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2021/22 AHSMC Part 1 Problem 4

Problem 4 of 16

Let g(k)=2k2g(k) = 2k^2 for any positive integer kk. Let p≥3p \ge 3 be a prime number. The number of positive integers dividing into g(g(p))g(g(p)) is

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Solution

Observe that g(g(p))=2(2p2)2=23p4g(g(p)) = 2(2p^2)^2 = 2^3 p^4. Thus the positive divisors of g(g(p))g(g(p)) are all of the form 2ipj2^i p^j for integers 0≤i≤30 \le i \le 3 and 0≤j≤40 \le j \le 4. Hence, g(g(p))g(g(p)) has 4×5=204 \times 5 = 20 positive divisors. The answer is (D).

University of Alberta