# 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 $$n^2$$:

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