Is there a non-brute-force way to find $x$ such that $45\leq x < 200$, $x\equiv 0 \pmod{5}$ , $x\equiv1 \pmod{8}$, $x\equiv1 \pmod{12}$? Like the title says, I'm wondering if there's a non-brute-force way to determine $x$ such that
$$45\le x<200,$$
$$x\bmod5\equiv0,$$
$$x\bmod8\equiv1,$$
$$x\bmod12\equiv1$$
I know I can simply enumerate all integers in $[45, 200)$ to get the answer (145), but I'm wondering if there's a more elegant solution.

From the last two congruences, $x\equiv 1 \pmod{24}$
Then $x\equiv 1 \equiv 25 \equiv 49 \equiv...$
Just keep going until you find something congruent to $0 \pmod{5}$ that is within your range of interest.
Or you could note that $25$ works, and then so will $25 + 120k$ for any integer $k$.  
