Sailors, coconuts and a monkey problem

From Rosetta Code
Task
Sailors, coconuts and a monkey problem
You are encouraged to solve this task according to the task description, using any language you may know.

Five sailors are shipwrecked on an island and collect a large pile of coconuts during the day.

That night the first sailor wakes up and decides to take his first share early so tries to divide the pile of coconuts equally into five piles but finds that there is one coconut left over, so he tosses it to a monkey and then hides "his" one of the five equally sized piles of coconuts and pushes the other four piles together to form a single visible pile of coconuts again and goes to bed.

To cut a long story short, each of the sailors in turn gets up once during the night and performs the same actions of dividing the coconut pile into five, finding that one coconut is left over and giving that single remainder coconut to the monkey.

In the morning (after the surreptitious and separate action of each of the five sailors during the night), the remaining coconuts are divided into five equal piles for each of the sailors, whereupon it is found that the pile of coconuts divides equally amongst the sailors with no remainder. (Nothing for the monkey in the morning.)


The task
  1. Calculate the minimum possible size of the initial pile of coconuts collected during the first day.
  2. Use a method that assumes an answer is possible, and then applies the constraints of the tale to see if it is correct. (I.e. no applying some formula that generates the correct answer without integer divisions and remainders and tests on remainders; but constraint solvers are allowed.)
  3. Calculate the size of the initial pile of coconuts if six sailors were marooned and went through a similar process (but split into six piles instead of five of course).
  4. Show your answers here.


Extra credit (optional)
  • Give some indication of the number of coconuts each sailor hides during the night.


Note
  • Of course the tale is told in a world where the collection of any amount of coconuts in a day and multiple divisions of the pile, etc can occur in time fitting the story line, so as not to affect the mathematics.
  • The tale is also told in a version where the monkey also gets a coconut in the morning. This is not that tale!


C.f



Bc[edit]

This script implements a solution in the coconuts function for a number of sailors > 1 and a number of monkeys between 1 and sailors-1. It also executes the coconuts function for some values of sailors/monkeys.

define coconuts(sailors, monkeys) {
print "coconuts(", sailors, ", ", monkeys, ") = "
if (sailors < 2 || monkeys < 1 || sailors <= monkeys) {
return 0
}
blue_cocos = sailors-1
pow_bc = blue_cocos^sailors
x_cocos = pow_bc
while ((x_cocos-blue_cocos)%sailors || (x_cocos-blue_cocos)/sailors < 1) {
x_cocos += pow_bc
}
return (x_cocos/pow_bc*(sailors^sailors)-blue_cocos)*monkeys
}
scale = 0
coconuts(1, 1)
coconuts(2, 1)
coconuts(3, 1)
coconuts(3, 2)
coconuts(4, 1)
coconuts(5, 1)
coconuts(5, 4)
coconuts(6, 1)
coconuts(101, 1)
Output:
$ time bc <coconuts_bc.in
coconuts(1, 1) = 0
coconuts(2, 1) = 11
coconuts(3, 1) = 25
coconuts(3, 2) = 50
coconuts(4, 1) = 765
coconuts(5, 1) = 3121
coconuts(5, 4) = 12484
coconuts(6, 1) = 233275
coconuts(101, 1) = 2731861967715741354199866657915606142014717766608\
81280465910305960827252944980667223385057449021203688309007889238399\
91099564447458450075226030128555294655577015766113909738825769262480\
452415909200510001

real    0m0.141s
user    0m0.031s
sys     0m0.062s

Befunge[edit]

Translation of: C

This is a translation of the second C solution. The output lists the number of sailors, the size of the original pile, and the final share each sailor receives the following morning.

>2+:01p9>`#@_00v
nvg10*g10:+>#1$<
#>\:01g1-%#^_:0v
-|:-1\+1<+/-1g1<
1>$01g.">-",,48v
^g10,+55<.,9.,*<
Output:
2 -> 11         1
3 -> 25         2
4 -> 765        60
5 -> 3121       204
6 -> 233275     13020
7 -> 823537     39990
8 -> 117440505  5044200
9 -> 387420481  14913080

Bracmat[edit]

Translation of: Tcl
(Though without the assert procedure.)
( ( divmod
= a b
. !arg:(?a.?b)&(div$(!a.!b).mod$(!a.!b))
)
& ( overnight
= ns nn result s q r
.  !arg:(?ns.?nn)
& :?result
& 0:?s
& whl
' ( !s+1:?s:~>!ns
& divmod$(!nn.!ns):(?q.?r)
& !r:1
& !q*(!ns+-1):?nn
& !result (!s.!q.!r.!nn):?result
)
& !s:>!ns
& divmod$(!nn.!ns):(?q.0)
& !result
)
& ( minnuts
= nsailors nnuts result sailor takes gives leaves
.  !arg:?nsailors
& 0:?nnuts
& whl
' ( 1+!nnuts:?nnuts
& ~(overnight$(!nsailors.!nnuts):?result)
)
& out$(!nsailors ": " !nnuts)
& whl
' ( !result:(?sailor.?takes.?gives.?leaves) ?result
& out
$ ( str
$ ( " Sailor #"
 !sailor
" takes "
 !takes
", giving "
 !gives
" to the monkey and leaves "
 !leaves
)
)
)
& out
$ ( str
$ ("In the morning, each sailor gets " !leaves*!nsailors^-1 " nuts")
)
)
& 4:?n
& whl
' ( 1+!n:~>6:?n
& out$("Solution with " !n " sailors:")
& minnuts$!n
)
)

Output:

Solution with  5  sailors:
5 :  3121
 Sailor #1 takes 624, giving 1 to the monkey and leaves 2496
 Sailor #2 takes 499, giving 1 to the monkey and leaves 1996
 Sailor #3 takes 399, giving 1 to the monkey and leaves 1596
 Sailor #4 takes 319, giving 1 to the monkey and leaves 1276
 Sailor #5 takes 255, giving 1 to the monkey and leaves 1020
In the morning, each sailor gets 204 nuts
Solution with  6  sailors:
6 :  233275
 Sailor #1 takes 38879, giving 1 to the monkey and leaves 194395
 Sailor #2 takes 32399, giving 1 to the monkey and leaves 161995
 Sailor #3 takes 26999, giving 1 to the monkey and leaves 134995
 Sailor #4 takes 22499, giving 1 to the monkey and leaves 112495
 Sailor #5 takes 18749, giving 1 to the monkey and leaves 93745
 Sailor #6 takes 15624, giving 1 to the monkey and leaves 78120
In the morning, each sailor gets 13020 nuts

