Ponder This Challenge - April 2026 - The Unlabeled Clock

We are given a strange analog clock - its dial has no markings, and although it has three hands that move continuously (not in steps), they are all of the same length and are indistinguishable. Quickly it becomes obvious that the clock is far from useless, since we can usually deduce the time from the cyclic triple of relative angles between its three hands.

However, there are still many moments in time that cannot be safely deduced from the state of the clock, because there are other moments in time that result in the same relative angles.

Here are some examples of both unique and undeducible clock configurations. Note that the first angle denotes the position of the first hand encountered clockwise from the unmarked 00:00:00 point, the second angle denotes the space between the first and second hands, and the third angle is between the second and third hands:

• 05:25:00 results in $150^\circ, 12.5^\circ, 197.5^\circ$ and this angle configuration is unique to it.
• Both 00:30:00 and 06:30:00 result in $15^\circ, 165^\circ, 180^\circ$ angles and cannot be distinguished.
• Both 03:00:00 and 09:00:00 result in $0^\circ, 90^\circ, 270^\circ$. These cases can be distinguished by checking where the two co-located hands are in respect to the third one - if they are located clockwise (09:00:00) or counterclockwise (03:00:00) to it.

Your goal: With the usual conventions of 60 seconds per minute (and full rotation), 60 minutes per hour (and full rotation), and 12 hours per full rotation, determine the number of moments in time that cannot be deduced.

A bonus "*" will be given for finding the "worst" convention, in the following sense: The clock still represents $12\cdot 60 \cdot 60=43200$ seconds, but the convention is now that there are $S$ seconds in a minute, $M$ minutes in an hour, and $H$ hours per rotation, such that $1<H< M< S$ and $S\cdot M\cdot H=43200$. What choice of $H,M,S\in \mathbb{N}$ yields the largest finite number of moments that cannot be deduced? Give both $(H,M,S)$ and the number of moments in time that cannot be deduced.

Solution

• Click here to view the solution

  The expected numerical solutions are:

  • 1541414
  • (30, 36, 40) => 3458432

  This time, the phrasing of the puzzle, even after corrections, remained confusing to many solvers and we apologize for this. However, most wrong answers arose from the initial thinking that one should consider time to be discrete, looking only at moment of the form HH:MM. The riddle explicitly noted the clock's hands move continuously, and hence simple enumeration of moments of the form HH:MM is not sufficient to solve the riddle; a different approach is required.

  We can model the current time as a value $t\in[0,1)$ and the locations of the clock's hands become

  • $H(t) = t$
  • $M(t) = 12t \text{ }(\text{mod } 1)$
  • $S(t) = 720t \text{ } (\text{mod } 1)$

  Let us define a three-component vector $A(t)=(H(t), M(t), S(t))$. Use the indices $1,2,3$, i.e. $A_1(t)=H(t)$, $A_2(t)=M(t)$, $A_3(t)=S(t)$.

  Now, an ambigous moment arises from two distinct times $t, u$ such that we can permute the elements of $A(t)$ and rotate them (i.e. add some $r$ modulo 1) and obtain $A(s)$. Hence, let $pi\in S_3$ be some permutation and $r\in [0,1)$ be the rotation value, and we consider the system of three modular equations:

  • $A_1(u) = A_{\pi(1)}(t)+r\text{ } (\text{mod } 1)$
  • $A_2(u) = A_{\pi(2)}(t)+r\text{ } (\text{mod } 1)$
  • $A_3(u) = A_{\pi(3)}(t)+r\text{ } (\text{mod } 1)$

  Subtracting the third equation from the first two, we obtain a system of two modular equations without $r$:

  • $A_1(u)-A_3(u) = A_{\pi(1)}(t)-A_{\pi(3)}(t) \text{ } (\text{mod } 1)$
  • $A_2(u)-A_3(u) = A_{\pi(2)}(t)-A_{\pi(3)}(t) \text{ } (\text{mod } 1)$

  Such modular equations can be solved using standard linear algebra. We demonstrate this in the case $\pi(1)=2, \pi(2)=3, \pi(3)=1$. In this case we obtain the modular equations

  • $H(u)-S(u) = M(t)-H(t) \text{ } (\text{mod } 1)$
  • $M(u)-S(u) = S(t)-H(t) \text{ } (\text{mod } 1)$

  i.e.

