Solution to Problem 127

Congratulations to this week's winner

Kenny Albright

Other correct solutions were sent in by Brandon Schwoerer, Nathan Pauli. Further solutions were submitted by alumnus Brian Laughlin ('81), and from Al Zimmermann, Sudipta Das, Nancy Schwarzkopf, Khanh Ngo, Burkart Venzke, Jonathan Lewin, Dane Brook, Mike Lynch, Robert T McQuaid, Ahron Teitelman, Francesc Suñol, Shekhar Joglekar, Steve Prowse, Hareendra Yalamanchili, K. Sankarasubramanian, Alexey Vorobyov, Philippe Fondanaiche.

Assuming that a and b are one digit numbers we can write a / b = .abbb... = a /10 + b(.0111...) = a /10 + b/90 = (9a + b)/90,
90a = 9ab + b² , from which we see that b is divisible by 3. Hence b = 3, 6, 9. A quick check shows the only solutions to be a = 1, b = 6, and a = 9, b = 9. Yes, really: 1 = 9/9 = .999...!!

Philippe Fondanaiche carried out the analysis for a and b with an arbitrary number of digits and discovered the only solutions to be those above together with a = 99 = b, a = 999 = b,...

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