We claim that in the two-hand case, there are ambiguous times, while in the three-hand
case, the time is unambiguous.
Consider a clock with two hands, where hand 2 makes an integer h revolutions for each
revolution of hand 1. Let t represent the number of revolutions hand 1 has made
(0 ≤ t < 1). Then hand 2 will have made ht revolutions; it will have appeared to have
made ht – [ht] revolutions, where [x] represents the greatest integer not exceeding x.
Suppose the hands are in a position at time t where interchanging the hands would place
them in a position at time u, (0 ≤ u < 1). Then t = hu – [hu] and u = ht – [ht]. This
means
t = hu – [hu]
= h2t – h[ht] – [h2t – h[ht]]
= h2t – h[ht] – [h2t] + h[ht] (since h[ht] is an integer)
= h2t – [h2t] .
Therefore,
[h2t] = t(h2 – 1).
(1)
The LHS of this expression is an integer, and so must the RHS, and so t = k/(h2 – 1) for
some integer 0 ≤ k < h2 – 1. In fact, (1) is true for any integer in this range, and so
t = k/(h2 – 1) and
u = hk/(h2 – 1) – [hk/(h2 – 1)].
Note that setting k = 1 results in two different times for t and u, and so this is one instance
of an ambiguous pair of times for two hands. For the case in the problem statement, we
have h = 12 (the minute hand makes 12 revolutions and the hour hand makes 1 revolution
in half a day). The ambiguous times corresponding to k = 1 are
t = 1/286 days ≈ 12:05:02.10 , and
u = 12/286 days ≈ 1:00:25.17.
For the three-hand case, suppose hand 2 makes h revolutions and hand 3 makes p
revolutions in the time hand 1 makes 1 revolution. Here, we suppose h and p are integers
greater than 1. Hands 2 and 3 can’t be pairwise ambiguous, as hand 1 is clearly in two
different positions at two different times t and u, (0 ≤ t,u < 1). If hands 1 and 2 are
pairwise ambiguous, then from the previous calculations,
t = k/(h2 – 1) and
u = hk/(h2 – 1) – [hk/(h2 – 1)] = k’/(h2 – 1)
where k and k’ are different integers between 0 and h2 – 2.
At time t, hand 3 points at pk/(h2 – 1) – [pk/(h2 – 1)] while at time u, hand 3 points at
pk’/(h2 – 1) – [pk’/(h2 – 1)]. If p shares no factors with (h2 – 1), then pk and pk’ are
different quantities modulo (h2 – 1), and so the third hand distinguishes the two times t
and u. In the original question, p = 720 and h = 12 or the reverse cause p not to share any
factors with h2 – 1.
If hands 1 and 3 are pairwise ambiguous then the argument that shows that the ambiguity
is broken by hand 2 can be written by switching the roles of h and p above.
The final case is when all three hands could cyclically change roles. Suppose at time t,
hand 1 points at t, hand 2 points at ht – [ht], and hand 3 points at pt – [pt], while at time u,
hand 1 points at ht – [ht], hand 2 points at pt – [pt], and hand 3 points at t. This means
u = ht – [ht], hu – [hu] = pt – [pt], and pu – [pu] = t.
We can combine these into two equations after algebra similar to the first part of the
problem:
[pht] = t(ph – 1),
[h2t] – [pt] = t(h2 – p)
(combining the first and third equations)
(combining the first and second equations.)
This means t is an integer multiple of 1/(ph – 1) and 1/(h2 – p) simultaneously, and
0 < t < 1. For the problem as stated, we have p = 720, h = 12 or the reverse. This
requires either t is an integer multiple of both 1/8639 and 1/576 or of both 1/8639 and
1/518388. But 8639 = 163 × 53 is relatively prime to both 576 and 518388, so no such t
exists and we’re done.