Video solution:

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Written solution:

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 $$k^2-70k+1000\le0\quad\Longrightarrow\quad (k-20)(k-50)\le0,$$ so $20\le k\le50$. The right inequality is $$70k-1000<k^2+2k+1\quad\Longrightarrow\quad k^2-68k+1001>0.$$ 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.