234 Answerszle appeared, gave an inferior solution in which it was necessary to view the
hour IX upside down and interpret it as XI. (Note that it is so interpreted in
the illustration accompanying the statement of the problem.) When a revised
second edition of the book was issued, this blemish was removed and the
solution shown here was substituted. Actually, there are twelve more legitimate
solutions. The reader is invited to see if he can find all of them without check-
ing my Scientific American columns for May and June, 1966, in which
all thirteen can be found.
It is assumed that the Roman numerals are permanently attached to the
rim of the clock face. Lines of breakage may separate the parts of an hour, as
shown in Dudeney's solution, but must not go around any numerals, separat-
ing them from the rim.-M. G.)
so. WHEN DID mE DANCING BEGIN?
The dancing must have begun at 598 ¥I43 minutes past ten, and the hands
were noticed to have changed places at 54 l31}l43 minutes past eleven.
- MISTAKING THE HANDS
The time must have been 5¥I I minutes past two oclock.- EQUAL DISTANCES
At 23 Vl3 minutes past three oclock.- RIGHT AND LEFT
It must be 41Vl3 minutes past three oclock.- AT RIGHT ANGLES
To be at right angles the minute hand must always be exactly fifteen minutes
either behind or ahead of the hour hand. Each case would happen eleven times
in the twelve hours-i.e., every 1 hour 5¥Il minutes. Starting from nine oclock,
the eighth addition will give the case 5 hours 43rlI minutes. In the other case,
starting from three oclock, the second addition gives 5 hours lOHy! I minutes.