MathDB
F(n)

Source: Cono Sur Olympiad, 1995, Problem #6

May 28, 2006
floor functionfunctionmodular arithmeticinequalitiesnumber theoryrelatively prime

Problem Statement

Let nn be a natural number and f(n)=2n−1995⌊n1000⌋f(n) = 2n - 1995 \lfloor \frac{n}{1000} \rfloor(⌊\lfloor ⌋\rfloor denotes the floor function). 1. Show that if for some integer rr: f(f(f...f(n)...))=1995f(f(f...f(n)...))=1995 (where the function ff is applied rr times), then nn is multiple of 19951995. 2. Show that if nn is multiple of 1995, then there exists r such that:f(f(f...f(n)...))=1995f(f(f...f(n)...))=1995 (where the function ff is applied rr times). Determine rr if n=1995.500=997500n=1995.500=997500
Back to ProblemsView on AoPS