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.