I need a way to find factorail for big numbers like 8479!

Thanks in advance

raju

A simple way.
Just multiply them out.

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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran

See Stirlings formula (equations 13 and 14 here)

RonL

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Giordano Bruno

That is only an approximation.

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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran

How many digits does 8479! have?

(Rhetorical question)

RonL

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Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno

Over 29,000.

25972686487041052159869258722936182200373790277711 48344797668281885988268302193747006721558615751528[...29429 digits...]00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000

I can find how many zeros is has .

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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran

i had this problems some weeks back. Stirlings function is probably the best for large factorials. But otherwise use this bit of code i wrote in java which is computationally efficient.


public double gammaln(float xx){
double x,y,tmp,ser;
double[] cof={76.18009172947146,-86.50532032941677,
24.01409824083091,-1.231739572450155,
0.1208650973866179e-2,-0.5395239384953e-5};
int j;
y=x=xx;
tmp=x+5.5;
tmp -= (x+0.5)*Math.log(tmp);
ser=1.000000000190015;
for (j=0;j <= 5;j++) {ser += cof[j]/++y;}
return -tmp+ Math.log(2.5066282746310005*ser/x);}


public double factorial(float n){
double value = gammaln(n + 1);
double answer = Math.exp(value);
return answer;}

This is based around a method of gamma approximation by C.lanczos. (Paper is called - (A precision approximation of the gamma function).

This method can also be used to find NON integer factorials

I think ur asking this for a question like "find the lowest common multiple" (LCM)

there is no need to use massive formulas unless the number is greater then 1trillion.

just use Prime factorisation.

eg. 4 = 2 power of 2
6 = 3 x 2
1001 = 7 x 11 x 13
etc...

take the number and find all possible prime numbers which go ito it. (see examples above)
then...

eg. find the lowest common multiple of 4, 2, 6
4 = 2 squared 2 = 2 and 6 = 2x3
therfore the number must contain 2 squared and 3.
= 2 squared x 3
= 12

find all prime factors of the numbers. and take the common ones out
eg. 2 squared and 2
only use 2 squared because 2 is within 2 squared

then multiply them together.


(i can go on but it would just be useless)

How does this relate to finding the factorial of a large number?

RonL

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Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno

it gives the base of a number in its prime factors,

then u can use all of theose prime numbers in different combinations to find the factors

This is incomprehensible, please give us an example

RonL

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Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno

hang on maybe some confusion. spanner were finding a method of carrying out the factorial function not the LCM. I dont see how in anyway ur method can find the factorials of large numbers.

I think that is what I was trying to indicate, indirectly so as not to cause offence .

RonL

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Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno