**Remark 1.** None of the Tables 1, 2, 3, or 4 need start with 1. If we start with any natural number \(r,\) \(1\le r\le 9\) for Tables 1 and 2, \(1\le r\le n\) for Table 3, or \(1\le r\le 4\) for Table 4, all the above results (with obvious translations) would remain the same.

**Remark 2.** Here is a simpler generalization of the problem. Again, suppose \(n^2>1\) trees are to be divided among \(n\) inheritors, where tree \(k\) yields \(k\) units of produce per year. Let us start out by writing \(n\) mutual derangements of the first \(n\) integers as in the following table.

1 | 2 | 3 | 4 | \(\cdots\) | \(n\) |

2 | 3 | 4 | 5 | \(\cdots\) | 1 |

3 | 4 | 5 | 6 | \(\cdots\) | 2 |

\(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\ddots\) | \(\vdots\) |

\(n\) | 1 | 2 | 3 | \(\cdots\) | \(n-1\) |

**Table 5**

Now, let us add \(0\) to all entries in row 1, \(n\) to all entries in row 2, \(2n\) to all entries in row 3, \(\dots,\) and \((n-1)n\) to all entries in row \(n.\) Each column of the new table so generated from Table 5 has the common sum \[[1+2+\cdots+n]+n[1+2+\cdots+(n-1)]=\frac{n(n^2+1)}{2},\] and the trees designated for inheritor \(j\) are the trees with labels in column \(j.\)

**Remark 3.** The generalized problem is equivalent to finding a solution of the following \(n\times n^2\) indeterminate system of equations with each variable a unique integer between 1 and \(n^2\):

\(x_{11}+x_{12}+\cdots+x_{1n}\) | \(={\sigma}_n\) | |||

\(x_{21}+x_{22}+\cdots+x_{2n}\) | \(={\sigma}_n\) | |||

\(\ddots\) | \(\vdots\) | |||

\(x_{n1}+x_{n2}+\cdots+x_{nn}\) | \(={\sigma}_n\) |