2020 AMC 10B #24

Problem:

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

Source: 2020 AMC 10B, Problem 24

Solution:

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

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

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 $\boxed{{\displaystyle 6}}$ values of $n$ that satisfy the given equation.