Problem
NT-B2-M11-P019 Period Exactly Three
#19
★★★★☆ Level 4 of 5
Find all \(n>1\), coprime to \(10\), for which the decimal expansion of \(\frac1n\) has period exactly \(3\).
Use \(10^3-1=999=3^3\cdot37\), and exclude period \(1\).
Period exactly \(3\) means that the order of \(10\) modulo \(n\) is \(3\). Hence \(n\mid10^3-1=999\), but \(n\nmid10-1=9\). Since \(999=3^3\cdot37\), the relevant divisors must contain the factor \(37\). They are \(37,3\cdot37,9\cdot37,27\cdot37\), namely \(37,111,333,999\).
For each of them, \(10^3\equiv1\pmod n\), while \(10\not\equiv1\pmod n\), because \(37\mid n\) but \(37\nmid9\). Thus the period is indeed \(3\). The answer is \(37,111,333,999\).
This checks whether the student can work with exact order, not only with divisibility of \(10^h-1\).