At what time immediately prior to six o'clock are the hands of the clock exactly opposite to each other?
Give the exact time in hours, minutes and seconds.

I take the total displacements of the hands as being t and 12t. Then they are opposite whenever t + 6 = 12t - 12k for some integer k. For example, if t = 6, and k = 5 we have the hands at 6 and 12. Solving the condition for t gives t = (6 + 12k)/11, so the hour hand is at 6/11, 18/11, 30/11, 42/11, 54/11, 66/11 (=6) when the minute hand is opposite. 54/11 = 4 9/11 = 4:54:32 8/11.
Edited on January 13, 2004, 1:11 am

Posted by Richard on 2004-01-12 23:58:20