Since a stopped clock already agrees with a correctly running clock twice per day,
So the answer (that requires the smallest clock hand speed) is
Since the answer must be symmetrical (as long as the difference in speeds remains the same, it doesn't matter if our clock is getting ahead or behind the correctly running one), we can get the other answer by ...
Let us denote the time by AB:CD:EF, where AB is between 00 and 23, CD and EF are each between 00 and 59, and of course A=F, B=E, C=D.
The middle two digits
Since C=D and CD ranges from 00 to 59, there are six possibilities: 00, 11, 22, 33, 44, 55.
The outer four digits
The first two digits (AB) range from 00 to 23. How many of these possibilities give ...
First produce 48 by putting the traces up/down/up/down/up in the freezer. Then use 18 of the ice cubes and put them in the corners of the botton 6 trays, produce 66 more ice cubes. Do this 2 more times and get 246 in total.
Possibly could make 264 if you can balance using 2 cubes only.
Let W, H, S be the men who come once every two, three, and seven days respectively.
Because W visits every 2 days, his first visit of the month must be on the 1st or 2nd. Similarly, H's first visit must be on the 1st, 2nd, or 3rd. But each pair of Monday, Wednesday, Friday are separated by at least 2 days, so W must first visit on the 1st and H on the 3rd.