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1000001!
By Anonymous on Thursday, March 25, 1999 - 04:37 pm:

I am a college student and I need some help!

I need to know what the Factorial of 1000001 (one million one) is. I have been working on this for some time now, but my math teacher still is not staisfied with my answer. My answer is:
1000001! = 8.263934135437012 E + 5565714

My teacher is wanting me to find out the exact digits of the mantissa (8263934135437012...) as well as where the zeroes begin and how many zeroes are included in the answer.

Any help would be greatly appreciated.

By Chris Jefferson (Caj30) on Friday, April 16, 1999 - 09:25 am:

Here is a challenge if anyone out there has computer skills. Would anyone like to start by exactly working out 1001!

By Alex Barnard (Agb21) on Wednesday, May 5, 1999 - 01:33 pm:



1001! is:

40278964733717086731724613635692698970509423907492
53471763437103403684509110276496126362526954563742
05280468598807393254690298539867803367460225153499
61453558842192859116083367874245135491592125229928
54569462713969958504379595406450196963727411427873
47450281325324373824456300226871609431497826989489
10952272579169116794569850928242153863296652337667
98918236969009820752231882794651940654891114985865
22997573307838057934994706212934291477882221464914
05874580817979513001896917560573982423724768451279
01696480137781586615203849163572855472196603375040
67910087936301580874662367543921288988208261944834
17836916980568248942050403833452938917784508967954
60750233058540061412562886338200799403953292515637
88399404652902154519302928365169452383531030755684
57850385148815409232357615031156932589119010592611
87616071002868279304729449132724208250789121587415
89850136017030887975452922434889688775883386977825
21590442368247894331380607214409743241869580741257
12923087398024810894070025239550801481840628104475
64594783139830113821372260474145316521647368313934
67078385848278150691528837894134807868969181565778
53058969122779932006398586962941995491077386355995
38328374931258525869323348477334798827676297868823
69302337741894230427226780050976580543565378753037
01182612199947525888664510727155837854953946845245
93296728611334955079882857173250037068541860372512
69317081925930941102783717661244469264917453642974
54210862877085881300821687927506971589017371302217
51430550976429258055277255676893874108456870904122
90225941722470713772340612581154995215962976677106
30794726792802138829785237854247603096781382687082
39764925768714349554665438389311198715040908077757
08690015938971244398767024424178790458509301154686
15020585500909148779008527016196482293321924010757
47543562989953271508977501771085759521631427816116
19176103125745449703967341424814921083600249711410
75659604585765252125561596349757155526386781721374
68172843066451093984443636560722213668172225585711
56655813446739265418546022258972331209759998725341
78314739395650710063443525180965644277812042000683
23913056897090916602712260306869786107237077572445
86657294576097772163940833843000997602897053915082
23365538566139627478146217470923489969157559834647
41082000337526945990059365493439921937093368896754
79141675960432489551466032591315784379603991781961
37173503809977812254720000000000000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000
000000000000000000000

By Chris Jefferson (Caj30) on Wednesday, May 5, 1999 - 02:30 pm:

Impressive!
However, before we've really started, 1000001! is going to have: 200000+40000+8000+1600+320+64+12+2=249998 zeroes. On a side note, 1001! had 249 zeroes. Can anyone see a pattern forming here??

It is the way you work them out. You work out the number of zeros of a! by doing [a!/5]+[a!/52]+[a!/53].... where [b] is the next integer below b. This is nearly a geometric series, first term a!/5, common ratio 1/5, so the sum to infinity will be (a!/5)/(1-1/5)=a!/4

Anyway, to work out 1000001!
It will vastly underestimate the number, but let's consider that it is in fact a thousand 1's, a thousand 1000's, a thousand 2000's,.... a thousand 9000's
So this underestimation puts 1000001! at:
(10001000(20001000)(30001000) .... (90001000)
=(21000*31000*41000...91000*1000 9000
=9!1000*10009000
is approx = 5.8*105559+9000*3
= 5.8*1032559
Remember, this is probably too small......
So, it's quite big then!

By Chris Jefferson (Caj30) on Wednesday, May 5, 1999 - 02:37 pm:

Oh no... messed that up in the middle just then, a quick check tells me that my friend way above me is correct. 8.263934135437012 * 105565714 is a good guess at the value. Whoops!

So, if you intend to work it out, you'll need a fair chuck of computer space to do it. It could be done though. Would anyone like to give it a try? A few people, some of whom can program would probably be best to do it. If you fancy a try, sign up below, along with your idea of how to.

Heres some ideas to get started:

Every one takes a group of numbers, say 2000-3000, multiplies them together and then someone multiplies these big numbers together.

Try instead to prime factorise 1000001!, then maybe multiply these together.

By Panna Lal Patodia (M152) on Wednesday, November 3, 1999 - 10:05 pm:

Here is a simple method to compute prime factors of any factorial:
"Say, you want to compute factorial of a number x. Take any prime number
(say p). Now, compute: m = [x/p]+[x/p^2]+.... till you encounter 1. Now,
note that x^p is the factor of x!. If you take p = 2,3,... all the prime
number till x, you got all the prime factors of x!."

Using above method, I have written a program in gp-pari which is as
follows:
{
fact(n,s=2,t=0,r=0,m=0,a="")=
if(t==0,t=n); forprime(i=s,t,m=0;r=n; while(r>1, r=floor(r/i);m+=r);
a = concat(concat(concat(concat(a,i),"^"),m)," * ")); return(a);
}

Using above program, I have computed all the prime factors between 2
to 100 of 1000001! which is as follows:

//Progam
? n=1000001;fact(n,2,100)

// Result
2^999993 * 3^499993 * 5^249998 * 7^166664 * 11^99998 * 13^83332 * 17^62497 *
19^55553 * 23^45453 * 29^35713 * 31^33332 * 37^27776 * 41^24998 * 43^23807 *
47^21737 * 53^19228 * 59^17240 * 61^16665 * 67^15150 * 71^14284 * 73^13887 *
79^12820 * 83^12194 * 89^11362 * 97^10416 * .....

If you are very particular to know the exact value, here is the result with
first 1001 significant digits (correct to its last digit):
8.26393995226292839361670847981876946380163911431185200904082105333926517441525
6120091614957841975284815580155817682799420053288415394786830137393276634275013
4152318618909200480576915758268336995995978510311212234500141184259408917606086
0821174405223188805731480150478348292958065090419764752269891861739882520822045
3391056680080210928911190657491278663627860022828472952322031305200737213830675
8893463471103248917739253539243953130853561384509272080258599966713101731031364
7455067347932611751140147289707931172536569038817425632515325947671158230676727
0337101993329869466429110135039948291574650111057146407647635491859825295407638
5653931658499939296925772280404187538337701847648337957138784584524734837773415
1939123159416137126620613403873847930720068737763777366621020120244273357856700
5948126505601702466398103746662848074658572433292044003318668892085418784806958
5881605307826023218733938263749478074557419645456861406325023218594325692286525
482571177244527877416592327733540922157910111909241968 * 10^5565714

The above result got by using the following formulas in gp-pari:
(Note that first line sets the precision to 1015 significant digits)
? \p 1015
? m=lngamma(1000002)/log(10)
? 10^frac(m)

To know where where the zeroes begin and how many zeroes are included
in the answer, we can use the following method:

We know 10 = 2*5. From the table given at the top, we find that
1000001! has 2^999993 and 5^249998 as its factors, obviously, the
number of zeros in the answer are 249,998 and it starts from
5,565,714 - 249,998 + 1 = 5,515,717th digit.

Good bye!
by P L Patodia, Bangalore, India.