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 <[EMAIL PROTECTED]>
Subject: Re: OT Last week's CarTalk puzzler
To: How to use Revolution <[email protected]>
Message-ID: <[EMAIL PROTECTED]>
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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