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# Distribution of 0 Digits in factorial series

Distribution of 0 Digits in factorial series is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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

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.

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.

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.

## 11l

Translation of: Python
`F facpropzeros(n, verbose = 1B)   V proportions = [0.0] * n   V (fac, psum) = (BigInt(1), 0.0)   L(i) 0 .< n      fac *= i + 1      V d = String(fac)      psum += sum(d.map(x -> Int(x == ‘0’))) / Float(d.len)      proportions[i] = psum / (i + 1)    I verbose      print(‘The mean proportion of 0 in factorials from 1 to #. is #..’.format(n, psum / n))    R proportions L(n) [100, 1000, 10000]   facpropzeros(n)`
Output:
```The mean proportion of 0 in factorials from 1 to 100 is 0.246753186.
The mean proportion of 0 in factorials from 1 to 1000 is 0.203544551.
The mean proportion of 0 in factorials from 1 to 10000 is 0.173003848.
```

### Base 1000 version

`F zinit()   V zc =  * 999   L(x) 1..9      zc[x - 1] = 2      zc[10 * x - 1] = 2      zc[100 * x - 1] = 2      L(y) (10.<100).step(10)         zc[y + x - 1] = 1         zc[10 * y + x - 1] = 1         zc[10 * (y + x) - 1] = 1    R zc F meanfactorialdigits()   V zc = zinit()   V rfs =    V (total, trail, first) = (0.0, 1, 0)   L(f) 2 .< 50000      V (carry, d999, zeroes) = (0, 0, (trail - 1) * 3)      V (j, l) = (trail, rfs.len)      L j <= l | carry != 0         I j <= l            carry = rfs[j - 1] * f + carry          d999 = carry % 1000         I j <= l            rfs[j - 1] = d999         E            rfs.append(d999)          zeroes += I d999 == 0 {3} E zc[d999 - 1]         carry I/= 1000         j++       L rfs[trail - 1] == 0         trail++       d999 = rfs.last      d999 = I d999 >= 100 {0} E I d999 < 10 {2} E 1       zeroes -= d999      V digits = rfs.len * 3 - d999      total += Float(zeroes) / digits      V ratio = total / f      I f C [100, 1000, 10000]         print(‘The mean proportion of zero digits in factorials to #. is #.’.format(f, ratio))       I ratio >= 0.16         first = 0      E I first == 0         first = f    print(‘The mean proportion dips permanently below 0.16 at ’first‘.’) meanfactorialdigits()`
Output:
```The mean proportion of zero digits in factorials to 100 is 0.246753186
The mean proportion of zero digits in factorials to 1000 is 0.203544551
The mean proportion of zero digits in factorials to 10000 is 0.173003848
The mean proportion dips permanently below 0.16 at 47332.
```

## C++

Translation of: Phix
`#include <array>#include <chrono>#include <iomanip>#include <iostream>#include <vector> auto init_zc() {    std::array<int, 1000> zc;    zc.fill(0);    zc = 3;    for (int x = 1; x <= 9; ++x) {        zc[x] = 2;        zc[10 * x] = 2;        zc[100 * x] = 2;        for (int y = 10; y <= 90; y += 10) {            zc[y + x] = 1;            zc[10 * y + x] = 1;            zc[10 * (y + x)] = 1;        }    }    return zc;} template <typename clock_type>auto elapsed(const std::chrono::time_point<clock_type>& t0) {    auto t1 = clock_type::now();    auto duration =        std::chrono::duration_cast<std::chrono::milliseconds>(t1 - t0);    return duration.count();} int main() {    auto zc = init_zc();    auto t0 = std::chrono::high_resolution_clock::now();    int trail = 1, first = 0;    double total = 0;    std::vector<int> rfs{1};    std::cout << std::fixed << std::setprecision(10);    for (int f = 2; f <= 50000; ++f) {        int carry = 0, d999, zeroes = (trail - 1) * 3, len = rfs.size();        for (int j = trail - 1; j < len || carry != 0; ++j) {            if (j < len)                carry += rfs[j] * f;            d999 = carry % 1000;            if (j < len)                rfs[j] = d999;            else                rfs.push_back(d999);            zeroes += zc[d999];            carry /= 1000;        }        while (rfs[trail - 1] == 0)            ++trail;        d999 = rfs.back();        d999 = d999 < 100 ? (d999 < 10 ? 2 : 1) : 0;        zeroes -= d999;        int digits = rfs.size() * 3 - d999;        total += double(zeroes) / digits;        double ratio = total / f;        if (ratio >= 0.16)            first = 0;        else if (first == 0)            first = f;        if (f == 100 || f == 1000 || f == 10000) {            std::cout << "Mean proportion of zero digits in factorials to " << f                      << " is " << ratio << ". (" << elapsed(t0) << "ms)\n";        }    }    std::cout << "The mean proportion dips permanently below 0.16 at " << first              << ". (" << elapsed(t0) << "ms)\n";}`
Output:
```Mean proportion of zero digits in factorials to 100 is 0.2467531862. (0ms)
Mean proportion of zero digits in factorials to 1000 is 0.2035445511. (1ms)
Mean proportion of zero digits in factorials to 10000 is 0.1730038482. (152ms)
The mean proportion dips permanently below 0.16 at 47332. (4598ms)
```

## Go

### Brute force

Library: Go-rcu

Timings here are 2.8 seconds for the basic task and 182.5 seconds for the stretch goal.