C[edit]

#include <stdio.h>
 
int valid(int n, int nuts)
{
int k;
for (k = n; k; k--, nuts -= 1 + nuts/n)
if (nuts%n != 1) return 0;
return nuts && !(nuts%n);
}
 
int main(void)
{
int n, x;
for (n = 2; n < 10; n++) {
for (x = 0; !valid(n, x); x++);
printf("%d: %d\n", n, x);
}
return 0;
}
Output:
2: 11
3: 25
4: 765
5: 3121
6: 233275
7: 823537
8: 117440505
9: 387420481

But it's faster to search backwards: if everyone receives some coconuts, see if we can backtrack to the original pile:

#include <stdio.h>
 
// calculates if everyone got some nuts in the end, what was the original pile
// returns 0 if impossible
int total(int n, int nuts)
{
int k;
for (k = 0, nuts *= n; k < n; k++) {
if (nuts % (n-1)) return 0;
nuts += nuts / (n-1) + 1;
}
return nuts;
}
 
int main(void)
{
int n, x, t;
for (n = 2; n < 10; n++) {
for (x = 1, t = 0; !(t = total(n, x)); x++);
printf("%d: %d\t%d\n", n, t, x);
}
return 0;
}
Output:

sailers: original pile, final share

2: 11   1
3: 25   2
4: 765  60
5: 3121 204
6: 233275       13020
7: 823537       39990
8: 117440505    5044200
9: 387420481    14913080

C#[edit]

Translation of: Java
class Test
{
static bool valid(int n, int nuts)
{
for (int k = n; k != 0; k--, nuts -= 1 + nuts / n)
{
if (nuts % n != 1)
{
return false;
}
}
 
return nuts != 0 && (nuts % n == 0);
}
 
static void Main(string[] args)
{
int x = 0;
for (int n = 2; n < 10; n++)
{
while (!valid(n, x))
x++;
System.Console.WriteLine(n + ": " + x);
}
}
}
2: 11
3: 25
4: 765
5: 3121
6: 233275
7: 823537
8: 117440505
9: 387420481

D[edit]

Translation of: Kotlin
 
import std.stdio;
 
void main() {
auto coconuts = 11;
 
outer:
foreach (ns; 2..10) {
int[] hidden = new int[ns];
coconuts = (coconuts / ns) * ns + 1;
while (true) {
auto nc = coconuts;
foreach (s; 1..ns+1) {
if (nc % ns == 1) {
hidden[s-1] = nc/ns;
nc -= hidden[s-1] + 1;
if (s==ns && nc%ns==0) {
writeln(ns, " sailors require a minimum of ", coconuts, " coconuts");
foreach (t; 1..ns+1) {
writeln("\tSailor ", t, " hides ", hidden[t - 1]);
}
writeln("\tThe monkey gets ", ns);
writeln("\tFinally, each sailor takes ", nc / ns);
continue outer;
}
} else {
break;
}
}
coconuts += ns;
}
}
}
 
Output:
2 sailors require a minimum of 11 coconuts
	Sailor 1 hides 5
	Sailor 2 hides 2
	The monkey gets 2
	Finally, each sailor takes 1

3 sailors require a minimum of 25 coconuts
	Sailor 1 hides 8
	Sailor 2 hides 5
	Sailor 3 hides 3
	The monkey gets 3
	Finally, each sailor takes 2

4 sailors require a minimum of 765 coconuts
	Sailor 1 hides 191
	Sailor 2 hides 143
	Sailor 3 hides 107
	Sailor 4 hides 80
	The monkey gets 4
	Finally, each sailor takes 60

5 sailors require a minimum of 3121 coconuts
	Sailor 1 hides 624
	Sailor 2 hides 499
	Sailor 3 hides 399
	Sailor 4 hides 319
	Sailor 5 hides 255
	The monkey gets 5
	Finally, each sailor takes 204

6 sailors require a minimum of 233275 coconuts
	Sailor 1 hides 38879
	Sailor 2 hides 32399
	Sailor 3 hides 26999
	Sailor 4 hides 22499
	Sailor 5 hides 18749
	Sailor 6 hides 15624
	The monkey gets 6
	Finally, each sailor takes 13020

7 sailors require a minimum of 823537 coconuts
	Sailor 1 hides 117648
	Sailor 2 hides 100841
	Sailor 3 hides 86435
	Sailor 4 hides 74087
	Sailor 5 hides 63503
	Sailor 6 hides 54431
	Sailor 7 hides 46655
	The monkey gets 7
	Finally, each sailor takes 39990

8 sailors require a minimum of 117440505 coconuts
	Sailor 1 hides 14680063
	Sailor 2 hides 12845055
	Sailor 3 hides 11239423
	Sailor 4 hides 9834495
	Sailor 5 hides 8605183
	Sailor 6 hides 7529535
	Sailor 7 hides 6588343
	Sailor 8 hides 5764800
	The monkey gets 8
	Finally, each sailor takes 5044200

9 sailors require a minimum of 387420481 coconuts
	Sailor 1 hides 43046720
	Sailor 2 hides 38263751
	Sailor 3 hides 34012223
	Sailor 4 hides 30233087
	Sailor 5 hides 26873855
	Sailor 6 hides 23887871
	Sailor 7 hides 21233663
	Sailor 8 hides 18874367
	Sailor 9 hides 16777215
	The monkey gets 9
	Finally, each sailor takes 14913080

Elixir[edit]

Translation of: Ruby

Brute Force[edit]

defmodule RC do
def valid?(sailor, nuts), do: valid?(sailor, nuts, sailor)
 
def valid?(sailor, nuts, 0), do: nuts > 0 and rem(nuts,sailor) == 0
def valid?(sailor, nuts, _) when rem(nuts,sailor)!=1, do: false
def valid?(sailor, nuts, i) do
valid?(sailor, nuts - div(nuts,sailor) - 1, i-1)
end
end
 
Enum.each([5,6], fn sailor ->
nuts = Enum.find(Stream.iterate(sailor, &(&1+1)), fn n -> RC.valid?(sailor, n) end)
IO.puts "\n#{sailor} sailors => #{nuts} coconuts"
Enum.reduce(0..sailor, nuts, fn _,n ->
{d, r} = {div(n,sailor), rem(n,sailor)}
IO.puts " #{inspect [n, d, r]}"
n - 1 - d
end)
end)
Output:
5 sailors => 3121 coconuts
  [3121, 624, 1]
  [2496, 499, 1]
  [1996, 399, 1]
  [1596, 319, 1]
  [1276, 255, 1]
  [1020, 204, 0]

6 sailors => 233275 coconuts
  [233275, 38879, 1]
  [194395, 32399, 1]
  [161995, 26999, 1]
  [134995, 22499, 1]
  [112495, 18749, 1]
  [93745, 15624, 1]
  [78120, 13020, 0]

Faster version[edit]

defmodule Sailor do
def coconuts(sailor), do: coconuts(sailor, sailor)
defp coconuts(sailor, nuts) do
if n = do_coconuts(sailor, nuts, sailor), do: n, else: coconuts(sailor, nuts+sailor)
end
 
defp do_coconuts(_sailor, nuts, 0), do: nuts
defp do_coconuts(sailor, nuts, _) when rem(nuts, sailor-1) != 0, do: nil
defp do_coconuts(sailor, nuts, i) do
do_coconuts(sailor, nuts + div(nuts, sailor-1) + 1, i-1)
end
end
 
Enum.each(2..9, fn sailor ->
IO.puts "#{sailor}: #{Sailor.coconuts(sailor)}"
end)
Output:
2: 11
3: 25
4: 765
5: 3121
6: 233275
7: 823537
8: 117440505
9: 387420481

Forth[edit]

Translation of: uBasic/4tH
: total
over * over 1- rot 0 ?do
over over mod if dup xor swap leave else over over / 1+ rot + swap then
loop drop
;
 
: sailors
1+ 2 ?do
1 begin i over total dup 0= while drop 1+ repeat cr i 0 .r ." : " . .
loop
;
 
9 sailors
Output:
2: 11 1
3: 25 2
4: 765 60
5: 3121 204
6: 233275 13020
7: 823537 39990
8: 117440505 5044200
9: 387420481 14913080  ok

Haskell[edit]

This program works by applying a function to increasing multiples of the number of sailors. The function takes a potential final number of coconuts (at the time the sailors awaken) and works backwards to get to the initial number of coconuts. At every step, it will abort the computation if the current number of coconuts can't arise as a result of splitting the previous pile.

import Control.Monad
import Data.Maybe
import System.Environment
 
-- Takes the number of sailors and the final number of coconuts. Returns
-- Just the associated initial number of coconuts and Nothing otherwise.
tryFor :: Int -> Int -> Maybe Int
tryFor s = foldr (>=>) pure $ replicate s step
where step n
| n `mod` (s - 1) == 0 = Just $ n * s `div` (s - 1) + 1
| otherwise = Nothing
 
-- Gets the number of sailors from the first command-line argument and
-- assumes 5 as a default if none is given. Then uses tryFor to find the
-- smallest solution.
main :: IO ()
main = do
args <- getArgs
let n = case args of
[] -> 5
s:_ -> read s
a = head . mapMaybe (tryFor n) $ [n, 2 * n ..]
print a

Examples:

$ ./coconuts
3121
$ ./coconuts 4
765
$ ./coconuts 6
233275

J[edit]

Here, we assume an answer which is less than 10000, and try each possibility, constraining ourselves to the list of answers which are valid. (As it happens, there's only one of those.)

   I.(=<.)%&5 verb def'4*(y-1)%5'^:5 i.10000
3121

These sailors must count coconuts extremely quickly.

When we do this with six sailors, it turns out that we have to assume a larger initial value:

   I.(=<.)%&6 verb def'5*(y-1)%6'^:6 i.1000000
233275 513211 793147

If it were not obvious which of the answers here was the minimum value we could additionally pick the smallest value from the list. But that would require typing two extra characters (for example, replace I. with i.&1), and most people already have so much trouble reading J that the extra code would surely be too much.

Java[edit]

Translation of: C
public class Test {
 
static boolean valid(int n, int nuts) {
for (int k = n; k != 0; k--, nuts -= 1 + nuts / n)
if (nuts % n != 1)
return false;
return nuts != 0 && (nuts % n == 0);
}
 
public static void main(String[] args) {
int x = 0;
for (int n = 2; n < 10; n++) {
while (!valid(n, x))
x++;
System.out.printf("%d: %d%n", n, x);
}
}
}
2: 11
3: 25
4: 765
5: 3121
6: 233275
7: 823537
8: 117440505
9: 387420481

JavaScript[edit]

ES5[edit]

Translation of: Python

( As in the recursive Python example )

(function () {
 
// wakeSplit :: Int -> Int -> Int -> Int
function wakeSplit(intNuts, intSailors, intDepth) {
var nDepth = intDepth !== undefined ? intDepth : intSailors,
portion = Math.floor(intNuts / intSailors),
remain = intNuts % intSailors;
 
return 0 >= portion || remain !== (nDepth ? 1 : 0) ?
null : nDepth ? wakeSplit(
intNuts - portion - remain, intSailors, nDepth - 1
) : intNuts;
}
 
// TEST for 5, 6, and 7 intSailors
return [5, 6, 7].map(function (intSailors) {
var intNuts = intSailors;
 
while (!wakeSplit(intNuts, intSailors)) intNuts += 1;
 
return intNuts;
});
})();
Output:
[3121, 233275, 823537]

ES6[edit]

Adding just a touch of curry to the coconuts. (See the Rosetta Code task: Currying)

(() => {
 
// wakeSplit :: Int -> Int -> Int -> Int
let wakeSplit = (intSailors, intNuts, intDepth) => {
let nDepth = intDepth !== undefined ? intDepth : intSailors,
portion = Math.floor(intNuts / intSailors),
remain = intNuts % intSailors;
 
return 0 >= portion || remain !== (nDepth ? 1 : 0) ?
null : nDepth ? wakeSplit(
intSailors, intNuts - portion - remain, nDepth - 1
) : intNuts;
};
 
 
//GENERIC FUNCTIONS
 
// curry :: ((a, b) -> c) -> a -> b -> c
let curry = f => a => b => f(a, b),
 
// until :: (a -> Bool) -> (a -> a) -> a -> a
until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
},
 
// succ :: Int -> Int
succ = x => x + 1;
 
 
// TEST for 5, 6, and 7 Sailors
return [5, 6, 7].map(intSailors => {
let intNuts = intSailors,
test = curry(wakeSplit)(intSailors);
 
return until(test, succ, intNuts);
 
});
})();
Output:
[3121, 233275, 823537]

jq[edit]

Works with: jq version 1.4

The first solution presented in this section is based on a simulation of the nighttime and daytime activities, and the second (much faster) solution works backwards.

If your jq does not have "until" defined, here is its definition:

def until(cond; next): def _until: if cond then . else (next|_until) end; _until;

Simulation[edit]

# If n (the input) is an admissible number of coconuts with respect to
# the night-time squirreling away of the coconuts by "sailors" sailors, then give 1 to the
# monkey, let one sailor squirrel away (1/sailors) coconuts, and yield the remaining number;
# otherwise, return false:
def squirrel(sailors):
def admissible: if . then (. % sailors) == 1 else . end;
 
if admissible then . - ((. - 1) / sailors) - 1
else false
end;
 
def nighttime(sailors):
reduce range(0; sailors) as $i (.; squirrel(sailors));
 
def morning(sailors):
if . then (. % sailors) == 0
else false
end;
 
# Test whether the input is a valid number of coconuts with respect to the story:
def valid(sailors): nighttime(sailors) | morning(sailors);

Five sailors: Find the minimum number of coconuts if there are 5 sailors -- start at 1 as there must be at least one to give to the monkey during the night:

1 | until( valid(5); . + 1)
Output:

3121

Six sailors: Find the minimum number of coconuts if there are 6 sailors -- start at 1 as there must be at least one to give to the monkey during the night:

1 | until( valid(6); . + 1)
Output:

233275

Working backwards[edit]

# If n (the input) is the number of coconuts remaining after
# the surreptitious squirreling away by one sailor,
# then emit the number of coconuts which that sailor originally
# saw if n is admissible, otherwise emit false:
def unsquirrel(sailors):
if . and (. % (sailors - 1) == 0)
then 1 + (sailors * (. / (sailors - 1)))
else false
end;
 
# If in the end each sailor received n coconuts (where n is the input), how many coconuts
# were there initially?
def backwards(sailors):
reduce range(0; sailors) as $i (. * sailors; unsquirrel(sailors));
 
def solve:
. as $sailors
# state: [ final_number_per_sailor, original_number_of_coconuts]
| [-1] | until( .[1]; .[0] += 1 | .[1] = (.[0] | backwards($sailors)) )
| "With \($sailors) sailors, there were originally \(.[1]) coconuts,"+
" and each sailor finally ended up with \(.[0])." ;
 
range(2;9) | solve
Output:
With 2 sailors, there were originally 3 coconuts, and each sailor finally ended up with 0.
With 3 sailors, there were originally 25 coconuts, and each sailor finally ended up with 2.
With 4 sailors, there were originally 765 coconuts, and each sailor finally ended up with 60.
With 5 sailors, there were originally 3121 coconuts, and each sailor finally ended up with 204.
With 6 sailors, there were originally 233275 coconuts, and each sailor finally ended up with 13020.
With 7 sailors, there were originally 823537 coconuts, and each sailor finally ended up with 39990.
With 8 sailors, there were originally 117440505 coconuts, and each sailor finally ended up with 5044200.

Kotlin[edit]

// version 1.1.2
 
fun main(args: Array<String>) {
var coconuts = 11
outer@ for (ns in 2..9) {
val hidden = IntArray(ns)
coconuts = (coconuts / ns) * ns + 1
while (true) {
var nc = coconuts
for (s in 1..ns) {
if (nc % ns == 1) {
hidden[s - 1] = nc / ns
nc -= hidden[s - 1] + 1
if (s == ns && nc % ns == 0) {
println("$ns sailors require a minimum of $coconuts coconuts")
for (t in 1..ns) println("\tSailor $t hides ${hidden[t - 1]}")
println("\tThe monkey gets $ns")
println("\tFinally, each sailor takes ${nc / ns}\n")
continue@outer
}
}
else break
}
coconuts += ns
}
}
}
Output:
2 sailors require a minimum of 11 coconuts
	Sailor 1 hides 5
	Sailor 2 hides 2
	The monkey gets 2
	Finally, each sailor takes 1

3 sailors require a minimum of 25 coconuts
	Sailor 1 hides 8
	Sailor 2 hides 5
	Sailor 3 hides 3
	The monkey gets 3
	Finally, each sailor takes 2

4 sailors require a minimum of 765 coconuts
	Sailor 1 hides 191
	Sailor 2 hides 143
	Sailor 3 hides 107
	Sailor 4 hides 80
	The monkey gets 4
	Finally, each sailor takes 60

5 sailors require a minimum of 3121 coconuts
	Sailor 1 hides 624
	Sailor 2 hides 499
	Sailor 3 hides 399
	Sailor 4 hides 319
	Sailor 5 hides 255
	The monkey gets 5
	Finally, each sailor takes 204

6 sailors require a minimum of 233275 coconuts
	Sailor 1 hides 38879
	Sailor 2 hides 32399
	Sailor 3 hides 26999
	Sailor 4 hides 22499
	Sailor 5 hides 18749
	Sailor 6 hides 15624
	The monkey gets 6
	Finally, each sailor takes 13020

7 sailors require a minimum of 823537 coconuts
	Sailor 1 hides 117648
	Sailor 2 hides 100841
	Sailor 3 hides 86435
	Sailor 4 hides 74087
	Sailor 5 hides 63503
	Sailor 6 hides 54431
	Sailor 7 hides 46655
	The monkey gets 7
	Finally, each sailor takes 39990

8 sailors require a minimum of 117440505 coconuts
	Sailor 1 hides 14680063
	Sailor 2 hides 12845055
	Sailor 3 hides 11239423
	Sailor 4 hides 9834495
	Sailor 5 hides 8605183
	Sailor 6 hides 7529535
	Sailor 7 hides 6588343
	Sailor 8 hides 5764800
	The monkey gets 8
	Finally, each sailor takes 5044200

9 sailors require a minimum of 387420481 coconuts
	Sailor 1 hides 43046720
	Sailor 2 hides 38263751
	Sailor 3 hides 34012223
	Sailor 4 hides 30233087
	Sailor 5 hides 26873855
	Sailor 6 hides 23887871
	Sailor 7 hides 21233663
	Sailor 8 hides 18874367
	Sailor 9 hides 16777215
	The monkey gets 9
	Finally, each sailor takes 14913080

Perl 6[edit]

Works with: rakudo version 2015.09

There is nowhere in the spec where it explicitly states that the sailors cannot equally share zero coconuts in the morning. Actually, The On-Line Encyclopedia of Integer Sequences A002021 considers the cases for 1 and 2 sailors equally sharing zero coconuts in the morning to be the correct answer.

