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. 

We can solve in a simple way without the Chinese Remainder Theorem.
Let $n = 3x + 2 = 5y + 3 = 7z + 2$.
Then, $5y + 3  \equiv 3x + 2 \equiv (\mod 3) \implies y \equiv 1(\mod 3)y$
Hence let $y = 3k + 1$ which gives $n = 5y + 3 = 15k + 8$.
Again $n = 15k + 8 \equiv 7z + 2 (\mod 7) \implies k \equiv 1 (\mod 7)$
Hence let $k = 7m + 1$. This gives $n = 15k + 8 = 105m + 23$
Thus in general the number $n$ is of the form $n = 105m + 23$. The smallest such number is when $m = 0$, we get $n = 23$.
