Prime denominators and cycle length

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 $\frac1{41}$.

Hints

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

Solutions

Recurring block theory gives $(10^L-1)/p\in\mathbb{Z}$, hence $10^L\equiv1\pmod p$; minimality gives least such $L$. By Fermat, $10^{p-1}\equiv1\pmod p$, so the order $L$ divides $p-1$. Since $37\mid(10^3-1)$ and not $10^1-1,10^2-1$, period is $3$. Since $41\mid(10^5-1)$ and not $10^1-1$, period is $5$.

Further Readings

• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/
• HSC Polys Ext 1: https://vumaths.com/booklets/hsc-polys-ext-1/
• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/

Connect with me

• YouTube - HSC Maths Extension 1+2: https://www.youtube.com/playlist?list=PLHSE0sAlTr2w
• Substack: https://vuhung16.substack.com/
• GitHub: https://github.com/vuhung16au/