2020 AMC 10B #24

Problem:

How many positive integers $n$ satisfy$$\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?$$(Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.)

Source: 2020 AMC 10B, Problem 24

Solution:

 Note that $n$ must be $50 \pmod{70}$ in order for the left hand side to be an integer. So, let $n=70x+50$ for some nonnegative integer $x$. We have $$\frac{70x+1050}{70} = \lfloor \sqrt{70x+50} \rfloor$$

$$\implies x+15 = \lfloor \sqrt{70x+50} \rfloor$$

From this, we know that $$(x+15)^2 \le 70x+50 < (x+16)^2$$ 

We split this into two inequalities, namely $(x+15)^2 \le 70x+50$ and $(x+16)^2 > 70x+50$. From the first inequality we get that $5 \le x \le 35$ must be true. From the second inequality, we get that one of $x< \approx 6.5$ and $x> \approx 31.5$ must be true. In the first case, namely $x<\approx 6.5$, the only integers that work are $x=5, 6$. In the second case, namely $x>\approx 31.5$, the only integers that work are $32, 33, 34, 35$. 

This gives us a total of $6$ possible integer values of $x$. Now, note that for each value of $x$, a new $n=70x+50$ is produced. Hence, we know that there are $\boxed{6}$ values of $n$ that satisfy the given equation.