6. Find the smallest natural number $n$ such that in the decimal representation of $\sqrt{n}$, two nines immediately follow the decimal point.
SOLUTION. Let $a$ be the nearest larger natural number to the irrational number $\sqrt{n}$. According to the problem, we then have $a-0.01 \leqq \sqrt{n}$. Since $a^2$ is a natural number greater than $n$, it must hold that

$$
(a-0.01)^2 \leqq n \leqq a^2-1
$$

After rearranging the inequality between the extreme terms, we get

$$
\frac{1}{50} a \geqq 1.0001 \quad \text{or } a \geqq 50.005 \text{.}
$$

Since the number $a$ is an integer, it follows that $a \geqq 51$. And since

$$
(51-0.01)^2 = 2601 - \frac{102}{100} + \frac{1}{100^2} \in (2599, 2600),
$$

the sought number is $n=2600$.

Note. A correct solution can also be recognized if it is solved using a calculator. If there are to be two nines after the decimal point, the number $n$ must be very close from the left to some square. Therefore, it is enough to try the numbers $\sqrt{3}, \sqrt{8}, \sqrt{15}$, etc., on a calculator. Since $51^2 = 2601$, we find that $\sqrt{2600} = 50.990195 \ldots$

A more laborious task would be to find the smallest number $n$ for which the two digits after the decimal point of the irrational number $\sqrt{n}$ are two eights, or two sevens, etc.

GUIDING AND ADDITIONAL PROBLEMS: