Question 15 Five monkeys divide a pile of peanuts. The first monkey divides the peanuts into five piles, eats one left over, and takes away one pile. The second monkey again divides the remaining peanuts into five piles, with exactly one left over, eats it, and takes away one pile. This continues until the fifth monkey comes to divide the remaining peanuts, and can also divide them into five piles with exactly one left over. How many peanuts were there in the pile at least?
From the problem, we know that the number of peanuts in each pile that each monkey gets is 1 more than a multiple of 5. Therefore,
$$
\begin{aligned}
N & =5 A_{1}+1,4 N_{1}=5 A_{2}+1,4 N_{2}=5 A_{3}+1,4 N_{3} \\
& =5 A_{4}+1,4 N_{4}=5 A_{5}+1,
\end{aligned}
$$

where $N$ represents the total number of peanuts, and $A_{1}, \cdots, A_{5}$ represent the number of peanuts in the pile each monkey takes away. Thus, we have
$$
\begin{array}{c}
4 A_{i}=5 A_{i+1}+1 \quad(i=1,2,3,4), \\
4\left(A_{i}+1\right)=5\left(A_{i+1}+1\right) \quad(i=1,2,3,4), \\
A_{5}+1=\frac{4}{5}\left(A_{4}+1\right)=\left(\frac{4}{5}\right)^{2}\left(A_{3}+1\right)=\cdots=\left(\frac{4}{5}\right)^{4}\left(A_{1}+1\right) .
\end{array}
$$

Since $A_{5}+1 \in \mathbf{Z}$, we have $\min \left(A_{1}+1\right)=5^{4}, A_{1}=5^{4}-1=624$.
$$
\min N=5 \times 624+1=3121 .
$$