"Puzzles"--the monkeys divide the peach

The original question is this: 5 monkeys picked 1 piles of peaches together, because they were too tired, they agreed to decide, first sleep and then points. After a long time, came the first monkey, it saw other monkeys did not come, will this pile of monkeys into five parts, the result of more than 1, will be more this eat, by the way take away the 1 piles. After a long time, the 2nd monkey came, it does not know that there are 1 companions have come, mistakenly think they are the 1th to, so the peach piled up on the ground, the average divided into five parts, found that there are 1 more, the same ate 1, take away 1 of the heap. The 3rd, 4th, and 5th monkeys are all like this ... Ask the 5 monkeys to pick at least how many peaches. How many peaches are left after the 5th monkey leaves. 1. Common Solution

Set these 1 piles of peaches at least X x, lend them 4 , called X+4 x+4.

5 Monkeys took A,b,c,d,e A, B, C, D, E Peaches (including one eaten), which were:

a=x+45b=15x (1−a) =425 (x+4) c=15 (1−a−b) =16125 (x+4) d=15 (1−a−b−c) =64625 (x+4) e=15 (1−a−b−c−d) =2563125 (x+4) \begin{ Split} &a=\frac{x+4}5\\ &b=\frac 15\times \left (1-a\right) =\frac4{25}\left (x+4\right) \ &c=\frac15\left (1-a-b\right) =\frac{16}{125}\left (x+4\right) \ &d=\frac15\left (1-a-b-c\right) =\frac{64}{625}\left (x+4\right ) \ \ &e=\frac15\left (1-a-b-c-d\right) =\frac{256}{3125} (x+4) \end{split}

E e is an integer, and 256 cannot be divisible by 3125, so X+4 x+4 is a multiple of 3125,

X+4=3125xk X+4=3125\times K
When K=1 k=1,