# Repeating decimal of $1/2009$ having a cycle (not necessarily minimal) with length of $\phi(2009)$

Explain why the Repeating decimal of $1/2009$ has a cycle (not necessarily minimal) with length of $\phi(2009)$

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Note $(10, 2009) = 1$, so (the class of) $10$ is an element of the group $G$ of units of $\mathbb{Z}_{2009}$. It follows from Lagrange's theorem that the period of $10$ in $G$ is a divisor of the order $\varphi(2009)$ of $G$.
Now you should know that this period is the same as the (length of the minimal) period of the decimal expansion of $1/2009$. (I will add an explanation below.) This proves your statement.
Suppose (the class of) 10 has period $n$ in $G$ as above. Then $10^n \equiv 1 \pmod{2009}$, that is, $10^n - 1 = 2009 \cdot k$ for some $k$. Thus $$\frac{1}{2009} = k \cdot \frac{1}{10^n - 1} = k \cdot \left( \frac{1}{10^n} + \frac{1}{10^{2n}} + \frac{1}{10^{3n}} + \dots \right),$$ that is, $1/2009$ is a periodic decimal, with period of length $n$.
+1 Undoubtedly you know this, but we can actually say a bit more. The factorization $2009=7^2\cdot41$ together with the facts that A) the order of $10$ in $\mathbb{Z}_{49}^*$ is a factor of $\phi(49)=42$, B) the order of $10$ in $\mathbb{Z}_{41}^*$ is a factor of $\phi(41)=40$ and the Chinese remainder theorem imply that the order of $10$ in $\mathbb{Z}_{2009}^*$ is a factor of $lcm(42,40)=840=\phi(2009)/2$. Not much more information, but if we want to milk everything out of this :-) – Jyrki Lahtonen Feb 4 '13 at 21:07