Magic square $29\times29$: Linear Congruences and Uniform Step Method

Question

This linear congruency I was given is part 1 to a 2 part question, I was able to get this. [ \begin{split} 14\cdot27(x+y)\equiv14\cdot16&\pmod{29}\Longrightarrow\\ x+y\equiv21&\pmod{29}\Longrightarrow\\ y\equiv21-x&\pmod{29} \end{split} ] The question also includes these congruences that ultimately creates the Magic Square using Uniform Step Method
\begin{split} x_j\equiv16+17\cdot j+23\cdot \left[\frac{j}{29} \right]\pmod{29}\\ y_j\equiv9+10\cdot j+4\cdot \left[\frac{j}{29} \right]\pmod{29}\\ \end{split} The part I am confused on the part where the question asks what the value of $j$ would be when placed in the cell with coordinates $(21,20)$ (the $x$-axis starts with $0$ and ends with $28$ and same rule applies to the $y$-axis) with these congruences and how the solved linear congruence has a relation to these congruences.

Answer 1

I think what you're being asked to do is to solve the system of simultaneous congruences $$ 21\equiv16+17j+23\left[{j\over29}\right]\pmod{29};\qquad 20\equiv9+10j+4\left[{j\over29}\right]\pmod{29} $$ I'll rewrite these as $$ 17j+23k\equiv5\pmod{29};\qquad10j+4k\equiv11\pmod{29};\qquad k=\left[{j\over29}\right] $$ By the usual methods of solving two linear equations in two unknowns, and of solving linear congruences in one variable, we get $j\equiv12\pmod{29}$, $k\equiv9\pmod{29}$. Combining these, we can take $j=9\cdot29+12=273$.