Maths Olympiad Prep

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Number theory Difficulty 4.5 AIME Find the answer

8. (HUN 1) For a given positive integer kk denote the square of the sum of its digits by f1(k)f_{1}(k) and let fn+1(k)=f1(fn(k))f_{n+1}(k)=f_{1}\left(f_{n}(k)\right). Determine the value of f1991(21990)f_{1991}\left(2^{1990}\right).

A number or a short expression. Spacing and $ signs are ignored.

Solution

8. Since 21990>02^{1990} > 0. Thus f3(21990)=r2f_{3}\left(2^{1990}\right)=r^{2} where r>1r > 1. Hence f1991(21990)=256f_{1991}\left(2^{1990}\right)=256.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.