If $\lfloor{\sqrt{x}}\rfloor=6$, how many possible integer values of $x$ are there?
To find the possible values of $x$, I need to think about what it means for $\lfloor{\sqrt{x}}\rfloor=6$. This means that the largest integer that is less than or equal to $\sqrt{x}$ is 6. In other words, $\sqrt{x}$ is between 6 and 7, not including 7. Squaring both sides, I get that $x$ is between 36 and 49, not including 49. So the possible integer values of $x$ are 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, and 48. There are 13 of them in total. 