OT Last week's CarTalk puzzler

[Charles Hartman]

Interesting. At first it looks straightforward:

1. If a factor is by definition an integer that when multipilied by  
another integer
	yields the number we're interested in as a product, then factors  
have to come
	in pairs. (It takes two to multiply.)
2. "Odd number of factors" is therefore a contradiction in terms,  
unless "factor" is
	shifted to mean "unique factor".
3. If a number has a pair of factors that are identical (so not  
unique, so they only
	"count" once), then it's the product of that factor (which provably  
can't be
	either 1 or the number itself), times that factor, which is the  
definition of a
	square.

So "having an odd number of factors" is a sufficient condition for  
being "a square".

But it doesn't seem to be a necessary condition. The factors of 36 --  
by the double definition you have to use in order to make sense of  
the statement of the problem -- are either
			1   36   2   2   3   3
or
			1   36   2   3
-- which counted one way amount to 6 and the other, 4, neither of  
which is conspicuously odd. It looks to me as though old Tom is  
wrong. Gee, does that ever happen?

Charles


On Nov 24, 2005, at 9:31 AM, Jim Hurley wrote:

>>
>> 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>
>> Content-Type: text/plain;	charset=US-ASCII;	delsp=yes; 	format=flowed
>>
>>
>> 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
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