If $AAA_4$ can be expressed as $33_b$, where $A$ is a digit in base 4 and $b$ is a base greater than 5, what is the smallest possible sum $A+b$?
To solve this problem, I need to find the values of $A$ and $b$ that satisfy the equation $AAA_4=33_b$. I can start by rewriting the equation in base 10, using the place value of each digit. For the left side, I have $AAA_4=A\cdot 4^2+A\cdot 4^1+A\cdot 4^0=A(16+4+1)=21A$. For the right side, I have $33_b=3\cdot b^1+3\cdot b^0=3(b+1)$. So I have $21A=3(b+1)$. To simplify this equation, I can divide both sides by 3, getting $7A=b+1$. Now I need to find a value of $A$ that is a digit in base 4, meaning it can only be 0, 1, 2, or 3. I also need to find a value of $b$ that is a base greater than 5, meaning it is at least 6. I can try different values of $A$ and see if they give me valid values of $b$. If $A=0$, then $b+1=0$, which is not possible. If $A=1$, then $b+1=7$, which gives $b=6$. This is a possible solution. If $A=2$, then $b+1=14$, which gives $b=13$. This is also a possible solution, but it is larger than the previous one. If $A=3$, then $b+1=21$, which gives $b=20$. This is also a possible solution, but it is larger than the previous one. So the smallest possible solution is $A=1$ and $b=6$. The smallest possible sum $A+b$ is then $1+6=7$.

