An old puzzle, extended precision - and logs

[Dieter]

(07-28-2016 09:26 AM)sa-penguin Wrote:  This requires you to solve: 100 (1 - (99/100)^30) - which ends up having 60 decimal places.

While a quick Google search will find a arbitrary precision calculator, I was intrigued by the original 1907 answer: while giving the right value, it claimed the use of logarithms was required.

No, logs are not required. But they can be used to evaluate 0,99^30.

(07-28-2016 09:26 AM)sa-penguin Wrote:  I can see that:

log (0.99)^30 = 30 * log (0.99)

That's the way with using logs.

(07-28-2016 09:26 AM)sa-penguin Wrote:  I just can't see how to calculate logs (and anti-logs) to the required number of decimal places. At best, in my youth, I saw tables of logarithms to 5 decimal places.

Do you really want a 60-digit result? Is this the result printed in the book?
Then there's another approach that can even be used with pencil and paper.

(07-28-2016 09:26 AM)sa-penguin Wrote:  Is there a "lost art" for using log tables, to get the extended precision required? Or is there some other secret?

I don't think this is a secret, but here is how you can do it:

Code:

  (99/100)^30 
= 99^30 / 100^30
= 99^30 / 10^60

      99x = 100x - x

    99*99 = 9900 - 99         = 9801   (=99^2)
  9801*99 = 980100 - 9801     = 970299   (=99^3)
970299*99 = 97029900 - 970299 = 96059601   (=99^4)
      ...   ...

Do this until 99^30, then shift the decimal point 60 digits left to get 0,7397...
1 minus this is 0,2602996... so the answer (times 100) is 26 cups.

Dieter