Show that if n is a product of m distinct primes, then the period of the decimal expansion of 1/n is the lowest common multiple of the periods of 1/p over all primes p|n.
I understand that the above fact is true, but I really don't understand how to go about proving it, and I have been stuck for days now.
Proofs have never been my strong point -- I have no idea where to start with this particular one. Thanks in advance.
The period is unchanged by the presence of a $2$ or a $5$, so we will assume that none of the prime divisors of $n$ is $2$ or $5$.
Then the period of $\frac{1}{n}$ the least positive integer $h$ such that $10^n \equiv 1\pmod{n}$. In other words, it is the order of $10$ modulo $n$.
Let $h_1, h_2, \dots, h_m$ be the orders of the primes divisors $p_1,p_2,\dots,p_m$ of $n$.
Let $h$ be any positive common multiple of the $h_i$. Let $h=q_ih_i$. Then $10^h=(10^{h_i})^{q_i}\equiv 1\pmod{p_i}$. It follows that $10^h\equiv 1\pmod{n}$.
So the order of $10$ modulo $n$ is less than or equal to the lcm of the $h_i$.
We need to show that nothing cheaper than the lcm will do. To show that, we show that if $e$ is the order of $10$ modulo $n$, then $e$ is divisible by the order of $10$ modulo $p_i$ for all $i$.
Let $e_i$ be the order of $10$ modulo $p_i$. We show that $e_i$ divides $e$. Use the division algorithm to express $e$ in the form $e=qe_i+r$, where $0\le r\lt e_i$. We will show that $r=0$.
We have $10^e\equiv 1\pmod{n}$, and therefore $10^e\equiv 1\pmod{p_i}$. Thus $$10^{qe_i+r}\equiv 1\pmod{p_i}.\tag{1}$$ But $$10^{qe_i+r}=(10^{e_i})^q 10^r\equiv 10^r \pmod{p_i}.\tag{2}$$ Since $e_i$ is the order of $10$ modulo $p_i$, and by (1) and (2) we have $10^r\equiv 1\pmod{p_i}$, with $r\lt e_i$, we must have $r=0$. This completes the proof.