Let's round each addend in the following expressions to $n=1,2,3, \ldots$ decimal places, and find the largest value of $n$ for which the rounded values of $K_{1}$ and $K_{2}$ are still equal. Also, examine the question for the case where the addends are rounded to the nearest unit, ten, hundred, and thousand.

$$
K_{1}=1842 \sqrt{2}+863 \sqrt{7}, \quad K_{2}=3519+559 \sqrt{6}
$$
a) We bring the factor in front of the root sign of the irrational addends under the root sign, and in calculating the square root of the resulting numbers, we go to the 6th decimal place. We will see that this is enough to determine the sought value of $n$.

$$
\begin{aligned}
1842 \sqrt{2} & =\sqrt{6785928}=2604.981381 \ldots \\
863 \sqrt{7} & =\sqrt{5213383}=2283.283381 \ldots \\
559 \sqrt{6} & =\sqrt{1874886}=1369.264766 \ldots
\end{aligned}
$$

The values of $K_{1}$ and $K_{2}$ calculated from the addends rounded to $n$ decimal places match for $n=1$ and $n=2$:

$$
\begin{aligned}
& 2605.0+2283.3=4888.3=3519+1369.3 \\
& 2604.98+2283.28=4888.26=3519+1369.26
\end{aligned}
$$

For $n=3$, they differ by 0.001, but for $n=4$, they match again; omitting the whole number digits:

$$
\begin{aligned}
& \ldots 981+\ldots 283=\ldots 264=\ldots+\ldots 265-\mathbf{0.001} \\
& \ldots 9814+\ldots 2834=\ldots 2648=\ldots+\ldots 2648
\end{aligned}
$$

For $n \geq 5$, $K_{2}$ is always larger; indeed, the excess for $n=5$ is 0.00001 (omitting the first three decimal places, $\ldots 38+\ldots 38=\ldots 76$, or ...77), which still necessitates the examination of the next case, but for every $n \geq 6$, it is at least 0.000002, because the 6th decimal place of the terms of $K_{1}$, even with rounding up, can only be 2 or 2, so their sum is at most 4, while the 6th decimal place of the second term of $K_{2}$, when rounded down, is 6. According to these, the sought value of $n$ is 4.

b) This peculiarity of calculations with rounded numbers is repeated in the case of rounding to the left of the decimal point. Rounding to the nearest unit, hundred, and thousand, $K_{1}=K_{2}=4888$, 4900, and 5000, respectively, but rounding to the nearest ten, $K_{1}=2600+2280=4880<K_{2}=3520+1370=4890$.

József Cseh (Heves, Gymn., III. o. t.)