The short story titled Coconuts, by Ben Ames Williams, appeared in the Saturday Evening Post on October 9, 1926.
The story tells about five men and a monkey who were shipwrecked on an island.
They spent the first night gathering coconuts.
During the night, one man woke up and decided to take his share of the coconuts.
He divided them into five piles.
One coconut was left over so he gave it to the monkey,
then hid his share and went back to sleep.

The third, fourth, and fifth man followed exactly the same procedure.
The next morning, after they all woke up,
they divided the remaining coconuts into five equal shares.
This time no coconuts were left over.

Solution to the Problem:

The smallest number of coconuts there could have been in the original pile is 3121.

        Time  Starting Pile =  Monkey + Share +   New Pile
        ----  -------------    ------   -----     --------
          1      3121       =       1 +   624 +       2496
          2      2496       =       1 +   499 +       1996
          3      1996       =       1 +   399 +       1596
          4      1596       =       1 +   319 +       1276
          5      1276       =       1 +   255 +       1020
          6      1020       =       0 +   204 +        816

 The 3,121 coconuts would be divided as follows:
 the monkey would get       5 coconuts
 the first man would get  828 coconuts
 the second man would get 703 coconuts
 the third man would get  603 coconuts
 the fourth man would get 523 coconuts
 the fifth man would get  459 coconuts

To try to see where this comes from, write down the equations that describe the problem.
We might want to use the letters A through G to represent the sizes of the pile of coconuts,
with A the beginning size, B through F the sizes after each sailor takes a share.
To preserve the pattern, we will let G be the number of coconuts left in the final pile after the monkey and the last sailor have taken theirs.

        B = A - 1 - ( A - 1 ) / 5 = (4/5) * ( A - 1 )
        C = B - 1 - ( B - 1 ) / 5 = (4/5) * ( B - 1 )
        D = C - 1 - ( C - 1 ) / 5 = (4/5) * ( C - 1 )
        E = D - 1 - ( D - 1 ) / 5 = (4/5) * ( D - 1 )
        F = E - 1 - ( E - 1 ) / 5 = (4/5) * ( E - 1 )
        G = F - 0 - ( F - 0 ) / 5 = (4/5) * ( F - 0 )

set talk on
f=1
do while f <= 1000
   g=(4*f)/5
   if int(g)=g
      * continue
      e=(5*f+1)/4
      if int(e)=e
         * continue
         d=(5*e+1)/4
         if int(d)=d
            * continue
            c=(5*d+1)/4
            if int(c)=c
               * continue
               b=(5*c+1)/4
               if int(b)=b
                  * continue
                  a=5*b+1
                  if int(a)=a
                      display memory
                      wait      
                  endif
               endif
            endif
         endif
      endif
   endif
   f=f+1
enddo

Sharina Broughton sent in the following computer program to solve the problem:

Solved using following ABAP code:

data total type i value 0.
data formonkey type i.
data count type i value 0.
data newtotal type i.
data subtotal type i.

do 10000 times.
  newtotal = total.
  do 6 times.
    count = count + 1.
    formonkey = newtotal mod 5.

    if formonkey = 1 AND count ne 6.
      subtotal = ( newtotal - 1 ) / 5.
      newtotal = newtotal - subtotal - 1.
    elseif formonkey = 0 AND count = 6.
      write: 'total is ', total.
    else.
      exit.
    endif.
    clear formonkey.
    clear subtotal.
  enddo.
  clear count.
  clear newtotal.
  total = total + 1.
enddo.