Problem
NT-B2-M11-P012 Period Exactly Two
#12
★★★☆☆ Level 3 of 5
Find all \(n>1\), coprime to \(10\), for which the decimal expansion of \(\frac1n\) has period exactly \(2\).
Period \(2\) means \(10^2\equiv1\pmod n\), but \(10\not\equiv1\pmod n\).
The order of \(10\) modulo \(n\) must be \(2\). Hence \(n\mid10^2-1=99\), but \(n\nmid10-1=9\). The divisors of \(99\) are \(1,3,9,11,33,99\). Excluding \(1\) and the divisors of \(9\), we get \(11,33,99\). For each of them, \(10^2\equiv1\pmod n\) while \(10\not\equiv1\pmod n\), so the period is exactly \(2\).
This reinforces the word “exactly”: a smaller period must be excluded.