Number of Digits in n!: Coding a Program for 500! Factorial

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The discussion centers around calculating the number of digits in the factorial of 500 (500!) using programming techniques, specifically in C. One participant describes using a logarithmic approach to compute the number of digits, summing the logarithms of integers from 1 to n, and arrives at an answer of 1135 digits. However, this conflicts with another source that states the answer is 1133. The discrepancy is attributed to potential rounding errors in the program. Participants mention using Mathematica for accurate calculations due to its arbitrary-precision arithmetic capabilities, which allows for quick computation of large factorials. Other methods discussed include using Stirling's approximation and programming in languages like Smalltalk and Python, which support arbitrary-length integers. The conversation highlights the importance of precision in calculations and the effectiveness of different programming tools for handling large numbers.
Gagan A
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I have to code a program in C that will give the correct number of digits of n! where n is upto 500.
I thought this way:
Get the value of log(n!) by using a for loop. (like log1 + log2 + log3... upto logn, log is to the base 10). Now the final answer will be (int)sum + 1. If I give 500! factorial the answer comes out to be 1135 acc. to my program, but the answer given is 1133. Am I right?
 
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I also get 1135.
 
500! =

Code:
122013682599111006870123878542304692625357434280319284219241358838584537315388\
199760549644750220328186301361647714820358416337872207817720048078520515932928\
547790757193933060377296085908627042917454788242491272634430567017327076946106\
280231045264421887878946575477714986349436778103764427403382736539747138647787\
849543848959553753799042324106127132698432774571554630997720278101456108118837\
370953101635632443298702956389662891165897476957208792692887128178007026517450\
776841071962439039432253642260523494585012991857150124870696156814162535905669\
342381300885624924689156412677565448188650659384795177536089400574523894033579\
847636394490531306232374906644504882466507594673586207463792518420045936969298\
102226397195259719094521782333175693458150855233282076282002340262690789834245\
171200620771464097945611612762914595123722991334016955236385094288559201872743\
379517301458635757082835578015873543276888868012039988238470215146760544540766\
353598417443048012893831389688163948746965881750450692636533817505547812864000\
000000000000000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000

This has 1135 digits.

- Warren
 
I too verified with the teacher, the answer is 1135. But how in heaven did you calculate the value of 500!, chroot? What is the maximum precision of the calculator you used?
 
Gagan A said:
I too verified with the teacher, the answer is 1135. But how in heaven did you calculate the value of 500!, chroot? What is the maximum precision of the calculator you used?
I think he used Methematica. It can handle most mathematical operations quite well.
It can give you the evaluation of 500!, say, in less than 1 second. :rolleyes:
------------------
The reason you got the wrong answer may be the rounding errors of the programme.
 
Gagan A,

I did indeed use Mathematica (VietDao29 probably recognized the tell-tale "\" characters at the end of each line in my output).

It uses arbitrary-precision arithmetic, so the maximum precision is that afforded by the entire memory of my computer (2 GB on this one). 500! is computed, indeed, in the blink of an eye, and probably only uses a few kilobytes of memory.

- Warren
 
Yes Mathemathica really wotks great for calculations like this. But there are other good way's of doing it as well. I have in the past also did stuff like this with the Smalltalk programming language (it isn't very good for doing mathemathics but it does enable arbitrary-precision arithmetic) which was also qouit fast. I once also made a programm in C that could work with large integers (it used strings) but it wasn't veary fast.
 
It may be 'cheating' within the context of your course (if they want you to do this a certain way), but you might want to look up Stirling's series. You can calculate the log of gamma (and hence the factorial) to the necessary accuracy to find the number of digits without resorting to a summation.
 
Python also supports arbitrary length integers:
Code:
>>> def fac(a):
    if a==1: return a
    return a * fac(a-1)
>>> len( str( fac( 500)))
1135
 
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