The last number of $365$ is $5$, therefore I’ve been told that $5$ is a factor of $365$, which it clearly is. This however does not work for other numbers, i.e., $9$ is not a factor of $8599$.
I’ve also been told to consider the sum of the numbers, i.e., $315: 3+1+5 = 9$, and since $9$ is divisible by $3$, $3$ is a factor of $315$, which it is. But this is severely limited because it only holds if the sum is a multiple of $3$, so using this for $389$ doesn’t work because $3+8+9 = 20$, which isn’t a multiple of $3$.
With this info in mind, is there a reliable way to factor numbers using inherent properties? If there is not, I am a bit worried that I will be unable to reliably factor any given number of a reasonable size. Thank you.
As noted in the comments, you can find plenty of similar divisibility rules in this Wikipeida article.
You are right to be worried, since factoring numbers is a hard problem. This is especially true for large integers with a small number of large prime factors. A good example are semiprimes, i.e., large integers $n=p_1\times p_2$ with exactly two very large prime components $p_1$ and $p_2$.
In fact, integer factorization is so hard, modern cryptographic methods rely on them.