This will test combinations of sailors and coconuts to see if they form a valid pairing. The first 6 are done using brute force, testing every combination until a valid one is found. For cases of 7 to 15 sailors, it uses a carefully crafted filter to drastically reduce the number of trials needed to find a valid case (to one, as it happens... :-) )

my @ones = flat 'th', 'st', 'nd', 'rd', 'th' xx 6;
my @teens = 'th' xx 10;
my @suffix = lazy flat (@ones, @teens, @ones xx 8) xx *;
 
# brute force the first six
for 1 .. 6 -> $sailors { for $sailors .. * -> $coconuts { last if check( $sailors, $coconuts ) } }
 
# finesse 7 through 15
for 7 .. 15 -> $sailors { next if check( $sailors, coconuts( $sailors ) ) }
 
sub is_valid ( $sailors is copy, $nuts is copy ) {
return 0, 0 if $sailors == $nuts == 1;
my @shares;
for ^$sailors {
return () unless $nuts % $sailors == 1;
push @shares, ($nuts - 1) div $sailors;
$nuts -= (1 + $nuts div $sailors);
}
push @shares, $nuts div $sailors;
return @shares if !?($nuts % $sailors);
}
 
sub check ($sailors, $coconuts) {
if my @piles = is_valid($sailors, $coconuts) {
say "\nSailors $sailors: Coconuts $coconuts:";
for ^(@piles - 1) -> $k {
say "{$k+1}@suffix[$k+1] takes @piles[$k], gives 1 to the monkey."
}
say "The next morning, each sailor takes @piles[*-1]\nwith none left over for the monkey.";
return True;
}
False;
}
 
multi sub coconuts ( $sailors where { $sailors % 2 == 0 } ) { ($sailors - 1) * ($sailors ** $sailors - 1) }
multi sub coconuts ( $sailors where { $sailors % 2 == 1 } ) { $sailors ** $sailors - $sailors + 1 }
Output:
Sailors 1: Coconuts 1:
1st takes 0, gives 1 to the monkey.
The next morning, each sailor takes 0
with none left over for the monkey.

Sailors 2: Coconuts 3:
...

Sailors 5: Coconuts 3121:
1st takes 624, gives 1 to the monkey.
2nd takes 499, gives 1 to the monkey.
3rd takes 399, gives 1 to the monkey.
4th takes 319, gives 1 to the monkey.
5th takes 255, gives 1 to the monkey.
The next morning, each sailor takes 204
with none left over for the monkey.

Sailors 6: Coconuts 233275:
1st takes 38879, gives 1 to the monkey.
2nd takes 32399, gives 1 to the monkey.
3rd takes 26999, gives 1 to the monkey.
4th takes 22499, gives 1 to the monkey.
5th takes 18749, gives 1 to the monkey.
6th takes 15624, gives 1 to the monkey.
The next morning, each sailor takes 13020
with none left over for the monkey.

Sailors 7: Coconuts 823537:
...

Sailors 15: Coconuts 437893890380859361:
...

Phix[edit]

The morning pile must be a multiple of sailors, so this only tries multiples of sailors! Needed an ugly kludge for solve(1), the limit of 1 billion suffices for solve(9), above that gets run-time type check errors as capacity of ints are blown anyway.

procedure solve(integer sailors)
integer m, sm1 = sailors-1
if sm1=0 then -- edge condition for solve(1) [ avoid /0 ]
m = sailors
else
for n=sailors to 1_000_000_000 by sailors do -- morning pile divisible by #sailors
m = n
for j=1 to sailors do -- see if all of the sailors could..
if remainder(m,sm1)!=0 then -- ..have pushed together sm1 piles
m = 0 -- (no: try a higher n)
exit
end if
m = sailors*m/sm1+1 -- add sailor j's stash and one for the monkey
end for
if m!=0 then exit end if
end for
end if
printf(1,"Solution with %d sailors: %d\n",{sailors,m})
for i=1 to sailors do
m -= 1 -- one for the monkey
m /= sailors
printf(1,"Sailor #%d takes %d, giving 1 to the monkey and leaving %d\n",{i,m,m*sm1})
m *= (sm1)
end for
printf(1,"In the morning each sailor gets %d nuts\n",m/sailors)
end procedure
 
solve(5)
solve(6)
Output:
Solution with 5 sailors: 3121
Sailor #1 takes 624, giving 1 to the monkey and leaving 2496
Sailor #2 takes 499, giving 1 to the monkey and leaving 1996
Sailor #3 takes 399, giving 1 to the monkey and leaving 1596
Sailor #4 takes 319, giving 1 to the monkey and leaving 1276
Sailor #5 takes 255, giving 1 to the monkey and leaving 1020
In the morning each sailor gets 204 nuts
Solution with 6 sailors: 233275
Sailor #1 takes 38879, giving 1 to the monkey and leaving 194395
Sailor #2 takes 32399, giving 1 to the monkey and leaving 161995
Sailor #3 takes 26999, giving 1 to the monkey and leaving 134995
Sailor #4 takes 22499, giving 1 to the monkey and leaving 112495
Sailor #5 takes 18749, giving 1 to the monkey and leaving 93745
Sailor #6 takes 15624, giving 1 to the monkey and leaving 78120
In the morning each sailor gets 13020 nuts

Python[edit]

You may want to read Solving the Monkey and coconuts problem to get more background on the evolution of the Python code.

Python: Procedural[edit]

def monkey_coconuts(sailors=5):
"Parameterised the number of sailors using an inner loop including the last mornings case"
nuts = sailors
while True:
n0, wakes = nuts, []
for sailor in range(sailors + 1):
portion, remainder = divmod(n0, sailors)
wakes.append((n0, portion, remainder))
if portion <= 0 or remainder != (1 if sailor != sailors else 0):
nuts += 1
break
n0 = n0 - portion - remainder
else:
break
return nuts, wakes
 
if __name__ == "__main__":
for sailors in [5, 6]:
nuts, wake_stats = monkey_coconuts(sailors)
print("\nFor %i sailors the initial nut count is %i" % (sailors, nuts))
print("On each waking, the nut count, portion taken, and monkeys share are:\n ",
',\n '.join(repr(ws) for ws in wake_stats))
Output:
For 5 sailors the initial nut count is 3121
On each waking, the nut count, portion taken, and monkeys share are:
  (3121, 624, 1),
  (2496, 499, 1),
  (1996, 399, 1),
  (1596, 319, 1),
  (1276, 255, 1),
  (1020, 204, 0)

For 6 sailors the initial nut count is 233275
On each waking, the nut count, portion taken, and monkeys share are:
  (233275, 38879, 1),
  (194395, 32399, 1),
  (161995, 26999, 1),
  (134995, 22499, 1),
  (112495, 18749, 1),
  (93745, 15624, 1),
  (78120, 13020, 0)

Python: Recursive[edit]

def wake_and_split(n0, sailors, depth=None):
if depth is None:
depth = sailors
portion, remainder = divmod(n0, sailors)
if portion <= 0 or remainder != (1 if depth else 0):
return None
else:
return n0 if not depth else wake_and_split(n0 - portion - remainder, sailors, depth - 1)
 
 
def monkey_coconuts(sailors=5):
"Parameterised the number of sailors using recursion including the last mornings case"
nuts = sailors
while True:
if wake_and_split(n0=nuts, sailors=sailors) is None:
nuts += 1
else:
break
return nuts
 
if __name__ == "__main__":
for sailors in [5, 6]:
nuts = monkey_coconuts(sailors)
print("For %i sailors the initial nut count is %i" % (sailors, nuts))
 
Output:
For 5 sailors the initial nut count is 3121
For 6 sailors the initial nut count is 233275

by solving Diophantine equation[edit]

The following is a more or less general solution for arbitrary number of sailors and varying numbers of coconuts the monkey gets. The monkey can be given more coconuts than there are sailors each turn. This is not part of task requirement.

# gives one solution of (x,y) for a x + by = c
def dioph(a, b, c):
aa,bb,x,y = a, b, 0, 1
 
while True:
q,a,b = a//b, b, a%b
x,y = y - q*x, x
if abs(a) == 1: break
 
if y*aa % bb != 1: y = -y
x,y = y*c, (c - aa*y*c)//bb
#assert(x*aa + y*bb == c)
return x,y
 
# rems: what monkey got each turn
# min_share: each sailor needs to get at least this many in the final round
def calcnuts(rems, min_share = 0):
n, r = len(rems) - 1, 0
c = (n - 1)**n
for x in rems: r,c = r + x*c, c//(n-1)*n
 
a, b = (n-1)**n, n**(n+1)
x, y = dioph(a, -b, r)
k = (min_share - y + a - 1)//a
return x + k*b, y + k*a
 
def distribute(nuts, monkey_nuts):
n = len(monkey_nuts) - 1
print("\n%d sailors, %d nuts:"%(n, nuts))
for r in monkey_nuts[:-1]:
p = (nuts - r)//n
print("\tNuts %d, hide %d, monkey gets %d" % (nuts, p, r))
nuts = p*(n - 1)
 
r = monkey_nuts[-1]
p = (nuts - r)//n
print("Finally:\n\tNuts %d, each share %d, monkey gets %d" % (nuts, p, r))
 
for sailors in range(2, 10):
monkey_loot = [1]*sailors + [0]
distribute(calcnuts(monkey_loot, 1)[0], monkey_loot)
 
# many sailors, many nuts
#for i in range(1, 5): print(10**i, calcnuts([1]*10**i + [0])[0])

Racket[edit]

Translation of: Python
#lang racket
 
(define (wake-and-split nuts sailors depth wakes)
(define-values (portion remainder) (quotient/remainder nuts sailors))
(define monkey (if (zero? depth) 0 1))
(define new-wakes (cons (list nuts portion remainder) wakes))
(and (positive? portion)
(= remainder monkey)
(if (zero? depth)
new-wakes
(wake-and-split (- nuts portion remainder) sailors (sub1 depth) new-wakes))))
 
(define (sleep-and-split nuts sailors)
(wake-and-split nuts sailors sailors '()))
 
(define (monkey_coconuts (sailors 5))
(let loop ([nuts sailors])
(or (sleep-and-split nuts sailors)
(loop (add1 nuts)))))
 
(for ([sailors (in-range 5 7)])
(define wakes (monkey_coconuts sailors))
(printf "For ~a sailors the initial nut count is ~a\n" sailors (first (last wakes)))
(map displayln (reverse wakes))
(newline))
Output:
For 5 sailors the initial nut count is 3121
(3121 624 1)
(2496 499 1)
(1996 399 1)
(1596 319 1)
(1276 255 1)
(1020 204 0)

For 6 sailors the initial nut count is 233275
(233275 38879 1)
(194395 32399 1)
(161995 26999 1)
(134995 22499 1)
(112495 18749 1)
(93745 15624 1)
(78120 13020 0)

REXX[edit]

uses a subroutine[edit]

Translation of: C
{from the 1st C example}
/*REXX program  solves  a  riddle  of 5 sailors, a pile of coconuts, and a monkey.      */
parse arg L H .; if L=='' then L=5 /*L not specified? Then use default.*/
if H=='' then H=6 /*H " " " " default.*/
/*{Tars is an old name for sailors.} */
do n=L to H /*traipse through a number of sailors. */
do $=0 while \valid(n,$); end /*perform while not valid coconuts. */
say 'sailors='n " coconuts="$ /*display number of sailors & coconuts.*/
end /*n*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
valid: procedure; parse arg n,nuts /*obtain the number sailors & coconuts.*/
do k=n by -1 for n /*step through the possibilities. */
if nuts//n\==1 then return 0 /*Not one coconut left? No solution. */
nuts=nuts - (1+nuts%n) /*subtract number of coconuts from pile*/
end /*k*/
return (nuts\==0) & \(nuts//n\==0) /*see if number coconuts>0 & remainder.*/

Programming note:   The parentheses in the last REXX (return) statement aren't necessary, but help for readability.

output   when using the default inputs:

sailors=5   coconuts=3121
sailors=6   coconuts=233275

uses in-line code[edit]

This REXX version is the same as the above version (but the defaults are different),
and it also eliminates the use of a subroutine, making it faster.

/*REXX program  solves  a  riddle of 5 sailors, a pile of coconuts, and a monkey.       */
 
do n=2 to 9 /*traipse through number of sailors. */
do $=0; nuts=$ /*perform while not valid # coconuts. */
do k=n by -1 for n /*step through the possibilities. */
if nuts//n\==1 then iterate $ /*Not one coconut left? No solution.*/
nuts=nuts - (1+nuts%n) /*subtract number of coconuts from pile*/
end /*k*/
if (nuts\==0) & \(nuts//n\==0) then leave /*is this a solution to the riddle ? */
end /*$*/
say 'sailors='n " coconuts="$ /*display number of sailors & coconuts.*/
end /*n*/ /*stick a fork in it, we're all done. */

output   when using the default inputs:

sailors=2   coconuts=11
sailors=3   coconuts=25
sailors=4   coconuts=765
sailors=5   coconuts=3121
sailors=6   coconuts=233275
sailors=7   coconuts=823537
sailors=8   coconuts=117440505
sailors=9   coconuts=387420481

shows the shares[edit]

Translation of: C
{from the 2nd C example}
/*REXX program  solves a  riddle  of 5 sailors, a pile of coconuts, and a monkey.       */
parse arg L H .; if L=='' then L=2 /*L not specified? Then use default.*/
if H=='' then H=9 /*H " " " " " */
/*{Tars is an old name for sailors.} */
do n=L to H /*traipse through the number of sailors*/
do $=1 until t\==0 /*perform while number coconuts not 0. */
t=total(n,$) /*perform while not valid coconuts. */
end /*$*/
say 'sailors='n " coconuts="t ' share='$
end /*n*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
total: procedure; parse arg n,nuts /*obtain the number sailors & coconuts.*/
nuts=nuts*n
nn=n-1 /*NN is used as calculation shortcut. */
do k=0 for n /*step through the possibilities. */
if nuts//nn\==0 then return 0 /*Not one coconut left? No solution. */
nuts=nuts + nuts%nn + 1 /*bump the number coconuts to the pile.*/
end /*k*/
return nuts /*see if number coconuts>0 & remainder.*/

output   when using the default inputs:

sailors=2   coconuts=11   share=1
sailors=3   coconuts=25   share=2
sailors=4   coconuts=765   share=60
sailors=5   coconuts=3121   share=204
sailors=6   coconuts=233275   share=13020
sailors=7   coconuts=823537   share=39990
sailors=8   coconuts=117440505   share=5044200
sailors=9   coconuts=387420481   share=14913080

Ruby[edit]

Brute Force[edit]

def valid?(sailor, nuts)
sailor.times do
return false if (nuts % sailor) != 1
nuts -= 1 + nuts / sailor
end
nuts > 0 and nuts % sailor == 0
end
 
[5,6].each do |sailor|
n = sailor
n += 1 until valid?(sailor, n)
puts "\n#{sailor} sailors => #{n} coconuts"
(sailor+1).times do
div, mod = n.divmod(sailor)
puts " #{[n, div, mod]}"
n -= 1 + div
end
end
Output:
5 sailors => 3121 coconuts
  [3121, 624, 1]
  [2496, 499, 1]
  [1996, 399, 1]
  [1596, 319, 1]
  [1276, 255, 1]
  [1020, 204, 0]

6 sailors => 233275 coconuts
  [233275, 38879, 1]
  [194395, 32399, 1]
  [161995, 26999, 1]
  [134995, 22499, 1]
  [112495, 18749, 1]
  [93745, 15624, 1]
  [78120, 13020, 0]

Faster version[edit]

Works with: Ruby version 2.1+
def coconuts(sailor)
sailor.step(by:sailor) do |nuts|
flag = sailor.times do
break if nuts % (sailor-1) != 0
nuts += nuts / (sailor-1) + 1
end
return nuts if flag
end
end
 
(2..9).each do |sailor|
puts "#{sailor}: #{coconuts(sailor)}"
end
Output:
2: 11
3: 25
4: 765
5: 3121
6: 233275
7: 823537
8: 117440505
9: 387420481

Implementation of Analysis on Discussion Page[edit]

NOTE: This example does not use a method that assumes an answer then applies the constraints of the tale to see if it is correct, and so technically does not perform the task.
 
def ng (sailors)
def _ng (sailors, iter, start)
n, g = [start], [start/sailors]
(1..iter).each{|s|
g[s],rem = n[s-1].divmod(sailors-1)
rem > 0 ? (return false) : n[s] = g[s]*sailors + 1
}
return [n,g]
end
n, start, step = [], sailors*(sailors-1), 1
(2..sailors).each{|s|
g=0; until n=_ng(sailors,s,start + g*step*sailors*(sailors-1)) do g=g+1 end
start,step = n[0][0], step*(sailors-1)
}
return n
end
 
Output:
 
(3..10).each{|sailors| puts "Number of sailors = #{sailors}"; p ng(sailors)}
 
Number of sailors = 3
[[6, 10, 16, 25], [2, 3, 5, 8]]
Number of sailors = 4
[[240, 321, 429, 573, 765], [60, 80, 107, 143, 191]]
Number of sailors = 5
[[1020, 1276, 1596, 1996, 2496, 3121], [204, 255, 319, 399, 499, 624]]
Number of sailors = 6
[[78120, 93745, 112495, 134995, 161995, 194395, 233275], [13020, 15624, 18749, 22499, 26999, 32399, 38879]]
Number of sailors = 7
[[279930, 326586, 381018, 444522, 518610, 605046, 705888, 823537], [39990, 46655, 54431, 63503, 74087, 86435, 100841, 117648]]
Number of sailors = 8
[[40353600, 46118401, 52706745, 60236281, 68841465, 78675961, 89915385, 102760441, 117440505], [5044200, 5764800, 6588343, 7529535, 8605183, 9834495, 11239423, 12845055, 14680063]]
Number of sailors = 9
[[134217720, 150994936, 169869304, 191102968, 214990840, 241864696, 272097784, 306110008, 344373760, 387420481], [14913080, 16777215, 18874367, 21233663, 23887871, 26873855, 30233087, 34012223, 38263751, 43046720]]
Number of sailors = 10
[[31381059600, 34867844001, 38742048891, 43046720991, 47829689991, 53144099991, 59048999991, 65609999991, 72899999991, 80999999991, 89999999991], [3138105960, 3486784400, 3874204889, 4304672099, 4782968999, 5314409999, 5904899999, 6560999999, 7289999999, 8099999999, 8999999999]]

Did someone ask the value for 100 sailors?

 
n = ng(100)
(0..100).each{|g| puts "#{n[0][100-g]}:#{n[1][100-g]}"}
 

The number of coconuts requires is as follows, the whole output is at Sailors, coconuts and a monkey problem/Ruby output 100

9899999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999901

Sidef[edit]

Translation of: Bc
func coconuts(sailors, monkeys=1) {
if ((sailors < 2) || (monkeys < 1) || (sailors <= monkeys)) {
return 0
}
var blue_cocos = sailors-1
var pow_bc = blue_cocos**sailors
var x_cocos = pow_bc
while ((x_cocos-blue_cocos)%sailors || ((x_cocos-blue_cocos)/sailors < 1)) {
x_cocos += pow_bc
}
return monkeys*(x_cocos / pow_bc * sailors**sailors - blue_cocos)
}
 
2.to(9).each { |sailor|
say "#{sailor}: #{coconuts(sailor)}";
}
Output:
2: 11
3: 25
4: 765
5: 3121
6: 233275
7: 823537
8: 117440505
9: 387420481

Tcl[edit]

This is a very straightforward implementation. The "overnight" proc attempts to fulfill the activities of the night, throwing an error (through "assert") if it cannot. "minnuts" keeps trying to call it with more nuts until it succeeds. On success, "overnight" will return a list which narrates the night's activities.

proc assert {expr {msg ""}} {    ;# for "static" assertions that throw nice errors
if {![uplevel 1 [list expr $expr]]} {
if {$msg eq ""} {
catch {set msg "{[uplevel 1 [list subst -noc $expr]]}"}
}
throw {ASSERT ERROR} "{$expr} $msg"
}
}
 
proc divmod {a b} {
list [expr {$a / $b}] [expr {$a % $b}]
}
 
proc overnight {ns nn} {
set result {}
for {set s 0} {$s < $ns} {incr s} {
lassign [divmod $nn $ns] q r
assert {$r eq 1} "Incorrect remainder in round $s (expected 1, got $r)"
set nn [expr {$q*($ns-1)}]
lappend result $s $q $r $nn
}
lassign [divmod $nn $ns] q r
assert {$r eq 0} "Incorrect remainder at end (expected 0, got $r)"
return $result
}
 
proc minnuts {nsailors} {
while 1 {
incr nnuts
try {
set result [overnight $nsailors $nnuts]
} on error {} {
# continue
} on ok {} {
break
}
}
puts "$nsailors: $nnuts"
foreach {sailor takes gives leaves} $result {
puts " Sailor #$sailor takes $takes, giving $gives to the monkey and leaves $leaves"
}
puts "In the morning, each sailor gets [expr {$leaves/$nsailors}] nuts"
}
 
 
foreach n {5 6} {
puts "Solution with $n sailors:"
minnuts $n
}
Output:
Solution with 5 sailors:
5: 3121
 Sailor #0 takes 624, giving 1 to the monkey and leaves 2496
 Sailor #1 takes 499, giving 1 to the monkey and leaves 1996
 Sailor #2 takes 399, giving 1 to the monkey and leaves 1596
 Sailor #3 takes 319, giving 1 to the monkey and leaves 1276
 Sailor #4 takes 255, giving 1 to the monkey and leaves 1020
In the morning, each sailor gets 204 nuts
Solution with 6 sailors:
6: 233275
 Sailor #0 takes 38879, giving 1 to the monkey and leaves 194395
 Sailor #1 takes 32399, giving 1 to the monkey and leaves 161995
 Sailor #2 takes 26999, giving 1 to the monkey and leaves 134995
 Sailor #3 takes 22499, giving 1 to the monkey and leaves 112495
 Sailor #4 takes 18749, giving 1 to the monkey and leaves 93745
 Sailor #5 takes 15624, giving 1 to the monkey and leaves 78120
In the morning, each sailor gets 13020 nuts

uBasic/4tH[edit]

Translation of: C

For performance reasons, we limit ourselves to seven sailors.

For n = 2 To 7
t = 0
For x = 1 Step 1 While t = 0
t = FUNC(_Total(n,x))
Next
Print n;": ";t;Tab(12); x - 1
Next
 
End
 
_Total Param(2)
Local(1)
 
[email protected] = [email protected] * [email protected]
[email protected] = [email protected] - 1
 
For [email protected] = 0 To [email protected]
 
If [email protected] % [email protected] Then
[email protected] = 0
Break
EndIf
 
[email protected] = [email protected] + 1 + [email protected] / [email protected]
Next
Return ([email protected])
Output:
2: 11       1
3: 25       2
4: 765      60
5: 3121     204
6: 233275   13020
7: 823537   39990

0 OK, 0:127

zkl[edit]

Translation of: Python
fcn monkey_coconuts(sailors=5){
nuts,wakes:=sailors,List();
while(True){
n0:=nuts; wakes.clear();
foreach sailor in (sailors + 1){
portion, remainder := n0.divr(sailors);
wakes.append(T(n0, portion, remainder));
if(portion <= 0 or remainder != (sailor != sailors).toInt()){
nuts += 1;
break;
}
n0 = n0 - portion - remainder;
}
fallthrough{ break }
}
 
return(nuts, wakes)
}
foreach sailors in ([5..6]){
nuts, wake_stats := monkey_coconuts(sailors);
println("For %d sailors the initial nut count is %,d".fmt(sailors, nuts));
println("On each waking, the nut count, portion taken, and monkeys share are:\n ",
wake_stats.concat("\n "));
}
Output:
For 5 sailors the initial nut count is 3,121
On each waking, the nut count, portion taken, and monkeys share are:
   L(3121,624,1)
   L(2496,499,1)
   L(1996,399,1)
   L(1596,319,1)
   L(1276,255,1)
   L(1020,204,0)
For 6 sailors the initial nut count is 233,275
On each waking, the nut count, portion taken, and monkeys share are:
   L(233275,38879,1)
   L(194395,32399,1)
   L(161995,26999,1)
   L(134995,22499,1)
   L(112495,18749,1)
   L(93745,15624,1)
   L(78120,13020,0)
Translation of: C
fcn total(n, nuts){
nuts *= n;
foreach k in (n){
if (nuts % (n-1)) return(0);
nuts += nuts / (n-1) + 1;
}
nuts;
}
 
println("sailers: original pile, final share");
foreach n,x in ([2..9],[1..]){
if(t := total(n, x)){
print("%d: %d\t%d\n".fmt(n, t, x));
break;
}
}
Output:
sailers: original pile, final share
2: 11	1
3: 25	2
4: 765	60
5: 3121	204
6: 233275	13020
7: 823537	39990
8: 117440505	5044200
9: 387420481	14913080