Mathematical Reflections 2011, Issue 6 - Problem J213

Problem:
For any positive integer $n$, let $S(n)$ denote the sum of digits in its decimal representation.
Prove that the set of all positive integers $n$ such that $n$ is not divisible by $10$ and $S(n) >
S(n^2 + 2012)$ is infinite.

Proposed by Preudtanan Sriwongleang.

Solution:

With the notation introduced, let $A = \{ n \in \mathbb{N} : n \neq 10k \textrm{ and } S(n)  > S(n^2+2012), k \in \mathbb{N}\}$. We prove that $A \supset \{5\cdot10^{m}-1, m \in \mathbb{N}\setminus\{0\}\} = \{49,499,\ldots,4\underbrace{9\ldots9}_{m},\ldots\}$, so that $A$ must be infinite. Clearly, $49,499 \in A$, since
$$49 \neq 10k, k \in \mathbb{N}, \qquad S(49)=13 > 12=S(4413)$$ and
$$499 \neq 10k, k \in \mathbb{N}, \qquad S(499)=22 > 12=S(251013).$$
Let $m > 2$. Then $5\cdot10^m-1 \neq 10k, k \in \mathbb{N}$ and $$(5\cdot 10^m - 1)^2+2012=25\cdot10^{2m}-10^{m+1}+2013=24\underbrace{9\ldots9}_{m-1}\underbrace{0\ldots0}_{m+1} + 2013,$$
from which $$S((5\cdot10^m-1)^2+2012)=2+4+9(m-1)+2+1+3=9m+3.$$
Therefore, $S(5\cdot10^m-1)=9m+4 > 9m+3=S((5\cdot10^m-1)^2+2012)$, i.e. $$5\cdot10^m-1 \in A \qquad \forall m \in \mathbb{N}\setminus\{0\},$$ which gives the desired conclusion.