AoPS - Problem

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Problem

Source: 2021 Irish Mathematical Olympiad P3

Tags: number theory, Digits, decimal representation

parmenides51

30.05.2021 20:30

For each integer $n \ge 100$ we define $T(n)$ to be the number obtained from $n$ by moving the two leading digits to the end. For example, $T(12345) = 34512$ and $T(100) = 10$ . Find all integers $n \ge 100$ for which $n + T(n) = 10n$ .

Pal702004

31.05.2021 21:27

$n=10\cdot \dfrac{10^{6k}-1}{91}$ Minimal solution

$n=109890$

pawssed

29.01.2024 02:32

can you please provide a solution how you got that? thanks

programjames1

29.01.2024 03:09

Let $y$ be the leading two digits, and $x$ the remaining $k$ , so $n = x + 10^{k}y, \qquad T(n) = 100x + y.$ We need $100x + y = T(n) = 9n = 9x + 9\cdot 10^{k}y\Longleftrightarrow x = \frac{10^{k}-1}{91}y.$ The order of $10$ modulo $91$ is six.