What is the order of the element $10 \in (\mathbb Z/p \mathbb Z)^*$? Let $p \geq 23$ be a prime number such that the decimal expansion (base 10) of $\frac{1}{p}$ is periodic with period $p-1$ . Let $(\mathbb Z/p \mathbb Z)^*$ denote the multiplicative group of integers modulo $p$. Then which of the following is true?
1) The order of the element $10 \in  (\mathbb Z/p \mathbb Z)^*$ is a proper divisor of $p-1$.
2) The order of the element $10 \in  (\mathbb Z/p \mathbb Z)^*$ is  $(p-1)/2$.
3) The element $10 \in  (\mathbb Z/p \mathbb Z)^*$  is a generator of the group $(\mathbb Z/p \mathbb Z)^*$
My Attempt : I know that the group $(\mathbb Z/p \mathbb Z)^*$ is cyclic. And the prime number $p$ is a divisor of $10^{p-1} -1$. I can not deduce anything else from the given question. Can anyone please help me to proceed?
 A: Gathering my comments above into an answer: We calculate $1\div 23$ the way I learned to do division in elementary school and compare to calculating powers of $10$ modulo $23$.
I start with how many times $1$ goes into $23$. The answer is none, with a remainder of $1$. So to begin with, we write
$$
1\div 23=0
$$
Next, we move one decimal place over, and make our remainder of $1$ into $10$. So we ask how many times $23$ goes into $10$. Once again the answer is $0$, this time with a remainder of $10$. And we write
$$
1\div23=0.0
$$
Next, we move another decimal place over and make our remainder into $100$. Once again we ask how many times does $23$ go into $100$? The answer is $4$, with a remainder of $8$. We write
$$
1\div 23=0.04
$$
And so on. We make $8$ into $80$, divide, get $3$ with a remainder of $11$. Next we make $11$ into $110$, divide, get $4$ with a remainder of $18$. So far, this has given us $$1\div 23=0.00434$$and this just continues indefinitely.
Now compare with computing powers of $10$ modulo $23$. We get
$$
\begin{array}{|c|c|}\hline
n&10^n\bmod 23\\\hline
0&1\\
1&10\\
2&8\\
3&11\\
4&18\\
\vdots&\vdots
\end{array}
$$
Do you recognize these numbers? Can you see how they are actually connected? (Think about how you might calculate the right-hand column, and see if you can recognize the steps.) And from this, can you answer the actual problem?
A: Now we have, $p$ is a prime $(\geq 23)$ and ${\mathbb{Z}_{p}}^{*}$ is a cyclic group with order $p-1$. Again by Fermat's Little theorem, $10^{p-1}\equiv 1 (\bmod \ p)$. So order of $10$ in group ${\mathbb{Z}_{p}}^{*}$ is $\leqslant p-1$. Suppose, $\circ (10) = m < p-1$$,$ then $10^{m}\equiv 1 (\bmod \ p)$. Since $p$ is a prime $(\geq 23)$, so $p\mid (1+10+\dots + 10^{m-1})$. Let $k$ be positive integer such that, $\frac{1}{p} = \frac{k}{1+10+\dots + 10^{m-1}}$. Now if $1\leq k< 10$, then $$\frac{k}{1+10+\dots + 10^{m-1}} = 0.\overline{\underset{\longleftarrow m-2 \longrightarrow}{00\dots 00}(k-1)(10-k)}.$$
Therefore, $\frac{1}{p}$ is periodic with period $m$, a big contradiction!. So option $(3)$ is the only correct option.
