If a and b are positive integers for which ab-6a+5b=15, what is the minimal possible value of |a-b|?
Substituting $b=6$ into $ab-6a+5b=15$ we get the absurdity $30=15$. Thus $b$ can never assume the value 6. By algebraic manipulation we can express $a$ in terms of $b$ as $a = (15-5b)/(b-6)$. Now, $15-5b > 0$ iff $b < 3$ and $b-6 > 0$ iff $b > 6$. Therefore, $a$ is positive only when $3 < b < 6$. Among these two plausible values of $b$ (4 and 5), only $b=5$ turns $a$ into a positive integer $a=10$. Thus, $|a-b|$ has only one value 5 and it must be the minimum.