What is the smallest positive integer such that when divided by 3 leaves a remainder of 1, when divided by 4 leaves a remainder of 2, and when divided by 5 leaves a remainder of 1?
+606 $x=1\pmod{3}, x=2\pmod{4}, x=1\pmod{5}$.
If $x=1\pmod{3,5}$, $x=1\pmod{\text{lcm}(3,5)=15}$.
Now $x=15k+1=4j+2$. Taking the equation modulo $4$, $1-k=2\pmod{4}\implies k=3\pmod{4}$. So minimum $x$ is $15*3+1=45+1=\boxed{46}$.
Nice one.
