Recurring decimals and modular structure

This problem sits at a beautiful intersection of number theory and the familiar world of decimal expansions. Understanding why the period of $\frac1p$ divides $p-1$ takes you right into the multiplicative order of $10$ modulo $p$, a concept that also underpins Fermat's Little Theorem. By working through the argument, you will see how modular arithmetic can reveal hidden structure in the repetend, including Midy's theorem – a classic result that often appears in HSC extension questions on sequences and series.

Problem Statement

Let $p$ be prime ($p\neq2,5$), and let $n$ be the period of $\frac1p$.

• Show $n\mid (p-1)$.
• If $n=2k$, prove $10^k\equiv-1\pmod p$ and the two half-blocks of the repetend sum to $10^k-1$.

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Hints

Factor $10^{2k}-1=(10^k-1)(10^k+1)$ and use minimality of $n$.

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Solutions

The period $n$ of the decimal expansion $\frac1p$ is the smallest positive integer such that $10^n \equiv 1 \pmod p$; this is because the repeating block corresponds to the integer $\frac{10^n-1}{p}$. Therefore $n$ is the multiplicative order of $10$ modulo $p$. By Fermat's Little Theorem, $10^{\,p-1}\equiv1\pmod p$, and since the order divides any exponent giving $1$, we must have $n\mid(p-1)$.

Now suppose the period is even, so $n=2k$. Because $n$ is the order, $p$ divides $10^{2k}-1$. The minimality of $n=2k$ tells us that $10^k-1$ is not divisible by $p$, so $p\nmid(10^k-1)$. Factoring $10^{2k}-1=(10^k-1)(10^k+1)$ forces $p$ to divide the other factor, thus $p\mid(10^k+1)$, which yields $10^k\equiv-1\pmod p$.

With $10^k\equiv-1\pmod p$ in hand, look at the repetend itself. Write the repeating block of length $2k$ as two $k$-digit blocks: let $A$ be the first half and $B$ the second half, so that

$$\frac1p = 0.\overline{A\,B} = \frac{A}{10^k} + \frac{B}{10^{2k}} + \cdots$$

Using $10^k\equiv-1\pmod p$ and the geometric series expansion, one obtains Midy's theorem:

$$A+B = 10^k-1.$$

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Takeaways

• The period of $\frac1p$ is the order of $10$ modulo $p$, so it must divide $p-1$ – a direct consequence of Fermat's Little Theorem.
• When the period is even, the repetend splits into two halves that sum to $10^k-1$ (a string of nines). This property, known as Midy's theorem, can be a handy shortcut for checking divisibility or for constructing fractions with prescribed repetends.
• This problem shows how modular arithmetic reveals structure in everyday decimal expansions, linking sequences, number theory, and calculus in one neat package.

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Further Readings

HSC Trigonometry, HSC Proofs, HSC Combinatorics, HSC Integrals