\overline{b_1{b}_2......b_2}$.
We show that r = $max\left(v_2\right(n);v_5(n\left)\right)$.
Let a and b be the numbers $a_1{a}_2........a_{r}andb=b_1{b}_2....b_s$ respectively.(Here $a_{1}and{b}_1$ can be both 0.) Then
$\frac{1}{n}=\frac{1}{{10}^r}(a+\sum _{k\ge 1}\frac{b}{({10}^s{)}^k})=\frac{1}{{10}^r}(a+\frac{1}{{10}^s-1})$
Thus we get ${10}^r({10}^s-1)=n({10}^s-1)a+b$. It shows that $r\ge max\left(v_2\right(n);v_5(n\left)\right)$. Suppose $r>max\left(v_2\right(n);v_5(n\left)\right)$. Then 10 divides b-a.Hence the last digits of a and b are equal: $a_r=b_s$. This means
$\frac{1}{n}=0:a_1{a}_2.......a_{r_1}\overline{b_2{b}_1{b}_2.....b_{s_1}}$.
This contradicts the definition of r. Therefore $r=max\left(v_2\right(n);v_5(n\left)\right)$.