Track / Stage 6 / 135 of 400 #1135 of 2000

Problem 1135

National olympiad, first round

Number theory Difficulty 6.2 Prove it

$1 \cdot 2$ Proof: For any natural number $k$ , there exist infinitely many natural numbers $t$ (in decimal notation) that do not contain the digit 0, such that the sum of the digits of $t$ and $kt$ are the same.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

[Proof] If the number $k$ is denoted as
$k=\overline{a_{n} a_{n-1} \cdots a_{0}},$

let's assume $a_{0} \geqslant 1, t$ represents the number composed of $m$ nines,
$t=\underset{m \uparrow}{99 \cdots 9}=10^{m}-1$

where $m>n$ . Then,
$k t=a_{n} a_{n-1} \cdots\left(a_{0}-1\right) 99 \cdots 9\left(9-a_{n}\right)\left(9-a_{n-1}\right) \cdots\left(9-a_{1}\right)(10$
$\left.-a_{0}\right) k t$ the sum of the digits of $k t$ is the same as the sum of the digits of $t$ , which equals 9 m.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.