user
Large Factorials and the Distribution of '0' in base 10 digits.

About the task

We can see that some features of factorial numbers (the series of numbers 1!, 2!, 3!, ...)
come about because such numbers are the product of a series of counting numbers, and so those
products have predictable factors. For example, all factorials above 1! are even numbers,
since they have 2 as a factor. Similarly, all factorials from 5! up end in a 0, because they
have 5 and 2 as factors, and thus have 10 as a factor. In fact, the factorial integers add
another 0 at the end of the factorial for every step of 5 upward: 5! = 120, 10! = 3628800,
15! = 1307674368000, 16! = 20922789888000 and so on.

Because factorial numbers, which quickly become quite large, continue to have another terminal 0
on the right hand side of the number for every factor of 5 added to the factorial product, one might
think that the proportion of zeros in a base 10 factorial number might be close to 1/5. However,
though the factorial products add another terminating 0 every factor of 5 multiplied into the product,
as the numbers become quite large, the number of digits in the factorial product expands exponentially,
and so the number above the terminating zeros tends toward 10% of each digit from 0 to 1 as the factorial
becomes larger. Thus, as the factorials become larger, the proportion of 0 digits in the factorial products
shifts slowly from around 1/5 toward 1/10, since the number of terminating zeros in n! increases only in
proportion to n, whereas the number of digits of n! in base 10 increases exponentially.

The task

Create a function to calculate the mean of the proportions of 0 digits out of the total digits found in each factorial
product from 1! to N!. This proportion of 0 digits in base 10 should be calculated using the number as printed as a base 10 integer.

Example: for 1 to 6 we have 1!, 2!, 3!, 4!, 5!, 6!, or (1, 2, 6, 24, 120, 720), so we need the mean of
(0/1, 0/1, 0/1, 0/2, 1/3, 1/3) = (2/3) (totals of each proportion) / 6 (= N), or 0.1111111...

Example: for 1 to 25 the mean of the proportions of 0 digits in the factorial products series of N! with N from 1 to 25 is 0.26787.

Do this task for 1 to N where N is in (100, 1000, and 10000), so, compute the mean of the proportion of 0 digits for each product
in the series of each of the factorials from 1 to 100, 1 to 1000, and 1 to 10000.

Stretch task

Find the N in 10000 < N < 50000 where the mean of the proportions of 0 digits in the factorial products from 1 to N
permanently falls below 0.16. This task took many hours in the Python example, though I wonder if there is a faster
algorithm out there.
assistant
Works, but depressingly slow for 10000.

```
sub postfix:<!> (Int $n) { ( constant factorial = 1, 1, |[\*] 2..* )[$n] }
sink 10000!; # prime the iterator to allow multithreading

sub zs ($n) { ( constant zero-share = (^Inf).race(:32batch).map: { (.!.comb.Bag){'0'} / .!.chars } )[$n+1] }

.say for (
     100
    ,1000
    ,10000
).map:  -> \n { "{n}: {([+] (^n).map: *.&zs) / n}" }

```

Output:

```
100: 0.24675318616743216
1000: 0.20354455110316458
10000: 0.17300384824186605

```