OT Last week's CarTalk puzzler
jhurley at infostations.com
Thu Nov 24 11:14:51 CST 2005
>Charles Hartman wrote:
>Interesting. At first it looks straightforward:
>1. If a factor is by definition an integer that when multipilied by
> 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
> either 1 or the number itself), times that factor, which is the
>definition of a
>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
> 1 36 2 3
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36; an odd number
of factors as the theorem predicts.
The factors of 32 are 1, 2, 4, 8, 16, and 32; an even number of
factors, again consistent with the theorem.
The only way I know to solve this problem, i.e. show that the number
of factors of a given number is odd, if and only if the number is a
perfect square, is to express the number as a product of primes.
For example 36 = 2^2 * 3^2
And 32 = 2^5
In case anyone wants to pursue this further, I will quit here. Beauty
is best enjoyed if you discover it for yourself.
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