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Solution for January 2010

The Problem:

Show that the product


is a power of 2.


Correct responses:

Correct solutions were submitted this month by

Berkay Anahtarci

Mei-Hui Fang

Remo Mantovanelli

Claudio Baiocchi

Philippe Fondanaiche

Claude Morin

Halima Bashier

Tom Fuzesy

Milan Pavic

Bojan Basic

Bruce Golfman

Shpetim Rexhepi

Luigi Bernardini

Patrick Gordon

Eric Reynaud

José Borges

Cornel Gruian

John T. Robinson

Lou Cairoli

David Jackrel

Heri Setiyawan

John Campbell

Magnus Jakobsson

Nutheti Shivdeep

Bernard Carpentier

Wolfgang Kais

A. Teitelman

Bernard Collignon

Stephen La Rocque

Jan van Delden
   (The Netherlands)

Olivier Cyr

Normand LaLiberté

Ray Van Raamsdonk
   (British Columbia)

Allen Druze

Daniel Lu

Leonardo Vicchi

  Karim Laaouini

            We will start with a proof that was common to many of the solutions sent to us this month.  We follow that by two approaches to a generalization that is no harder than the original problem, and we end with two further generalizations.


The solution:

The original problem.


which is a power of 2, as desired.

A generalization.  Ten of our correspondents observed that the number 2009 could be replaced by 4n + 1 in the statement of our problem; specifically, for all integers n ≥ 1,

product                                    (1)

in the original problem we have n = 502.  Bernardini, Colignon, Golfman, Gruian, Morin, and van Delden applied a version of the argument we presented above; namely,


as claimed.

Mathematical induction.  Basic, Fondanaiche, La Liberté, and Vicchi let P(n) stand for the product on the left side of equation (1), and they used induction to prove that P(n) = 22n. P(1) is simply 5 – 1 = 22, so assume that P(k) = 22k; we are to prove that P(k + 1) = 22(k+1).


where we inserted the term 2k + 2 in both the numerator and the denominator.  Note that the expression in the square brackets is P(k), which we have assumed to equal 2k. Thus,

inductive step

as was to be proved.

Comments.  We thank our colleague Richard McIntosh for suggesting this month's problem.  He recently discovered equation (1) and several related identities in his investigation of Bernoulli and Euler numbers that will be incorporated into a forthcoming article.  In fact, a version of equation (1) holds also for 4n – 1, namely

equation 1, n ≥ 1.                                   (2)

Furthermore, there is a similar identity with a plus sign in the product, namely

equation 2, n ≥ 1.                                      (3)

Identities (2) and (3) can both be proved using the same methods that worked for (1).





Math Central is supported by the University of Regina and the Pacific Institute for the Mathematical Sciences.



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