Coconuts problem

Question

I am trying to solve the following problem and I would appreciate any help on where I am wrong

Problem

Five men crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.

That night each man took a turn watching for rescue searchers while the others slept. The first watcher got bored so he decided to divide the coconuts into five equal piles. When he did this, he found he had one remaining coconut. He gave this coconut to a monkey, took one of the piles, and hid it for himself. Then he jumbled up the four other piles into one big pile again.

To cut a long story short, each of the five men ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.

What is the smallest number of coconuts there could have been in the original pile?

My approach

I let $o$ be the original number of coconuts

We know that when $o$ is divided into 5 piles there is one left, so we can say that $o=5a+1$ where $a$ represents the number of coconuts in the equal piles that were split.

I then know that the 4 piles saved where again split equally and there was one left so writing an equation for each of the 5 splits we have the following $$o=5a+1$$ $$4a=5b+1$$ $$4b=5c+1$$ $$4c=5d+1$$ $$4d=5e+1$$

So I begin substituting back to the original $$\begin{align} o&=\frac{5}{4}(5b+1) \\ o&=\frac{5^2}{4}b+\frac{5}{4} \\ o&=\frac{5^3}{4^2}c+\frac{5^2}{4^2}+\frac{5}{4} \\ o&=\frac{5^4}{4^3}d+\frac{5^3}{4^3}+\frac{5^2}{4^2}+\frac{5}{4} \\ o&=\frac{5^5}{4^4}e+\frac{5^4}{4^4}+\frac{5^3}{4^3}+\frac{5^2}{4^2}+\frac{5}{4} \end{align}$$

Now at this point I try to see if im on the right way and I check the solution which is somehow $$ o = (\frac{5^5}{4^4}) (e + 1) − 4$$

By substituting a value for e I know my equation and this is not equal, so what did I do wrong and how did it get here?

Answer 1

You forgot one in the first step: $$\begin{align}o&=5a+1\\ &=\frac54(5b+1)+1\\ &=\frac{5^2}4b+\frac54+1.\end{align} $$ Finally note that $1+a+a^2+a^3+a^4=\frac{a^5-1}{a-1}$ to simplify the sum of powers in the end (or, use four blue coconuts).