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2014/15 AHSMC Part 1 Problem 7

Problem 7 of 16

Let f(n)=2n3f(n) = 2n^3 for any positive integer nn. For any odd number m≥3m \ge 3, the largest positive integer kk such that 2k2^k divides f(f(f(m)))f(f(f(m))) is

Answer choices

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Solution

We have f(m)=2m3f(m) = 2m^3, f(2m3)=24m9f(2m^3) = 2^4 m^9 and f(24m9)=213m27f(2^4 m^9) = 2^{13} m^{27}. The answer is (D).

University of Alberta