Problem
NT-B2-M11-P016 The Period Equals the Product
Find all reduced proper fractions \(\frac{a}{b}\) whose decimal expansion is an infinite repetition of the decimal notation of the number \(ab\). For example, \(\frac13=0.\overline3\).
If \(ab\) has \(k\) digits, then \(0.\overline{ab}=\frac{ab}{10^k-1}\).
Let \(P=ab\), and suppose \(P\) has \(k\) digits. Then \(0.\overline{P}=\frac{P}{10^k-1}=\frac{ab}{10^k-1}\). The condition gives \(\frac{a}{b}=\frac{ab}{10^k-1}\). Since \(a>0\), cancel \(a\): \(\frac1b=\frac{b}{10^k-1}\), so \(b^2=10^k-1\).
For \(k=1\), this gives \(b=3\). For \(k\ge2\), \(10^k-1\equiv3\pmod4\), but no square is congruent to \(3\pmod4\). Therefore \(b=3\). Since the fraction is proper and reduced, \(a=1\) or \(a=2\). The answer is \(\frac13\) and \(\frac23\).
The difficulty is interpreting the period correctly: it is the notation of the product, not the product of digits.