What is the number that, when divided by $3$, $5$, $7$, leaves remainders of $2$, $3$, $2$, respectively? What is the number?
The LCM of the divisors is 105. I think this has something to do with the Chinese remainer theorem, but I am not sure how to apply this knowledge. 

Let $N$ be the number so we have$$N=3n_1+2=7n_2+2\Rightarrow n_1=7k\Rightarrow N=21k+2$$ Besides
 $$21k+2=5n_3+3\iff 21k-5n_3=1\qquad (*)$$ Since $(k,n_3)=(1,4)$ is a solution of $(*)$ the general solution is $$\begin{cases}k=5t+1\\n_3=21t+4\end{cases}$$ It follows $$N=21(5t+1)+2=23+105t\text{ where } t\in\Bbb Z$$
