Analog clock with same hands - sometimes one can't tell time There is an accurate analog clock, however both hands are the same size and shape.
How many moments during a day a person can not conclude current time from the position of the hands?
This is from a competition, I was told it is a difficult problem.
 A: Let's measure time in units of $12$ hours, measure angles as a fraction of $2\pi$ and consider a period from midnight to noon.  Time will run from $0$ to $1$.  The hour hand position is $t$.  The minute hand position is $12 t \pmod 1$, the fractional part of $12t$  The times when you can't tell the time is when you can interchange the hands and get a legal time.  This would be when you consider the minute hand to be the hour hand, the hand you think is the hour hand, which is at $12t \pmod 1$ and the hand you think is a minute hand is the hour hand, so $t=144t \pmod 1$.  Noon and midnight are solutions to this equation, but you know what time it is. There will be $143$ times between $0$ and $1$ where this is satisfied, so $286$ times per day.
A: Let's slightly generalize the problem by considering clocks where the hour hand makes one round in $n$ hours (for our normal clocks we have $n=12$; however e.g. for this clock you have $n=24$).
Let's measure the time in hours. The minute hand makes one cycle every hour, so it gives the position modulo $1$; that is, the minute hand at times $t_1$ and $t_2$ is in the same position if and only if
$$t_1 \equiv t_2\ (\mod 1) \tag 1$$
On the other hand, the hour hand makes one cycle every $n$ hours, that is, the hour hand at times $t_1$ and $t_2$ is in the same position if and only if
$$t_1 \equiv t_2\ (\mod n) \tag 2$$
To avoid dealing with different mods, we can scale equation $(1)$ to also be modulo $n$:
$$nt_1 \equiv nt_2\ (\mod n) \tag{1a}$$
We can see that this rescaling is allowed by going back to the definition of the modulo operation: $x \equiv y\ (mod m)$ means there exists an integer $k$ so that $x=y+km$. Obviously we can multiply that equation with an arbitrary factor $s$ without changing its validity; thus we get $(sx) = (sy)+k(sm)$, or $sx\equiv sy\ (\mod sm)$.
Note that consistently calculating modulo $n$ nicely reflects the fact that after $n$ hours both hands are again in the same position.
Now two times $t_1$, $t_2$ are indistinguishable is the following two conditions are fulfilled:


*

*At time $t_2$ the hands are exactly reversed from time $t_1$, that is,
$$\begin{align}
    nt_1 \equiv t_2\ (\mod n) \tag 3\\
    nt_2 \equiv t_1\ (\mod n) \tag 4
    \end{align}$$
Equation $(3)$ says that the minute hand at time $t_1$ points in the same direction as the hour hand at time $t_2$, and equation $(4)$ the same with the two times exchanged. Note that, unlike in the integer case, we cannot replace $nt_k$ with $0$ in the modulo equation.

*The two times to distinguish are not the same,
$$t_1 \not\equiv t_2\ (\mod n) \tag 5$$


Now to proceed, we would like to multiply both sides of $(4)$ with $n$ without adjusting the modulo part. However, may we do that? Let's go back to the definition: Equation $(4)$ says there exists an integer $k$ such that $nt_2 = t_2 + kn$. The new equation would say that there exists an integer $k'$ such that $n^2t_2 = nt_1 + k'n$. Obviously if $k$ is the integer solving the first equation, then $k'=nk$ is an integer (because $n$ is) and solves the second equation, so whenever the second condition holds, also the first does. However the reverse i not true; we might get additional solutions, so we need to check for that afterwards. This is similar to squaring both sides of an ordinary equation, where you also may get additional solutions which you might want to get rid of afterwards.
So let's do the transformation, and keep in mind that we will have to check equation (4) again for any solution we may find using the new equation. We thus have:
$$n^2t_2 \equiv nt_1\ (\mod n) \tag 6$$
From (3) and (6) we therefore get the condition for $t_2$
$$n^2t_2 \equiv t_2\ (\mod n)$$
or after bringing everything on the same side:
$$(n^2t_2 - 1)t_2 \equiv 0\ (\mod n)$$
The latter says that $(n^2-1)t_2$ must be a multiple of $n$, or
$$t_2 = \frac{nk}{n^2-1},\quad k\in\mathbb Z \tag 7$$
Now for $k=n^2-1$, we get $t_2=n$, which is equivalent to $t_2=0$, therefore this gives $n^2-1=(n+1)(n-1)$ different solutions for equation $(7)$.
From exchange symmetry between $t_1$ and $t_2$, we also get
$$t_1 = \frac{nk'}{n^2-1},\quad k'\in\mathbb Z \tag 8$$
However we might have too many solutions since we used non-equivalent transformations; therefore we have to check which of the solutions solve equations $(3)$ and $(4)$.
Inserting $(7)$ and $(8)$ into $(3)$ and $(4)$ gives
$$\begin{align}
\frac{n^2k}{n^2-1} &\equiv \frac{nk'}{n^2-1}\ (\mod n) \tag 9\\
\frac{n^2k'}{n^2-1} &\equiv \frac{nk}{n^2-1}\ (\mod n) \tag {10}
\end{align}$$
To simplify those relations, we rescale again, this time with the factor $(n^2-1)/n$, so the equations get modulo $n^2-1$:
$$\begin{align}
nk &\equiv k'\ (\mod n^2-1) \tag {11}\\
nk' &\equiv k\ (\mod n^2-1) \tag {12}
\end{align}$$
Note that $(11)$ and $(12)$ are now integer modulo equations; since further $n$ and $n^2-1$ are relatively prime, this time multiplying both sides of $(11)$ by $n$ is an equivalence relation; furthermore, obviously $n^2\equiv 1\ (\mod n^2-1)$. Therefore equations $(11)$ and $(12)$ are equivalent, and thus all values of $k$ indeed give valid solutions.
However we have to still check equation $(5)$. Remember that $(12)$ means that for some $m\in\mathbb Z$ we have $nk' = k + (n^2-1)m$.
Inserting $t_1$ and $t_2$, we get the condition
$$\frac{nk}{n^2-1} \equiv \frac{k+(n^2-1)m}{n^2-1}\ (\mod n)$$
which we can again rescale to
$$nk \equiv k+(n^2-1)m\ (\mod n(n^2-1))$$
and then rewrite into
$$(n-1)k \equiv (n+1)(n-1)m\ (\mod n(n+1)(n-1))$$
or after rescaling again
$$k \equiv (n+1)m\ (\mod n(n+1))$$
which means there exists an $m'$ so that $k = (n+1)m + n(n+1)m' = (n+1)m''$.
So equation $(5)$ is fulfilled whenever $k$ is an integer multiple of $n+1$. This happens $(n^2-1)/(n+1) = (n-1)$ times.
Therefore per round trip of the hour hand, that is per $n$ hours, there are $(n^2-1)-(n-1) = n(n-1)$ moments where you cannot tell the current time from the indistinguishable hands.
For the normal clock, $n=12$ and therefore there are $12\cdot 11=132$ such moments per $12$ hours, or $264$ moments per day.
For a 24h clock, $n=24$, and therefore there would be $552$ such moments.
