Prime denominators and cycle length

When we write a fraction like $\frac{1}{p}$ as a decimal, the digits often fall into a repeating block — and the length of that block satisfies a beautiful number-theoretic property. This problem connects decimal expansions, modular arithmetic, and Fermat’s little theorem, giving you a powerful way to predict the period without doing the long division. By working through the steps, you’ll consolidate your understanding of orders modulo a prime, a topic that appears in both the Extension 1 and Extension 2 HSC courses.

Problem Statement

Let $p$ be prime, $p\neq2,5$, and $L$ be period length of $\frac1p$.

• Show $L$ is the least positive integer with $10^L\equiv1\pmod p$.
• Deduce $L\mid(p-1)$.
• Find periods for $\frac1{37}$ and $\frac1{41}$.

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Hints

Use Fermat's theorem and first occurrence of $p\mid(10^n-1)$.

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Solutions

We start by linking the period length $L$ to a divisibility condition. The standard theory of recurring decimals tells us that the block of length $L$ occurs because the remainder $10^L$ after multiplying and subtracting is $1$, giving the fractional part exactly $1/p$ shifted. This translates directly into the integer condition $(10^L-1)/p\in\mathbb{Z}$, which is precisely the congruence $10^L\equiv1\pmod p$. The minimality of $L$ as the period length guarantees that no smaller positive exponent yields this congruence — $L$ is the least positive integer satisfying the relation.

To deduce that $L$ divides $p-1$ we bring in Fermat’s little theorem. Since $p$ does not divide $10$, we know $10^{p-1}\equiv1\pmod p$. The order of an element modulo $p$ — here the order of $10$ — always divides any exponent that returns $1$. Our $L$ is exactly that order, so we must have $L\mid(p-1)$. This step reveals why the period of $\frac{1}{p}$ is always a factor of $p-1$, a delightful general fact.

Now we can check the specific primes. For $p=37$, we test small $n$ to find the least positive integer with $37\mid(10^n-1)$.
$10^1-1=9$ is not a multiple of $37$.
$10^2-1=99$ is not a multiple of $37$.
$10^3-1=999$; since $37\times27=999$, we have $37\mid(10^3-1)$. No smaller $n$ works, so the period $L=3$.
Similarly, for $p=41$ we examine $10^n-1$:
$10^1-1=9$ is not divisible by $41$.
Checking further, $10^5-1=99999$ and $41\times2439=99999$, so $41\mid(10^5-1)$. Because no smaller $n$ gives divisibility, the period is $5$.

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Takeaways

• The period length of $\frac{1}{p}$ (for a prime $p\neq2,5$) is exactly the order of $10$ modulo $p$, i.e. the smallest $L$ with $10^L\equiv1\pmod p$.
• Fermat’s little theorem forces $L$ to divide $p-1$, giving a divisor upper bound on the period.
• For a concrete prime, you can find $L$ by testing powers of $10$ until $10^n-1$ becomes divisible by $p$; the smallest such $n$ is the period.

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Further Readings

HSC Integrals, HSC Inequalities, HSC Polys Ext 1, HSC Vectors