OT Last week's CarTalk puzzler

[Jim Hurley]

>Mark Wieder wrote:
>
>Jim-
>
>Friday, November 25, 2005, 6:45:22 AM, you wrote:
>
>I think it's simpler than that. (warning - it's Friday morning and the
>caffeine hasn't really taken effect yet)
>
>Consider any number in terms of its factors. By definition, factors
>come in pairs:
>
>the factors of 15 are 1 x 15, 3 x 5
>the factors of 17 are 1 x 17
>the factors of 18 are 1 x 18, 2 x 9, 3 x 6
>
>The *only* exceptions to this are perfect squares, unless you want to
>count the same factor twice:
>
>the factors of 16 are 1 x 16, 2 x 8, and 4
>
>Therefore, every switch will be toggled an even number of times and
>turned off except for the perfect squares.
>
>--
>-Mark Wieder
>  mwieder at ahsoftware.net
>


Yes, Mark. But I think that is essentially Charles solution. It is 
certainly simpler. (But not as rich, IMHO :))

For some reason I didn't get Issue 99 of the Run Rev list. I presume 
that is where my alternate approach appears, and what you are 
responding to.


Jim