jmc

Since the expression $\lfloor \sqrt{x}\rfloor$ stands for the greatest integer that is less than or equal to $x$, the smallest possible value of $x$ that could satisfy the equation is $6^2$, or $36$. The next integer greater than $6$ is $7$, so the smallest integer (greater than $36$) that would not satisfy $\lfloor \sqrt{x}\rfloor =6$ must be $7^2$, or $49$. Therefore, any integer that is in the range $36\le x<49$ could be considered a possible integer value of $x$. Since there are 13 numbers in this range, our final solution is $\boxed{13}$.