  • $u-720u = 12t-t \text{ } (\text{mod } 1)$
  • $12u-720u = 720t-t \text{ } (\text{mod } 1)$

  Which reduces to finding solutions to the linear system $\left(\begin{array}{cc} 11 & 719\\ 719 & 708 \end{array}\right)\left(\begin{array}{c} t\\ u \end{array}\right)=\left(\begin{array}{c} n\\ m \end{array}\right)$

  For $n,m\in\mathbb{Z}$.

  This allows collection of all the possible ambiguous times. Situations like 03:00:00 and 09:00:00 still needs to be carefully ruled out.

  The solution can be found manually; the goal of the "*" part was to encourage automation that will allow systematic checks for all (H,M,S) systems. However, some of those system yields infinite number of ambiguous moments; the intention was to find the maximum finite number.

Solvers

• *Lazar Ilic (2/4/2026 1:08 PM IDT)
• *Bertram Felgenhauer (3/4/2026 12:26 AM IDT)
• *Jack Saleeby (3/4/2026 1:00 AM IDT)
• *King Pig (3/4/2026 4:12 AM IDT)
• Prashant Wankhede (3/4/2026 6:35 AM IDT)
• *Paul Lupascu (3/4/2026 6:51 AM IDT)
• *Guangxi Liu (3/4/2026 8:56 AM IDT)
• Ahmet Yüksel (3/4/2026 10:03 AM IDT)
• *Kang Jin Cho (3/4/2026 10:16 AM IDT)
• Alex Fleischer (3/4/2026 4:19 PM IDT)
• Stéphane Higueret (3/4/2026 11:46 PM IDT)
• Stephen Ebert (4/4/2026 11:29 AM IDT)
• *Pitiwat Chimplee (5/4/2026 8:21 AM IDT)
• Rahid Zaman (5/4/2026 3:58 PM IDT)
• Nickita (5/4/2026 10:08 PM IDT)
• *Sanandan Swaminathan (6/4/2026 7:42 AM IDT)
• *Daniel Chong Jyh Tar (6/4/2026 9:55 AM IDT)
• Hakan Summakoğlu (6/4/2026 11:49 AM IDT)
• *Jan Ondras (7/4/2026 2:36 AM IDT)
• *Jean-françois HERMANT (7/4/2026 3:16 AM IDT)
• *Shouky Dan & Tamir Ganor (7/4/2026 11:07 AM IDT)
• *Hao Li (8/4/2026 11:50 PM IDT)
• *Vladimir Volevich (9/4/2026 3:20 PM IDT)
• *Latchezar Christov (10/4/2026 12:58 PM IDT)
• Fatih Keşan (10/4/2026 1:22 PM IDT)
• Chern Arthas (12/4/2026 8:59 PM IDT)
• *Daniel Bitin (12/4/2026 11:37 PM IDT)
• *Phil Naveen (13/4/2026 4:13 AM IDT)
• *Alper Halbutogullari (13/4/2026 7:18 AM IDT)
• *Franciraldo Cavalcante (13/4/2026 3:54 PM IDT)
• *Junyu Zhou (13/4/2026 6:46 PM IDT)
• *Rethna Pulikkoonattu (17/4/2026 3:49 AM IDT)
• Guy Daniel Hadas (25/4/2026 3:13 PM IDT)
• *Justin Mazenauer (27/4/2026 5:22 PM IDT)
• *Martin Thorne (27/4/2026 6:38 PM IDT)
• *Lorenz Reichel (28/4/2026 1:28 PM IDT)
• Jan Fricke (29/4/2026 12:30 PM IDT)
• *Liubing Yu (29/4/2026 6:58 PM IDT)
• Nyles Heise (29/4/2026 9:19 PM IDT)
• Karl D’Souza (29/4/2026 10:49 PM IDT)
• Lorenzo Gianferrari Pini (29/4/2026 11:15 PM IDT)
• *Nadir S. (30/4/2026 9:33 AM IDT)
• *Luke Eilers (30/4/2026 4:01 PM IDT)
• *Chris Shannon (30/4/2026 6:48 PM IDT)
• *Tim Walters (30/4/2026 7:29 PM IDT)
• *Alexander Bielik (30/4/2026 11:28 PM IDT)
• *Motty Porat (30/4/2026 11:28 PM IDT)
• Oscar Volpatti (1/5/2026 8:12 AM IDT)
• David Greer (1/5/2026 3:08 PM IDT)
• Radu-Alexandru Todor (3/5/2026 1:26 PM IDT)
• *Li Li (5/5/2026 3:46 PM IDT)
• *Evan Semet (5/5/2026 5:04 PM IDT)
• *Armin Krauß (6/5/2026 9:19 AM IDT)
• *David F.H. Dunkley (7/5/2026 3:54 AM IDT)