In America, temperature is measured in Fahrenheit. This is a uniform scale where the freezing point of water is $32^{\circ} \mathrm{F}$ and the boiling point is $212^{\circ} \mathrm{F}$.

Someone gives the temperature rounded to the nearest whole Fahrenheit degree, which we then convert to Celsius and round to the nearest whole degree again. What is the maximum possible difference between the resulting value and the original temperature in Celsius degrees?
I. solution. According to the data, $180{ }^{\circ} \mathrm{F}=100{ }^{\circ} \mathrm{C}$, i.e., $1{ }^{\circ} \mathrm{F}=\frac{5}{9}{ }^{\circ} \mathrm{C}$.

In the first rounding, the maximum deviation is $\frac{1}{2}{ }^{\circ} \mathrm{F}=\frac{5}{18}{ }^{\circ} \mathrm{C}$. In the second, the maximum deviation is $\frac{1}{2}{ }^{\circ} \mathrm{C}$. Nevertheless, the maximum deviation is not $\frac{1}{2}{ }^{\circ} \mathrm{C}+\frac{5}{18}{ }^{\circ} \mathrm{C}=\frac{14}{18}{ }^{\circ} \mathrm{C}$.

Indeed, whole Fahrenheit degrees cannot fall precisely between two whole Celsius degrees. According to the conversion, an entire Fahrenheit degree can only correspond to a fraction on the Celsius scale, where the denominator in the appropriate fractional form is 9. Therefore, in the rounding of Celsius degrees, the deviation is at most $\frac{4}{9}{ }^{\circ} \mathrm{C}$ instead of $\frac{1}{2}{ }^{\circ} \mathrm{C}$.

Thus, the deviation from the original is at most $\frac{4}{9}{ }^{\circ} \mathrm{C}+\frac{5}{18}{ }^{\circ} \mathrm{C}=\frac{13}{18}{ }^{\circ} \mathrm{C}$.

II. solution. Let $(x)^{*}$ denote the number $x$ rounded to the nearest integer, and let $(x)_{*}=x-(x)^{*}$. Then $(x)_{*} \in[-0.5 ; 0.5]$. Let the temperature in Fahrenheit be $f$, then in Celsius it is

\[
\frac{f-32}{180} \cdot 100=\frac{5 f-160}{9}
\]

The question is, what is the maximum absolute value of $\frac{5 f-160}{9}-\left(\frac{5(f)^{*}-160}{9}\right)^{*}$. This can be transformed as follows:

\[
\begin{gathered}
\left|\frac{5\left((f)^{*}+(f)_{*}\right)-160}{9}-\left(\frac{5(f)^{*}-160}{9}\right)^{*}\right|= \\
=\left|\frac{5(f)^{*}-160}{9}-\left(\frac{5(f)^{*}-160}{9}\right)^{*}+\frac{5(f)_{*}}{9}\right|= \\
=\left|\left(\frac{5(f)^{*}-160}{9}\right)_{*}+\frac{5(f)_{*}}{9}\right| \leq\left|\left(\frac{5(f)^{*}-160}{9}\right)_{*}\right|+\left|\frac{5(f)_{*}}{9}\right|
\end{gathered}
\]

The maximum of the first term is $\frac{4}{9}$, because the numerator is an integer and the denominator is 9. The second term is at most $\frac{5 \cdot \frac{1}{2}}{9}=\frac{5}{18}$, so the sum is at most $\frac{4}{9}+\frac{5}{18}=\frac{13}{18}$. The sought difference is thus certainly not greater than $\frac{13}{18}$.

For $f=32.5$, equality holds, so the maximum deviation in Celsius is $\frac{13}{18}$.