Talk:Distribution of 0 Digits in factorial series
Can you please explain the following example?
Example: for 1 to 6 we have 1!, 2!, 3!, 4!, 5!, 6!, or (1, 2, 3, 24, 120, 720), so (2 zero digits) / (8 total digits) = 0.25.
Where does the number 8 come from?
number of trailing zeros for factorials
In the Rosetta Code task preamble, it states:
- Because factorial numbers, which quickly become quite large, continue to have another terminal 0 on the right hand side of the number for every 5 increase in the factorial base number, one might think ...
This is factually true, but understated. In fact, another trailing zero is added for every increase in five, and for every power of five as well.
- for every increase in 5, an additional trailing zero is added.
- for every increase in 52, an additional trailing zero is added.
- for every increase in 53, an additional trailing zero is added.
- for every increase in 54, an additional trailing zero is added.
- for every increase in 55, an additional trailing zero is added.
/*REXX program computes the number of trailing zeros in the factorial for any given N. */
numeric digits 100 /*just in case the user gets ka─razzzy.*/
parse arg N . /*obtain an optional argument from C.L.*/
if N=='' | N=="," then N= 1000000 /*Not specified? Then use the default.*/
z= 0 /*number of trailing zeroes (so far). */
v= 5 /*initialize V to five. */
do while v<=N /*calculate number of trailing zeroes. */
z= z + N % v /*bump Z if multiple power of five.*/
v= v * 5 /*calculate the next power of five. */
end /*while*/ /* [↑] % is REXX's integer division.*/
if z==0 then z= 'no' /*use gooder English for the message. */
say commas(N)'! has ' commas(z) " trailing zeroes."
exit z /*stick a fork in it, we're all done. */
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
Here is the output for the number of trailing zeros in the factorial of one million.
- output when using the input of: 10000000
1,000,000! has 249,998 trailing zeroes.
True. So the number of terminal zeros actually goes up as the summation over [k from 1 to rounded up(log base 5 of n)] of (n / 5^k), ie n/5 + n/25 + n/125 .... Interesting, though the nonterminal part of n! still dominates the result eventually. --Wherrera (talk) 18:27, 10 June 2021 (UTC)