Hi David. That's a great technique. Thanks.

If B1:B365 contains =ROW()...

Perhaps we could eliminate the B1:B365 cells with this Array formula...

=PRODUCT(ROW(INDIRECT("A1:A365"))/100)

Another variation of that same equation is:

=EXP(GAMMALN(366)-365*LN(100))

but it's only good for a few digits, as GammaLn is not very accurate.

Thanks. :)
Dana DeLouis


David Biddulph wrote:
You can use Excel to get an rough (15 sig fig) approximation to FACT(365) by
trimming down by a couple of orders of magnitude in each row. If B1:B365
contains =ROW(), you can use an array formula =PRODUCT(B1:B365/100) to get
2.51041E+48 or as number with no decimal places to see that the significant
figures are 251041286755588, and then the number would need to be multiplied
by 10^(2*365), so 1E+730 would change the exponent term from E+48 to E+778
(as Dana gave below).
--
David Biddulph


"Dana DeLouis" wrote in message
...
Jitka wrote:
Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks
Hi. As a side note, a math program indicates it is 779 digits long.
Here's an approximate value:

2.510412867555873*10^778

As others have mentioned, it's too large for excel.

(If it helps, the Natural Log of 365! is...)

Log(365.!) = 1792.3316495780505
- - -
Dana DeLouis