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Q(3^k) \ge Q(3^{k+1} for infinite n, whereas Q(n) is sum of decimal digits of n

Source: 1999 German Federal - Bundeswettbewerb Mathematik - BWM - Round 2 p2

January 27, 2020
number theorysum of digitsDigitsinequalities

Problem Statement

For every natural number nn, let Q(n)Q(n) denote the sum of the decimal digits of nn. Prove that there are infinitely many positive integers kk with Q(3k)≥Q(3k+1)Q(3^k) \ge Q(3^{k+1}).
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