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Tenth IMAR Mathematical Competition

Number theory Difficulty 5.3 AIME, harder Prove it Romania

For every non-negative integer $n$ , let $s_n$ be the sum of the digits in the decimal expansion of $2^n$ . Is the sequence $(s_n)_{n \in \mathbb{N}}$ eventually increasing?

Solution

Suppose, if possible, that the sequence is eventually increasing, say from some rank $n_0$ on. Fix a non-negative integer $m$ such that $6m \ge n_0$ to write
$\begin{cases} s_{6m+1} \ge s_{6m} + 1, \\ s_{6m+2} \ge s_{6m+1} + 2, \\ s_{6m+3} \ge s_{6m+2} + 4, \\ s_{6m+4} \ge s_{6m+3} + 8, \\ s_{6m+5} \ge s_{6m+4} + 7, \\ s_{6m+6} \ge s_{6m+5} + 5, \end{cases}$
and deduce thereby that $s_{6m+6} \ge s_{6m} + 27$ , so
$s_{6m+6n} \ge s_{6m} + 27n, \quad n \in \mathbb{N}. \quad (*)$
On the other hand, the number of non-vanishing digits in the decimal expansion of $2^{6m+6n}$ does not exceed $\lceil(6m + 6n)\log_{10} 2\rceil < 2m + 2n$ , so $s_{6m+6n} \le 18m + 18n$ , contradicting (*) for $n$ large enough. The conclusion follows.

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