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Repunits And 2008 (Posted on 2007-10-06) Difficulty: 3 of 5
Can the 2008th repunit be a perfect square in any integer base x ≥ 2?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

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re: Repunits Comment 2 of 2 |
(In reply to Repunits by Gamer)

"Notice that x-1 times the repunit with n ones in base x equals x^(n+1)-1. So the repunit with n ones in base x equals x^(n+1)/(x-1)"

But then, in base x = 10, with one 1, this gives 1 = (10^2)/9. With two 1's, 11 = (10^3) / 9.

Actually, in base x, a repunit of n 1's = (x^n - 1) / (x - 1).

  Posted by Charlie on 2007-10-07 02:35:47

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