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   No one breaks any coconuts, so all the variables are integers.

   Initial pile:    X
   First man takes: (X - 1)/5   First man leaves:  (X - 1)4/5 = A
   Second man gets: (A - 1)/5   Second man leaves: (A - 1)4/5 = B
   Third man gets:  (B - 1)/5   Third man leaves:  (B - 1)4/5 = C
   Fourth man gets: (C - 1)/5   Fourth man leaves: (C - 1)4/5 = D
   Fifth man gets:  (D - 1)/5   Fifth man leaves:  (D - 1)4/5 = E
   Each man gets:   E/5 = F

   Substituting:

   5F = E = (D - 1)4/5          25F = 4D - 4             25F + 4 = 4D

   25F + 4 = 4(C - 1)4/5        125F + 20 = 16C - 16     125F + 36 = 16C

   125F + 36 = 16(B - 1)4/5     625F + 180 = 64B - 64    625F + 244 = 64B

   625F + 244 = 64(A - 1)4/5    3125F + 1220 = 256A - 256
                                3125F + 1476 = 256A

   3125F + 1476 = 256(X - 1)4/5    15625F + 7380 = 1024X + 1024
                                   15625F + 8404 = 1024X


   1024X = 15625F + 8404   "That leaves us two unknowns in one equation."


   8404 and 1024X are both divisible by two (twice), so 15625F must be
also, but 15625 is not, so F must be.  Say F = 4G.

   1024X = 15625(4G) + 8404

   256X = 15625G + 2101

   2101 is odd, and 256X is even, so 15625G must be odd, so G must be
odd.  Say G = 2H + 1 (where H >= 0).

   256X = 15625(2H + 1) + 2101 = 15625(2H) + 15625 + 2101
   256X = 15625(2H) + 17726

   128X = 15625H + 8863   Similarly, H = 2J + 1.

   128X = 15625(2J + 1) + 8863 = 15625(2J) + 15625 + 8863
   128X = 15625(2J) + 24488

   64X = 15625J + 12244   Here, J = 4K.

   64X = 15625(4K) + 12244

   16X = 15625K + 3061   Here, K = 2M + 1.

   16X = 15625(2M + 1) + 3061 = 15625(2M) + 15625 + 3061
   16X = 15625(2M) + 18686

   8X = 15625M + 9343   Here, M = 2N + 1.

   8X = 15625(2N + 1) + 9343 = 15625(2N) + 15625 + 9343
   8X = 15625(2N) + 24968

   4X = 15625N + 12484   Here, N = 4P.

   4X = 15625(4P) + 12484

   X = 15625P + 3121

   This formula gives all the solutions.  The smallest one is where P is
zero, and X is 3121.


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© 2008 Steven M. Schweda.