NT-B2-M11-P006 When One Repunit Divides Another

For the converse, divide \(n\) by \(m\) with remainder and consider \(10^n-1\) modulo \(10^m-1\).

If \(m\mid n\), then \(10^m-1\mid10^n-1\), hence \(R_m\mid R_n\).

Conversely, suppose \(R_m\mid R_n\). Then \(10^m-1=9R_m\) divides \(9R_n=10^n-1\). Write \(n=qm+r\), where \(0\le r0\), then \(10^r-1\) is positive and smaller than \(10^m-1\), so it cannot be divisible by \(10^m-1\). Therefore \(r=0\), so \(m\mid n\).

A core problem for the entire repunit topic; students should solve it themselves.