`package main import (    "fmt"    big "github.com/ncw/gmp"    "rcu") func main() {    fact  := big.NewInt(1)    sum   := 0.0    first := int64(0)    firstRatio := 0.0        fmt.Println("The mean proportion of zero digits in factorials up to the following are:")    for n := int64(1); n <= 50000; n++  {        fact.Mul(fact, big.NewInt(n))        bytes  := []byte(fact.String())        digits := len(bytes)        zeros  := 0        for _, b := range bytes {            if b == '0' {                zeros++            }        }        sum += float64(zeros)/float64(digits)        ratio := sum / float64(n)        if n == 100 || n == 1000 || n == 10000 {            fmt.Printf("%6s = %12.10f\n", rcu.Commatize(int(n)), ratio)        }         if first > 0 && ratio >= 0.16 {            first = 0            firstRatio = 0.0        } else if first == 0 && ratio < 0.16 {            first = n            firstRatio = ratio                   }    }    fmt.Printf("%6s = %12.10f", rcu.Commatize(int(first)), firstRatio)    fmt.Println(" (stays below 0.16 after this)")    fmt.Printf("%6s = %12.10f\n", "50,000", sum / 50000)}`
Output:
```The mean proportion of zero digits in factorials up to the following are:
100 = 0.2467531862
1,000 = 0.2035445511
10,000 = 0.1730038482
47,332 = 0.1599999958 (stays below 0.16 after this)
50,000 = 0.1596200546
```

### 'String math' and base 1000

Translation of: Phix

Much quicker than before with 10,000 now being reached in 0.35 seconds and the stretch goal in about 5.5 seconds.

`package main import (    "fmt"    "rcu") var rfs = []int{1} // reverse factorial(1) in base 1000var zc = make([]int, 999) func init() {    for x := 1; x <= 9; x++ {        zc[x-1] = 2     // 00x        zc[10*x-1] = 2  // 0x0        zc[100*x-1] = 2 // x00        var y = 10        for y <= 90 {            zc[y+x-1] = 1      // 0yx            zc[10*y+x-1] = 1   // y0x            zc[10*(y+x)-1] = 1 // yx0            y += 10        }    }} func main() {    total := 0.0    trail := 1    first := 0    firstRatio := 0.0    fmt.Println("The mean proportion of zero digits in factorials up to the following are:")    for f := 2; f <= 10000; f++ {        carry := 0        d999 := 0        zeros := (trail - 1) * 3        j := trail        l := len(rfs)        for j <= l || carry != 0 {            if j <= l {                carry = rfs[j-1]*f + carry            }            d999 = carry % 1000            if j <= l {                rfs[j-1] = d999            } else {                rfs = append(rfs, d999)            }            if d999 == 0 {                zeros += 3            } else {                zeros += zc[d999-1]            }            carry /= 1000            j++        }        for rfs[trail-1] == 0 {            trail++        }        // d999 = quick correction for length and zeros        d999 = rfs[len(rfs)-1]        if d999 < 100 {            if d999 < 10 {                d999 = 2            } else {                d999 = 1            }        } else {            d999 = 0        }        zeros -= d999        digits := len(rfs)*3 - d999        total += float64(zeros) / float64(digits)        ratio := total / float64(f)        if ratio >= 0.16 {            first = 0            firstRatio = 0.0        } else if first == 0 {            first = f            firstRatio = ratio        }        if f == 100 || f == 1000 || f == 10000 {            fmt.Printf("%6s = %12.10f\n", rcu.Commatize(f), ratio)        }    }    fmt.Printf("%6s = %12.10f", rcu.Commatize(first), firstRatio)    fmt.Println(" (stays below 0.16 after this)")    fmt.Printf("%6s = %12.10f\n", "50,000", total/50000)}`
Output:
```Same as 'brute force' version.
```

## jq

Works with jq

The precision of jq's integer arithmetic is not up to this task, so in the following we borrow from the "BigInt" library and use a string representation of integers.

Unfortunately, although gojq (the Go implementation of jq) does support unbounded-precision integer arithmetic, it is unsuited for the task because of memory management issues.

