Prove that there exists a power of three that ends with 001

Solution


|Contact| |Front page| |Contents| |Up| |Store|

Copyright © 1996-2012 Alexander Bogomolny

Prove that there exists a power of three that ends with 001

As in a related problem, let 3n and 3m (n > m) have the same remainder when divided by 1000. Thus 3n - 3m = 3m(3n-m - 1) is divisible by 1000. Since 1000 and 3m have no common factors, 1000 is bound to divide the second factor (3n-m - 1). This exactly means that 3n-m ends with 001!


Related material
Read more...

  • 17 rooks on a Chessboard
  • Chinese Remainder Theorem
  • Pigeonhole in a Chessboard
  • Pigeonhole in Concurrent Cuts
  • Monochromatic Rectangle in a 2-coloring of the Plane
  • Polynomial and Integer Division
  • Pigeonhole Principle (subsets)
  • Pigeonhole in Integer Differences
  • Four Numbers, Six Differences, GCD of the Products

  • |Contact| |Front page| |Contents| |Up| |Store|

    Copyright © 1996-2012 Alexander Bogomolny

     40616208

    A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
    Sites for teachers
    Sites for parents
    Terms of use
    Awards
    Interactive Activities

    CTK Exchange
    CTK Wiki Math
    CTK Insights - a blog
    Math Help
    Games & Puzzles
    What Is What
    Arithmetic
    Algebra
    Geometry
    Probability
    Outline Mathematics
    Make an Identity
    Book Reviews
    Stories for Young
    Eye Opener
    Analog Gadgets
    Inventor's Paradox
    Did you know?...
    Proofs
    Math as Language
    Things Impossible
    Visual Illusions
    My Logo
    Math Poll
    Cut The Knot!
    MSET99 Talk
    Old and nice bookstore
    Other Math sites
    Front Page
    Movie shortcuts
    Personal info
    Privacy Policy

    Guest book
    News sites

    Recommend this site

    Sites for parents

    Education & Parenting

    Search:
    Keywords:

    Google
    Web CTK
    Supported by
    3wVentures