Integer long division

Integer long division
You are encouraged to solve this task according to the task description, using any language you may know.

Write a function that prints the result of the division of two integers with infinite precision (only limited by the available memory), stopping before the period starts repeating itself. Return also the length of this period (0 if there is no period).

Demonstrate it with the division 1/149, whose result is 0.0067114093959731543624161073825503355704697986577181208053691275167785234899328859060402684563758389261744966442953020134228187919463087248322147651, where the last 148 digits repeat endlessly.

The result could be stored as a string or simply output to the screen.
Note that the division of any two numbers will always produce a period, but if the numerator is an exact multiple of the denominator, or if the denominator contains only the factors 2 and 5, the period will be 0. In the remaining cases, these possible 2's and 5's of the denominator produce a leading number of digits in the quotient, but have no effect on the period.

References

• [1] Wikipedia article on Long Division

Common Lisp

<lang lisp> (defun $/ (a b)

"Divide a/b with infinite precision printing each digit as it is calculated and return the period length"

($/ 1 17) => 588235294117647 ; 16

 (assert (and (integerp a) (integerp b) (not (zerop b))))
 (do* (c 
      (i0 (1+ (max (factor-multiplicity b 2) (factor-multiplicity b 5)))) ; the position which marks the beginning of the period
      (r a (* 10 r)) ; remainder
      (i 0 (1+ i)) ; iterations counter
      (rem (if (= i i0) r -1) (if (= i i0) r rem)) ) ; the first remainder against which to check for repeating remainders
     ((and (= r rem) (not (= i i0))) (- i i0))
   (multiple-value-setq (c r) (floor r b))
   (princ c) ))

(defun factor-multiplicity (n factor) "Return how many times the factor is contained in n"

(factor-multiplicity 12 2) => 2

(do* ((i 0 (1+ i)) (n (/ n factor) (/ n factor)) )

            ((not (integerp n)) i)

() ))

</lang>

Output:

($/ 1 149) 
00067114093959731543624161073825503355704697986577181208053691275167785234899328859060402684563758389261744966442953020134228187919463087248322147651
148

Raku

It's a built-in. <lang perl6>for 1/1, 1/3, 1/7, 1/17, 10/13, 3227/555, 1/149, 1/5261 -> $rat {

   printf "%8s - Period is %-5s: %s%s\n", $rat.nude.join('/'), .[1].chars, .[0], (.[1].comb RZ~ "\c[COMBINING OVERLINE]" xx *).join
     given $rat.base-repeating

}</lang>

Output:

     1/1 - Period is 0    : 1
     1/3 - Period is 1    : 0.̅3
     1/7 - Period is 6    : 0.̅1̅4̅2̅8̅5̅7
    1/17 - Period is 16   : 0.̅0̅5̅8̅8̅2̅3̅5̅2̅9̅4̅1̅1̅7̅6̅4̅7
   10/13 - Period is 6    : 0.̅7̅6̅9̅2̅3̅0
3227/555 - Period is 3    : 5.8̅1̅4̅4
   1/149 - Period is 148  : 0.̅0̅0̅6̅7̅1̅1̅4̅0̅9̅3̅9̅5̅9̅7̅3̅1̅5̅4̅3̅6̅2̅4̅1̅6̅1̅0̅7̅3̅8̅2̅5̅5̅0̅3̅3̅5̅5̅7̅0̅4̅6̅9̅7̅9̅8̅6̅5̅7̅7̅1̅8̅1̅2̅0̅8̅0̅5̅3̅6̅9̅1̅2̅7̅5̅1̅6̅7̅7̅8̅5̅2̅3̅4̅8̅9̅9̅3̅2̅8̅8̅5̅9̅0̅6̅0̅4̅0̅2̅6̅8̅4̅5̅6̅3̅7̅5̅8̅3̅8̅9̅2̅6̅1̅7̅4̅4̅9̅6̅6̅4̅4̅2̅9̅5̅3̅0̅2̅0̅1̅3̅4̅2̅2̅8̅1̅8̅7̅9̅1̅9̅4̅6̅3̅0̅8̅7̅2̅4̅8̅3̅2̅2̅1̅4̅7̅6̅5̅1
  1/5261 - Period is 1052 : 0.̅0̅0̅0̅1̅9̅0̅0̅7̅7̅9̅3̅1̅9̅5̅2̅1̅0̅0̅3̅6̅1̅1̅4̅8̅0̅7̅0̅7̅0̅8̅9̅9̅0̅6̅8̅6̅1̅8̅1̅3̅3̅4̅3̅4̅7̅0̅8̅2̅3̅0̅3̅7̅4̅4̅5̅3̅5̅2̅5̅9̅4̅5̅6̅3̅7̅7̅1̅1̅4̅6̅1̅6̅9̅9̅2̅9̅6̅7̅1̅1̅6̅5̅1̅7̅7̅7̅2̅2̅8̅6̅6̅3̅7̅5̅2̅1̅3̅8̅3̅7̅6̅7̅3̅4̅4̅6̅1̅1̅2̅9̅0̅6̅2̅9̅1̅5̅7̅9̅5̅4̅7̅6̅1̅4̅5̅2̅1̅9̅5̅4̅0̅0̅1̅1̅4̅0̅4̅6̅7̅5̅9̅1̅7̅1̅2̅6̅0̅2̅1̅6̅6̅8̅8̅8̅4̅2̅4̅2̅5̅3̅9̅4̅4̅1̅1̅7̅0̅8̅8̅0̅0̅6̅0̅8̅2̅4̅9̅3̅8̅2̅2̅4̅6̅7̅2̅1̅1̅5̅5̅6̅7̅3̅8̅2̅6̅2̅6̅8̅7̅7̅0̅1̅9̅5̅7̅8̅0̅2̅6̅9̅9̅1̅0̅6̅6̅3̅3̅7̅1̅9̅8̅2̅5̅1̅2̅8̅3̅0̅2̅6̅0̅4̅0̅6̅7̅6̅6̅7̅7̅4̅3̅7̅7̅4̅9̅4̅7̅7̅2̅8̅5̅6̅8̅7̅1̅3̅1̅7̅2̅4̅0̅0̅6̅8̅4̅2̅8̅0̅5̅5̅5̅0̅2̅7̅5̅6̅1̅3̅0̅0̅1̅3̅3̅0̅5̅4̅5̅5̅2̅3̅6̅6̅4̅7̅0̅2̅5̅2̅8̅0̅3̅6̅4̅9̅4̅9̅6̅2̅9̅3̅4̅8̅0̅3̅2̅6̅9̅3̅4̅0̅4̅2̅9̅5̅7̅6̅1̅2̅6̅2̅1̅1̅7̅4̅6̅8̅1̅6̅1̅9̅4̅6̅3̅9̅8̅0̅2̅3̅1̅8̅9̅5̅0̅7̅6̅9̅8̅1̅5̅6̅2̅4̅4̅0̅6̅0̅0̅6̅4̅6̅2̅6̅4̅9̅6̅8̅6̅3̅7̅1̅4̅1̅2̅2̅7̅9̅0̅3̅4̅4̅0̅4̅1̅0̅5̅6̅8̅3̅3̅3̅0̅1̅6̅5̅3̅6̅7̅8̅0̅0̅7̅9̅8̅3̅2̅7̅3̅1̅4̅1̅9̅8̅8̅2̅1̅5̅1̅6̅8̅2̅1̅8̅9̅6̅9̅7̅7̅7̅6̅0̅8̅8̅1̅9̅6̅1̅6̅0̅4̅2̅5̅7̅7̅4̅5̅6̅7̅5̅7̅2̅7̅0̅4̅8̅0̅8̅9̅7̅1̅6̅7̅8̅3̅8̅8̅1̅3̅9̅1̅3̅7̅0̅4̅6̅1̅8̅8̅9̅3̅7̅4̅6̅4̅3̅6̅0̅3̅8̅7̅7̅5̅8̅9̅8̅1̅1̅8̅2̅2̅8̅4̅7̅3̅6̅7̅4̅2̅0̅6̅4̅2̅4̅6̅3̅4̅0̅9̅9̅9̅8̅0̅9̅9̅2̅2̅0̅6̅8̅0̅4̅7̅8̅9̅9̅6̅3̅8̅8̅5̅1̅9̅2̅9̅2̅9̅1̅0̅0̅9̅3̅1̅3̅8̅1̅8̅6̅6̅5̅6̅5̅2̅9̅1̅7̅6̅9̅6̅2̅5̅5̅4̅6̅4̅7̅4̅0̅5̅4̅3̅6̅2̅2̅8̅8̅5̅3̅8̅3̅0̅0̅7̅0̅3̅2̅8̅8̅3̅4̅8̅2̅2̅2̅7̅7̅1̅3̅3̅6̅2̅4̅7̅8̅6̅1̅6̅2̅3̅2̅6̅5̅5̅3̅8̅8̅7̅0̅9̅3̅7̅0̅8̅4̅2̅0̅4̅5̅2̅3̅8̅5̅4̅7̅8̅0̅4̅5̅9̅9̅8̅8̅5̅9̅5̅3̅2̅4̅0̅8̅2̅8̅7̅3̅9̅7̅8̅3̅3̅1̅1̅1̅5̅7̅5̅7̅4̅6̅0̅5̅5̅8̅8̅2̅9̅1̅1̅9̅9̅3̅9̅1̅7̅5̅0̅6̅1̅7̅7̅5̅3̅2̅7̅8̅8̅4̅4̅3̅2̅6̅1̅7̅3̅7̅3̅1̅2̅2̅9̅8̅0̅4̅2̅1̅9̅7̅3̅0̅0̅8̅9̅3̅3̅6̅6̅2̅8̅0̅1̅7̅4̅8̅7̅1̅6̅9̅7̅3̅9̅5̅9̅3̅2̅3̅3̅2̅2̅5̅6̅2̅2̅5̅0̅5̅2̅2̅7̅1̅4̅3̅1̅2̅8̅6̅8̅2̅7̅5̅9̅9̅3̅1̅5̅7̅1̅9̅4̅4̅4̅9̅7̅2̅4̅3̅8̅6̅9̅9̅8̅6̅6̅9̅4̅5̅4̅4̅7̅6̅3̅3̅5̅2̅9̅7̅4̅7̅1̅9̅6̅3̅5̅0̅5̅0̅3̅7̅0̅6̅5̅1̅9̅6̅7̅3̅0̅6̅5̅9̅5̅7̅0̅4̅2̅3̅8̅7̅3̅7̅8̅8̅2̅5̅3̅1̅8̅3̅8̅0̅5̅3̅6̅0̅1̅9̅7̅6̅8̅1̅0̅4̅9̅2̅3̅0̅1̅8̅4̅3̅7̅5̅5̅9̅3̅9̅9̅3̅5̅3̅7̅3̅5̅0̅3̅1̅3̅6̅2̅8̅5̅8̅7̅7̅2̅0̅9̅6̅5̅5̅9̅5̅8̅9̅4̅3̅1̅6̅6̅6̅9̅8̅3̅4̅6̅3̅2̅1̅9̅9̅2̅0̅1̅6̅7̅2̅6̅8̅5̅8̅0̅1̅1̅7̅8̅4̅8̅3̅1̅7̅8̅1̅0̅3̅0̅2̅2̅2̅3̅9̅1̅1̅8̅0̅3̅8̅3̅9̅5̅7̅4̅2̅2̅5̅4̅3̅2̅4̅2̅7̅2̅9̅5̅1̅9̅1̅0̅2̅8̅3̅2̅1̅6̅1̅1̅8̅6̅0̅8̅6̅2̅9̅5̅3̅8̅1̅1̅0̅6̅2̅5̅3̅5̅6̅3̅9̅6̅1̅2̅2̅4̅1̅0̅1̅8̅8̅1̅7̅7̅1̅5̅2̅6̅3̅2̅5̅7̅9̅3̅5̅7̅5̅3̅6̅5̅9

