Problem 10.8. For what least natural $a$ does the numerical interval ( $a, 3 a$ ) contain exactly 50 perfect squares?
Answer: 4486.

Solution. Let's choose the largest natural $n$ such that $n^{2} \leqslant a$. Then

$$
n^{2} \leqslant a\frac{49}{\sqrt{3}-1}=\frac{49(\sqrt{3}+1)}{2}>\frac{49(1.7+1)}{2}>\frac{132}{2}=66
$$

From the fact that $n+1>66$, it follows that $n \geqslant 66$ and $a \geqslant 66^{2}$.

If on the corresponding interval for $n=66$, $\left[66^{2} ; 67^{2}\right)$, there are suitable values of $a$, then the smallest of them will be the answer to the problem. In this case, the conditions $66^{2}=4356 \leqslant a<4489=67^{2}$ must be satisfied. Additionally, the conditions $13456=116^{2}<3 a \leqslant 117^{2}=13689$ must be satisfied, which are equivalent to $4485 \frac{1}{3}<a \leqslant 4563$. Clearly, the smallest $a$ that fits all these conditions is $a=4486$. It fits: on the corresponding interval $(a ; 3 a)$, there are only perfect squares $67^{2}, 68^{2}, \ldots, 116^{2}$, and there are exactly 50 of them.

## 11th Grade