From BigInt.jq

` # multiply two decimal strings, which may be signed (+ or -)def long_multiply(num1; num2):   def stripsign:    .[0:1] as \$a    | if \$a == "-" then [ -1, .[1:]]     elif \$a == "+" then [  1, .[1:]]     else [1, .]    end;   def adjustsign(sign):     if sign == 1 then . else "-" + . end;   # mult/2 assumes neither argument has a sign  def mult(num1;num2):      (num1 | explode | map(.-48) | reverse) as \$a1    | (num2 | explode | map(.-48) | reverse) as \$a2    | reduce range(0; num1|length) as \$i1        ([];  # result         reduce range(0; num2|length) as \$i2           (.;            (\$i1 + \$i2) as \$ix            | ( \$a1[\$i1] * \$a2[\$i2] + (if \$ix >= length then 0 else .[\$ix] end) ) as \$r            | if \$r > 9 # carrying              then                .[\$ix + 1] = (\$r / 10 | floor) +  (if \$ix + 1 >= length then 0 else .[\$ix + 1] end )                | .[\$ix] = \$r - ( \$r / 10 | floor ) * 10              else                .[\$ix] = \$r              end         )        )     | reverse | map(.+48) | implode;   (num1|stripsign) as \$a1  | (num2|stripsign) as \$a2  | if \$a1 == "0" or  \$a2 == "0" then "0"    elif \$a1 == "1" then \$a2|adjustsign( \$a1 * \$a2 )    elif \$a2 == "1" then \$a1|adjustsign( \$a1 * \$a2 )    else mult(\$a1; \$a2) | adjustsign( \$a1 * \$a2 )    end; `

` def count(s): reduce s as \$x (0; .+1); def meanfactorialdigits:   def digits: tostring | explode;   def nzeros: count( .[] | select(. == 48) ); # "0" is 48    . as \$N   | 0.16 as \$goal   | label \$out   | reduce range( 1; 1+\$N ) as \$i ( {factorial: "1", proportionsum: 0.0, first: null };        .factorial = long_multiply(.factorial; \$i|tostring)        | (.factorial|digits) as \$d        | .proportionsum += (\$d | (nzeros / length))         | (.proportionsum / \$i) as \$propmean	| if .first	  then if \$propmean > \$goal then .first = null else . end	  elif \$propmean <= \$goal then .first = \$i	  else .	  end)    | "Mean proportion of zero digits in factorials to \(\$N) is \(.proportionsum/\$N);" +       (if .first then " mean <= \(\$goal) from N=\(.first) on." else " goal (\(\$goal)) unmet." end); # The task:100, 1000, 10000 | meanfactorialdigits`
Output:
```Mean proportion of zero digits in factorials to 100 is 0.24675318616743216; goal (0.16) unmet.
Mean proportion of zero digits in factorials to 1000 is 0.20354455110316458; goal (0.16) unmet.
Mean proportion of zero digits in factorials to 10000 is 0.17300384824186707; goal (0.16) unmet.
```

## Julia

`function meanfactorialdigits(N, goal = 0.0)    factoril, proportionsum = big"1", 0.0    for i in 1:N        factoril *= i        d = digits(factoril)        zero_proportion_in_fac = count(x -> x == 0, d) / length(d)        proportionsum += zero_proportion_in_fac        propmean = proportionsum / i        if i > 15 && propmean <= goal            println("The mean proportion dips permanently below \$goal at \$i.")            break        end        if i == N            println("Mean proportion of zero digits in factorials to \$N is ", propmean)        end    endend @time foreach(meanfactorialdigits, [100, 1000, 10000]) @time meanfactorialdigits(50000, 0.16) `
Output:
```Mean proportion of zero digits in factorials to 100 is 0.24675318616743216
Mean proportion of zero digits in factorials to 1000 is 0.20354455110316458
Mean proportion of zero digits in factorials to 10000 is 0.17300384824186707
3.030182 seconds (297.84 k allocations: 1.669 GiB, 0.83% gc time, 0.28% compilation time)
The mean proportion dips permanently below 0.16 at 47332.
179.157788 seconds (3.65 M allocations: 59.696 GiB, 1.11% gc time)
```

