Disclaimer: I'm not a mathematician and for the most part I do not understand what, exactly, it takes to mathematically prove anything. That is where this question comes from.
Why can't the Collatz Conjecture be proven by showing that all even numbers less than 11 do "resolve" (via the collatz division and multiplication) to 1. From there, couldn't it be shown that all even numbers have a multiple factor that is less than 10, thus proving the case for starting with an even number? Which just leaves the odd-numbered case. But for odd numbers, mulitplying by 3 and adding one always results in an even number, which can be proven to resolve to 1 via the even numbered part of the proof.
Obviously I'm missing something but i don't have the words and labels to say what it is to investigate further.
I apologize in advance if this question doesn't belong on this SE and thank you in advance for any insight.
EDIT: I haven't actually thought about this in a little while before writing it up here and have obviously goofed in a couple of spots.
One thing to correct is that every even number has a factor that is of the form 2^x and that any number 2^x resolves to 1 in a straight line.
If we knew that every Collatz orbit eventually hits a number less than $10$ (is that what you mean?) then we'd be done immediately. So we don't know that! More generally if we knew that every Collatz orbit eventually hits a number less than $N$ for any $N$ then we'd have reduced the Collatz conjecture to checking $N$ cases, and we'd be done as long as $N$ was not enormous; apparently the conjecture has been verified up to $N = 2^{68}$ or so. We don't know this either! This is the entire difficulty of the problem.
In other words, despite having verified the conjecture up to $N = 2^{68}$, for all we know it's possible that some extremely large number has a Collatz orbit that never dips below, say, $2^{100}$, and so would be a counterexample we haven't tested computationally yet.