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2020/21 AHSMC Part 1 Problem 9

Problem 9 of 16

For a positive integer NN, let S(N)S(N) denote the sum of the digits of NN. Consider the sequence a1=2a_1 = 2, an=(S(an−1))2−1a_n = \bigl(S(a_{n-1})\bigr)^2 - 1 for n≥2n \ge 2. The term a2020a_{2020} is equal to

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Solution

The sequence is {2,3,8,63,80,63,80,…}\{2, 3, 8, 63, 80, 63, 80, \ldots\} and repeats with a period of 22. For even values of n≥4n \ge 4 we have an=63a_n = 63, hence a2020=63a_{2020} = 63. The answer is (D).