### Base 1000 version

Translation of: Pascal, Phix
`function init_zc()    zc = zeros(Int, 999)    for x in 1:9        zc[x] = 2       # 00x        zc[10*x] = 2    # 0x0        zc[100*x] = 2   # x00        for y in 10:10:90            zc[y+x] = 1         # 0yx            zc[10*y+x] = 1      # y0x            zc[10*(y+x)] = 1    # yx0        end    end    return zcend function meanfactorialzeros(N = 50000, verbose = true)    zc = init_zc()    rfs =      total, trail, first, firstratio = 0.0, 1, 0, 0.0     for f in 2:N        carry, d999, zeroes = 0, 0, (trail - 1) * 3        j, l = trail, length(rfs)        while j <= l || carry != 0            if j <= l                carry = (rfs[j]) * f + carry            end            d999 = carry % 1000            if j <= l                rfs[j] = d999            else                push!(rfs, d999)            end            zeroes += (d999 == 0) ? 3 : zc[d999]            carry ÷= 1000            j += 1        end        while rfs[trail] == 0            trail += 1        end        # d999 = quick correction for length and zeroes:        d999 = rfs[end]        d999 = d999 < 100 ? d999 < 10 ? 2 : 1 : 0        zeroes -= d999        digits = length(rfs) * 3 - d999        total += zeroes / digits        ratio = total / f        if ratio >= 0.16           first = 0           firstratio = 0.0        elseif first == 0            first = f            firstratio = ratio        end        if f in [100, 1000, 10000]            verbose && println("Mean proportion of zero digits in factorials to \$f is \$ratio")        end    end    verbose && println("The mean proportion dips permanently below 0.16 at \$first.")end meanfactorialzeros(100, false)@time meanfactorialzeros() `
Output:
```Mean proportion of zero digits in factorials to 100 is 0.24675318616743216
Mean proportion of zero digits in factorials to 1000 is 0.20354455110316458
Mean proportion of zero digits in factorials to 10000 is 0.17300384824186707
The mean proportion dips permanently below 0.16 at 47332.
4.638323 seconds (50.08 k allocations: 7.352 MiB)
```

## Mathematica/Wolfram Language

`ClearAll[ZeroDigitsFractionFactorial]ZeroDigitsFractionFactorial[n_Integer] := Module[{m},  m = IntegerDigits[n!];  Count[m, 0]/Length[m]  ]ZeroDigitsFractionFactorial /@ Range // Mean // NZeroDigitsFractionFactorial /@ Range // Mean // NZeroDigitsFractionFactorial /@ Range // Mean // NZeroDigitsFractionFactorial /@ Range // Mean // NZeroDigitsFractionFactorial /@ Range // Mean // N fracs = ParallelMap[ZeroDigitsFractionFactorial, Range, Method -> ("ItemsPerEvaluation" -> 100)];means = Accumulate[[email protected]]/Range[Length[fracs]];len = LengthWhile[[email protected], # < 0.16 &];50000 - len + 1`
Output:
```0.111111
0.267873
0.246753
0.203545
0.173004
47332```

## Nim

Library: bignum
`import strutils, std/monotimesimport bignum let t0 = getMonoTime()var sum = 0.0var f = newInt(1)var lim = 100for n in 1..10_000:  f *= n  let str = \$f  sum += str.count('0') / str.len  if n == lim:    echo n, ":\t", sum / float(n)    lim *= 10echo()echo getMonoTime() - t0`
Output:
```100:    0.2467531861674322
1000:   0.2035445511031646
10000:  0.1730038482418671

(seconds: 2, nanosecond: 857794404)```

Library: bignum

At each step, we eliminate the trailing zeroes to reduce the length of the number and save some time. But this is not much, about 8%.

