For given $ab\leq n$, do there exist $a'\geq a$ and $b'\geq b$ such that $a'b'=n$? For example, given $a=3$, $b=3$, and $n=14$, no such $a',b'$ exists. On the other hand, for $a=3$, $b=3$, and $n=12$, we can use $a'=3$ and $b'=4$.
Is there a simple formula that can help determine the answer to this question, rather than simply searching through possible combinations of $a'$ and $b'$?
 A: There is no set formula to achieve this, AFAIK. What you can do is focusing on one variable, like $a'$, and finding the higest possible value such that the other variable is still in the allowed range:
$$a \le a' \le \left\lfloor\frac{n}b\right\rfloor \tag{1}\label{eq1}$$
and then try every $a'$ in that range if it divides $n$.
One optimization method would be to test $n$ for 'large' prime factors $p$, but in our case 'large' already starts at $p=5$. Because if $n$ is divisible by some prime $p$, at least one of $a'$ and $b'$ need to be divisible by $p$. 
So instead of testing all $a'$ that satisfy $\eqref{eq1}$, you test all $a'$ that satisfy $\eqref{eq1}$ and the additional condition that $p|a'$. But since you don't know if $a'$ or $b'$ will be divisble by $p$, you have to test all $b'$ with $p|b'$ and
$$b \le b' \le \left\lfloor\frac{n}a\right\rfloor \tag{2}\label{eq2}$$
as well. But since you need to test only every $p$-th number, that should normally still be an advantage over the method without considering $p$.
A: In reverse, it's related to the formula for markup and margin. No possibility of such a factorization exists, with $ab=n$.  In fact: $$\sqrt{n}<a\land \sqrt{n}<b \implies n<ab<a'b'$$
Which is then a contradiction. One way would be:$${n\over ab}=(1+b_\text{markup})(1+a_\text{markup})$$ where you use percentage markup that can be integers when multiplied by b and a, respectively. But that's still more of a brute force approach.
