Problem
NT-B2-M11-P008 Repunits Divisible by \(37\)
#8
★★★☆☆ Level 3 of 5
Find all positive integers \(n\) for which the repunit \(R_n=11\ldots1\) with \(n\) ones is divisible by \(37\).
Notice that \(R_3=111=3\cdot37\), and use divisibility of repunits.
Since \(R_3=111=3\cdot37\), if \(3\mid n\), then \(R_3\mid R_n\), so \(37\mid R_n\).
Conversely, \(37\mid R_n\) is equivalent to \(10^n\equiv1\pmod{37}\), because \(9\) is invertible modulo \(37\). The order of \(10\) modulo \(37\) is \(3\): \(10^3=1000\equiv1\pmod{37}\), while \(10\not\equiv1\) and \(10^2=100\not\equiv1\pmod{37}\). Therefore \(3\mid n\).
The answer is all multiples of \(3\).
This connects repunits with the order of \(10\), preparing students for period problems.