`import strutils, std/monotimesimport bignum let t0 = getMonoTime()var sum = 0.0var first = 0var f = newInt(1)var count0 = 0for n in 1..<50_000:  f *= n  while f mod 10 == 0:    # Reduce the length of "f".    f = f div 10    inc count0  let str = \$f  sum += (str.count('0') + count0) / (str.len + count0)  if sum / float(n) < 0.16:    if first == 0: first = n  else:    first = 0 echo "Permanently below 0.16 at n = ", firstecho "Execution time: ", getMonoTime() - t0`
Output:
```Permanently below 0.16 at n = 47332
Execution time: (seconds: 190, nanosecond: 215845101)```

## Pascal

Doing the calculation in Base 1,000,000,000 like in Primorial_numbers#alternative.
The most time consuming is converting to string and search for zeros.
Therefor I do not convert to string.I divide the base in sections of 3 digits with counting zeros in a lookup table.

`program Factorial;{\$IFDEF FPC} {\$MODE DELPHI} {\$Optimization ON,ALL} {\$ENDIF}uses  sysutils;type  tMul = array of LongWord;  tpMul = pLongWord;const  LongWordDec = 1000*1000*1000;  LIMIT = 50000;var  CountOfZero : array[0..999] of byte;  SumOfRatio :array[0..LIMIT] of extended;  procedure OutMul(pMul:tpMul;Lmt :NativeInt);// for testingBegin  write(pMul[lmt]);  For lmt := lmt-1  downto 0 do    write(Format('%.9d',[pMul[lmt]]));  writeln;end; procedure InitCoZ;//Init Lookup table for 3 digitsvar  x,y : integer;begin  fillchar(CountOfZero,SizeOf(CountOfZero),#0);  CountOfZero := 3; //000  For x := 1 to 9 do  Begin    CountOfZero[x] := 2;     //00x    CountOfZero[10*x] := 2;  //0x0    CountOfZero[100*x] := 2; //x00    y := 10;    repeat      CountOfZero[y+x] := 1;      //0yx      CountOfZero[10*y+x] := 1;   //y0x      CountOfZero[10*(y+x)] := 1; //yx0      inc(y,10)    until y > 100;  end;end; function getFactorialDecDigits(n:NativeInt):NativeInt;var  res: extended;Begin  result := -1;  IF (n > 0) AND (n <= 1000*1000) then  Begin    res := 0;    repeat res := res+ln(n); dec(n); until n < 2;    result := trunc(res/ln(10))+1;  end;end; function CntZero(pMul:tpMul;Lmt :NativeInt):NativeUint;//count zeros in Base 1,000,000,000 numbervar  q,r : LongWord;  i : NativeInt;begin  result := 0;  For i := Lmt-1 downto 0 do  Begin    q := pMul[i];    r := q DIV 1000;    result +=CountOfZero[q-1000*r];//q-1000*r == q mod 1000    q := r;    r := q DIV 1000;    result +=CountOfZero[q-1000*r];    q := r;    r := q DIV 1000;    result +=CountOfZero[q-1000*r];  end;//special case first digits no leading '0'  q := pMul[lmt];  while q >= 1000 do  begin    r := q DIV 1000;    result +=CountOfZero[q-1000*r];    q := r;  end;  while q > 0 do  begin    r := q DIV 10;    result += Ord( q-10*r= 0);    q := r;  end;end; function GetCoD(pMul:tpMul;Lmt :NativeInt):NativeUint;//count of decimal digitsvar  i : longWord;begin  result := 9*Lmt;  i := pMul[Lmt];  while i > 1000 do  begin    i := i DIV 1000;    inc(result,3);  end;  while i > 0 do  begin    i := i DIV 10;    inc(result);  end;end; procedure DoChecks(pMul:tpMul;Lmt,i :NativeInt);//(extended(1.0)* makes TIO.RUN faster // only using FPU?Begin  SumOfRatio[i] := SumOfRatio[i-1] + (extended(1.0)*CntZero(pMul,Lmt))/GetCoD(pMul,Lmt);end; function MulByI(pMul:tpMul;UL,i :NativeInt):NativeInt;var  prod  : Uint64;  j     : nativeInt;  carry : LongWord;begin  result := UL;  carry := 0;  For j := 0 to result do  Begin    prod  := i*pMul+Carry;    Carry := prod Div LongWordDec;    pMul := Prod - LongWordDec*Carry;    inc(pMul);  end;   IF Carry <> 0 then  Begin    inc(result);    pMul:= Carry;  End;end; procedure getFactorialExact(n:NativeInt);var  MulArr : tMul;  pMul : tpMul;  i,ul : NativeInt;begin  i := getFactorialDecDigits(n) DIV 9 +10;  Setlength(MulArr,i);  pMul := @MulArr;  Ul := 0;  pMul[Ul]:= 1;  i := 1;  repeat    UL := MulByI(pMul,UL,i);    //Now do what you like to do with i!    DoChecks(pMul,UL,i);    inc(i);  until i> n;end; procedure Out_(i: integer);begin  if i > LIMIT then    EXIT;  writeln(i:8,SumOfRatio[i]/i:18:15);end; var  i : integer;Begin  InitCoZ;  SumOfRatio:= 0;  getFactorialExact(LIMIT);  Out_(100);  Out_(1000);  Out_(10000);  Out_(50000);  i := limit;  while i >0 do  Begin    if SumOfRatio[i]/i >0.16 then      break;    dec(i);  end;  inc(i);  writeln('First ratio < 0.16 ', i:8,SumOfRatio[i]/i:20:17);end.`
Output:
```     100 0.246753186167432
1000 0.203544551103165
10000 0.173003848241866
50000 0.159620054602269
First ratio < 0.16    47332 0.15999999579985665
Real time: 4.898 s  CPU share: 99.55 % // 2.67s on 2200G freepascal 3.2.2```

## Perl

Library: ntheory
`use strict;use warnings;use ntheory qw/factorial/; for my \$n (100, 1000, 10000) {    my(\$sum,\$f) = 0;    \$f = factorial \$_ and \$sum += (\$f =~ tr/0//) / length \$f for 1..\$n;    printf "%5d: %.5f\n", \$n, \$sum/\$n;}`
Output:
```  100: 0.24675
1000: 0.20354
10000: 0.17300```

## Phix

Using "string math" to create reversed factorials, for slightly easier skipping of "trailing" zeroes, but converted to base 1000 and with the zero counting idea from Pascal, which sped it up threefold.

