What a big difference a small change can make

First, we must determine the time that each clock shows. Since Clock 1 is correct, it shows the time $8 \! : \! 20$ . Clock 2 gains one minute every hour, so since $20 \frac{1}{3}$ hours past since $12$ AM, Clock 2 shows $8 \! : \! 40 \frac{1}{3}$ . In the same fashion, we can find that Clock 3 shows the time $7 \! : \! 59 \frac{2}{3}$ .

Next, we find the non-convex angle between the hour and minute hands for each of these times. We will derive a general formula for the angle first, then plug in our values:

Let $H$ be the hour and $M$ be the minute. Every minute, the minute hand moves $6^{\circ}$ counterclockwise, so after M minutes, the minute hand moved $6M$ degrees from the pointing up position. Now, for every minute that passes, the hour hand moves $\frac{1}{2}^{\circ}$ , and for every hour that passes the hour hand moves $30^{\circ}$ . Thus, the hour hand is $30H + \frac{1}{2}M$ degrees from the pointing up position. The angle between the hour and minute hands is the absolute value of the difference between these two angles.

$\begin{aligned} \text{Angle between the hour and minute hands} &= \left | 30H + \frac{1}{2}M - 6M \right | \\ &= \left | 30H - \frac{11}{2} M \right | \\ &= \frac{1}{2} | 60H - 11M |. \end{aligned}$

Now, we can find the angles. For Clock 1, we have

$\frac{1}{2} | 60H - 11M | = \frac{1}{2} |60(8) - 11(20) | = 130^{\circ}.$

For Clock 2, we have

$\frac{1}{2} | 60H - 11M | = \frac{1}{2} \left | 60(8) - 11( 40 \frac{1}{3} ) \right | = \frac{109}{6}^{\circ}.$

For Clock 3, we have

$\frac{1}{2} | 60H - 11M | = \frac{1}{2} \left |60(7) - 11(59 \frac{2}{3}) \right | = \frac{709}{6}^{\circ}.$

Thus,

$\left \lfloor A + B + C \right \rfloor = \left \lfloor 130 + \frac{109}{6} +\frac{409}{3} \right \rfloor = \boxed{266} .$

$\text{Note that 1 minute movement by minute arm M is 6 degrees rotation,} \\ \text{and by the hour arm H, it is 6/12= 0.5 degree rotation.} \\ \text{So the CHANGE in angle between the two arms, } \\ \text{ for one minute by M, will be } 6 - 0 .5=\color{#3D99F6}{ 5.5 degrees.} \\ ~~~\\\text{The start and end has 20.33 hours.} \\ \text {So Clock 2 will gain 20.33 minutes and Clock 3 will loss 20.33 minute.} \\~~~\\ \text{ At 8:00, all clocks will have the minute arm at 0 degrees} \\ \text{ and hour arm at 240 degrees clockwise. } \\~~\\Clock ~1 \\ \text{It will show 8:20 . So M has moved by 20 minutes. So} \\ \text{the distance between H and M has reduced by } 20 * 5.5=110^o.\\ \therefore~ angle~ between~ H~ and ~M ~is~240-110=\color{#D61F06}{130^o} .\\~~~~~\\Clock 2. \\ \text{It will show 8:20 +20.33= 8:40.33. M has moved 40.33 minutes. So}\\ \text{the distance between H and M has reduced by } 40.33 * 5.5=221.833^o.\\ \therefore~ angle~ between~ H~ and ~M ~is~240-221.833=\color{#D61F06} {18.166^o} . \\~~~~~~\\Clock 3. \\ \text{It will show 8:20 -20.33= 7.59.66. M has moved - 0.33 minutes. So}\\ \text{the distance between H and M has reduced by } - 0.33* 5.5=- 1.86^o.\\ \therefore~ angle~ between~ H~ and ~M ~is~240-(-1.86)={241.86^o} . But ~we \\ \text{need angle less than 180. So the angle is } 360- 241.86=\color{#D61F06}{118.1933^o}. ~~\\~~\\\text{Angle between H and M is always +tive for this problem.} \\ \lfloor 130+18.166+118.1933 \rfloor =~~~~~~~~~\Large \color{#EC7300}{266}$