# Al-Maghribî’s Mecca Problem Meets Sudoku – Further Remarks on the Mecca Problem

Author(s):
Ilhan M. Izmirli (George Mason University)

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 $n^2$:

 $x_{11}+x_{12}+\cdots+x_{1n}$ $={\sigma}_n$ $x_{21}+x_{22}+\cdots+x_{2n}$ $={\sigma}_n$ $\ddots$ $\vdots$ $x_{n1}+x_{n2}+\cdots+x_{nn}$ $={\sigma}_n$

Ilhan M. Izmirli (George Mason University), "Al-Maghribî’s Mecca Problem Meets Sudoku – Further Remarks on the Mecca Problem," Convergence (August 2016)