RE: [MLUG - DISCUSSION] beginner slide rule wanted

[Mike Miller <EMAIL:PROTECTED>]

On Wed, 2 Jul 2003, Jonathan King wrote:

> On Wed, 2 Jul 2003, Mike Miller wrote:
>
> > On Wed, 2 Jul 2003, Mike Miller wrote:
> >
> > > > So every year I walk students through the Tower of Hanoi Problem with a
> > > > small number of disks, establish that the number of moves it takes for
> > > > the n disk problem is 2^n - 1, and then ask for people to guess how long
> > > > the 64 disk problem takes if you can do 1 move per second.
> > >
> > > I do it as (2^10)^6 * 2^4 which more than 16*(10^3)^6 = 1.6*10^19 seconds.
> > > But how long is that?  There are 3600 seconds per hour and 24 hours per
> > > day so there are about 8.6*10^5 seconds per day and about 3.7*8.6*10^7 or
> > > 3.2*10^8 seconds per year.  So that means (1.6/3.2)*10^(19-8) = 5 x 10^10
> > > days to solve.  That is 50 billion.
> >
> > And I screwed up at the end.  Meant to say 5 x 10^10 *years* to solve, or
> > 50 billion years.
>
> Alas, that wasn't your only error.  3600*24 < 10^5.

Right, I should have had 8.6*10^4, and that screwed me up.


> Repeat after me:  there are pi*10^7 seconds in a year, approximately.

That's a good one.


> > Doing it with a computer, I get 58.4 billion years, which doesn't
> > bother me because I knew I was low and I was shooting mostly for the
> > order of magnitude.
>
> Well, it bothers me since your computer has slipped a place. :-)

Here's what really happened:  I did it in Octave with "format long" (don't
ask me why, because it was a bad idea).  Then I misread the very long
number.

So, this problem really involves two numerical tricks:  knowing that 2^10
is about 10^3, which everyone should know! (but they don't) and knowing
that there are about 3.15 x 10^7 seconds in a year, which is a good thing
to know, but also not known to many.  So I can see why your students don't
get it.

Now, soon, anyone with access to Google will be able to see how badly I
performed on your test.  I hope they won't fire me!

Mike
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