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.
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.
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