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Let \(p(n)\) be the number of digits when \(8^{n}\) is written is base 6, and let \(q(n)\) be the number of digits when \(6^{n}\) is written in base 4. For example, \(8^{2}\) in base 6 is 144, hence \(p(2) = 3\). Then \(lim_{n \rightarrow \infty} \) \({p(n)q(n) \over n^2}\) equals :

(A) 1

(B) \(\frac{4}{3}\)

(C) \(\frac{3}{2}\)

(D) 2

Updated Tue, 21 May 2019 12:56 pm

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