olymmath

ISL 2009 C8

    For any integer n≥2n \geq 2, we compute the integer h(n)h(n) by applying the following procedure to its decimal representation. Let rr be the rightmost digit of nn. (1) If r=0r=0, then the decimal representation of h(n)h(n) results from the decimal representation of nn by removing this rightmost digit 0 . (2) If 1≤r≤91 \leq r \leq 9 we split the decimal representation of nn into a maximal right part RR that solely consists of digits not less than rr and into a left part LL that either is empty or ends with a digit strictly smaller than rr. Then the decimal representation of h(n)h(n) consists of the decimal representation of LL, followed by two copies of the decimal representation of R−1R-1. For instance, for the number n=17,151,345,543n=17,151,345,543, we will have L=17,151,R=345,543L=17,151, R=345,543 and h(n)=17,151,345,542,345,542h(n)=17,151,345,542,345,542. Prove that, starting with an arbitrary integer n≥2n \geq 2, iterated application of hh produces the integer 1 after finitely many steps.

    Dataset text NuminaMath olympiads. Quoted with source.

    Official text: IMO 2009 Shortlist booklet, page 9.

    Solutions

    Official solution Solution 1 by the IMO Problem Selection Committee

    The terms of the source let olymmath link to this solution but not copy it. Read the solution (IMO official site).

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