99. Find the smallest square that starts with six twos.
99. Since the number 222222 is not a perfect square, the decimal representation of the unknown number has the form $222222 a_{7} a_{8} \ldots a_{n}$, where $a_{7}, a_{8}, \ldots, a_{n}$ are some unknown digits.

First, assume that the number of digits $n$ of the unknown number is even: $n=2 k$. We will now extract the square root of this number according to the usual rules:

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-145.jpg?height=378&width=567&top_left_y=1413&top_left_x=346)
'The fifth digit of the result is 0, since $x_{1}$ can only be 4 or 5 and, therefore, less than 9; for the same reason, the sixth digit, if it is the last, is 5).

The remainder is zero if $a_{9}=4, a_{10}=0, a_{11}=2, a_{12}=5$; $x_{1}=4, x_{2}=7, x_{3}=1$, from which we easily deduce $a_{8}=6+1=7, a_{7}=(7+9)-10=6$. Thus, the smallest number with an even number of digits that satisfies the condition of the problem is $222222674025=471405^{2}$.

Similarly, consider the case when $n=2 k+1$ is odd:

$$
\begin{aligned}
& \sqrt{222222 a_{7} a_{5} a_{9} \ldots a_{2 k} a_{2 k+1}}=149071 \ldots \\
& 24 \left\lvert\, \frac{1}{122}\right. \\
& \frac{4 \mid 96}{289 \mid 2622} \\
& \frac{9 \mid 2601}{29807 \mid 21} 2 a_{7} a_{8} a_{8}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-146.jpg?height=150&width=436&top_left_y=691&top_left_x=226)

Since the number formed by the digits $x_{1}, x_{2}$ is less than $33=$ ( $=119-86$ ) and not more than $43(=129-86)$, the sixth digit of the root is 1, and the process of extracting the root does not stop here but continues further. Therefore, the smallest number with an odd number of digits that satisfies the condition of the problem is at least thirteen digits, i.e., exceeds the number 222222674025. Thus, the number we are looking for is 222222674025.