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.