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An old puzzle, extended precision - and logs - sa-penguin - 07-28-2016 09:26 AM I found an old [1907] puzzle book, Canterbury Puzzles One of the puzzles involved a cask holding 100 cups of wine. Every night, for 30 nights, a young monk steals a cup - and tops up the cask with a cup of water. The puzzle is to determine exactly how much wine the monk stole over those 30 nights. 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. I can see that: log (0.99)^30 = 30 * log (0.99) 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. Is there a "lost art" for using log tables, to get the extended precision required? Or is there some other secret? RE: An old puzzle, extended precision - and logs - Dieter - 07-28-2016 12:16 PM (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. 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: 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 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 RE: An old puzzle, extended precision - and logs - sa-penguin - 07-29-2016 11:48 AM (07-28-2016 12:16 PM)Dieter Wrote: I don't think this is a secret, but here is how you can do it:Very elegant! A re-reading of the answer mentions a "simplified process of calculation". Probably the same as you have shown. Quote:and after the thirtieth theft there would remain in the cask the thirtieth power of 99 divided by the twenty-ninth power of 100. RE: An old puzzle, extended precision - and logs - Dieter - 07-30-2016 09:27 AM (07-28-2016 09:26 AM)sa-penguin Wrote: log (0.99)^30 = 30 * log (0.99) I just noticed that of course this has to read log(0,99^30) = 30 · log(0,99) Dieter RE: An old puzzle, extended precision - and logs - ColinJDenman - 07-30-2016 04:10 PM Just as an aside to the discussion, my father, a mariner had a large volume of 10-figure tables of all kinds, for navigation and other nautical purposes. Sadly, I lost them in a house move some years ago. Presumably larger precision tables would have been more commonplace in the pre-digital era. RE: An old puzzle, extended precision - and logs - Dieter - 07-30-2016 07:56 PM (07-30-2016 04:10 PM)ColinJDenman Wrote: Just as an aside to the discussion, my father, a mariner had a large volume of 10-figure tables of all kinds, for navigation and other nautical purposes. Sadly, I lost them in a house move some years ago. Presumably larger precision tables would have been more commonplace in the pre-digital era. The well known "Abramowitz & Stegun" (Handbook of Mathematical Functions) includes 10-digit tables of logs, powers and roots. The ln table has 15 digits, e^x has 18 and sin/cos even 23 digits. At least if you have a printed version or a PDF where the tables have not been removed. ;-) Dieter RE: An old puzzle, extended precision - and logs - ColinJDenman - 07-31-2016 02:25 PM (07-30-2016 07:56 PM)Dieter Wrote: At least if you have a printed version or a PDF where the tables have not been removed. ;-) Thank you for the suggestion -- another wasted afternoon guaranteed ;-)) Purely for old times sake. RE: An old puzzle, extended precision - and logs - StephenG1CMZ - 08-07-2016 08:12 AM (07-29-2016 11:48 AM)sa-penguin Wrote:(07-28-2016 12:16 PM)Dieter Wrote: I don't think this is a secret, but here is how you can do it:Very elegant! Earlier the problem referred to cups, but here we have " 26.03 pints". Is that a typo or is the size of a cup in pints known? I am guessing they are pint-sized cups. RE: An old puzzle, extended precision - and logs - sa-penguin - 08-07-2016 08:14 AM I think he was stealing with a cup that held one pint. RE: An old puzzle, extended precision - and logs - Thomas Klemm - 07-01-2022 05:28 AM (07-29-2016 11:48 AM)sa-penguin Wrote:Quote:By a simplified process of calculation, I have ascertained that the exact quantity of wine stolen would be: We can use the following program with Free42 Decimal: Code: 00 { 3-Byte Prgm } 30 ENTER -.01 R/S -2.602996266117195772699849076832851e-1 RE: An old puzzle, extended precision - and logs - John Keith - 07-01-2022 11:37 AM (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. Such as the HP 50g in exact mode: Code:
RE: An old puzzle, extended precision - and logs - JoJo1973 - 07-01-2022 01:34 PM Not to mention newRPL, of course: Code: « |