How many decimal places of accuracy does the following approximate equality hold?

$$
495 \sqrt{2}-388 \sqrt{3} \approx 28
$$

that is, how many zeros can we write after 28 following the decimal point, so that the resulting number is still the rounded value of the correct left-hand side?
I. Solution. We can answer the question by calculating the two terms on the left side in parallel to more and more decimal places until we get different decimal digits. The individual terms arise from square roots and multiplications. We can only give the result of the square root in decimal form approximately, and multiplication would increase the error of the approximation, so it is advisable to bring the factors 495 and 388 under the square root first. Thus:

$$
\begin{aligned}
& 495 \sqrt{2}=\sqrt{490050}=700,035713374 \ldots \\
& 388 \sqrt{3}=\sqrt{451632}=672,035713336 \ldots
\end{aligned}
$$

The difference is 28.000 00003 ..., so we can write at most seven 0s after the 28.

Ferenc Szigeti (Kunszentmárton, József A. g. II. o. t.)

Remark. We took for granted that we would eventually get different digits in the two square roots. This conjecture is correct; we can see in advance that the value of our expression is not exactly 28. Indeed, its square

$$
2 \cdot 495^{2}+3 \cdot 388^{2}-2 \cdot 495 \cdot 388 \sqrt{6}
$$

is an irrational number, it cannot be equal to $28^{2}$, which is a rational number.

II. Solution. We determine the deviation $x$ of the left side from 28 without taking square roots, by estimation. The

$$
x=495 \sqrt{2}-(388 \sqrt{3}+28)
$$

expression's numerator is rationalized (writing a 1 in the denominator) in two steps:

$$
\begin{gathered}
x=\frac{495^{2} \cdot 2-(388 \sqrt{3}+28)^{2}}{495 \sqrt{2}+(388 \sqrt{3}+28)}=\frac{2(18817-10864 \sqrt{3})}{(495 \sqrt{2}+(388 \sqrt{3}+28)}= \\
=\frac{2\left(18817^{2}-3 \cdot 10864^{2}\right)}{[495 \sqrt{2}+(388 \sqrt{3}+28)](18817+10864 \sqrt{3})} .
\end{gathered}
$$

Here the value of the numerator is 2. In the denominator, "large" sums stand. Since we are satisfied with 2 significant digits for $x$, it is enough to determine 3 significant digits in the factors of the denominator. Always taking a lower approximation for the denominator, we get an upper approximation for $x$. With $\sqrt{2} \approx 1.41$ and $\sqrt{3} \approx 1.73$, the value of the first parenthesis is greater than $697+671+28=1396$, and the second is greater than $18817+18794=37611$, so the denominator is greater than $1390 \cdot 37600$, thus more than $522 \cdot 10^{6}$, and finally $x$ is less than $39 / 10^{9}$. - If, however, we take upper approximations for the square roots, we get a lower approximation for $x$: with $\sqrt{2} \approx 1.42$ and $\sqrt{3} \approx 1.74$, the denominator is less than $1410 \cdot 37800$, even less than $533 \cdot 10^{6}$, so $x$ is greater than $37 / 10^{9}$. Thus, the first significant digit of $x$ stands in the 8th place and is less than 5, so we can write seven 0s.

Péter Tekulics (Szeged, Radnóti M. g. II. o. t.)

Remark. The latter solution is more elaborate because it involves less mechanical calculation, and in the former, we do not see in advance how far we need to calculate. However, the first solution does show that the question can be answered quite simply.