Recurring decimals and modular structure

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 $10^k-1$.

Hints

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

Solutions

From order arguments, $n$ is the multiplicative order of $10$ mod $p$, so $n\mid(p-1)$. If $n=2k$, then $p\mid(10^{2k}-1)$. Since $2k$ is minimal, $p\nmid(10^k-1)$, hence $p\mid(10^k+1)$, so $10^k\equiv-1\pmod p$. Writing repetend as two $k$-digit blocks $A,B$ gives Midy's theorem:

Further Readings

• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/
• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Combinatorics: https://vumaths.com/booklets/hsc-combinatorics/
• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/

Connect with me

• LinkedIn: https://www.linkedin.com/in/nguyenvuhung/
• Website - Vu's Maths Hub: https://vumaths.com/
• YouTube - HSC Maths Extension 1+2: https://www.youtube.com/playlist?list=PLHSE0sAlTr2w