Re: [math-fun] [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result
At 03:18 PM 1/27/2012, you wrote:
998 starts 3rd row from bottom, 2 digits from righthand end ...
that's 799 at the end of that row. thanks
On 1/27/12, Ray Tayek <rtayek@ca.rr.com> wrote:
At 11:34 AM 1/27/2012, you wrote:
SetAccuracy[1/998001, 4000] gives the same problem (no 998) for me.
i used java.
import java.math.*; public class Main { public static void main(String[] args) { BigDecimal n = BigDecimal.valueOf(998001); MathContext mc = new MathContext(3010); BigDecimal x = BigDecimal.ONE.divide(n, mc); String s = x.toString(); for (int i = 0; i < 10; i++) System.out.println(s.substring(2 + 300 * i, 2 + 300 * (i + 1))); } }
000001002003004005006007008009010011012013014015016017018019020021022023024025026027028029030031032033034035036037038039040041042043044045046047048049050051052053054055056057058059060061062063064065066067068069070071072073074075076077078079080081082083084085086087088089090091092093094095096097098099
100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199
200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299
300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399
400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499
500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599
600601602603604605606607608609610611612613614615616617618619620621622623624625626627628629630631632633634635636637638639640641642643644645646647648649650651652653654655656657658659660661662663664665666667668669670671672673674675676677678679680681682683684685686687688689690691692693694695696697698699
700701702703704705706707708709710711712713714715716717718719720721722723724725726727728729730731732733734735736737738739740741742743744745746747748749750751752753754755756757758759760761762763764765766767768769770771772773774775776777778779780781782783784785786787788789790791792793794795796797798799
800801802803804805806807808809810811812813814815816817818819820821822823824825826827828829830831832833834835836837838839840841842843844845846847848849850851852853854855856857858859860861862863864865866867868869870871872873874875876877878879880881882883884885886887888889890891892893894895896897898899
900901902903904905906907908909910911912913914915916917918919920921922923924925926927928929930931932933934935936937938939940941942943944945946947948949950951952953954955956957958959960961962963964965966967968969970971972973974975976977978979980981982983984985986987988989990991992993994995996997999000
--- co-chair http://ocjug.org/
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--- co-chair http://ocjug.org/
799 is three digits from the end. If you start two digits from the end, you get the 99 part of the 799 followed by the 8 part of the 800 in the next row. Voila 998! On 1/27/2012 8:31 PM, Ray Tayek wrote:
At 03:18 PM 1/27/2012, you wrote:
998 starts 3rd row from bottom, 2 digits from righthand end ...
that's 799 at the end of that row.
thanks
At 05:37 PM 1/27/2012, you wrote:
799 is three digits from the end. If you start two digits from the end, you get the 99 part of the 799 followed by the 8 part of the 800 in the next row. Voila 998!
yes. and all of the rest seem to be in sequence. i expected the last line to end in 997998999 instead of 997999000 thanks
On 1/27/2012 8:31 PM, Ray Tayek wrote:
At 03:18 PM 1/27/2012, you wrote:
998 starts 3rd row from bottom, 2 digits from righthand end ...
that's 799 at the end of that row.
thanks
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--- co-chair http://ocjug.org/
On Fri, Jan 27, 2012 at 8:06 PM, Ray Tayek <rtayek@ca.rr.com> wrote:
i expected the last line to end in 997998999 instead of 997999000
It does end in 997 998 999 but then the next "three-digit" number is 1000, and after carrying the 1 you can see why it should be 997 999 000. Why can't we invent some three-digit bins that can hold numbers with more than three digits? --Joshua Zucker
If we want 998, we can use 1/99980001 (we will obtain more exactly 0998). But this time 9998 will disappear for the same reason. 1/9801, 1/998001, 1/99980001, 1/9999800001, .... = no 98, no 998, no 9998, no 99998, ... Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Joshua Zucker Envoyé : samedi 28 janvier 2012 06:06 À : math-fun Objet : Re: [math-fun] [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result On Fri, Jan 27, 2012 at 8:06 PM, Ray Tayek <rtayek@ca.rr.com> wrote:
i expected the last line to end in 997998999 instead of 997999000
It does end in 997 998 999 but then the next "three-digit" number is 1000, and after carrying the 1 you can see why it should be 997 999 000. Why can't we invent some three-digit bins that can hold numbers with more than three digits? --Joshua Zucker _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
At 02:01 AM 1/28/2012, you wrote:
If we want 998, we can use 1/99980001 (we will obtain more exactly 0998). But this time 9998 will disappear for the same reason.
so the 998 really is not "in order" with the rest of the 3 digit numbers? thanks
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Joshua Zucker Envoyé : samedi 28 janvier 2012 06:06 À : math-fun Objet : Re: [math-fun] [EXTERNAL] Re: Fun with math: Dividing one by 998001 yields a surprising result
On Fri, Jan 27, 2012 at 8:06 PM, Ray Tayek <rtayek@ca.rr.com> wrote:
i expected the last line to end in 997998999 instead of 997999000
It does end in 997 998 999 but then the next "three-digit" number is 1000, and after carrying the 1 you can see why it should be 997 999 000. Why can't we invent some three-digit bins that can hold numbers with more than three digits? ...
--- co-chair http://ocjug.org/
Really, this is so simple. You are adding numbers in columns 000 001 002 ... 000001002... What happens around 1000? 997 998 999 1000 1001 1... 997999000001002... The carry from the 1000 term flips 999 to 1000, and the carry from that 1000 flips 998 to 999. The effect is that the 999 shifts 3 digits left and overwrites the 998, which is lost. On 1/28/2012 8:43 PM, Ray Tayek wrote:
At 02:01 AM 1/28/2012, you wrote:
If we want 998, we can use 1/99980001 (we will obtain more exactly 0998). But this time 9998 will disappear for the same reason.
so the 998 really is not "in order" with the rest of the 3 digit numbers?
thanks
At 05:10 AM 1/29/2012, you wrote:
Really, this is so simple.
You are adding numbers in columns
000 001 002 ... 000001002...
What happens around 1000?
997 998 999 1000 1001 1... 997999000001002...
The carry from the 1000 term flips 999 to 1000, and the carry from that 1000 flips 998 to 999. The effect is that the 999 shifts 3 digits left and overwrites the 998, which is lost.
i got it! thanks
On 1/28/2012 8:43 PM, Ray Tayek wrote:
At 02:01 AM 1/28/2012, you wrote:
If we want 998, we can use 1/99980001 (we will obtain more exactly 0998). But this time 9998 will disappear for the same reason.
so the 998 really is not "in order" with the rest of the 3 digit numbers? ...
--- co-chair http://ocjug.org/
Which made me curious... What is the smallest value a(d) such that every d-digit sequence occurs in the decimal expansion of 1/d (after the decimal point). For small d >= 1 my program says 17, 109, 1019, 10007, 100019, ... but it seems stuck on 6 digits, so perhaps there is an overflow issue. All primes slightly above 10^d, as one might expect, but not necessarily the smallest such primes.
participants (4)
-
Christian Boyer -
David Wilson -
Joshua Zucker -
Ray Tayek