When the two hands are indistinguishable, the clock face no longer reveals which hand performs which role. Some images still determine one time; others connect two different instants.
Identical Hands
Riddle statement
An analog clock runs continuously and perfectly, but its two hands are identical in length, thickness, and appearance. You cannot tell which is the hour hand and which is the minute hand.
Consider the twelve-hour interval beginning at 12:00:00 noon —included—and ending at 24:00:00, the following midnight —excluded—.
The reading is ambiguous at an instant if the same visible arrangement of the two hands also corresponds to a different instant in the same cycle when their roles are exchanged.
Count instants, not distinct images: if an ambiguous image occurs at two different times, both instants count.
At how many instants in the cycle is the reading ambiguous?
Show solution
Solution
Let h be the position of the hour hand, measured as a fraction of one revolution from twelve. During the cycle:
0 ≤ h < 1.
The minute hand moves twelve times faster, so its position m satisfies:
m ≡ 12h (modulo 1).
Now mentally exchange the roles of the two hands. For the old minute hand to act as the hour hand and the old hour hand to act as the minute hand, we must also have:
h ≡ 12m (modulo 1).
Substituting the first relation into the second gives:
h ≡ 144h (modulo 1),
so:
143h is an integer.
The candidate positions are therefore:
h = k/143, where k = 0, 1, …, 142.
This gives 143 candidate instants in the cycle.
Some are not ambiguous, however. When the hands coincide, exchanging them does not produce a different time. The condition h = m, together with m ≡ 12h, gives:
11h is an integer.
Thus there are 11 meetings:
h = j/11, where j = 0, 1, …, 10.
All of them belong to the candidate list because j/11 = 13j/143. Removing them leaves:
143 − 11 = 132 ambiguous instants.
To see why these represent 66 images, label a candidate instant by k. The instant obtained by exchanging the hands corresponds to:
k' ≡ 12k (modulo 143).
This transformation is an involution because 12² = 144 ≡ 1 (modulo 143). The 11 meetings are its fixed points; the remaining 132 instants are paired two by two:
132 ÷ 2 = 66 distinct ambiguous images.