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Number theory Difficulty 7.8 National Olympiad, round 2 Prove it Baltic Way

Prove that there are infinitely many positive integers $n$ , which are not divisible by $10$ and such that $s(n^2) < s(n) - 5$ where $s(n)$ is the sum of digits of $n$ .

Solution

All integers of the form $499\ldots99$ satisfy the condition. Indeed, if $n = 4\underbrace{99\ldots99}_{k} = 5 \cdot 10^k - 1$ then
$n^2 = 25 \cdot 10^{2k} - 10^{k+1} + 1 = 24 \underbrace{99\ldots9}_{k-1} \underbrace{00\ldots00}_{k} 1.$
In such a case $s(n) = 4 + 9k$ , but $s(n^2) = 7 + 9(k - 1) = 9k - 2$ .

Solution:
Consider a sequence $10^{3m} - 10^{2m} - 1$ . Similarly to the original solution it is easy to check that if $m$ increases by $1$ then $s(n)$ increases by $27$ , but $s(n^2)$ increases by $18$ only.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.