What is the number of digits (IntegerLength) of the factorial of 12 345 678 987 654 321? The number of zeros at the end of this factorial was calculated and it is huge: exactly 3 086 419 746 913 569 zeros. The calculation of the number of digits for this factorial if done by simple multiplication, will take about 199 years in my current machine which has a I7 with 6 cores and 64GB of memory. Yes, I know, this number will not fit in my memory, even though I have a SSD with 1TB extra memory which will also not be able to store this gigantic number. So, any ideas?

PS: the program to calculate the number of zeros at the end of a factorial is:

Fnz[n_] := Module[{f, s, p, z},
  p = 5;
  f = IntegerDigits[n, 5];
  s = Total[f];
  z = (n - s)/(p - 1);
  Print[n, "! has ", z, " zeros at the end."];

Here the result:


12345678987654321! has 3 086 419 746 913 569 zeros at the end.

  • 2
    $\begingroup$ Thanks for the question. One of my pet peeves, however, is people describing their problems as "hard", when what they mean is "I can't work out how to do this" - which is implied, because you posted the question here! :) $\endgroup$ Commented Dec 30, 2015 at 18:12
  • $\begingroup$ Patrick, factoring a large number n (with say 300 digits) is also a hard math problem even though everybody knows how to do it. Just try every prime until you reach the Sqrt[n], right? Same thing here. You see, I could calculate the number of zeros at the end of the number WITHOUT having it in my hand to count the zeros. That´s is what I´m talking about. $\endgroup$
    – Giorgio
    Commented Dec 30, 2015 at 18:31
  • 1
    $\begingroup$ Factoring is in general difficult, but factoring a factorial is not :) $\endgroup$ Commented Dec 30, 2015 at 18:53
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    $\begingroup$ @Giorgio One possibility: buy 3.10^4 - 1 additional 1To SSD drives. More seriously, did you check Chip Hurst's answer (which gives another result)? $\endgroup$
    – anderstood
    Commented Dec 30, 2015 at 19:19
  • 1
    $\begingroup$ What do you mean by "program"? You can always pack the code in my answer into it's own function. Something like Ndf[x_] := Ceiling[LogGamma[N[x + 1]]/Log[10]], and now Ndf[12345678987654321] outputs 193299018111544064. $\endgroup$
    – Greg Hurst
    Commented Dec 30, 2015 at 19:47

2 Answers 2


We can exploit the built in LogGamma:

x = 12345678987654321;
Ceiling[LogGamma[N[x + 1]]/Log[10]]

Edit, Addressing precision:

We have naively for $n > 1$, that $n! < n^n$. Taking logs of both sides gives the (not very tight) bound $\log\Gamma(x + 1) < x \log(x)$ for $x > 1$.

This means if we want the number of digits of $n!$ for some large $n$, asking for $\log_{10}(n \log(n))$ number of digits in $\log\Gamma$ will be more than enough.

Here it is in a function.

Ndf[x_] := 
  With[{p = 20 + Max[1, Ceiling[Log10[x Log[x]]]]},
    Ceiling[LogGamma[N[x + 1, p]]/Log[10]]

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    $\begingroup$ Correct, but at the edge of what can be done with machine precision. $\endgroup$ Commented Dec 30, 2015 at 19:23
  • $\begingroup$ Yep, I suppose that is worth mentioning. $\endgroup$
    – Greg Hurst
    Commented Dec 30, 2015 at 19:24
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    $\begingroup$ 12345678987654321+1 can just barely be represented exactly as an IEEE double-precision float, but only because it's even! If it were odd, that last 1 bit would be lost or rounded up. There is not even the potential of a loss of precision if you just do Ceiling[LogGamma[12345678987654321 + 1]/Log[10]]. $\endgroup$
    – Mark Adler
    Commented Dec 30, 2015 at 22:28
  • $\begingroup$ @MarkAdler That's a good point about using exact inputs, the only issue is that there is a gray area where Mathematica will try to auto evaluate LogGamma, but it would require more memory than most computers have. For example In[2]:= MemoryConstrained[LogGamma[10^10], 10^8] Out[2]= $Aborted. It looks like it stops auto evaluating around 10^14. $\endgroup$
    – Greg Hurst
    Commented Dec 31, 2015 at 0:17
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    $\begingroup$ I don't know what the gray area is, but this doesn't fall into it. With the Ceiling, Mathematica uses less than 4K of memory to do the exact calculation. $\endgroup$
    – Mark Adler
    Commented Dec 31, 2015 at 0:22

I wondered if Chip's answer was exactly correct, given Daniel's comment about machine precision. So I did it a little differently with higher precision in a way that gives good confidence in the result. (It turns out that the machine precision answer is correct.)

LogGamma can be expanded around infinity thusly:

Series[LogGamma[z], {z, Infinity, 4}]

to give:

$$z \left(-\log \left(\frac{1}{z}\right)-1\right)+\frac{1}{2} \left(\log \left(\frac{1}{z}\right)+\log (2 \pi )\right)+\frac{1}{12 z}-\frac{1}{360 z^3}+O\left(\left(\frac{1}{z}\right)^5\right)$$

For large $z$ that should converge very rapidly. In fact, the $1\over z$ terms can be dropped, since $O\left(1\over z\right)$ will be way less than one digit. For really big z, like this one, this is a sufficient approximation for LogGamma to count digits (unless you happen to be right on the hairy edge of an integer):

$$\left(z-{1\over 2}\right)\log z-z+{1\over 2}\log\left(2\pi\right)$$

(Simplified a little.) You can add the $1\over 12z$ if you like, but it won't make any difference here.

Evaluating that for this case, I indeed get:

(z - 1/2) Log[z] - z + 1/2 Log[2 \[Pi]] /. 
  z -> 12345678987654321 + 1 // Ceiling[#/Log[10]] &

Looking at the digits after the decimal:

(z - 1/2) Log[z] - z + 1/2 Log[2 \[Pi]] /. 
  z -> 12345678987654321 + 1 // N[#/Log[10], 20] &

So it's not on the hairy edge since the ...63.73 was rounded up to ...64, and in fact adding the $1\over z$ terms don't change the result.

Interestingly, that formula is not the log of Stirling's formula, and the log of Stirling's formula does not give the correct answer. (It's low by eight digits.)

  • $\begingroup$ Since the number of digits in $n!$ is bounded above by $2n\log_{10}(n)$, we could always do Ceiling[LogGamma[N[x + 1, Ceiling[2x Log10[x]]]]/Log[10]] and be assured that our answer is correct. (I may edit my post with something similar later today.) $\endgroup$
    – Greg Hurst
    Commented Dec 30, 2015 at 21:47
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    $\begingroup$ If you use the 2nd version of Stirling's approximation at en.wikipedia.org/wiki/Stirling%27s_approximation which has corrections of order $\ln(n)$, n! ~ Sqrt[2*Pi*n] (n/E)^n, it comes out right, though: Ceiling@N[(Log[Sqrt[2 \[Pi] n]] + n (Log[n] - 1))/Log[10] /. n -> 12345678987654321, 20] yields 193299018111544064. $\endgroup$
    – evanb
    Commented Dec 30, 2015 at 22:27

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