Argh!

I was caught up in binary thinking.

-dtd

> -----Original Message-----
> From: Barry Lake [mailto:[log in to unmask]]
> Sent: Monday, December 04, 2000 9:20 AM
> To: Dave Darnell; [log in to unmask]
> Subject: Re: Math Help
>
>
> At 8:12 AM -0700 12/4/00, Dave Darnell wrote:
> >?? I'm getting non-powers of the base:
> >
> >ASTRO/DARNELL/PUB(1):comeau 127,8
> >127 = 7+56+64
> >Written as 177 in base 8
> >ASTRO/DARNELL/PUB(1):calc 8*8*8
> >512, $200, %1000
> >ASTRO/DARNELL/PUB(1):calc 8*8
> >64, $40, %100
> >ASTRO/DARNELL/PUB(1):calc 8*7
> >56, $38, %70
> >ASTRO/DARNELL/PUB(1):
> >
> >56 is not a power of 8
>
> But 56 is a multiple of a power of 8.
>
> Try running the script with the third parameter provided:
>
>   :powersum 127,8,detail
>   127 = (7*1)+(7*8)+64
>
> Base 2 is a special case. At any level or power you can only
> have either
> zero or one multiple of that power added into the sum. If you
> have two,
> then you've moved up to the next power.
>
> For example, 3 would be "1+2" and 5 might be "1+(2*2)". But
> since 2*2 is a
> power of two (as well as a multiple of a power of two), we have to say
> "1+4". In the case of base 8, or any other base higher than
> two, it is not
> possible represent all numbers as sums of single multiples of
> the powers of
> the base. In other words, your misunderstanding is that you
> are thinking
> about base 8 as if it were base 2.
>
>
>