Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Find the answer

For $n \in \mathbf{N}_{+}$ , define
$S(n)=\left[\frac{n}{10^{[18 n]}}\right]+10\left(n-10^{[i \mid n]}\left[\frac{n}{10^{\left[1 / B^{n}\right]}}\right]\right),$

where $[x]$ denotes the greatest integer not exceeding the real number $x$ . Then, among $1,2, \cdots, 2012$ , the number of positive integers $n$ that satisfy $S(S(n))=n$ is

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

6. 108.

Let $t=10^{[\lg n]}$ . Then
$S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right) \text {. }$

Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$ , and $\left[\frac{n}{t}\right]$ is the first digit of $n$ .
We will discuss the cases below.
(1) If $n$ is a one-digit number, all satisfy the requirement, totaling 9.
(2) If $n=\overline{x y}, S(n)=\overline{y x}, S(S(n))=\overline{x y}$ , totaling 81 (excluding those with $y=0$ ).
(3) If $n=\overline{x y z}, S(n)=\overline{y z x}, S(S(n))=\overline{z x y}$ , thus, $x=y=z$ , totaling 9.
(4) If $n=\overline{x y z w}, S(S(n))=\overline{z w x y}$ , thus, $w=y, z=x$ , totaling 9.
Therefore, the number of $n$ that satisfy the requirement is 108.

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