As a mere
arithmetical problem this question
presents no difficulty. In order that the hands shall all point to
twelve o'clock at the same time, it is necessary that B shall gain at
least twelve hours and that C shall lose twelve hours.
As B gains a
minute in a day of twenty-four hours, and C loses a minute in precisely
the same time, it is evident that one will have gained 720 minutes
(just twelve hours) in 720 days, and the other will have lost 720
minutes in 720 days.
Clock A keeping perfect time, all three clocks
must indicate twelve o'clock simultaneously at noon on the 720th day
from April 1, 1898.
What day of the
month will that be?
I published this
little puzzle in 1898 to
aware of the fact that 1900 would not be a leap year.
It was surprising
how many were then ignorant on the point.
Every year that can be
divided by four without a remainder is bissextile or leap year, with
the exception that one leap year is cut off in the century. 1800 was
not a leap year, nor was 1900.
On the other hand, however, to make the
calendar more nearly agree with the sun's course, every fourth hundred
year is still considered bissextile.
Consequently, 2000, 2400, 2800,
3200, etc., will all be leap years. May my readers live to see
therefore find that 720 days from noon of April 1, 1898, brings us to
noon of March 22, 1900.