We need the approximate value of $\sqrt{17}$ rounded to eight decimal places. Can we obtain this from the following square roots rounded to seven decimal places?

$$
\begin{array}{ll}
\sqrt{17}=4.1231056, & \sqrt{425}=20.6155281 \\
\sqrt{68}=8.2462113, & \sqrt{612}=24.7386338 \\
\sqrt{153}=12.3693169, & \sqrt{833}=28.8617394 \\
\sqrt{272}=16.4924225, & \sqrt{1088}=32.9848450
\end{array}
$$

Could we get $\sqrt{17}$ rounded to five decimal places if the given values were rounded to four decimal places (i.e., 4.2131, ... 32.9848)?
Starting from 17, and further from 17 being a multiple of $4=2^{2}$, and from the square roots of the multiples of $3^{2}, 4^{2}, \ldots, 8^{2}$, rounded, that is, the multiples of 1, 2, ..., 8 of the first data, disregarding the rounding. The rounding of a number $x$ to the 7th decimal place, denoted as $x^{*}$, deviates by at most half the value of the 7th decimal place, which is $0.5 / 10^{7} = 5 / 10^{8}$, meaning $x$ lies between the two numbers obtained by subtracting and adding $5 / 10^{8}$ to $x^{*}$. For example,

$$
4.12310555 \leq \sqrt{17} \leq 4.12310565
$$

In other words, for $\sqrt{17}$, we get a smaller and a larger finite decimal fraction (a so-called lower and upper bound), each ending in 5, with a difference of $1 / 10^{7}$. (In principle, an equality could stand in place of both inequalities, depending on the rounding instruction followed when only one digit needs to be omitted and that digit is exactly 5; in our case, however, equality cannot stand anywhere because $\sqrt{17}$ is irrational, while the two bounding numbers are rational.)

For our calculation, it would be sufficient to have a pair of inequalities similar to (1), where the right and left sides are 9-decimal numbers ending in 5, with a difference of $1 / 10^{8}$.

Based on our data, we can write similar double inequalities for the double, triple, ..., and octuple of $\sqrt{17}$, and from these, by dividing by 2, 3, ..., 8, we get inequalities for $\sqrt{17}$ itself, where the difference between the right and left sides is $1 / 2, 1 / 3, \ldots, 1 / 8$ of $1 / 10^{7}$. For example, from

$$
20.61552805 < \sqrt{425} = 5 \sqrt{17} < 20.61552815
$$

and

$$
24.73863375 < \sqrt{612} = 6 \sqrt{17} < 24.73863385
$$

by dividing by 5 and 6, respectively,

$$
\begin{aligned}
& 4.12310561 \leq \sqrt{17} < 4.12310563 \\
& \underline{4.123105625 < \sqrt{17} < 4.1231056} 41 \ldots
\end{aligned}
$$

From the underlined inequality, we can conclude more than desired: the 8-decimal rounded value of $\sqrt{17}$ is 4.12310563. (We chose the two examples above after writing all the double inequalities because the left side of (3) is the largest among the left sides, and the right side of (2) is the smallest among all the right sides.)

Accordingly, our first data was rounded down, the second was rounded up, as the right side of (2) indicates: $\sqrt{68} = 2 \sqrt{17} < 2 \cdot 4.12310563 = 8.24621126$, and this leads to the data by rounding up. Similarly, the third and sixth data were rounded up, the latter being exactly twice the third; the fourth, fifth, and eighth data were rounded down, and this alternation made it possible to write the desired value. (We cannot determine the direction of rounding for the seventh data, because from the underlined parts of (2) and (3), multiplying by 7, we get $28.861739375 < 7 \sqrt{17} < 28.861739410$.

If the data were rounded to four decimal places, we could not determine the fifth decimal place from them, because all data would be exactly $1, 2, \ldots, 8$ times the first. We could only determine the fifth decimal place if either $10 \sqrt{17}$ were among the data, or the four-decimal rounded approximate value of the intermediate $9 \sqrt{17}$ were not exactly 9 times the first data (from the 7-decimal data, we see that this would indeed be the case).

Györgyi Gallyas (Budapest, Szilágyi Erzsébet Secondary School II.o. t.)

Note. Some sought the 8th and 5th decimal places of the square root in the last digit of the 8-decimal and 5-decimal rounded values, respectively; these are two different things. - Some calculated the average of the data; this is common in physical measurements or where an average is explicitly sought.