Sailor,Monkey,Coconut answer in elaborate In Sailor, Monkey, Coconut Problem
Can anyone tell me how  adding 56  gives me another solution??I understand that cocount is divided into 5 piles.But how is 56 give me another solution?why wouldn't I add the cocounuts that are being given to monkey? 
 A: If the heap has $k\cdot 5^m$ additional coconuts when a sailor wakes up (with $m\ge 1$) then his secret share will be bigger by $k\cdot 5^{m-1}$ coconuts and the heap he leaves for the next round is larger by $4k\cdot 5^{m-1}$ coconuts. Hence if we start with $5^6$ additional coconuts, the heap after the five sailors take their secret shares will be bigger by $4^5\cdot 5^1$ coconuts. This is still (though just barely) divisible by $5$, hence there will still be an extra coconut for the monkey in the end (as well as within each previous round).
A: Let $x_0^0$ be a solution of the Sailor,Monkey,Coconut problem.
consider $x_1^0=x_0 + 5^6$. the claim is that $x_1^0$ is another solution. Let's see why:
Sailor 1  divided $x_0^0$ in five equal parts and one was left. $x_0 = 5 x_0^1 + 1$.
Sailor 2  divided $4x^1_0$ in five equal parts and one was left. $4x_0^1 = 5 x_0^2 + 1$.
Sailor 3  divided $4x^2_0$ in five equal parts and one was left. $4x_0^2 = 5 x_0^3 + 1$.
Sailor 4  divided $4x^3_0$ in five equal parts and one was left. $4x_0^3 = 5 x_0^4 + 1$. 
Sailor 5  divided $4x^4_0$ in five equal parts and one was left. $4x_0^4 = 5 x_0^5 + 1$. 
In the morning  the group of sailors divided $4x^5_0$ in five equal parts and one was left. $4x_0^5 = 5 x_0^6 + 1$. 
The important is that at each step $x_0^i$ is an integer and $4x_0^i = 5 x_0^{i+1} + 1$, If you consider $x_1^0 = x_0^0 + 5^6 $ you will see that each division operation  this relation is preserved.
A: Given that we have $x$ coconuts to begin with, the pile of coconuts $P_k$ remaining after $k$ repetitions of the process can be shown to equal
$$
P_k=\frac{4^kx-4\cdot 5^k+4^{k+1}}{5^k}
$$
It is simple to check that this formula makes sense after one round, namely
$$
P_1=\frac{4x-4\cdot 5+4^2}{5}=\frac{4x-4}{5}=\frac45(x-1)
$$
and then one can check using induction and the principle that $P_{k+1}=\frac45(P_k-1)$ that it holds in general. Considering the numerator of the formula, one sees that
$$
4^k x-4\cdot 5^k+4^{k+1}\equiv 4^k(x+4)\mod{5^k}
$$
and each $P_k$ is only an integer if the numerator is congruent to $0$ modulo the denominator $5^k$. Thus we must have $4^k(x+4)\equiv 0\mod{5^k}$. This requires $x$ to be congruent to $-4$ modulo $5^k$. So $x$ must have the form
$$
x=5^k q-4
$$
With $x=5^6q-4$ this holds for all $k=1,2,...,6$ and so all the piles $P_1,P_2,...,P_6$ will be integers throughout the process. Hence solutions differ by exactly $5^6$, and the smallest non-negative solution is $x=5^6-4=15621$, the next is $x=5^6\cdot 2-4=31246$ and so on.
