Project Euler Problem 891
A round clock only has three hands: hour, minute, second.
Solution
Answer: 1541414
Let the hand positions (modulo one full turn) at time $t\in[0,12\text{ h})$ be
$$H=\frac{t}{43200},\qquad M=\frac{12t}{43200},\qquad S=\frac{720t}{43200}.$$
A clock reading is determined only up to:
- rotation of the whole clock, and
- permutation of the three indistinguishable hands.
So an ambiguous moment is a time $t$ for which there exists another time $t'\neq t$ and a nontrivial permutation $\pi$ such that
$$(H',M',S') \equiv (H,M,S)_\pi + c \pmod 1$$
for some rotation $c$.
Subtracting the first equation from the other two removes the unknown rotation.
For each nontrivial permutation we obtain a linear system modulo $1$, whose determinant gives the number of solutions in one 12-hour cycle.
There are five nontrivial permutations.
1. Swap $M,S$: $(H,M,S)\mapsto(H,S,M)$
$$\begin{cases} 11t' -719t \equiv 0\ 719t' -11t \equiv 0 \end{cases}$$
Determinant:
$$11^2-719^2=-516840.$$
Hence there are $516840$ solutions for $t$.
But $t'=t$ occurs when $M=S$, i.e.
$$708t\equiv0,$$
giving $708$ non-ambiguous fixed points.
Contribution:
$$516840-708=516132.$$
2. Swap $H,M$: $(M,H,S)$
Determinant:
$$15697.$$
Fixed points are exactly $H=M$, giving $11$ points.
Contribution:
$$15697-11=15686.$$
3. 3-cycle $(H,M,S)\mapsto(M,S,H)$
Determinant:
$$509173.$$
Only fixed point is $t=0$, so:
$$509173-1=509172.$$
4. 3-cycle $(H,M,S)\mapsto(S,H,M)$
This produces the same solution set as case 3, so it contributes no new moments.
5. Swap $H,S$: $(S,M,H)$
Determinant:
$$501143.$$
Fixed points are exactly $H=S$, giving $719$ points.
Contribution:
$$501143-719=500424.$$
The four distinct families are pairwise disjoint except for the fixed point $t=0$, which has already been excluded.
Therefore the total number of ambiguous moments is
$$516132+15686+509172+500424 =1541414.$$
So the exact answer is
Answer: 1541414