1 million factorial

ril https://www.quora.com/What-is-an-efficient-algorithm-to-find-the-factorial-of-huge-numbers-which-lie-in-the-range-100000-1000000

just by using bc

rewrote a factorial function not to use recursive calling.

define g(x) {
    answer=1;
    for(i=1;i<x+1;i++) {
        answer *= i;
    }
    return answer;
}

g(1000000);

real 578m28.905s
user 450m56.192s
sys 12m33.006s
(with poor computing power (centOS virtual machine on windows7. i5 processor))

5,565,709 digits starting

82639316883312400623766461031726662911353479789638730451677758855633796110356450844465305113114639733516068042108785885414647469506478361823012109754232995901156417462491737988838926919341417654578323931987280247219893964365444552161533920583519938798941774206240841593987701818807223169252057737128436859815222389311521255279546829742282164292748493887784712443572285950934362117645254493052265841197629905619012120241419002534128319433065076207004051595915117186613844750900755834037427137686877042093751023502633401248341314910217684549431273636399066971952961345733318557782792616690299056202054369409707066647851950401003675381978549679950259346666425613978573559764142083506…

ending with 249,998 zeros(about 4.49% of the whole digits)

more easy way to get the number of trailing zeros(from here)

Every multiple of 5 contributes a zero.
Every multiple of 25 contributes a second zero
Every multiple of 125 contributes a third zero
Every multiple of 625 contributes a fourth zero
etc.

floor(1000000/5) + floor(1000000/25) + ... floor(1000000/5^8)
= 200,000 + 40,000 + 8,000 +1,600 +320 + 64 +12 + 2
= 249,998 trailing zeros

someone can get this in about only 10sec!! here
He's saying he used FFT

ref

http://www.luschny.de/math/factorial/FastFactorialFunctions.htm

http://www.luschny.de/math/factorial/csharp/FactorialPrimeSwing.cs.html

http://answers.google.com/answers/threadview/id/509662.html

http://en.wikipedia.org/wiki/Elliptic_curve_factorization