Christmas Streak 64/88: Clock #3

In a given day, the minute hand does $24$ full rotations, and the hour hand does $2$ full rotations, meaning that the two hands line up $24-2=22$ times in a day. From one instance when the hands line up to the next instance, exactly $360-1=359$ positive, integral values of $\alpha$ will be attained. Hence, the answer is $22\times 359=\boxed{7898}$ .

Starting at midnight, after $m$ minutes, the minute hand has swept out $6m$ degrees, and the hour hand has swept out $m/2$ degrees. So $\alpha$ is an integer if and only if $6m-m/2=11m/2$ is an integer. (Note: I'm not saying that $\alpha = 11m/2,$ just that they are either both integers or both not.)

There are $1440$ minutes in a day, so at midnight of the next day, $11m/2 = 7920.$ So $11m/2$ takes every integer value between $0$ and $7920$ once in that $24$ -hour period. But we must throw out the values where $11m/2$ is a multiple of $360,$ since these correspond to $\alpha = 0$ and we want $\alpha$ to be a positive integer. So we throw out $0,360,720,\ldots,7920.$ Our final answer is just $7920 - 7920/360 = \fbox{7898}.$