TITLE: Is there a better way to factor $375007$ with out testing first $612$ primes? No calculators please
QUESTION [5 upvotes]: Is there a better way to factor $375007$ with out testing first $612$ primes ? 

I know this factors to $31\times 12097$ by testing the primes $2,3,5,\ldots,31$. 
Is there any other clever way to work this ? I have tried Fermat's factorization by writing the number as $x^2-y^2$ but it is also taking too many iterations because the factors differ by large magnitude. 
Also I have been trying to factor it by changing the base to 10^2 :  $37x^2 + 50x+7 = (ax+b)(cx+d)$ and other bases but no success yet.

REPLY [2 votes]: The last prime I need to know to factorize that number is $113$, the $30$th prime number.  This is because $\sqrt{12097}\approx110$.  After $110$, I know that any higher value to divide $12097$ would force its counterpart to be smaller:  $\frac{12097}{113}\approx107$.  Thus, if a number higher than $110$, like $113$, divided $12097$, I would have already found its counterpart.  Once I hit $113$, I know that $12097$ is prime, because I've crossed the square root barrier.