Day Clock

From Sunday noon to the next Sunday noon, the hour hand will pass the day hand $2 \cdot 7 - 1 = 13$ times over a period of $24 \cdot 7 = 168$ hours. Therefore, the day hand and the hour hand will be superimposed every $\frac{168}{13}$ hours, and if $n$ is the number of times that the hands are superimposed since Sunday noon, and $h$ is the number of hours since Sunday noon, then $h = \frac{168}{13}n$ .

Friday is $5$ days after Sunday, so Friday noon is $5 \cdot 24 = 120$ hours after Sunday noon, and Friday morning would start $120 - 12 = 108$ hours after Sunday noon. Therefore, the superimposed hands must occur between $h = 108$ hours and $h = 120$ hours, so $108 < \frac{168}{13}n < 120$ , which means $8\frac{5}{14} < n < 9\frac{2}{7}$ , and since $n$ is an integer, $n = 9$ (which means Friday morning is the ninth time in the week that the hands are superimposed).

Since $n = 9$ for the superimposed hands on Friday morning, the number of hours since Sunday noon is $h = \frac{168}{13}n = \frac{168}{13} \cdot 9 = 116\frac{4}{13}$ . Since Friday starts at $108$ hours after Sunday noon, the hands are superimposed $116\frac{4}{13} - 108 = 8\frac{4}{13}$ hours into Friday morning. And since $\frac{4}{13}$ of an hour is $\frac{4}{13} \cdot 60 \approx 18$ minutes, the hands are superimposed at approximately 8:18 am on Friday morning, which would be entered as $\boxed{818}$ with the given instructions.

Let 1 unit be equal to $\frac{\pi}{6}$ radians. I set up the equation (1 unit/hour)t = [(12 unit)/((24)(7) hour)]t+12k. This simplifies to $t=\frac{1}{14}t+12k.$ I manipulated the equation to say $\frac{1}{14}t=\frac{12}{13}k.$ Note that $\frac{1}{14}t$ is the number of units the day hand moves in $t$ hours. We want this number of units to occur on a Friday, which is between 4.5 and 5.5 days from noon Sunday. Converting days the day hand moves to units gives $4.5\cdot\frac{12}{7}$ units and $5.5\cdot\frac{12}{7}$ units. So, we want $4.5\cdot \frac{12}{7}<\frac{1}{14}t<5.5\cdot\frac{12}{7}.$ Since $\frac{1}{14}t=\frac{12}{13}k,$ we can write $4.5\cdot\frac{12}{7}<\frac{12}{13}k<5.5\cdot\frac{12}{7} \Leftrightarrow 8\frac{5}{14}<k<9\frac{3}{7}.$ The only integer $k$ inside this interval is $k=9.$ Solving the equation, $t=\frac{1}{14}t+12k,$ with $k=9$ results in $t=\frac{14}{13}\cdot 108=8\frac{4}{13}.$ Convert $\frac{4}{13}$ to minutes by multiplying by $60$ , and you will get the final answer 8:18.