# 1190.19 – Two Erratic Clocks

Two Friends, Arthur and Robert, were curators at a museum. Both of them were born in the month of May, one in 1932 and the other a year later. Each was in charge of a beautiful antique clock. Both of the clocks worked pretty well, condidering their ages, but one of them lost ten seconds an hour and the other gained ten seconds an hour.

On one bright day in January, the two friends set both clocks right at exactly 12:00 noon. "You realize," said Arthur, "that if we don't adjust them the clocks will start drifting apart and they won't be together again until . . . the very day you will be 47 years old. Am I right?"

Robert made a quick calculation. "That's right!" he said.

Who is older, Arthur or Robert?

Solution

Since one clock is losing time at the same rate that the other is gaining, you can imagine that one will say 11:00 when the other says 1:00, then 10 and 2, then 9 and 3, until they both say 6:00. That's the next time the clocks will be together.

So, how long does it take a clock to lose six hours, if it loses ten seconds per hour? Well, in six hours it will lose one minute. To lose 60 minutes, i.e. 1 hour, it will take 60x6 hours, that's 360 minutes, or 15 days. To lose SIX hours will take 90 days.

Now, this 90-day span begins in January and ends in May. The only way that can happen is for the clocks' time span to begin on January 31 and end on May 1st, and even that can't happen unless it's not a leap year.

So May 1st is Robert's 47th birthday, and we know that it's not a leap year. Robert was born either in 1932 or 1933. 47 years after 1932 is 1979, not a leap year, and 47 years after 1933 is 1980, which is indeed a leap year. So Robert was born in 1932 and he is the older. The clocks were set on January 31, 1979; they were together again on May 1, 1979, Robert's 47th birthday.