```with javascript_semantics
sequence rfs = {1}  -- reverse factorial(1) in base 1000

function init_zc()
sequence zc = repeat(0,999)
for x=1 to 9 do
zc[x] = 2       -- 00x
zc[10*x] = 2    -- 0x0
zc[100*x] = 2   -- x00
for y=10 to 90 by 10 do
zc[y+x] = 1         -- 0yx
zc[10*y+x] = 1      -- y0x
zc[10*(y+x)] = 1    -- yx0
end for
end for
return zc
end function
constant zc = init_zc()

atom t0 = time(),
total = 0
integer trail = 1,
first = 0
for f=2 to iff(platform()=JS?10000:50000) do
integer carry = 0, d999,
zeroes = (trail-1)*3,
j = trail, l = length(rfs)
while j<=l or carry do
if j<=l then
carry = (rfs[j])*f+carry
end if
d999 = remainder(carry,1000)
if j<=l then
rfs[j] = d999
else
rfs &= d999
end if
zeroes += iff(d999=0?3:zc[d999])
carry = floor(carry/1000)
j += 1
end while
while rfs[trail]=0 do trail += 1 end while
-- d999 := quick correction for length and zeroes:
d999 = rfs[\$]
d999 = iff(d999<100?iff(d999<10?2:1):0)
zeroes -= d999
integer digits = length(rfs)*3-d999

total += zeroes/digits
atom ratio = total/f
if ratio>=0.16 then
first = 0
elsif first=0 then
first = f
end if
if find(f,{100,1000,10000}) then
string e = elapsed(time()-t0)
printf(1,"Mean proportion of zero digits in factorials to %d is %.10f (%s)\n",{f,ratio,e})
end if
end for
if platform()!=JS then
string e = elapsed(time()-t0)
printf(1,"The mean proportion dips permanently below 0.16 at %d. (%s)\n",{first,e})
end if
```
Output:
```Mean proportion of zero digits in factorials to 100 is 0.2467531862 (0s)
Mean proportion of zero digits in factorials to 1000 is 0.2035445511 (0.2s)
Mean proportion of zero digits in factorials to 10000 is 0.1730038482 (2.3s)
The mean proportion dips permanently below 0.16 at 47332. (1 minute and 2s)
```

(stretch goal removed under pwa/p2js since otherwise you'd get a blank screen for 2 or 3 minutes)

### trailing zeroes only

Should you only be interested in the ratio of trailing zeroes, you can do that much faster:

```with javascript_semantics
atom t0 = time(),
f10 = log10(1),
total = 0
integer first = 0
for f=2 to 50000 do
f10 += log10(f)
integer digits = ceil(f10),
zeroes = 0,
v = 5
while v<=f do
zeroes += floor(f/v)
v *= 5
end while
total += zeroes/digits
atom ratio = total/f
if ratio>=0.07 then
first = 0
elsif first=0 then
first = f
end if
if find(f,{100,1000,10000}) then
printf(1,"Mean proportion of trailing zeroes in factorials to %d is %f\n",{f,ratio})
end if
end for
string e = elapsed(time()-t0)
printf(1,"The mean proportion dips permanently below 0.07 at %d. (%s)\n",{first,e})
```
Output:
```Mean proportion of trailing zeroes in factorials to 100 is 0.170338
Mean proportion of trailing zeroes in factorials to 1000 is 0.116334
Mean proportion of trailing zeroes in factorials to 10000 is 0.081267
The mean proportion dips permanently below 0.07 at 31549. (0.1s)
```

## Python

`def facpropzeros(N, verbose = True):    proportions = [0.0] * N    fac, psum = 1, 0.0    for i in range(N):        fac *= i + 1        d = list(str(fac))        psum += sum(map(lambda x: x == '0', d)) / len(d)        proportions[i] = psum / (i + 1)     if verbose:        print("The mean proportion of 0 in factorials from 1 to {} is {}.".format(N, psum / N))     return proportions  for n in [100, 1000, 10000]:    facpropzeros(n) props = facpropzeros(47500, False)n = (next(i for i in reversed(range(len(props))) if props[i] > 0.16)) print("The mean proportion dips permanently below 0.16 at {}.".format(n + 2)) `
Output:
```The mean proportion of 0 in factorials from 1 to 100 is 0.24675318616743216.
The mean proportion of 0 in factorials from 1 to 1000 is 0.20354455110316458.
The mean proportion of 0 in factorials from 1 to 10000 is 0.17300384824186707.
The mean proportion dips permanently below 0.16 at 47332.
```

The means can be plotted, showing a jump from 0 to over 0.25, followed by a slowly dropping curve:

`import matplotlib.pyplot as pltplt.plot([i+1 for i in range(len(props))], props) `

### Base 1000 version

Translation of: Go via Phix via Pascal
`def zinit():    zc =  * 999    for x in range(1, 10):        zc[x - 1] = 2        # 00x        zc[10 * x - 1] = 2   # 0x0        zc[100 * x - 1] = 2  # x00        for y in range(10, 100, 10):            zc[y + x - 1] = 1           # 0yx            zc[10 * y + x - 1] = 1      # y0x            zc[10 * (y + x) - 1] = 1    # yx0     return zc def meanfactorialdigits():    zc = zinit()    rfs =     total, trail, first = 0.0, 1, 0    for f in range(2, 50000):        carry, d999, zeroes = 0, 0, (trail - 1) * 3        j, l = trail, len(rfs)        while j <= l or carry != 0:            if j <= l:                carry = rfs[j-1] * f + carry             d999 = carry % 1000            if j <= l:                rfs[j-1] = d999            else:                rfs.append(d999)             zeroes += 3 if d999 == 0 else zc[d999-1]            carry //= 1000            j += 1         while rfs[trail-1] == 0:            trail += 1         # d999 is a quick correction for length and zeros        d999 = rfs[-1]        d999 = 0 if d999 >= 100 else 2 if d999 < 10 else 1         zeroes -= d999        digits = len(rfs) * 3 - d999        total += zeroes / digits        ratio = total / f        if f in [100, 1000, 10000]:            print("The mean proportion of zero digits in factorials to {} is {}".format(f, ratio))         if ratio >= 0.16:            first = 0        elif first == 0:            first = f     print("The mean proportion dips permanently below 0.16 at {}.".format(first))   import timeTIME0 = time.perf_counter()meanfactorialdigits()print("\nTotal time:", time.perf_counter() - TIME0, "seconds.") `
Output:
```The mean proportion of zero digits in factorials to 100 is 0.24675318616743216
The mean proportion of zero digits in factorials to 1000 is 0.20354455110316458
The mean proportion of zero digits in factorials to 10000 is 0.17300384824186707
The mean proportion dips permanently below 0.16 at 47332.

