I need to find a positive integer $n$ that satisfies two conditions: $n \equiv 3 \pmod{4}$ and $n \equiv 2 \pmod{5}$. These are two congruences in different moduli, so I can try to use the Chinese remainder theorem to solve them. But $11 \equiv 1 \pmod{5}$, not $2$, so it doesn't work. The next one is $15$, but $15 \equiv 0 \pmod{5}$, so it also doesn't work. The next one is $19$, and $19 \equiv 4 \pmod{5}$, so it still doesn't work. The next one is $23$, and $23 \equiv 3 \pmod{5}$, so it doesn't work either. The next one is $27$, and $27 \equiv 2 \pmod{5}$, so this one works! I have found a value of $n$ that satisfies both conditions, and it is the smallest one greater than $10$, so the answer is $n = 27$. 
When we say that Ray is climbing up the stairs $m$ at a time, we mean that he starts on the floor (step $0$) then jumps to step $m$ and then to $2m$ and so on until the number of steps to the top is less than $m$. Ray climbs up a flight of stairs of $n$ steps in two ways. When he does it $4$ steps at a time, there are $3$ steps left at the top. When he does it $5$ steps at a time, there are $2$ steps left at the top. What is the smallest possible value of $n$ that is greater than $10$?