Wren

This is based on the Python code here. <lang ecmascript>var divide = Fn.new { |m, n|

   if (m < 0) Fiber.abort("m must not be negative")
   if (n <= 0) Fiber.abort("n must be positive.")
   var quotient = (m/n).floor.toString + "."
   var c = 10 * (m % n)
   var zeros = 0
   while (c > 0 && c < n) {
       c = c * 10
       quotient = quotient + "0"
       zeros = zeros + 1
   }
   var digits = ""
   var passed = {}
   var i = 0
   while (true) {
       if (passed.containsKey(c)) {
           var prefix = digits[0...passed[c]]
           var repetend = digits[passed[c]..-1]
           var result = quotient + prefix + "(" + repetend + ")"
           result = result.replace("(0)", "").trimEnd(".")
           var index = result.indexOf("(")
           if (index == -1) return [result, "", 0]
           result = result.replace("(", "").replace(")", "")
           for (i in 0...zeros) {
               if (repetend[-1] == "0") {
                   result = result[0..-2]
                   repetend = "0" + repetend[0..-2]
               } else break
           }
           return [result + "....", repetend, repetend.count]
       }
       var q = (c / n).floor
       var r = c % n
       passed[c] = i
       digits = digits + q.toString
       i = i + 1
       c = 10 * r
   }

}

var tests = [ [1, 1], [1, 3], [1, 7], [1, 17], [10, 13], [3227, 555], [1, 149] ] for (test in tests) {

   var a = test[0]
   var b = test[1]
   var res = divide.call(a, b)
   System.print("%(a)/%(b) = %(res[0])")
   System.print("Repetend is '%(res[1])'")
   System.print("Period is %(res[2])\n")

}</lang>

Output:

1/1 = 1
Repetend is ''
Period is 0

1/3 = 0.3....
Repetend is '3'
Period is 1

1/7 = 0.142857....
Repetend is '142857'
Period is 6

1/17 = 0.0588235294117647....
Repetend is '0588235294117647'
Period is 16

10/13 = 0.769230....
Repetend is '769230'
Period is 6

3227/555 = 5.8144....
Repetend is '144'
Period is 3

1/149 = 0.0067114093959731543624161073825503355704697986577181208053691275167785234899328859060402684563758389261744966442953020134228187919463087248322147651....
Repetend is '0067114093959731543624161073825503355704697986577181208053691275167785234899328859060402684563758389261744966442953020134228187919463087248322147651'
Period is 148