Total time: 648.3583232999999 seconds.
```

## Raku

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
```

## REXX

`/*REXX program computes the mean of the proportion of "0" digits a series of factorials.*/parse arg \$                                      /*obtain optional arguments from the CL*/if \$='' | \$=","  then \$= 100 1000 10000          /*not specified?  Then use the default.*/#= words(\$)                                      /*the number of ranges to be used here.*/numeric digits 100                               /*increase dec. digs, but only to 100. */big= word(\$, #);  != 1                           /*obtain the largest number in ranges. */                                do i=1  for big  /*calculate biggest  !  using 100 digs.*/                                != ! * i         /*calculate the factorial of  BIG.     */                                end   /*i*/if pos('E', !)>0  then do                        /*In exponential format?  Then get EXP.*/                       parse var !  'E'  x       /*parse the exponent from the number.  */                       numeric digits    x+1     /*set the decimal digits to  X  plus 1.*/                       end                       /* [↑]  the  +1  is for the dec. point.*/ title= ' mean proportion of zeros in the (decimal) factorial products for  N'say '     N     │'center(title, 80)              /*display the title for the output.    */say '───────────┼'center(""   , 80, '─')         /*   "     a   sep   "   "     "       */   do j=1  for #;  n= word(\$, j)                  /*calculate some factorial ranges.     */  say center( commas(n), 11)'│' left(0dist(n), 75)...    /*show results for above range.*/  end   /*j*/ say '───────────┴'center(""   , 80, '─')         /*display a foot sep for the output.   */exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?/*──────────────────────────────────────────────────────────────────────────────────────*/0dist:  procedure; parse arg z;        != 1;         y= 0                     do k=1  for z;    != ! * k;     y= y   +   countstr(0, !) / length(!)                     end   /*k*/        return y/z`
output   when using the default inputs:
```     N     │       mean proportion of zeros in the (decimal) factorial products for  N
───────────┼────────────────────────────────────────────────────────────────────────────────
100    │ 0.2467531861674322177784158871973526991129407033266153063813195937196095976...
1,000   │ 0.2035445511031646356400438031711455302985741167890402203486699704599684047...
10,000   │ 0.1730038482418660531800366428930706156810278809057883361518852958446868172...
───────────┴────────────────────────────────────────────────────────────────────────────────
```

