Path: ...!news.mixmin.net!weretis.net!feeder8.news.weretis.net!reader5.news.weretis.net!news.solani.org!.POSTED!not-for-mail
From: Physfitfreak <Physfitfreak@gmail.com>
Newsgroups: comp.os.linux.advocacy
Subject: Re: Prog Challenge
Date: Fri, 8 Mar 2024 13:06:57 -0600
Message-ID: <usfnkh$13n9o$1@solani.org>
References: <17ba8df1ac0783cb$104093$1054558$802601b3@news.usenetexpress.com>
 <use0es$12qgm$1@solani.org>
 <17bacfe9ea8cf1b9$13872$3298354$802601b3@news.usenetexpress.com>
 <17bad1c8e19520b7$8$821545$802601b3@news.usenetexpress.com>
 <17bad47b58d27fdc$49069$1507308$802601b3@news.usenetexpress.com>
 <17bad5a0f333db18$120726$3000918$802601b3@news.usenetexpress.com>
 <usfk5n$13lc3$1@solani.org>
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
Injection-Date: Fri, 8 Mar 2024 19:06:57 -0000 (UTC)
Injection-Info: solani.org;
	logging-data="1170744"; mail-complaints-to="abuse@news.solani.org"
User-Agent: Mozilla Thunderbird
Cancel-Lock: sha1:OWOPLpn452G1Ei+ICod7xleQy8g=
Content-Language: en-US
X-Antivirus: Avast (VPS 240308-0, 3/7/2024), Outbound message
X-Antivirus-Status: Clean
X-User-ID: eJwFwYEBwDAEBMCVqH9iHCL2H6F3NFe/AaeDy83UqKtd20WeHYWgU6beGp5zb0Tkt++DKaIpkzVWp1v89A9mxhXz
In-Reply-To: <usfk5n$13lc3$1@solani.org>
Bytes: 5990
Lines: 136

On 3/8/2024 12:07 PM, Physfitfreak wrote:
> On 3/8/2024 10:04 AM, Nuxxie wrote:
>> On Fri, 08 Mar 2024 15:43:54 +0000, Nuxxie wrote:
>>
>>>
>>> 1/37252902384619140625
>>>
>>>
>>> The decimal expansion will begin with 11 digits followed by a repeating
>>> porion of length 54,494,498,496.
>>>
>>> This number would need 64 GiB of memory, but we could write it out to a
>>> file using Maxima or bc or GMP.
>>>
>>
>> Someone please write out this number to a file.  As 1 byte per character
>> it should be about 55 GiB in size.
>>
>> Then verify that the decimals start with 11 digits and then repeat
>> after 54,494,498,496 + 11 digits.  Just verify that the first 100 or
>> so digits are the same.
>>
>> Y'all have big, fat mouths.  Let's see if that can translate into
>> serious computing action.
>>
>> The first to do it will receive a BIG PRIZE.
>>
> 
> 
> Again, you're turning the spoon twice around your head before placing it 
> in your mouth to eat your soup.
> 
> If you're after _proving_, mathematically, that a method will give the 
> repeating decimals, that's another matter. And I'm not good at that and 
> am not interested in stuff and degree of precise logic that 
> mathematicians are accustomed to and apply in their proofs. My 
> background is not math. That's why.
> 
> But if you're only interested in the answer, then method of nines is the 
> simplest and fastest way of getting it.
> 
> If you haven't seen it, then what kind of numerical methods class did 
> you attend? You must be way younger than me. I'm thinking the generation 
> after me with access to much better computers didn't need all that nitty 
> gritty to get by, and therefore profs simply omitted a bulk of such 
> material from the courses.
> 
> With an almost two lines of program, one can find a number consisting 
> only of digits of 9, for which the denominator of the fraction is a 
> divisor. Then with one more line of code, the fraction is rewritten with 
> that denominator consisting of 9's. The numerator thus obtained will be 
> the decimals that get repeated. As simple as that.
> 
> How many lines of code is that? 3 ? At most 4.
> 
> Example. Say we want to find the repeating decimals of 14/33 without 
> actually dividing the two numbers to visually find that out, but by 
> knowing in advance what those repeating decimals will be.
> 
> 1- find the smallest number consisting of only digits of 9 that 33 is a 
> divisor for:
> 
> 33 x 3 = 99
> 
> 2. apply this factor of 3 to the fraction:
> 
> 14/33 x 3/3 = 42/99
> 
> 3. therefore the repeating decimals are 42.
> 
> That's how you have your soup :)
> 
> The cases involving other numbers appearing in the decimals before 
> repeating decimals begin are quite similar. You first convert the 
> fraction to different pieces to separate the repeating part of decimals 
> from the rest. This turns the term with repeating decimals to have a 
> bunch of zeros coming before the repeating part begins. Then you use a 
> slightly different form of method of nines (called method of nines with 
> zeros) to find the repeating decimals.
> 
> 
> 
> 

Method of nines have some more detail than just that. The example and 
method I mentioned above applies only to cases where numerator and 
denominator are of the same order of magnitude. If denominator is of a 
magnitude of order n more than the numerator, then n zeros appear in the 
repeating decimals before nonzeros in them begin. Example:

1/13 (i.e. there will be one repeating zero before non-zero repeating 
decimals begin)

1. find the nines: (a two liner loop)

13 x 76923 = 999999

2. apply the factor:

1/13 x 76923/76923 = 76923/999999

3. therefore the repeating decimals are 076923 (not 76923).

Now let's take the example you brought earlier:

1/31

1. There will be one repeating zero before non-zero digits of repeating 
decimals begin.

2. get the nines:

31 x 32258064516129 = 999999999999999

3. apply the factor:

1/31 x 32258064516129/32258064516129 = 32258064516129/999999999999999

4. therefore the repeating decimals will be 032258064516129


Note that the zeros coming before non-zero digits begin, are not a case 
of zeros after radix that are _not_ part of the repeating decimals. The 
latter cases are handled by method of nines with zeros which is slightly 
different from method of nines, and I won't go into it. Those interested 
can dig it themselves.








-- 
This email has been checked for viruses by Avast antivirus software.
www.avast.com