2020 AMC 10B Problem 24

Let $$k=\left\lfloor\sqrt n\right\rfloor.$$ The equation gives $n=70k-1000.$ Also, by the definition of the floor function, $$k^2\le n<(k+1)^2.$$ Substituting $n=70k-1000,$ we get $$k^2\le 70k-1000<(k+1)^2.$$ The left inequality is $$\begin{aligned} &k^2-70k+1000\le0 \\ &\quad \Longrightarrow (k-20)(k-50)\le0, \end{aligned}$$ so $20\le k\le50.$ The right inequality is $$\begin{aligned} &70k-1000<k^2+2k+1 \\ &\quad \Longrightarrow k^2-68k+1001>0. \end{aligned}$$ The roots of $k^2-68k+1001$ are $34\pm\sqrt{155},$ which are approximately $21.55$ and $46.45.$ Thus, together with $20\le k\le50,$ the possible integer values are $$k=20,21,47,48,49,50.$$ There are $6$ such values. Thus, C is the correct answer.