Maximization of a ratio Edit: Removed solved in title, because I realize I need someone to check my work.
Ok, so the problem is a lot more straight forward than I originally approached it (which  was a false statement -- so it was excluded).
Question:
Let R,S, x $\in$ N with x $\le$ R*S and $0 \lt$ R $\le$ S. Next, define B as a multiplicative factor of x - c with c $\ge 0$ and B $\le$ S such that $\frac{x - c}{B} = A \le R$ and A $\in$ N. What value of B maximizes A?

Consider the expression $\frac xB$. For B to be a multiplicative factor of x - c, c must equal  the remainder between B and x. Therefore, we can rewrite c as c = x - dB, where d is the unique natural number satisfying both dB $\le$ x and (d + 1)B $\gt$ x. Substituting,
A = $\frac{x - (x - dB)}{B}$ = d.
However, d depends on B, so find $B \in N$ with $B \le$ S such that $d_B \le R$ and $d_{B-1} \gt R$. Following these two sets of inequalities will maximize A.
