Puzzle Solution: The Case of the Smithsonian Clocks


The Case of the Smithsonian Clocks

We must first determine the number of days it will take for the two clocks to come together. Since one clock is losing time at the same rate as the other is gaining, then the next time they will be together is when the slow clock has lost six hours and the fast clock has gained six hours. (Both clocks will then read six o’clock; of course they will both be wrong, but together.)

Now, ten seconds an hour is four minutes (240 seconds) a day, which is one hour in 15 days, or six hours in 90 days. And so the clocks will come together exactly 90 days after the day in January on which they were set right. Also, they will come together on Robert’s 47th birthday, which is in May.

How can 90 days be fitted between a day in January and a day in May? Consulting a calendar, we see that there are exactly 90 days between January 31 and May I providing it is not a leap year! In a leap year the shortest possible time between any day in January and any day in May is 91 days (since February 29 falls in between). This proves that Robert’s 47th birthday cannot fall in a leap year, and therefore Robert could not have been born in 1933 (since 47 years after 1933 is 1980, a leap year). Therefore Robert must have been born in 1932, and it is Arthur who was born in 1933. And so Robert is the older (incidentally, the year of the story-the year in which the clocks were set right-must be 1979).

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