## Rust

Translation of: Phix
`fn init_zc() -> Vec<usize> {    let mut zc = vec![0; 1000];    zc = 3;    for x in 1..=9 {        zc[x] = 2;        zc[10 * x] = 2;        zc[100 * x] = 2;        let mut y = 10;        while y <= 90 {            zc[y + x] = 1;            zc[10 * y + x] = 1;            zc[10 * (y + x)] = 1;            y += 10;        }    }    zc} fn main() {    use std::time::Instant;    let zc = init_zc();    let t0 = Instant::now();    let mut trail = 1;    let mut first = 0;    let mut total: f64 = 0.0;    let mut rfs = vec!;     for f in 2..=50000 {        let mut carry = 0;        let mut d999: usize;        let mut zeroes = (trail - 1) * 3;        let len = rfs.len();        let mut j = trail - 1;        while j < len || carry != 0 {            if j < len {                carry += rfs[j] * f;            }            d999 = carry % 1000;            if j < len {                rfs[j] = d999;            } else {                rfs.push(d999);            }            zeroes += zc[d999];            carry /= 1000;            j += 1;        }        while rfs[trail - 1] == 0 {            trail += 1;        }        d999 = rfs[rfs.len() - 1];        d999 = if d999 < 100 {            if d999 < 10 {                2            } else {                1            }        } else {            0        };        zeroes -= d999;        let digits = rfs.len() * 3 - d999;        total += (zeroes as f64) / (digits as f64);        let ratio = total / (f as f64);        if ratio >= 0.16 {            first = 0;        } else if first == 0 {            first = f;        }        if f == 100 || f == 1000 || f == 10000 {            let duration = t0.elapsed();            println!(                "Mean proportion of zero digits in factorials to {} is {:.10}. ({}ms)",                f,                ratio,                duration.as_millis()            );        }    }    let duration = t0.elapsed();    println!(        "The mean proportion dips permanently below 0.16 at {}. ({}ms)",        first,        duration.as_millis()    );}`
Output:
```Mean proportion of zero digits in factorials to 100 is 0.2467531862. (0ms)
Mean proportion of zero digits in factorials to 1000 is 0.2035445511. (1ms)
Mean proportion of zero digits in factorials to 10000 is 0.1730038482. (149ms)
The mean proportion dips permanently below 0.16 at 47332. (4485ms)
```

## Wren

### Brute force

Library: Wren-big
Library: Wren-fmt

Very slow indeed, 10.75 minutes to reach N = 10,000.

`import "/big" for BigIntimport "/fmt" for Fmt var fact = BigInt.onevar sum = 0System.print("The mean proportion of zero digits in factorials up to the following are:")for (n in 1..10000) {    fact = fact * n    var bytes = fact.toString.bytes    var digits = bytes.count    var zeros  = bytes.count { |b| b == 48 }    sum = sum + zeros / digits    if (n == 100 || n == 1000 || n == 10000) {        Fmt.print("\$,6d = \$12.10f", n, sum / n)    }}`
Output:
```The mean proportion of zero digits in factorials up to the following are:
100 = 0.2467531862
1,000 = 0.2035445511
10,000 = 0.1730038482
```

### 'String math' and base 1000

Translation of: Phix

Around 60 times faster than before with 10,000 now being reached in about 10.5 seconds. Even the stretch goal is now viable and comes in at 5 minutes 41 seconds.

`import "/fmt" for Fmt var rfs =   // reverse factorial(1) in base 1000 var init = Fn.new { |zc|    for (x in 1..9) {        zc[x-1] = 2         // 00x        zc[10*x - 1] = 2    // 0x0        zc[100*x - 1] = 2   // x00        var y = 10        while (y <= 90) {            zc[y + x - 1] = 1       // 0yx            zc[10*y + x - 1] = 1    // y0x            zc[10*(y + x) - 1] = 1  // yx0            y = y + 10        }    }} var zc = List.filled(999, 0)init.call(zc)var total = 0var trail = 1var first = 0var firstRatio = 0System.print("The mean proportion of zero digits in factorials up to the following are:")for (f in 2..50000) {    var carry = 0    var d999 = 0    var zeros = (trail-1) * 3    var j = trail    var l = rfs.count    while (j <= l || carry != 0) {        if (j <= l) carry = rfs[j-1]*f + carry        d999 = carry % 1000        if (j <= l) {            rfs[j-1] = d999        } else {            rfs.add(d999)        }        zeros = zeros + ((d999 == 0) ? 3 : zc[d999-1])        carry = (carry/1000).floor        j = j + 1    }    while (rfs[trail-1] == 0) trail = trail + 1    // d999 = quick correction for length and zeros    d999 = rfs[-1]    d999 = (d999 < 100) ? ((d999 < 10) ? 2 : 1) : 0    zeros = zeros - d999    var digits = rfs.count * 3 - d999    total = total + zeros/digits    var ratio =  total / f    if (ratio >= 0.16) {        first = 0        firstRatio = 0    } else if (first == 0) {        first = f        firstRatio = ratio    }    if (f == 100 || f == 1000 || f == 10000) {        Fmt.print("\$,6d = \$12.10f", f, ratio)    }}Fmt.write("\$,6d = \$12.10f", first, firstRatio)System.print(" (stays below 0.16 after this)")Fmt.print("\$,6d = \$12.10f", 50000, total/50000)`
Output:
```The mean proportion of zero digits in factorials up to the following are:
100 = 0.2467531862
1,000 = 0.2035445511
10,000 = 0.1730038482
47,332 = 0.1599999958 (stays below 0.16 after this)
50,000 = 0.1596200546
```