58. Clock with identical hands. It is known that, when determining the time without a clock, no one makes an error of more than six hours.

A clockmaker inserted two identical hands into the clock, so that it is impossible to distinguish the small hand from the large one. What will be the greatest error that threatens the owner of the clock?
58. As is known, the time shown by the clock is fully determined by the coordinate of the hour hand on the clock face (Fig. 127); the minute hand has only an auxiliary significance, namely, it creates something like a vernier on the clock face, which allows the accuracy of determining the coordinate of the hour hand to be increased twelvefold.

If we denote the coordinate of the hour hand on the clock face by $\xi$, and the coordinate of the minute hand by $\eta$, we get:

$$
\eta-12\{\xi\}=0, \quad 0 \leqslant \xi<12, \quad 0 \leqslant \eta<12
$$

where $\{\xi\}$ denotes the fractional part of the number $\xi$.

Suppose both clock hands are identical. Let $x$ be the coordinate of one hand, and $y$ be the coordinate of the other.

There can be three cases.

I. If

$$
x-12\{y\} \neq 0
$$

then necessarily

$$
y-12\{x\}=0
$$

and the clock shows the time $x$.

II. If

$$
y-12\{x\} \neq 0
$$

then necessarily

$$
x-12\{y\}=0
$$

and the clock shows the time $y$.

III. If both equalities

$$
y-12\{x\}=0 \text { and } x-12\{y\}=0
$$

hold simultaneously, then with identical clock hands, we cannot read the time, because each hand can be taken for the hour hand, and thus the time $x$ and the time $y$ are equally possible; there are 143 such exceptional positions of the hands.

The corresponding numbers $x$ and $y$ are the coordinates of 143 points of intersection of the graphs of the functions $y=12\{x\}$ and $x=12\{y\}$ (Fig. 128). These points lie on 23 lines defined by the equations

$$
\begin{gathered}
y=x+\frac{12}{13} k \\
(k=0, \pm 1, \ldots, \pm 11)
\end{gathered}
$$

If we now denote by

$$
r=|x-y|
$$

the error we will make by taking one of the two clock readings ($x$ and $y$) instead of the other, then according to equation (1) we will have:

$$
r=\frac{12}{13}|k| \quad(k=0, \pm 1, \ldots, \pm 11)
$$

Discarding, according to the problem's condition, errors greater than 6 hours, we get the maximum error for $k=$ $= \pm 6$. It is $r=\frac{72}{13}=5 \frac{7}{13}$, i.e., 5 hours 32 minutes $18 \frac{6}{13}$ seconds. (At what time does this error threaten us?) Of course, in this reasoning, we assumed that the clock owner reads the coordinates of the hands without error.

126