OT Last week's CarTalk puzzler

[Jim Hurley]

>
>Message: 10
>Date: Wed, 23 Nov 2005 21:19:31 -0500
>From: Charles Hartman <charles.hartman at conncoll.edu>
>Subject: Re: OT Last week's CarTalk puzzler
>To: How to use Revolution <use-revolution at lists.runrev.com>
>Message-ID: <51A8F1BE-8874-493D-888F-0668BC162A59 at conncoll.edu>
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>
>On Nov 23, 2005, at 6:07 PM, Jim Hurley wrote:
>
>>
>>  All those numbers are called perfect squares. And only they have an 
>>  odd number of factors, because one of the factors is the square 
>>  root of the number in question. For example, nine has three 
>>  factors, 1 and 9 and 3.  [I confess, I can't see how this follows. 
>>  Jim]
>
>
>Well, because 9 has four factors -- 1, 3, 3, and 9 -- two of which 
>are assigned to the same chain-puller, who however only pulls the 
>chain once.
>
>Charles
>


Charles,

I expressed myself badly. What I meant was that I didn't see how this 
one example proved the theorem. A proof needs to show how the theorem 
follows for all perfect squares and only for perfect squares, i.e. it 
must be both a necessary  and sufficient condition.

Jim