7746=1
4509=2
6003=1
3669=3
7681=3
8898=7
9243=1
8285=4
5439=1
2282=2
1886=?

Those numbers don't equal the values you claim. Do you mean

$f(7746) = 1$
$f(4509) = 2$
...
...
...
$f(2282) = 2$
$f(1886) = ?$

for some function $f$?

If that is what you mean, then there are an infinite number of functions that satisfy that. Without context, it is impossible to know which answer is correct.

All the info I have, is what I have supplied. I am wondering if it is some kind of sick joke. My boss said if I can solve it, he will give me a prize. I have been trying to find some kind of pattern for hours. I am not even sure the answer is mathematical. It could be a riddle.

Here is a pattern:

Those are all equivalences modulo 1. Of course, modulo 1, all integers are equivalent, so you can pick any integer you want to be the answer, and it will fit that pattern. If your boss doesn't like it, he/she should have given you more specifications.

If there's any lesson to be learned from this problem it's that you shouldn't spend a lot of time essentially guessing at answers to a totally under specified problem.

The answer could be anything. It could be the last digit of the weight of that many argon atoms expressed in base 13. Who knows.

By the way, in case you don't understand the pattern I pointed out, if two integers, $a$ and $b$, are equivalent modulo some integer $n$, we write:

$a \equiv b \pmod{n}$

That means $\dfrac{a-b}{n}$ is an integer.

Here is an example:
$a \equiv b \pmod{2}$ if and only if both $a$ and $b$ are even or if both are odd.

So, $7746 \equiv 1 \pmod{1}$ means $\dfrac{7746-1}{1}$ is an integer. Because the difference of any two integers is an integer, and any integer divided by 1 is still an integer, $a \equiv b \pmod{1}$ is true for any integers $a$ and $b$. Hence the "pattern" I pointed out for your list of equations.