What is the smallest whole number $b$ such that 62 can be expressed in base $b$ using only three digits?
To express 62 in base $b$ using only three digits, I need to find $b$ such that $62 = a_2b^2 + a_1b + a_0$, where $a_2$, $a_1$, and $a_0$ are digits between 0 and $b-1$. I can try some small values of $b$ and see if they work. If $b = 2$, then 62 in base 2 is 111110, which is too long. If $b = 3$, then 62 in base 3 is 2102, which is also too long. If $b = 4$, then 62 in base 4 is 322, which is exactly three digits. This might be the answer, but I need to check if there is a smaller $b$ that works. If $b = 5$, then 62 in base 5 is 222, which is also three digits. But 4 is smaller than 5, so I don't need to check any larger $b$. Therefore, the smallest $b$ that works is 4. 