(related to Problem: The Three Clocks)
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 $720$th 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 on March $22, 1900.$
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