A bag of coconuts is to be shared among five children and a monkey. One of the five children, without the others knowing, divided the coconuts in the bag into five equal parts. This way, he got one coconut as a remainder, which he gave to the monkey, and then left with one-fifth of the coconuts. The second child also divided the remaining coconuts into five equal parts, and he too got one as a remainder, which he gave to the monkey, and left with one-fifth of the coconuts. The other three children did the same. Finally, the five children came together and divided the remaining coconuts into five equal parts, and once again got one coconut as a remainder, which they gave to the monkey.

What is the minimum number of coconuts that were originally in the bag?
Let $x$ denote the number of coconuts each child received after the last division. Thus, before the last division, there were $(5 x+1)$ coconuts in the bag. The fifth child gave one coconut to the monkey and took one-fifth of the remaining coconuts, leaving $(5 x+1)$ coconuts in the bag. Therefore, before this, there must have been $1+\frac{5}{4}(5 x+1)$ coconuts in the bag. Similarly, we find that before the fourth child divided the coconuts, there were

$$
1+\frac{5}{4}\left(1+\frac{5}{4}(5 x+1)\right)
$$

coconuts in the bag.

Continuing this, we find that originally there were

$$
1+\frac{5}{4}\left\{1+\frac{5}{4}\left[1+\frac{5}{4}\left(1+\frac{5}{4}\left(1+\frac{5}{4}(5 x+1)\right)\right)\right]\right\}
$$

coconuts in the bag. Performing the indicated operations and using the identity

$$
a^{6}-b^{6}=(a-b)\left(a^{5}+a^{4} b+a^{3} b^{2}+a^{2} b^{3}+a b^{4}+b^{5}\right)
$$

we get:

$$
\begin{aligned}
& \frac{5^{6} x+5^{5}+5^{4} \cdot 4+5^{3} \cdot 4^{2}+5^{2} \cdot 4^{3}+5 \cdot 4^{4}+4^{5}}{4^{5}}= \\
= & \frac{5^{6} x+\frac{5^{6}-4^{6}}{5-4}}{4^{5}}=\frac{5^{6}(x+1)}{4^{5}}-\frac{4^{6}}{4^{5}}=\frac{5^{6}(x+1)}{4^{5}}-4
\end{aligned}
$$

From the problem statement, it follows that this must be a natural number. For this to be true, the first term of the difference must be an integer, meaning $5^{6} \cdot(x+1)$ must be divisible by $4^{5}$. The first factor of the product is relatively prime to $4^{5}$, so $4^{5}$ can only be a divisor of the product if it is a divisor of the second factor, $(x+1)$. Thus, $(x+1)$ must be at least $4^{5}$. The number of coconuts in the bag is the smallest when $x$ is the smallest possible. Therefore, the bag originally contained at least

$$
\frac{5^{6} \cdot 4^{5}}{4^{5}}-4=15621
$$

coconuts.

If there were exactly 15621 coconuts in the bag, then the described division could have been carried out:

The first child took

The second child took

The third child took

The fourth child took

The fifth child took

Finally, each received. $\quad \frac{5115}{5}=1023$ coconuts.