THE THREE CLOCKS.
(
Money Puzzles)
On Friday, April 1, 1898, three new clocks were all set going precisely
at the same time--twelve noon. At noon on the following day it was found
that clock A had kept perfect time, that clock B had gained exactly one
minute, and that clock C had lost exactly one minute. Now, supposing
that the clocks B and C had not been regulated, but all three allowed to
go on as they had begun, and that they maintained the same rates of
progress without stopping, on what date and at what time of day would
all three pairs of hands again point at the same moment at twelve
o'clock?
Answer:
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 see how many people were 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 them. We therefore find
that 720 days from noon of April 1, 1898, brings us to noon of March 22,
1900.
Informational
